Thank you both! I think I'm fine with diodes then, if that's all we need.Normally the ground is the lowest reference, which is why we can do what we normally do with other circuits. But since there is a potential that is lower than this reference in this case, we have to look at the situation differently, as we could find the two currents in the two resistors. But as the second was greater than the first, the current had to come from somewhere (again current flows from higher potential to lower potential, so that's why some current was drawn from there). In the normal case, where the lowest potential is the ground, the currents are flowing towards this direction, and so we can look at the problem normally.
As for the whole earthing thing. If there is some current that is leaking from the ground in this case, why don't we have to take that into consideration for other circuits? Or do we? Like, for simple circuit analysis where there is a circuit with a battery, a resistor, and ground at some point, can we not calculate the current simply because there is the ground??? I think ground just confuses me, I thought its only purpose was as a reference point :/
could someone help me with the first twolet I_{B1} be the voltage drawn from the battery 1 and I_{B2} be the voltage drawn from battery 2
in a force distance graph for satellite, is the kinetic energy greater at a greater distance or is the kinetic energy greater at a smaller distancekinetic energy is greater at a smaller distance.
General questions about voltage;I'm not sure about the voltage thing but since voltage is relative, I think it is possible. But I'm prettyy sure its not necessary in the course. (not in the study design) I might get back to this later :)
Why can you have negative voltage (AC) ? Doesn't that imply we can have "negative" energy? Wouldn't it be best if we have negative and positive current for saying which way the current flow?
Also;
I lost a mark on a SAC with the question being about explainging how seatbelt and crumpling reduce the severity of the drivers injury, it was a 3 mark question. What type of exam "style" answers should I have. The teacher noted on it "how does the seat belt increase the time of collision?", how would I undergo answering that for an 'exam' quality answer?
Thanks :)
Ok, thanks for that. I will ask my teacher what her opinion of what the answer "should" be (even though I find people on this forum more knowledgable and reliable with answer).Split this post and moved it to the Technical Score Discussion boards
Also, another question (this is more a general education question).
Does the only thing that matters with SACs is your ranking? Please only answer if you are 100% sure due to reading a report from VCAA or talking to an examiner?
If so, then does that means there is no different between averaging 90% and 100% on SACs if you go to a low scoring school?
Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.I'm not entirely sure on this one but I'll give you my thoughts...
^help anyone? thanks in advance
Jasper has a transistor radio with ﬂat batteries. Suggest a reason to explain why the sound is distorted when he tunes in to a strong radio signal.
^help anyone? thanks in advance
The input signal may be too large, causing the amplifier to be limited by the supply voltage.
What does a negative gain amplifier actually achieve?
I know it inverts the Vout graph but how does that have a meaningful purpose? What would be the difference in a microphone that is hooked up to an amplifier with a negative gain rather than a positive one?
Thanks :)
Hey guys how would you calculate the value of R?
Thanks ;D
ANS: 309.1ohms
For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:
Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.Ahhh yeah I had a mental blank lol
Hopefully you can see how the solutions got the answer from here!
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
Question about the 'gain' in amplifiers
First off, can you have a negative gain? Does that occur when the gradient is negative?
Secondly, my interpretation of the gain is the factor in which the input voltage is multiplied to get the output voltage. However, in Qs 3a in Test 2 on electronic & photonics part in the atar notes book, it doesn't work because the gradient is negative.
Also, my original understanding with circuits, if you have two resistors in a circuit, the resistor with the highest resistance will have the highest power output. But this doesn't seem the case with resistors in parallel. Could someplease please confirm this and try to explain it a bit.
Thanks
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
Thanks for that, I had questions on those topics for a practice SAC and our class haven't cover much of this stuff in detail if at all.
Another question relating to springs.
Pinball game has a ball with mass 0.15kg which is launch by a compressed spring at 4.2m/s that has been compressed by 20 cm. If the spring is "ideal" what is the spring constant k in N/m?
Since k is N/m, would you use constant acceleration formulas to find the acceleration, multiply it by mass (since F=ma) and divide by distance (k=F/x)? Would you have to worry about the constant acceleration question or are you meant to substitute the final speed for acceleration?
Just looking at the 2012 Exam 2 paper and confused with question 1. Basically, there is a point P and it asks for you to draw an arrow at P indicating the direction of the magnetic field. There is a solenoid which will produce a field (comparable to Earth's) going right, and there is an arrow pointing upwards that says "magnetic north."
The answers say that the arrow drawn should be "to the right and up the page at an angle of approximately 45°." However, in saying that "magnetic north" is upwards, does that not mean that the field lines due to the Earth should be going downwards? If magnetic north is up, magnetic south is down, and field lines go from north to south. I would have drawn an arrow "to the right and *DOWN* the page at an angle of approximately 45°."
Basically, in saying that magnetic north is up, does that not mean that geographical north is downwards?
how would i find the tension in each cable?As the structure is in rotational equilibrium you can sum the moments (I think you refer to it as torque in yr 12 physics?) about the point of rotation and equate them to zero.
It depends on how the angle is defined. The formula is only used when the angle is measured between the heads of the vectors r and F (it's a vector equation). In this case, 60 degrees is the angle between the ladder and the horizontal, while here theta is the angle between r and mg, in this case the ladder and the VERTICAL.Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
The young's modulus for a rope of length 30m, cross-sectional area 30cm2, which stretches 2m under a load of 1000N is? ANS: 5 x 10^6 Nm-2
edit: accidentally clicked modify instead of quote, my bad (2/cos(c))
Is the torque formula examinable material for our core topics, or is it only required for a particular detailed study because I haven't seen it in the Heinemann textbook :/
Also, are you guys gonna reinforce the Unit 3 knowledge-through some mid-year exams- or continue on with the course, as revision?
Also, What would be the tension in the cables of the cranes? Thankyou
hey just a question why is it 0.003 wouldnt it be so ?
Hey guys,Okay so the forces are identical, but what is the difference here? Something has to be different between you and the ball, because clearly one is moving while the other is stable. The difference is mass; you weigh a lot more than the ball you've kicked and according to the formula: a = F/m, you are going to experience negligible acceleration while the ball is going to experience much greater acceleration, because if 'm' increases, then acceleration of an object will decrease. Therefore, you are going to 'absorb' a reaction force of 100 N by only experiencing negligible acceleration, but this force will cause the ball to go 'flying'.
i am having a little trouble getting my head around Newton's third law. so it's the whole equal and opposite reaction thing.
so if an object exerts a force on the another, that object while exert the same force of same magnitude in the opposite direct. what i don't understand is how object are still able to move (i understand how the forces are acting on different and can't be cancelled out).
for example if you kick a ball with 100N, the ball exerts 100N back on you. Where does that *extra* force to make the ball go flying??
thank you guys
What would be the direction of the magnetic force and how do we work it out?
impulse/momentum question here:
as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.
thanks guys
you have it wrong my friend. By increasing time in the car case you increase your impulse:
the equation is ΣF= ∆P/∆T. So if we increase the time the Net force decreases.
say ∆P=4 and ∆T=1 then the net force is 4N. but if we increase the change in time to, lets say, 2, then the Net force is 2N.
similarly, say we have a force of 5N and ∆T=1 then the impulse will be 5. however if we increase the change in time to, lets say, 2, then the impulse is 10.
You understand broda?
impulse/momentum question here:
as we know in cars, there are crumple zones which increases the amount of time the change in velocity occurs so from forumla
. what i need help with understanding is for cars more time has resulted in less impulse, but why when talking about baseball/golf and following through more time results in greater impulse.
thanks guys
More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.
Could someone please explain why when the secondary coil in a transformer is connected to a circuit with an open switch (i.e. one with no load) the energy used by the primary coil is zero? Wouldn't the power being dissipated by the primary coil always just be IxV regardless of what's going on in the secondary circuit? This is in relation to question 9 of chapter 10.6 in Heinemann.
Thanks
To create an induced current we do not always need loops and magnets. Whenever a changing magnetic flux encounters a conducting material an induced current will occur. These currents are often called eddy currents and may result in lost energy in electrical machinery. The eddy currents produced in a moving conductor will themselves be subject to the IlB force.Eddy currents are present in the core of transformers too.
p369 of the Heinemann textbook
i know the top of the beam experiences compression and the bottom tension, but what about the centre? is it a little of both?In theory, the beam experiences no bending stresses at its centre. Although I believe in most cases there will be a very small amount, which is negligible.
Because VCE physics is inadequate and doesn't give you the full picture. Period.Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.
Jks.
The force is actually given by a cross product: Force = iL x B
where L is the current length vector; it has magnitude equal to the length of the wire and has the same direction as the wire
and B is the magnetic field vector which has a magnitude and a direction.
The magnitude of this force is given by i*|L||B| |sin theta| where theta is the angle between L and B. As you've seen from VCE physics, if the current wire is parallel to the field, no force results.
Likewise, if it's perpendicular, the sine function returns the maximum value of 1. Hence the force is maximum when the external magnetic field is perpendicular to the conductor.
Read up on wiki if you're not sure about cross products in general.
Oh how I can relate to that! VCE physics is especially vague in explaining Electronics & Photonics.Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense
Basically, dot/scalar product is cos and cross product is sin. You use a few of each in VCE physics torque, force (in magnetism), work etc.
I think you mean "you're meant to use a few of each in VCE physics torque, force, work etc".They use them without knowing that they are using them :P
The dot product is "too difficult" for VCE physics. Funny. Maths was made primarily for physics, so why are we omitting it like this. Hmph.
They're looking for the max instantaneous emf induced in the coil, not the average. You'll need a derivative here.Thanks :)
Frequency of 100 Hz => 0.01 s = period
Let's denote the rotation speed by w rad/s. Then the angle made between the magnetic field and the coil changes, so the flux is going to be B*A*cos(wt)
As it's a uniform rotation rate, the coil rotates by 2pi radians in one period. Therefore the rotation rate is 2pi/T = 2pi rad/(0.01s) = 200pi rad/s = w
So now we can differentiate the flux with respect to time. This yields -w*BAsin(wt). The maximum of this is simply wBA, which by Faraday's law is the maximum emf induced at any time.
But from before, w = 200 pi rad/s, so plugging in all the values yields emf = 200 pi rad/s * 8*10^-4 T * 40*10^-4 m^2 = 64000pi * 10^-8 Wb/s = 2.01^-3 V
Someone check my working.
I think the only other approach would be to memorize the formula... typical VCE Physics...
I was surprised this question came up in the Heinemann book, coz I dont think the exercises showed us how to find the max. EMF; is it just me or is Electric Power a hard topic (especially since the Heinemann book doesn't help)
Thanks again!
Formula for dark fringes is d sin theta = (n+1/2) wavelength where the first fringe is at n = 0
So the fourth dark fringe for the known wavelength has a path difference of 9/2*600 nm
This corresponds to the fifth dark fringe, which has a path difference of (11/2)*w
9/2*600 = 11/2 * w
w = 5400/11 nm
So...his maximum speed is 2.5a
Maintains this for a time of 10.4 - 2.5 = 7.9 seconds
Distance travelled = 7.9 * 2.5a + 1/2*at^2 = 1/2*a*6.25 + 7.9*2.5a
22.875a = 100
a = 4.37 m/s^2
Max speed = 2.5a = 10.9 m/s
im a bit confused because, if he's maintaing the vmax, shouldnt the acceleration be zero instead of 4.37? and within that 7.9 seconds, he travelled a distance less than 100m because he took 10.4s to finish, so i dont understand why t=7.9 and dist=100 were subbed together into one eqn, to me this implies that he ran 100m in the 7.9s
sorry if this is confusing but im having trouble understanding it all
i gotta question:
Why is the EMF induced in a DC generator a maximum when the coil is parallel to the magnetic field, and a minimum when perpendicular to the field?
We are doing a prac at school with a circular motion kit, I've stumbled on a task that i have no clue how to go about:
"Given Fnet = mv^2/r
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.
Clue: what is another way of expressing v^2/r and how can this be kept constant even when radius is changing. "
Not sure how to star, any pointers would be appreciated.
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
can you tell me which question it is? is it from hinemann?Well question 5 and questions 8 of 10.4 involve peak induced voltage and magnetic flux; and yes, it's from Heinemann.
Find the average RMS voltage and multiply by to get the peak voltage.
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
Faraday's law: emf=-d(phi)/dt
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
This question has popped up in the textbook a couple of times an I can't seem to get the right answer:
How would you calculate the maximum induced voltage? I know that the maximum rate of change of magnetic flux occurs when the plane of the coil is parallel to the magnetic field lines but I can't seem to translate this into Faraday's Law.
Thanks :)
P.S. (I've got a SAC on this tomorrow and this is the only thing I'm unsure about so please answer quickly ;)).
Find the average RMS voltage and multiply by to get the peak voltage.
Faraday's law: emf=-d(phi)/dt
If you draw the graph, it should look like a cosine graph (fyi flux is the dot product of area and magnetic field, that's why we have the cos function). After drawing the graph, i think it should be clear that the gradient of the graph is greatest/lowest when magnetic flux=0 (if you know calculus, this is easy to prove). Greatest/lowest gradient (which is basically the biggest/smallest value of -d(phi)/dt) gives you the maximum value for the magnitude of emf.
Yeah, that seems to make sense, but I don't really know how to apply it to the questions I've mentioned above. :-\
How would you go about finding the average RMS voltage of a generator? I can find the average induced voltage but I don't think this is equivalent to the average RMS voltage.
Also, I might as well post the related question(s) from the textbook:
5. 100 turns, area=20cm^2, B=5.0mT, coil rotates at a rate of 15degrees per millisecond. What is the peak value of the induced EMF for this coil?.
ANS: 0.263V.
8. Peak voltage=8.0kV, N=1000 turns, each coil has a radius of 10cm, magnetic field strength=B, frequency=50Hz. Calculate the strength of the magnetic field required to produce a peak voltage of 8.0kV.
ANS: B=0.81T.
I guess sometimes you just have to use calculus, huh.
But that's too difficult for the delicate minds of VCE physics students.
Well you seem to be one of those guys who has no respect for the VCE design (for all subjects) even though you'll be one of the people who simply ace VCE.
May I ask why?
Also is it wrong that I never fully understood motion in physics (which we was the first thing we did in term 1) until I began kinematics in spesh a few weeks ago?
And do you know if we're allowed to use I and J systems in physics?
nliu1995, I am very intrigued by you at the moment.
Never seen anyone who seems to actually be capable of virtually destroying VCE with minimal effort.
But what? Spesh in year 10? In year 10 I was still learning what a parabola was lol...
I am simply speechless...
Man what the hell, although deep down know you're telling the truth, I simply cannot believe what you are saying.
Year 5... Pokemon Emerald/FireRed dominated my life throughout that year?
I only learnt what that f(-b/2a) thing meant last year lol.
v = radius * angular velocity
= rw
v^2/r = (rw)^2/r = w^2*r
Or v^2/r = (2pi*r/t)^2/r = 4pi*r/t^2
The second form is probably more useful, so if you keep the same distance and period you'll be fine. I don't know about the practicalities of that though.
Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
I'm just a more outspoken member of the "dissatisfied with VCE" group. There are heaps of people like me. They just don't want to be so vocal for fear of offending people.
...
lol well, I spent too much on maths in earlier life. Parabolas was year five for me. I still remember fumbling around to understand what this f(-b/2a) meant in primary school.
Determine a relationship between F and m so everything else must be kept constant. Think about how best to keep v^2/r constant while collecting your f vs m data.Thanks, so i will select a time to make the period for every test i do and also raidus.. however do you know how i would go about keeping period the same w/o increasing speed (because it increases my r)
To be honest, I had no idea what he wanted, as simply investigating the relationship between F and m is...trivial. F = ma arises from a definition of force in Newtonian mechanics and with a constant mass. So why would anyone measure the relationship between F and m? It's just measuring your experimental capabilities.
EDIT (in spoiler):I'm guessing you meant it tongue in cheek, but anyway :P it depends if your teacher is expecting you to get the correct results from the experiment though, a well designed practical sac would allow you to still do pretty well if you're able to explain how your data turned out to be messed up.SECRET TO SUCCEEDING IN ANY PRACJust copy what the smart people in your class are doing :P Or worse comes to worse make up/tweak your data results after the prac then write some cohesive BS about errors and stuff :D
Or just blame the teacher hahahaha. that always goes down well..
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.
If the wave model was correct, there would be a delay before electrons were emitted, as it would require time for the electron to build up energy. There would also be electrons emitted for all frequencies of light, but the energy would take longer to build up. It also predicts that higher intensities would produce more current. These have been shown to be false through experiment. The particle model can explain all of these so it is more appropriate than the wave model. Since the wave model couldn't explain it and the particle model can, the particle model was reintroduced.
Erm...higher intensities DO produce more current. Another piece of evidence was the fact that intensities had no effect on the energies of the emitted electrons.True. I'll edit the post.
QuestionA reverse bias voltage can be used to find the stopping voltage (ie. when current=0A). In doing so, the kinetic energy of the electrons (since voltage=energy/coulomb(1.609*10^19 electrons I believe). From this, we can determine the work function of the metal (amount of energy to liberate
Explain the role of a reverse bias voltage and an ammeter in the photoelectric effect
Why is momentum measured in Ns for some questions? I know the units all add up but I thought that you just keep it as kgms^-1I presume it's from Newton's second Law, which in its true form is (I'm not sure if you do methods so I don't know if you are familiar with differential notation - force is the rate of change of momentum with respect to time)
Do we need to know about the thermal motion of electrons (I think that's what it's called)? A question came up about it in the second atarnotes exam from the study guide in regards to an incandescent globe
In the examiners report of last years second exam, in relation to the last question it stated "a common misconception was that the electrons moved around the orbit in a wave pattern"
If someone explain to me how this is wrong that would be great
The electrons don't orbit in a 'wave' pattern - electrons are the wave. This is probably a bit tricky for most people to understand, since it's difficult to comprehend something being a particle and a wave at the same time.
The electron exhibits both particle-like properties and wave-like properties. They have a rest mass, like a particle, but it's easiest to explain the orbitals of electrons using a wave model for electrons. The electrons can only form orbits when the 'wavelength' of the electron fits exactly into the circumference of their orbit, and they form a 'standing wave' structure. The electron cannot take any other orbital, or else it will 'destructively interfere' with itself, and so it can only exist at certain orbits and at certain energy levels. If you do chemistry, this is what causes the shells and sub-shells.
So the electron's orbit is not a wave-pattern, but instead we find the orbits by treating the electron as a wave.
Some Questions related to light and matter...
How does the medium (air vs water) influence the Fringe Spacing?
Electrons of known energy are fired into Mercury vapour. The energy of the scattered electrons is then measured. When electrons of energy from 0eV to 4.8eV are fired into the mercury vapour, the energy of the scattered electrons equal the energy of the incident electrons. At 4.8eV the energy of some of the scattered electrons falls to zero. Which of the following statements best explains this observation?
The answer is...
Inelastic collisions within the atoms can occur for electrons of energy 4.8eV, but not at lower energies.
Could someone please explain this to me.
For electronics...
If a diode with a turn on voltage of 0.7V is in parallel with a 6.0V battery as well as a resistor, what would be the voltage drop across the resistor? Would it matter whether the diode is in forward bias or reverse bias ?
And for electric power...
Also, are we meant to know how to work the EMF generated from a moving object in a constant magnetic field using EMF=BLv as it was in the STAV 2013 practice exam?
Thanks
As for electric power...it's just a formula, another what, two lines on your cheat sheet? Can't be that bad.
That's literally 7 characters (EMF=Blv)
The word commutator takes up more space than that lol
TSFX Exam 2 2010, Light and Matter question 3. Basically, photoelectric effect measured on a metal with blue light and with ultraviolet light. The graph of current vs. potential difference shows that blue light produces a higher current for the forward potentials.
Question 3 asks "which of the lights had a greater intensity?" with options "blue," "ultraviolet" or "unable to determine." The answers say "unable to determine" because the intensity of the light isn't the only determining factor in the current - higher kinetic energy will also make a higher current, and this depends on the frequency of the incident light.
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
I chose "blue", because the ultraviolet light will have higher energy photons - E=hf. It would need more photons to equal the photocurrent of the ultraviolet light, let alone produce a larger photocurrent. Right?
Yep along with thousands of other physics students.
I've got a fair few friends who have physics as their last exam as well.
We're all planning on keeping our cool together after they say pens down in the physics exam (we don't want to be reported to VCAA or anything haha) then we're gonna go absolutely ballistic outside.
Planning on getting smashed on the footy oval, hopefully it's a nice day as well considering it'll be halfway through November!
for me its gonna be a full day of sleep and PS4 session.
Just got some electricity questions from the VCAA 2007 exam
Questions: http://imgur.com/a/UhAma
Answers: http://imgur.com/wab3ASk
1) How come the graph needed to be inverted?
2) How come it clips at 3.0 V?
Thanks
Edit: Also, it's fine for explanation questions to be answered in dot points, right? Or are we required to write full sentences?
Lucky. For me, after the physics exam, all I can say is three down, three to go.
Oh well, another thing to add to my cheat sheet when nliu answers haha
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
In sorry to disappoint for I'm not nliu (@ nliu you have no idea what autocorrect on my iPod suggested for your nick LOL)
1) Anonymiza youre actually doing a paper from an older study design when students where actually required to know how things worked D; D; D; IKR THE SHOCK
2) Most simply put, what you have us a single transistor amplifier. Because you have an NPN transistor, when a large input signal comes in through the base, a "smaller" output voltage signal I recorded. Note if it was a double transistor amplifier then it would be non inverting
3) I really don't want to go into capacitors specifically, only because the textbook brushes on n-p and p-n junctions (not in stuy design) but no capacitors or the saturation point or cutoff point of transistors.
it's suffices to say you WON'T get a question like this. it's nice to know how an amplifiers work so you can answer questions like this, but VCE physics just doesnt require this standard.
Aww English Language is the day after and you still have Uni exams?
But look at the bright side, you'll be able to say three 50s down (the way you demolish those spesh, chem and physics questions... far out lol), one 45+ to go and the uni stuff doesn't really even count haha.
I believe both are acceptable, I usually just write or whatever the gain is if it is inverting, otherwise just the number.
Again, I'm not strictly certain on this, but I usually do dotted vertical lines. I doubt they are pedantic enough to pick up on either of your two questions though :)
In the VCAA 2009 paper there was a 3-mark question:Assessment Report Comment:SpoilerAt the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."Is a response like that enough for the whole 3 marks??SpoilerYoung’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Also in general, for some 2-mark questions such as "What is the power dissipated in R_{1}?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
Personally I would add what the particle model would predict for the experiment, and perhaps mention that nature of the interference pattern actually observed as well (i.e. alternating dark and light fringes). I do think that some examiners would only award you two marks for the response given there by the examiner's report...
Why not just write down the working to be safe? Usually you'll only need one or two very short lines =) You'll also minimize the chance of making a silly mistake, as well as enabling the examiner to give you one of the two marks if you get the answer wrong but have valid working.
In the VCAA 2009 paper there was a 3-mark question:Assessment Report Comment:SpoilerAt the time of Young's double-slit experiment there were two competing models of the nature of light. Explain how Young's experiment supported one of these models compared with the other."Is a response like that enough for the whole 3 marks??SpoilerYoung’s experiment demonstrated interference effects, so his work supported the wave model of light. The particle model did not explain the interference effect.
Also in general, for some 2-mark questions such as "What is the power dissipated in R_{1}?"
Obviously it's expected that there's some working out involved. But say you did the working out straight into your calculator, and got the right answer (no working out written down). Would you still obtain the full marks? (like for multiple choice questions)
Explain the wave model of light and what it predicts (interference pattern of light and dark) | 1 Mark |
Explain the particle model of light and what it predicts (no such pattern, just one light band) | 1 Mark |
Young's DS Experiment supported the wave model of light because a fringe interference pattern was observed on the screen | 1 Mark |
Correct formula with wrong values or wrong conversion substituted in | 0 Marks |
Correct formula with right values substituted in but wrong answer | 1 Mark |
Correct formula with right values substituted in and correct answer | 2 Mark |
No/wrong formula and wrong answer | 0 Marks (no duh) |
Wrong formula/working but somehow correct answer | 1 Mark (you showed bad physics in your working) |
No formula and correct answer | 1 Mark I'm pretty sure for 2013 because for short answer questions it says: |
"In questions worth more than 1 mark appropriate working should be shown." |
You have to remember that most assessment reports are brief because they don't want students to memorise answers from the reports. So, they just touch on important points
"In questions worth more than 1 mark appropriate working should be shown."
where does it say that?
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it."You should clean your room"
I had always thought the fact that is explicitly states "should" rather than the "must" of Methods and Spesh exams suggests that they want you to but they can't actually mark you down for it.
"You should clean your room"
"You must clean you room"
Either way you get yelled at if you don't do it :P
But seriously, better safe than sorry. Plus, what if you did the question all on your calculator get the answer wrong. Don't even get any marks for method or working. It's just safer this way :))
I always showed full working last year, but if someone on AN wants to try it with no working at all this year be my guest and let us know how it went :) (no, this is not a personal challenge at you nliu :P)
in the 2013 trial paper is says:
(http://i.imgur.com/9KVgf3d.png)
I've seen those questions before in past exams. Given the cutting of the physics course down, it's possible that you're not asked to know this.
In case you do, something on the lines of the thermal vibrations of the electrons creates an electromagnetic wave, light, would probably suffice.
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
To nliu and Alwin and potentially others, I got my hands on the VCE Physics Exam marking principles today. No working, correct answer = full marks regardless of number of available marks.So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.
Some other absolute gems:
"If the student changes the units in the answer box and his answer is correct, give full marks."
"Scientific notation – 2 E 9 is acceptable"
So if there are 3-4 marks devoted to a question, we can full mark it simply by writing the answer? Source please.
Why would anyone put just an answer when there are multiple marks allocated?I'm the type of person that will show working even if the questions is worth 1 mark, there's no way I'm ever gonna write an answer without the working shown. I was just curious about whether the examiners have concrete rules for allocating marks
If you get the answer wrong and have no working out, then you get zero, it's as simple as that.
However, if you have the incorrect answer but have working out is certainly possible to obtain marks.
In a subject where you hear numerous stories of people mistyping things into their calculators and obtaining an incorrect final answer, I really don't understand why you would simply write an answer and nothing else, especially considering it takes, what, an extra 5 seconds to write? ???
In order to replace the fuse as safely as possible, which of the following is the best precaution for Joan toThis is just my thought process.
take?
A. stand on a rubber mat
B. switch off the mains supply
C. disconnect the transformer from the mains supply
D. remove the load from the transformer
Why C instead or B?
Photonics (detailed study) question,Bumping this
Could someone please explain the what "numerical aperture" and "acceptance angle". I can usually get MC questions related to them right by applying the formula but I have no genuine idea what the numbers even mean.
Cheers
Our equation for the extent of diffraction is where w is the width of the slit.
We can see that as we increase w, becomes smaller, thus less diffraction occurs.
As we decrease w, more diffraction occurs as is now bigger.
This also explains why we need to use very thin slits/apertures for this experiment, as if they are too big any diffraction is not evident.
what about the intensity of the pattern? and what effect does it have on the fringe spacing?
Dont be smartass alright. It was a genuine question.Sorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
i just got the physics marking scheme from my teacher. it says "where questions dont specify that the working must be shown, if the correct answer is in the box the student receives full marks". Maybe you should get some suffice information yourself before absolutely classifying other peoples question invalid and, from the sound of your tone, stupid.
offtopicSorry lolipopper if I made you feel like you'd asked a "stupid question" or anything, I didn't mean it in that way. I was just joking with you, about the "read the instructions first" part I put in the picture. I haven't been on this thread in a while, been rather stressed for Indo oral exam so again I apologise for being or seeming a bit short with you.
I was just showing the source of my "information". I can only talk about the pre-2012 exam marking scheme, and as Tim...blahhh has confirmed from more accurate sources Re: Physics [3/4] Question Thread! it turns out I was wrong. Not going to blame the practise exam, it's just that it turns out I was wrong and I'm sorry if my response was inconsiderate.
Good luck to everyone with physics :)
With questions talking about the extent of diffraction: do we assume that max. diffraction occurs when wavelength/width ratio ~ 1 or when this ratio is a large value?
The corresponding constant DC voltage that provides the same average power output.
what is a good answer to the following:
-what is modulation and demodulation?
Honestly mate, I don't fully understand half the crap we're learning in physics and I got 96% on that sample VCAA 2013 exam.
All you need to do is sub in values.
We know speed = distance/time so therefore time = distance/speed = 0.02/0.04 = 0.5, all values directly from the question subbed into our formulas from our A3 cheat sheet.
Then sub time = 0.5 into the EMF equation and wallah, you get the answer.
VCAA 2010 Q16:
Basically a moving trolley with a spring on its front collide into a stationary trolley.
SpoilerI think this is quite a difficult question to grasp conceptually, and difficult to answer without simply parroting the line that momentum is always conserved. But essentially what is happening is that forces are acting on the spring in both directions, and the spring is exerting a force on both trolleys. Because momentum is a vector quantity, that is, it has direction, exerting equal forces in opposite directions results in no change in momentum. The problem is it is a little more complicated than that, as the spring would not exert equal forces on both the trolleys, but essentially what would happen is that it would exert a slightly greater force on the stationary trolley than the moving trolley behind it, and the 'unbalanced' force actually contributes to the change in momentum of the spring itself. I expect that VCAA might expect you to briefly touch on some of these points and simply mention that momentum is always conserved.
I've talked about both forces on the spring and forces on the trolley, and used these interchangeably as they'll be of the same magnitude as they're Newton third law pairs, but it can make it a bit confusing.
The square doesn't need to be in the middle; it just needs to be in a position such that the entire square is in the magnetic field. This is because the magnetic field is uniform.
Remember, the formula flux = BA refers to the area that is exposed to the magnetic field.
Now, the maximum voltage induced would then be due to the change in flux from when the square first enters the field to when it is fully immersed in the field. The side length is 2 cm, and the speed of motion is 4 cm/s, so the time we want is 0.5 s.
Just one loop only.
The change in flux is 3.7*10^-3 T * (2 cm)^2 = 14.8 * 10^-7 T m^2 (note change in units)
So dividing by the time, our max voltage is 29.6* 10^-7 V, so 2.96 uV which is the unrounded form of the answer given.
The trick is identifying what the time means. The time is the period of time over which the change in flux occurs. Note that when the square is in the middle of the magnetic field, the entire square will have remained in the field for a period of time, so the flux isn't changing; the voltage is zero in that case.
Another example of how physics is becoming ridiculous; you don't even need any understanding of the mechanics of the formulas to get the marks.
But of course, the force accelerates the trolley, which increases its velocity, reducing its speed relative to the spring, reducing its compression, which I think is what you're trying to say?
The answer is A, but shouldnt it be the other way around? SO negative current at start then positive? (because of the EMF being (-) change in flux?) ThanksI don't think it matters, remember the motion has to be relative.
How do you do this question?Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.Spoiler(http://i.imgur.com/RXyjDio.png)
What happens between the two?
How do you do this question?Spoiler(http://i.imgur.com/RXyjDio.png)
What happens between the two?
I don't think it matters, remember the motion has to be relative.Remember when there are two South poles to draw a dotted line (similar to an asymptote in maths) between the two. Then draw your field lines going from the two south poles and behaving asymptotically with the line you have drawn. I am pretty sure you can also draw some conventional lines from the north pole to the south pole as well. It's pretty hard to describe so I am sorry if you do not understand. Maybe someone can upload a picture.
its explained in the pic
The difference in the questions is that the question in the 2008 VCAA exam states that the field can be considered 0 outside the poles, while it cannot be assumed so in the 2011 paper. In terms of realism, the curved graph is a more accurate representation, but sometimes for simplicity's sake, we make assumptions like no field outside the poles.
When drawing a circuit, how do we know when the switch has to be opposite a thermistor or LDR rather than a variable resistor (or vice versa)?
you have to see when the heater must be turned on. If it is to be turned on when the voltage increases then against the element whose resistance increases (Voltage increases across it as well) and vice versa.
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.
The answer you have gotten is correct. One upside of physics is that there often aren't very many technicalities to a question, it's just a matter of plugging numbers in.
Lets be honest here, this is VCAA style physics.
The only thing that has some degree of 'better watch out for that' I find is light.
Motion is essentially 'I've got these letters, let me look at my cheatsheet to see which formula has them.'
Electronics is 'can I read a graph, a formula and use V=IR.'
Light actually takes some learning, but only because VCAA tend to ask more theory orientated questions.
This satellite question is driving me crazy!!
Q8 of motion in the 2013 sample exam by VCAA
R=1.35*10^7m
G=6.67*10^-11
M (earth)=5.98*10^24kg
m (satellite)=525kg
What is the period?
Both Vicphysics and Itute answer say it is1.53*10^4s but whenever I do it I always get 1.56*10^4s. It is so close yet so far away, could someone please test this question, I know how to do it but it just isn't working.
Does it matter where you place the diode in a series circuit?Current would first pass through the resistor, resulting in a drop, and THEN it'll be stopped by the diode in reverse-biased.
For example, you have the positive end of supply connected to reverse biased diode then diode connected to resistor then resistor connected back to supply.
The entire voltage supply drops across the diode as no current flows through it hence no drop across resistor.
But how about if the position of the diode and the resistor are swapped? Would the same thing happen?
What happens to the current that flows across the resistor first?
But what I'm asking is what if the current flows through the resistor first, then the reverse biased diode stops the current flow?For the current to flow, you would need a complete circuit (ie doesn't flow in one half then stops)
how to do 1b from 2012 Exam 2? About the orientation the magnet will take?
bumb :)
Figure out the where the 'north' and 'south' end of the solenoid is. Your smaller magnet will act as a compass needle which(without any other magnetic fields) points at magnetic north. Placing the magnet near point Q and allowing it freely rotate means that the north end of the magnet will rotate to face the south end of the solenoid.Spoiler(the answer is C)
also just to be 100% sure, we can take in 2 double sided a4 sheets into the exam yeh? :)
stuck together? how do i do that? you can't bound 2 A4 sheets can you? would stapling them together work?
"when the switch was closed a magnetic field built up to the left. To oppose this, the induced current must produce a magnetic field to the left."
Yeah must have been a typo, without looking at the question id say the flux increased in the left direction, and so to oppose this, a current was induced that had a flux in the right direction, then do the right hand rule to find current flow (guessing its a solenoid question or something similar)
hope this helps :)
Highly doubt there would be an error in VCAA solutions. If something's wrong they will revise it.
Okay, so when the switch is closed, the current is increasing, creating an increasing magnetic field. The field's direction points towards the left. At the second coil, the direction of the magnetic field is also pointing towards the left. The current induced in the coil will be such that it opposes the change, hence we want to induce a field towards the right. Creating a field pointing towards the right requires a current to flow from Y to X.
The current only flows momentarily, as once the current maintains a steady value there will no longer be any change in flux and hence no induced current.
So the answer is B.
North pole moves to left=>induced magnetic field points to the the right=>use right hand rule, current is from q to p in the EXTERNAL circuit, but p to q through the ammeter.
Having some difficulties with lenzs law.
A loop is sitting in a magnetic field that is going out of the page, it is quickly pulled to the left. Which direction will the current flow (clockwise or anticlockwise)?
Looking at the left hand side of the loop, you can use the right hand slap rule to determine the direction of the current( fingers going out of the page, palm to the right to oppose the force to the left, current will be clockwise). However, if you do the same analysis on the other side, the thumb would be pointing the same way so current will be anticlockwise. Which one is correct and why?
I guess?Ah okay, what I was trying to ask was whether *only* mentioning interference would suffice, or is it better to also talk about diffraction.
The thing is that it's a double slit diffraction experiment that you're being asked about. They've given you two slits, why only talk about single slit diffraction? It's clear that the question is directing you to talk about interference patterns, and if it asks you specifically what Young's double-slit experiment shows, the most notable thing about it is the interference pattern it shows. If you wanted an experiment to show simple diffraction of light, you probably wouldn't use Young's double-slit experiment.
I'm not sure what point you're trying to make about electrons and de Broglie wavelengths. The fact that electrons can diffract, and also interfere, speaks about the wave nature of matter, not the nature of light.
For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.
*begins rant about VCE physics
Umm, no, there will be more diffraction if wavelength > slit.
Could someone verify this?
I'm a bit worried now after reading this
And I don't understand any of nlius explanation
I haven't read anything outside of nliu1995's points, but they seem to make physical sense.
Basically, if you have a wavelength/slit ratio greater than 1, you don't get any minima and maxima forming, but instead just have a complete diffraction of the light source, which I'm assuming means an equal distribution of light intensity in all directions.
I should clarify that, obviously you can't have more than 90 degrees of diffraction, which occurs when wavelength approaches the slit width. If wavelength is greater than the slit width, you will get 90 degrees of diffraction as well. So if wavelength is greater than slit width, you will get the maximum possible amount of diffraction occurring.
For diffraction through a single slit, the angular positions of the intensity minima are given by
where m is a constant (not zero), d is the slit width and theta is the angle the line from the slit to the minima makes with a line parallel to the slit opening.
As you can see, if the wavelength approaches the slit width, the first angle approaches ninety degrees, which means the first intensity minima spreads out completely. If the wavelength exceeds the slit width, then you will not have any minima and the wave will spread completely.
*begins rant about VCE physics
Numerical approximation of diffraction pattern from a slit of width equal to wavelength of an incident plane wave in 3D spectrum visualization: | Numerical approximation of diffraction pattern from a slit of width equal to five times the wavelength of an incident plane wave in 3D spectrum visualization | Numerical approximation of diffraction pattern from a slit of width four wavelengths with an incident plane wave. The main central beam, nulls, and phase reversals are apparent. |
(http://upload.wikimedia.org/wikipedia/commons/thumb/4/46/Wavelength%3Dslitwidthspectrum.gif/220px-Wavelength%3Dslitwidthspectrum.gif) | (http://upload.wikimedia.org/wikipedia/commons/thumb/0/0a/5wavelength%3Dslitwidthsprectrum.gif/220px-5wavelength%3Dslitwidthsprectrum.gif) | (http://upload.wikimedia.org/wikipedia/commons/thumb/3/3c/Wave_Diffraction_4Lambda_Slit.png/220px-Wave_Diffraction_4Lambda_Slit.png) |
nliu, they kinda sorta mildly touch this in Synchrotron detail study, but I agree. Explained muchhhh more in depth in A level (eg singapore) or highschool equivalent in china and other asian countries.
Anyways, for those interested from page 496:
A collimated beam of X-rays incident upon a layer of atoms will be scattered. Bragg’s law states that the beams will interfere constructively, producing maxima when the following relationship is satisfied:
2dsinθ = nλ where d is the distance between layers of atoms (m), θ is the angle the X-ray beam makes with the surface, λ is the wavelength of incident X-ray photons (m) and n is the number of the maxima occurring, 1, 2, 3, etc.
What nliu is trying to say is that as d (which can be considered as the slit width) and λ (wavelength) are approaching the same width, ie d/λ -> 1, the equation becomes:
2λsinθ = nλ -> sinθ= n/2.
n=1 is the centre band, so we look for n=2, the band next to it:
sinθ= n/2 -> sinθ= 2/2 -> θ= 90 degrees.
This implies that the wave has defracted so completely that the next band occurs at right angles to the path, ie parallel to the screen.
So, using a "similar" version of Bragg's Law (sadly barely touched upon in VCE) the max defraction occurs when the ratio of wavelength to slit with approaches one.
So essentially for this exam (which is all I care about):
- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction
Correct?
So essentially for this exam (which is all I care about):
- lamda/width < 1 = not a lot of diffraction
- lamda/width = 1 = max diffraction
- lamda/width > 1 = max diffraction
Correct?
See, in the diagram with w=d, there is a beam that spreads across the entire wall, whereas with the third one, the most intense part of the wave is quite limited in range.You referring to the ones in my post?
What happens to magnetic field (B) when the current increases, and which equation tells us thisMisread question.
Magnetic field strength is directly proportional to the current.
What equation tells you that?
Biot-Savart law:
Oh wait, not part of course. My bad.
So we can see that as the current increases, the magnetic field needs to increase or the time needs to be reduced.
This is only for an induced current, not for currents ttat create magnetic fields
Wait, so if a question talks about doubling the current how am I meant to know the effect on the field?
Also, I came across a question involving two parallel current carrying wires, where the distance between them was reduced by a factor of 2, In the answers it said field varies inversely with distance squared. Again how am I meant to know this?
It's a parabola; the particle's horizontal velocity component is constant but the vertical component increases in magnitude. Just draw something like that and I think you'll be fine.
thanks but are these types in the course, because it similar to relative motion using frames of references and stuff?I think it is sort of on the course since you have to be aware that for projectile motion, the horizontal component of the velocity is constant
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.
Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)
I'm so confused.
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
http://www.youtube.com/watch?v=Iuv6hY6zsd0
Henreezy, the reaction force rule is given by
F(a on b)= -F (b on a) that is, newton's 3rd law
From this, we can sort of sub in and solve so we get
F(earth's gravity on object)= -F(object's gravity on earth)
So what actually occur is the gravitation force from the ball actually pulls the earth up but since the earth is so massive, nothing is observed on the earth's behalf.
For young's double slit experiment, does it matter if the light source is coherent or not to achieve interference? I did an exam practice by ITUTE which said that using torch light would not allow interference to occur but in this video by veritasium, he achieved interference with sunlight
http://www.youtube.com/watch?v=Iuv6hY6zsd0
Question 4 b) VCAA 2012 Exam 1:
Is the tension force referred to as the 'gravitational force - upwards' due to the fact that the reaction is due to gravity? I don't know and that confuses me.
Action: Gravity 'pulling' it downwards
Reaction: Gravity 'pushing' it upwards (results in 'tension' from string?)
I'm so confused.
Can someone help me with the concept in part of b of this question?
One (vce level reply) would be to say that electrons need to form stationary waves to be stable so the orbit circumference has to be a whole number multiple of the wavelength of the electron's de broglie wavelength.
If an electron orbited at radius such that the circumference was not a whole number multiple, destructive interference would occur and the electron would 'drop' to a lower stable orbit.
Thus, only certain energy levels, orbits, are possible for the electrons of the hydrogen atom... or something along those lines
inb4 nliu comments how they're not levels so much, but bands but that's beyond vce
Is diffraction stronger when wavelength equals slit width or when wavelength is greater than slit width?
definitely noticeable diffraction at 1 but vcaa will either give you something obviously small like 0.000001 if theres no diffraction
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA
For light and matter, do we need to know the required voltage to accelerate an electron? I have been given heaps of question related to it in private companies exams (particularly atar notes) but not VCAA
It's not hard... but I'm pretty sure that that sounds like a detail study (synchrotron irrc) question not a core studies question...
Pretty sure it's not something you need to know.
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy. Use E=Vq if you ever come across it.
Technically, in the photoelectric effect, if you have electron energy=hf-W, then the stopping voltage would be 1/q*(hf-W) which explains the gradient of the curve. It's not entirely out of the course.
The definition of voltage is the potential energy difference per charge. So, a 1 V potential energy difference would give a 1 C charge 1 J of energy.Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?
Thanks nilu, I understand that part it is just the random formulas out there for accelerating electrons from close to rest to form a diffraction pattern similar to x-rays, then work out what voltage is required to accelerate them. Just want to know if they are on the course?
I have a fairly good idea in deriving the formulas but if I do have to know it, it would be easier to just plug and play VCE style
-Identify rms voltage as an AC voltage which produces the same power in a resistive component as a DC voltage of the same magnitude
-explain the production of atomic absorption and emission spectra, including those from metal vapours?
I don't understand these 2 dot points, could someone elaborate further?
hey guys, did we have to know the skin effect and optical fibres and how they work? I don't remember learning about it at all and haven't seen any on the exam.
It's been put into one of the detailed studies :) not in the core section of the exam
Hi guys, on the 2011 Exam 2 Electric Power Q.11, I've been talking with my friend from school and we can't figure the graph out. We looked at the VCAA solution and we get what they're saying but we were thinking something else - I put our train of thought in the attachment below.I'm a little confused but I think I see where you're coming from. The only thing is that the flux doesn't reduce when the magnet is inside the loop. Flux still "runs through" the inside of a magnet, and the field lines are still there when this magnet is in the centre of the loop.
Pretty much it's that since the flux will reduce once the magnet is actually inside the loop shouldn't that cause some emf generation, spiking in the opposite direction? The attachment will make what I'm saying clearer hopefully (even if it is nonsense lol)
Any advice would be great, thanks :)
If anyone is here :)
PLeaseee:
A beam of red light of frequency 4 × 10^14 Hz is found to deliver 7.54 × 10^19 photons per second.
What would be the power of the beam?
If anyone is here :)
Can the year 11's moving into year 12 use this thread next year? Or do we have to make our own, new thread?
A question says: "A spring has stiffness 20 N/m and is hang vertically. When a mass is attached, it stretches by 0.2 m. Assuming that the spring has no mass, what is the value of the mass."
What I did was that:
change(Elastic E)=Change(Gravitational E)
1/2kx^2=mgh
1/2(20)(0.2)^2=m(10)(0.2)
m=0.2 kg
But the ans says to use kx=mg, and the ans is 0.4 kg.
I am thinking that since the F due to gravity on the mass is constant, the 1/2kx^2 should not be used. Is that the reason why my method gave the wrong ans?
Can I pls have help with the following questions?
1) An astronaut standing on the moon experiences a gravitational force of attraction of 160N. He moves away from the surface of the moon to an altitude where the gravitational force is 40N.
a) How far from the centre of the moon is this new location in terms of the radius of the moon?
b) The astronaut now travels to another location at a height of three moon radii above the surface. Calculate the gravitational force at this altitude
Cheers ;)
Yes, KE final = KE initial + change in GPE, then solve for speed.
That only works if the person does not add further energy. If it is in the vertical plane, tension at the bottom will be greater than the tension at the top because of the weight force acting against the centripetal force. Drawing force diagrams with the directions of the force will help you understand it.
At the top: centripetal force = tension + weight
At the bottom: centripetal force = tension - weight
There's not enough information. You haven't been told about the speeds and angles of each missile.
Can someone help with this question please? I'm not sure how to rearrange the circuit. The answer is supposed to be 83.3 ohms but from the circuit I drew I get 116.7 ohms.
Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?
Help would be much appreciated. Thank you :)
I have a few questions:
1. Describe the origin of the centripetal force that causes an object to follow a circular path.
2. I want to know whether I've approached this correctly.
Saturn has at least eighteen natural satellites, two of which are Titan and Tethys.
mass of Titan = 1.35 x 10^23 kg
radius of Titan = 2.6 x 10^6 m
period of Titan's orbit = 1.38 x 10^6 s
radius of Titan's orbit = 1.22 x 10^9 m
mass of Tethys = 7.4 x 10^20 kg
radius of Tethys orbit = 2.9 x 10^8 m
Calculate the gravitational field strength of the surface of titan.
g = GM/r^2.
G = 6.67 x 10^-11
M = 1.35 x 10^23 kg
r = 2.6 x 10^6 m
Is that value of r correct? Which radius do I use? How do I know which one to use in a case like this?
Help would be much appreciated. Thank you :)
For the first part to your question: the centripetal force is always supplied by a real force. This could include the tension in a string or the frictional force between the tyres of a car and the road.
Let's first take the example of the tension in the string scenario. Say if I have a bucket of water attached to a string, as I begin to swing it in the circular path the tension will act towards the center of the circle, which is by consequence the centripetal force.
The second example is one that has popped up a few times on exams. So consider a 'normal' road; that is, one that hasn't been rained on or has much due so it's pretty dry. If a car wanted to go in a circular path around this dry road it would be quite possible. This is because there is a frictional force between the surface of the road and the car tyres which acts towards the center the circular path, and this too is the centripetal force. Often in calculations on horizontal surfaces you will denote this as the net force.
Caution: One of the reasons why you can't maintain such motion on an icy track is because there is little friction between the car tyres and the surface of the road. So as you speed up it becomes difficult to control the car and the car will leave the circular path at a tangent to the path.
Yes that is the correct value for the radius, the gravitational field strength is calculated purely on the object in question. Any gravitational field on the surface of a planet in the solar system (Earth) will also have external gravitational fields acting upon them as is the nature of matter in the known universe. The question is a bit ambiguous, however I believe what it's asking is what you have done. :)
Does anyone have any good ideas for EPI on electronics? I want to do somthing awesome/interesting :)Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.
Maybe you could investigate factors that affect a modulated signal. Like for example my friend was telling me that the radio in his car always gets distorted when a motorcycle is near his car, maybe because of the fact those 'drumming' noises are interfering with the modulated signal being formed by the carrier and input signal.
If we were to do another EPI I would base it on that, looks so fun.
Good luck Rishi :)
Yeah that does sound fun Rod. But we haven't learnt anything about modulated signals at school. So would I still be able to do it?Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.
Yeah you should be able to, it's unit 3 area of study 2 stuff, but confirm with your teacher just in case.
All my teacher when through was 'a low frequency input signal in increased by a carrier signal, and then we get our modulated signal.....'. And showed us one diagram. So yeah lol he didn't go through it with us a single bit but we still understand it.
It's just a suggestion, I would probably do it.
Best of luck!
how would you do these questions
Calculate the energy in electronvolts of
a)an alpha particle with 8.5*10^-12 J of energy
b)a beta particle with 6.4*10^-11 J of energy
c)a gamma ray with 4.7*10^-11 J of energy
ok so I got my physics EPI project info today and I found out that I have to design a circuit using either resistors, led or diodes. We had to buy this short circuit kit from jaycar and our EPI can only be made using the components which are spring connectors, battery holders, and all the basic electronic components.What does 'all the basic electronic components' actually entail?
I have no idea what to investigate. Any ideas would be greatly greatly appreciated :)
A massive thanks :D Pls help me
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)
I asked for clarification just because it didn't seem like much sorry :).
Maybe a row of LED's with different brightness? Honestly I can't think of much you can do with that (at this level).
The effect of doubling resistance on brightness or something? Or different configurations of resistors eg series/parellel
Hey guys,
bit stuck on this question:
In a laboratory class at school, Lee is given a spring with a stiffness of 15.4 N m–1 and unstretched length of 0.40 m. He hangs it vertically, and attaches a mass to it, so that the new length of the spring is 1.10 m when the mass is stationary. Assuming the spring has no mass, what was the value of the mass he attached?
Thanks!
I got 0.18kg. Do you have the answers?
it's 1.08.. :/
I'm getting a different answer of 1.72...Looks alright?
The mass will pull the spring down with a force equal to its mass * gravity.
Therefore the force exerted on the spring is
We also know that
Therefore
Substituting in and rearranging gives 1.72 which is wrong.
(I've left out all the negative signs so this isn't strictly true.)
Looks alright?
If we define downwards as being positive, our change in x (of the spring) will be a positive value as well as our acceleration of gravity so I don't think we'll encounter any negative answers.
Diodes, resistors, LEDS, globes (it has more things but we are allowed to only use these)
3Wm^-2 =3000mWm^-2
Does anyone know any good youtube channels to watch vce physics 1/2 videos?
Is anyone going to the TSFX Physics lecture this weekend???nah, are you?
nah, are you?
Quick question:Spoiler(http://i.imgur.com/usMS3pS.png)
Generally with these graphs, to find gain must you find the gradient (rise/run) or can you just go Vout/Vin for any value? Because I was taught the latter but for this question that is wrong (must use gradient).
It would be illogical to have the two signal waves looking like what the answer has shown you because they would simply cancel each other out upon wave summation. Thus, the 'modulated wave' and the carrier wave would look exactly the same. Which is why I don't understand what they are trying to say there...
So how come there are "two" signal waves?
Silly mistakes will be the death of me in this subject. Argh.
Anyone else doing Materials and Structures for their detailed study? It actually looks like a decent detailed study. We then have to do data analysis on materials and structures for this SAC.
Is unit 4 Phys more difficult or easier than U3?
We're doing materials and structures for our detailed study as well.. although, I would much prefer to do sound. It 'sounds' more interesting. hehe lol
Ok first things first, just do me a favour and completely ignore the answers because I think it is wrong. Either that, or I just don't understand it.
Now, this is an example of amplitude modulation, not frequency modulation, I think we gathered that much.
Secondly, I think the carrier wave is a monoamplitude sinusoidal wave with the same frequency as the mondulated wave.
Thus, the signal wave would have to be the other component of the modulated wave. And the only other wave that can summate with the carrier wave I just described to give you the modulated wave shown would be the positive sin looking graph in the answers (not the negative sin graph).
Why? because only a maximum on the signal wave would boost the positive and negative amplitudes of the carrier wave to give you that first section of that modulated wave. and only a negative minimum on the signal wave would squeeze the positive and negative amplitudes of the carrier wave to give you that section at 2ms of the modulated wave (that part which looks like a noose had been wrung tightly over the wave)
can someone help me with this question 886?
thanks in advance
for units 1/2 this year, i scored 92% on my radioactivity sac, 100% on Flight (dual highest) and just recently electricity 98% (highest). Despite this, i have chosen to drop physics, im not enjoying it but i am getting really good marks, if you were in my position what would you do? do you think ive made the right decision to drop out of it for Accounting.
how and what is the answer to this question, from electricity sac.They need to be a multiple of the elementary charge, which is approximately (I can't remember how many sig figs you do it to in physics but it doesn't matter here).
Thanks
how is this derived?already forgot about this. had to refer to my year 11 textbook. Sv is the unit for dose equivalent while Gy is the unit for absorbed dose.
Hi guys,
I'm really stuck with a voltage divider question, I thought I had them down but this one is confusing me.
It's question 32 and 33 of A+ Notes for physics.
I realise the photo isn't clear - Change in Vout = 6V and change in Vin = 0.4V
So here's the answers from the book and my answers afterwards:
Answer to Q30 is 0.6V and 4V - no worries.
Answer to Q31 is 15V - wtf? Isn't it negative 15V because the amplifier is inverting?
Answer to Q32 is 2.5V - That's reading from the graph but when you do Vout = Vgain * Vin = -15 * 0.7 = 10.5V, it doesn't agree with the graph. Can anyone explain why this is the case?
Answer to Q33 is 5.5V - Same problem as Q32.
Thanks,
Stewart
When we're calculating the co-efficient of friction, formula Fr= m x N
where m is the co-efficient of friction,
What unit do we attribute to N, the normal reaction force? Do we use the mass provided or do we convert it into newtons?
So how should I approach this question...
I get a result that's 10x more than the actual answer :/
As we are told the system is moving left, let left be positive. Let the tension in the left rope be and the tension in the right rope be .
Draw the forces the acting on each block.
For M1: mg is down and T1 is up
For M2: mg is down, N is up, T1 is left, T2 is right, friction is right (as we are told block is moving to the left)
For M3: mg is down, T2 is up
Since acceleration of whole system is the same, we can apply Newton's second law to each block individually.
For M1:
--- Eq. 1
For M2:
You also know that
Thus:
--- Eq. 2
For M3:
--- Eq. 3
Adding Eq. 1 and Eq 3. gives:
--- Eq. 4
Adding Eq. 2 and Eq 4. gives:
(negative sign shows that it's decelerating)
Hope that helps.
I'm sorry, i'm not sure what you did wrong but apparently that wasn't what the answer is...I'll give this a shot. Hopefully my explanation makes sense.
Solution for the question was 5.6 ms ^-2
Thankyou for your help guys
but in my book it says displacement=final position-initial position
in this case it would be 30-50=-20 which would be south
Can anyone explain how this works
- - - - - >
<- - -
So he moves 50km right (north) then 30km south. His initial position is 0, then +50, then he moves back 30km so his final position is +20km.Can anyone help with this questionWell the instantaneous velocity would require you to find the gradient at point t=35s. So that requires differentiation - which is basically impossible because you have no formula to differentiate!
Thanksseeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement). You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.
how would you do this question
A lamp purchased in the USA is designed to operate at optimum efficiency when it is connected to an AC supply that has a peak voltage of 170 V and a frequency of 60 Hz. The lamp has an operating resistance of 100 ohmsJust wondering what the answer was?
What is the peak power produced by this lamp?
I know the peak power = I peak x V peak but this isn't working for me
Thanks :)
Just wondering what the answer was?
289 WYep, I am not sure what you did wrong but your formula was right. Pp=VpIp
Yep, I am not sure what you did wrong but your formula was right. Pp=VpIp
we have Vp so we need to find Ip. Ip= Vp/R =170/100 =1.7
Pp=1.7*170 = 289W
seeing as you are going from v/t to pos/t graph this time. ANTIdifferentiation is required (area under the graph=displacement). You can figure out the v-t graph formulas this time as they are linear so it is possible to antidiff.
Thanks for your help but i still dont understand what will be my position points for my position/time grapgh and how do i work them outOk, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
Ok, I just realised that you don't even need to antidiff any formulas. You can just count the squares - its the same thing...area under the graph and all.
So at t=1 the area under the graph is 1m/s x 1 s = 1m (seconds cancel out to give you metres - the position!). t=2 ----> 1m/s x 2s = 2m. t=3 ----> 3 and a half squares so 3.5m So on and so forth.
Just find the area under the graph at each second and plot those values on the Pos/time graph. Then connect the dots. Do you understand how that works?
yes i do thanks so much :)
What is the difference between a slip-ring and a split ring communtator?
Which is better and why?
The acceleration is given by the gradient, I think. So, if you can find the gradient of the line (generally found by drawing a straight line at the position you want, ie. 10s and for the second question 40s) then you can use the rise over run formula to find the gradient which gives you the acceleration.
eg. the acceleration at 10s I would say is 20/20 = 1m/s^2.
This is described here (scroll down to finding the gradient of a curve): http://www.mathsrevision.net/gcse-maths-revision/algebra/gradients-and-graphs
Does that make sense? Hope it helps,Stewart
how would you do these questions
What is the acceleration of the train 10 s after starting?
What is the acceleration of the train 40 s after starting?
Oh and Rishi, for your split/slip ring question. The full definitions can be found in my cheat sheet under the section Electric Power. And neither is "better" per se, rather they both have different functions.
SLIP rings are used to "collect" the AC current generated from generators while SPLIT rings facilitate the constant rotation of motors.
You should try to understand how these rings achieve those feats though, and not just memorise the answers :)
With velocity-time graphs, acceleration can be calculated from the gradient.Hello there,
So:
(a) 20m/s / 10s = 2m/s/s
(b) 40m/s / 40s = 1m/s/s
We can validate this because we can see that the graph eventually plateus; an indication of the fact that the object is travelling at constant speed (at which gradient = 0).
how would you draw an acceleration–time graph for the busHave you read the chapter?
Have you read the chapter?
You find the accelation of each linear portion of the graph (you can find it using rise over run as this v-t graph is linear) and graph the acceleration against time.
Q1 Hint: What does the gradient of a velocity-time graph tell you?SpoilerAcceleration
Q2,3,4 Hint: What does the area under a velocity-time graph tell you?SpoilerDisplacement
If you still need help just post again
can someone please help me?
An electron moving north enters a magnetic field
that is directed vertically upwards.
If the electron’s motion was inclined upwards
at an angle, as well as travelling north, what
would be the path of the electron?
Hello knightrider,
When asking questions and for help, it is better if you let us know what you are struggling with specifically, rather than just saying you need help. For example, you could tell us how you attempted the problem and that way we would know where a mistake may have been made and point you in the right direction.
I remember that question with the bus and Anna from last year, I had to ask my teacher.
You can usually find the worked solutions by googling "heinemann physics 11 worked solutions chapter X" where X is obviously the chaper you're up to.
I hope this helps.
Cheers,
Stewart
hi guys,
For these 2 equations how do you know when to use the one with the plus 1/2 and the one with -1/2
What does the plus and minus mean and how would you know to use which formula
Apparently the second one is not officially recognised (maybe only by my school), but it is quite useful in some situations.
They are both used under one main condition, that is, when acceleration is constant. You can notice that one involves u (initial velocity) and the other involves v (final velocity). So if you have one quantity missing out of x, u, t, a, then you use the equation x=ut+1/2at^2. If you have one quantity missing out of x, v, t, a, then you use the equation x=vt-1/2at^2.
Hope that helps :)
Hi how would you do the following question
I have worked out the initial speed to be 20m/s and the maximum height to be 20m
how would you find the speed of the cork as it returned to its
starting point?
Well, technically you don't even have to do any working out... i mean, the cork decelerates on the way up and accelerates by the same amount on the way down. It is bound to have the same velocity at the end of the flight as it did at the beginning.
You've invoked the conservative nature of the gravitational force, and a rigorous proof of that isn't going to be given before second year university :P
Technically, it should reach the starting height slower than its initial speed because of the non-conservative nature of air resistance. In addition, PB I think you've made a typo there that I'll highlight. I trust you can see what is wrong with it :P
why no force is produced when the wire is parallel to the magnetic field?
thank you
This is just proving the conservation of momentum (if initial total masses are equal to final total masses, then total initial speeds are equal to total final speeds)
VCAA won't give anything this vague, it will always have actual figures for you to plug into equations
Hi everyone 8) ,A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Can someone please help me out with this question for light:
Estimate the size of the smallest object that can be clearly imaged by a microscope that uses visible light. Explain this limitation.
A microscope works by shining light through a slide and the specimen absorbs some frequencies of light, producing an image that enters the lens and into your eye (or something like that).
Now, we all know that light whose wavelength is greater than the width of a gap will diffract significantly. The same thing goes if the wavelength is greater than the width of an object! Fortunately, this doesn't affect our day to day lives because every-day objects are definitely not smaller than the wavelengths of the visible light spectrum. If not, there will be crazy diffractions going everywhere, and nobody will be able to see anything.
However when we are dealing with extremely tiny objects, whose width is actually smaller than the wavelength of visible light, significant diffraction starts to occur, and (as you can imagine) starts to create blurry images.
So there you go. Now go and find the smallest wavelength of visible light, and the object of interest should be no smaller than that in order to be clearly imaged by a light microscope.
a ball is thrown vertically upward from a platform 16m above the ground and has a initial velocity of 24.5m/s. Find:(a) v^2=u^2+2ax
a) max. height the ball reaches
b) time taken to reach max height
c) time taken to reach ground from max height
d) total time the ball is in the air
e) speed of ball when it returns to the platform
f) speed of ball when it hits the ground
thanks you and please
Ask your teacher why he/she thinks that. The magnetic field lines are leaving at the marked N terminal, so I see nothing wrong with the diagram given.
What he says is that any point is north relative to a point further down the line (in the direction of the field), which is a south. Basically that labelling a north and south is relative, so INSIDE the solenoid, the field starts at a north and goes to a south, thus doing the opposite to the diagram. He argues that you should never label the poles of a solenoid as they are relative (he's a VCAA assesor, and took a mark off my friend in a test for labelling the poles). However when I asked him a question on lenz's law, with a bar magnet going into a loop, he said to determine the direction of current, you need a north end produced by the coil that opposes the north coming towards it - to do this he used the north and south relative to inside the solenoid. Which, when looking over some questions, I found to be wrong.
The problem is, you can't say that north and south are relative terms. They're not. Magnetic field lines form closed loops, so by his logic you can keep going more and more and more north, which is ludicrous.
Inside the solenoid, you go from a south pole to a north pole, not the other way around, exactly like a bar magnet. I don't know what he's thinking, but honestly I don't follow his logic.
As for Lenz's law and a bar magnet going into the loop, if it's the north pole of the bar magnet facing the loop, you have a stronger magnetic field going into the loop so the resulting current should produce a magnetic field which opposes the original magnetic field. And vice versa.
How would you do this question
A man of mass 70 kg steps forward out of a boat and
onto the nearby river bank with a velocity, when he
leaves the boat, of 2.5 m s−1 relative to the ground.
The boat has a mass of 400 kg and was initially at
rest. With what velocity relative to the ground does
the boat begin to move?
Conservation of momentum. Initially both the man and the boat are at rest. Then, the man pushes on the boat to start moving. Have a think about it.
Thanks would this be right :)The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
using m1u1+m2u2=m1v1+m2v2
(70*0)+(400*0)=(70*2.5)+(400*v2)
0=175+400v2
v2=-0.44metres per second what does the negative mean in the answer
What accelerating potential difference would be required to give an alpha particle a de Broglie wavelength of 2.0nm?
mass of alpha is 6.67x10^-27kg
I attached the solution and am just unsure why they doubled the coulomb of charge. The way i tried to work it out is find the velocity using de broglie's wavelength formula which gave me 49.7m/s. Then use this to find the kinetic energy which was 8.24x10^-24J. Once i did this i used the formula W=qV and subbed in 8.24x10^-24=(1.6x10^-19)V which gave me the answer of 5.1x10^-5V required.
what is the importance of frequency in determining the non-ideal behaviour of transformers. are there any other parameters?
question: in a DC motor why does the armature continue to spin even when at one point on force (the forces cancel out) are acting on it?
I know it has something to do with momentum but I don't quite get it....
Lol, I know I'm not doing VCE anymore but I'm doing a VCE equivalent (kinda) Physics subject at university and I'm getting confused on the question attached. My answer for the second situation doesn't make sense so I'm assuming I'm doing it wrong. Thanks for the help. :)
The negative simply means the boat is moving in the opposite direction to the man. If you define the man's velocity to be positive 2.5m/s, the boat's velocity will be 0.44m/s in the other direction - this is because the man's movement pushes the two apart, they are not in the same direction.
Thanks Zealous :)
Can you help with these questions?
A small research rocket of mass 250 kg is launched
vertically as part of a weather study. It sends out
50 kg of burnt fuel and exhaust gases with a velocity
of 180 m s−1 in a 2 s initial acceleration period.
a)What is the velocity of the rocket after this initial
acceleration?
b)What upward force does this apply to the rocket?
c)What is the net upward acceleration acting on the
rocket?
Sorry, I'm not Zealous but I hope you'll accept my help too :P
To me, the trick to this question is that the mass of the rocket changes from 250 to 200kg (50 kg of fuel is sent out)
So, I would use the 'average' mass of 225kg in my calculations. (there are more scientific methods to find the average mass than just choosing the middle value, but they're too bothersome and not required for VCE)
Part a
Conservation of momentum tells us that the momentum before the launch is equal to that after the launch. In maths form,
Now initially we know that the rocket was stationary and we're given all the information in the question (just don't forget to use the average mass!). Note that I take up as positive
Part b
To find the upwards force, we can use the impulse as we know the momentum and we are given the time
Part c
To find the net acceleration we just use Newton's 2nd law remembering to use the average mass again
Work (W) = Force (F) * Displacement (x)
As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.
For the first box:
x=1.5m
W=(100)(1.5) = 150N
You will have to lift each box an extra 0.3m
Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Work (W) = Force (F) * Displacement (x)
As we are lifting boxes, the force is made up by the weight force = mg
Hence, each box, when lifted, has a force of (10)(10)=100N pulling it down.
For the first box:
x=1.5m
W=(100)(1.5) = 150N
You will have to lift each box an extra 0.3m
Total work done = (100)(1.5+1.8+2.1+2.4+2.7) = (100)(10.5) = 1050N = 1.1*10^3 N
Your answer is correct, but be careful of the working.
For instance, if you were to actually try to exert a force equal to the weight force on the box, the net force on the box would be zero. Problem. In reality, we pull up with a greater force than the weight force to get the box moving, and then we reduce the force when we want to slow it down, so the force isn't constant. W = Fx only works for constant forces.
What you should actually do is note that the kinetic energy at the beginning and the end is zero, so the work done is the change in gravitational potential energy. The working out is the same, the reasoning isn't.
Thx lzxnl and jhardwickvce :)
But remember jhardwickvce that work is measured in joules or newton metres
in your working out you wrote newtons at the end :)
By the way jhardwickvce and lzxnl how different is physics 1/2 to 3/4 physics and how are you guys finding it/found it
I know I'm not jhardwickvce nor lzxnl but I hope you don't mind me answering this question.
Motion in 3/4 builds upon what was learnt in 1/2 very well. Some new concepts involve circular motion, more intricate block/ramp systems and newtons's law of gravity.
Electricity involved familiar stuff too. Some additions included diodes, thermistors, photodiodes, light-dependent resistors in circuits. A common question is to figure out where to place a cooling/heating element in a circuit.
Light is a different story entirely (at least from what I learned in 1/2). In 1/2 we did a lot of ray tracing and calculations based on focal lengths of lenses/mirrors, but in 3/4 the content shifted towards wave/particle duality of light and the experiments that proved such light properties (photoelectric effect, interference patterns). I found this topic the driest and most repetitive.
Electromagnetism was not something we were taught in 1/2 so concepts like flux and field took a bit of time to get comfortable with, but otherwise it's fine.
Can someone please help me calculate this question for structures and materials:
A winch uses a steel cable to lift a large piece of machinery
from a loading dock. While supporting this load the length
of the steel cable increases to 1.001 times its original
value. What is the tensile strain on the cable when
supporting this load? Express your answer as a
percentage 8)
Is the yield strength the same as the elastic limit of a material? If not what is the difference? 8) :-\
Hey everyone
In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/
Thanks
Hey everyone
In the physics exam, can we use formulas that are not in the study design? Because my tutor said we can only use formulas if we know how to derive them :/
Thanks
Which formulas would you even want to use? I never felt the need to use extra formulas explicitly.
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.
You sure those aren't on the formula sheet? Maybe my memory isn't working properly for me here but I swear the range is at least.
In any case, deriving it wouldn't take you more than half a minute.
any help is greatly appreciated :)
an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
any help is greatly appreciated :)
an object, projected vertically upwards with a speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
V= u -10t
t = -(v-u)/10
t = u-v/10 *2
t= u-v/5
Not sure if this is your answer but that's my jab! :-)
Not constant acceleration; there's air resistance here
Your working missed a nuance in the question. The question says 'speed v'. But it's returning downwards, so its final velocity is actually -v, not v.
The angle used in the formula is the angle between the force and the lever. In this case, as the angle with the ground is 65, the angle with the lever is 25, so the torque is given byOooh I see, thank you so much!
which means that E is correct. Because it appears that the cosine of the angle is used, but you need to make sure you use the right angle. The cosine of the angle with the ground happens to be equal to the sine of the angle between the weight force and the lever, which is what is actually used in the calculation.
Why is F_{box on trailer} > F_{trailer on box}?Spoiler(http://i.imgur.com/syISyRb.png)
I could be wrong, but I think you're not looking at the diagram correctly. The arrow from the trailer upwards is actually from the bottom of the box (where the trailer makes contact with the box), and the arrow downwards is from the bottom of the trailer downwards, so in reality the proportion of upwards to downward arrow size is the same. It would have to be according to Newton's 3rd Law of Motion...
for the gain of an amplifier, (even if it is an inverting amplifier), should it always be a positive value?I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.
I quite sure that as long as you take the gradient of the Vin/Vout graph, whether negative or positive VCAA will accept it (from the assessment reports I've seen). I usually just take the magnitude anyway though - negative gain just sounds awkward.
thank you guys!
just another one:
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
but the ans says i can't? why?
can i ask another one?Slip Rings maintain contact with the rotating coil in an AC generator as a way to transfer the current out from a coil into a circuit for use.
what is the purpose of slip rings in AC generator?
The lines diverge because the magnetic field (B) gets smaller the further away from the loop you go.Thanks for the response.
Just on that, wouldn't the magnetic field strength have no effect on the direction?
Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator
edit: that is with g accurate to two significant figures... if you use g = 10 m/s^2 you get 1.01 m. I never did 3/4 physics, is it common to use g = 10 m/s^2? ???
Can someone please help me with question 12 for structures...I have looked at the worked solutions on itute and i don't understand how you can neglect the torque of one of the beams when answering this question :/Torque is calculated as - the force multiplied by the distance from a "pivot point".
The ans is B...
Torque is calculated as - the force multiplied by the distance from a "pivot point".
When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to . So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.
Torque is calculated as - the force multiplied by the distance from a "pivot point".
When we solve this question, we look at it almost like a seesaw, where Y is the centre and we take torque about the point Y. So the torque created by the beam Y is equal to . So we can neglect the force of the beam Y as we are calculating torque about this point, so the distance is 0 and the force isn't working to create any torque. Then we can just look at the force from the mass at Z and the force from the mass of the beam to solve Q12.
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?
thanks in advance :)
Can someone please clarify this concept?As I understand it depends on what the wheels are trying to do;
The forward driving force is provided by the friction exerted by the ground on the rear wheel. It is the action – reaction pair to the friction force of the wheel pushing backwards on the ground
Retarding forces acting on the bicycle include air resistance (drag) and friction on the wheels (especially when the brakes are applied).
So what does friction do? How is part of the driving force and the retarding force? So if the bicycle is travelling where there is no air resistance, it won’t ever stop? Because the retarding force is trying to stop it, but instead it actually helps it to move, but as it moves the friction on the wheels try to stop it, and then the cycle repeats itself?
Why is it that no work is done when the component is in a perpendicular direction to the objects movement?Work is what I like to call - "useful energy". The consumed energy that actually contributed towards an object's propagation in a certain direction.
thanks in advance :)
Hi everyone
Just did 2006 VCAA, and I found motion section so hard, much harder than any neap, insight etc commercial exam motion section. How is everyone else feeling about motion 06? Will the motion section be of that difficulty in 14? Does the difficulty stay the same from 06-13?
thanks
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marksThank god.
all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)
should be fine!
based on grade distributions that paper was the hardest one evar, A+ cutoff was 79% for unit 3 2006 and 0 people got full marks
all unit 3 and 4 exams except that from 2004-2012 have A+ cutoffs between 85% and 96% with the years towards 85% being considered the harder papers (they try to go for ~90%) (with the exception of unit 3 2004 which was 81%) (I havent added 2013 because this is my stats from last year)
should be fine!
How can you tell how many people got full marks for particular examsUsually it says in the assessors report. For the 2013 one it didnt, but for years like 2009 where 100s and 100s people got 100% they did
How can you tell how many people got full marks for particular exams
can someone please help me with this question?
I guess im stuck between those three big words: weight = mg correct, force due to gravity
weightlessness? - whats that? actually having weight = 0, i.e. m=0 or g=0, doesnt truly happen at any times here
aparent weightlessness - when normal = 0 yep
Yes some older physics/specialist reports seemed to have this information, but I'm going based off the spreadsheet that my physics teacher gave me towards the end of last year which was his info on grade distributions and everything so idk he may have gotten the information from some other source related to being a VCE physics teacher. he had all sorts of info like that.Hey silver sorry for the questions!
Where would you find the A+ cutoffs and things like that on the vcaa website for physics
Hey silver sorry for the questions!
Would you reccomend me going through the qs for motion for 06? Or leave it?
thank you! silverpixeli
but at the top of the flight, wouldn't it still experience centripetal acceleration/therefore centripetal force downwards in that situation? in the same direction as its weight force? but its weight force is zero? unknow? im confused....
so for that question I would ans it like this?
the plane has a weight force because it has mass and experiences 9.8N (g) on the Earth.
weightlessness can never happen, cause for this to happen it requires g=0, which is like never? so it does not apply to any situation?
apparent weightlessness occurs bcause there is no normal reaction force acting on the plane because only the weight force is needed? but how do we know this?
Hi all, I was doing the 2008 VCAA exam 1. In the electronics and photonics section it shows a npn transistor and stuff like that. That's no longer in our course right? Just want to confirm. I don't even know how to go about the question.
Can someone please help me with this question from the 2006 vcaa exam 2?
I am unsure how they found the change in time. I presume they converted the 4m/s using a distance but I can't interpret what the distance would be. :-[
Hey guys, quick question
How do you know whether to use the formula net force=Fn+Fg, or net force=Fn-Fg?
Thanks!
Can someone please help me with part ii to this question from the 2007 exam 2?
I understand the method to work it out but i always get confused when the resistance illustrated is the total resistance or you have to add it? How do i know what the total resistance is for these types of questions?
hey I was wondering whether pn junctions and those type of things are in our study design. I remember my teaching doing something brief on them mid-year but every practice exam I've done I've not witnessed anything to do with them. Theyre also in my textbook - n and p type semiconductors and all that stuff.
Can someone please help me with this question.
Whats the difference between elastic and isolated systems?
Isolated = no external forcesYeah I think he means elastic collisions. Elastic collisions (in terms of physics 3&4) are collisions in which no kinetic energy is lost. (Ek(initial) = Ek(final)). And then inelastic is when kinetic energy is lost.
Elastic systems? I haven't heard that term used. If you mean elastic as in collisions, you can have an isolated system in which collisions are not elastic, namely that of light and matter, or a collision that involves heat transfer. Friction can still act in an isolated system.
This is a stupidly simple question, but I've got myself confused, when you find the 'direction' of a projectile, which angle to you take?
Usually when the object is moving up or down, we take the acute angle that the direction has with the horizontal.
Hope that makes sense, and best of luck with the exams.
Resistance here is ~19kOhm right?
Answers says it is 1kOhm
careful of the scale of the illumination axis, the question gives you but the graph is in
but yeah if it was the point you thought, you are reading the graph correctly and ~19 would have been correct
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
I guess you could also talk about diffraction as a wave property?
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy?
It seems REALLY brief, but it actually does summarise essentially why Young's model supported the wave theory of light. You should, however, explain why it suggests interference.
I guess you could also talk about diffraction as a wave property?
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?
Just need some clarification thanks. :D
Is this seriously an answer that would give full marks?Spoiler(http://i.imgur.com/PtIPW2h.png)
(http://i.imgur.com/rNvoTSd.png)
For these questions, I mentioned that interference is a typical property of waves. Young's double-slit experiment demonstrated that light interferes constructively and destructively to produce light and dark bands respectively. Thus, Young's experiment supports the wave-model.
You should also explain why it doesn't support the particle model - for any of these questions, you have to explain why it 'disproves' the other too. (You might have, but just going off what you wrote ^^ :) )
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
Quick question: electrons can only make transitions upward (higher energy levels) from ground state right? So if there's n=3 at 4eV and n=4 at 7eV, an electron won't be able to absorb 3eV and take that transition in energy (or it can't go from n=2 straight to ionisation)? Is this because we expect the electron to stabilise by dropping back to ground state first before it can accept any more energy?
Just need some clarification thanks. :D
Sigh. This gets complicated :P you have all these issues about which energy is it more favourable for it to relax to, or will it preferentially absorb energy...don't worry about that xP
Fluorescence is essentially when excited electrons go via intermediate energy levels back down to the ground state. So to be honest, I don't exactly know how particles determine which energy level to go to. Need to ask a quantum physicist there :P
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
So for a 3 mark question, would we be expected to say for instance:
1 mark - whether it supports the wave or particle model
1 mark - why it supports the particle model (or vice versa)
1 mark - why it doesn't support the wave model (or vice versa)
??
Perfect. :)
Looks good to me, I'd say;
The experiment supports the wave model for light, as the pattern that is observed on the screen can be explained by diffraction and interference, both wave phenomena. It contradicts the particle model, which does not account for diffraction or interference, instead predicting two bright bands of light directly behind the slits.
Since the question seems to be geared towards a comparison, I'd chuck in a quick diagram of the two expectations if I had time! That's probably more than needed for 3/3, but it can't hurt to give more than required (as long as it's all correct)
Not sure if this has been mentioned yet, but VCAA don't provide super detailed solutions (enough to get the full 3 marks) to worded questions because everyone would just copy it down onto their cheat sheet.
Easiest fix ever: remove the bloody cheat sheet and make people actually remember stuff. Like, what, 99% of the other subjects?
Also, are incandescent globes on the course? I've been doing pre-2006 exams and in the sections for Light and Matter, it's got questions on incandescent lights and I'm not sure whether or not they're still on the course. :/ thanks!
what mass should i use 33, 44 or 77??Depends on how you want to approach the question.
N = mg - ma?
Explain why adding a soft iron core increases the strength of an electromagnet.
please help :)
Explain why adding a soft iron core increases the strength of an electromagnet.
please help :)
STAV 2013
can someone please help me with question 3,4 and 14 in the attachment? all multiple choice questions
thank you
hey speedy! :)
question 3 is above question 4 in the same attachment.
ans to question 4 is B?
hey guys
for significant diffraction to occur, does lambda/slit width have to be greater than or equal to one or does it need to have the same order of magnitude? I havent been able to get a definite answer, what do you guys think?
For complete diffraction, you need your slit to be smaller than the wavelength.
Are there any main difference's between a split ring commutator and slit ring commutator that we should note? So far on my cheat sheet I have only put that split rings are for DC and produce a DC output, and slip rings are for AC and produce an AC output... :-\Split ring reverse direction of current every half turn but slip ring maintains it.
Split ring reverse direction of current every half turn but slip ring maintains it.
Can anyone confirm?
For the cheat sheet, do we have to stick the two A4 pieces together?
Got a few questions:
What should I chuck on my cheat sheet for modulation? I don't think we touched on it during the year but the exams seem to look to ask questions on it. I was thinking of just putting what an information signal/unmodulated signal/modulated signal/carrier wave looks like.
If anyone is doing structures & materials; What's the difference between strain energy and toughness? Are they both the same thing or is one multiplied by volume? Also, when reading a f-x graph do we calculate the area under the graph (similar to a stress-strain graph) for strain energy?
does anybody have a suitable definition for single slit interference? I have it such that when light passes through a slit it diffracts and the waves from each side interfere with each other creating bright/dark bands but am still a bit unclear.
furthermore would the purpose of modulation be to transmit information more effectively since higher frequency waves diffract less (hence why we impose the signal wave onto the carrier wave)
It's not like you really need a definition for this and I haven't tried to define it myself before so I'll see what I come up with here.
Huygen's wave model of light assumes that in a light wave, every point on the edge of the light wave acts as a new site of propagation of the light wave. In other words,you can think of light as originated from every single point on a light wave. Therefore, in single slit diffraction, you can think of light as radiating from every single point in the slit. Considering light being emitted from different parts of the slit and considering their path differences allows for analysis of the single slit diffraction phenomenon.
Heck no. Hence why I said you don't need a definition of this.
Hey guys, need help with question 22.c
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/2013physics-cpr-w.pdf
Here is my working out:
....
-----------------------------------------
Okay so I got 2 out of the possible 3 marks for this. Where have I gone wrong? According to the assesors report my path difference was wrong for the bright band, it should have been 2/. Why 2? THanks
And just another quick q sorry;
23.B
How does momentum influence the wavelength of an electon? ANd is 'fringe spacing' just referring to the diffraction pattern? Thanks
The path difference is 2 whole wavelengths. The path difference to the first bright band is 1 whole wavelength and the path difference to the second bright band is 2 whole wavelengths. Recall the formula , we are looking at the 2nd bright band so n=2, and we get: .Oh I see! So the central maxima does not count as a bright band? Thanks!
The formula for de Broglie matter wavelength is . So the wavelength of an electron (matter) is inversely proportional to the momentum of the electron. Basically, the faster an electron is travelling (greater momentum), the smaller it's de Broglie wavelength and conversely, the slower an electron (and lower momentum), the larger it's de Broglie wavelength.
You're right with "fringe spacing" - the bands in the circular diffraction pattern (or rings) are refered to as fringes, and their spacing depends on the wavelength of electrons/light. It's a bit like taking Young's Double Slit interference pattern and converting it into a circle so you've got bright and dark sections from the inside to outside.
Oh I see! So the central maxima does not count as a bright band? Thanks!Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.
Well, the central maximum is still a bright band, but you don't use it in the PD formula because there isn't any path difference. The central band is exactly half way between the two slits, so waves from both slits would've travelled the exact same distance, there's no difference in the paths they've taken so they'll interfere constructively.Thanks again
It becomes a central maximum (brighter than the other bands) because the intensity of light is inversely proportional to the square of the distance (), so the brightest band would be in the center (where the waves have travelled the least) and then the pattern will get dimmer as you go outwards (and the light travels further).
2b and 6c2b:
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf
Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
Thanks again
Sorry for bothering but would it be okay if you please explaind two more
2b and 6c
http://www.vcaa.vic.edu.au/Documents/exams/physics/2013/physics_examrep13.pdf
Don't get 6c AT ALL. I get how they have assumed the spring potential to be 0 N as incorrect, but how do I prove that it is not 0?
What is an oscilloscope?
Thanks :)
Is this correct to say about transformers?: The current in the primary coil creates a magnetic field which is strengthened by the iron core, therefore threading the secondary coil and inducing a current proportional to the amount of turns in the coil.I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
I don't even think you need to mention an iron core unless they've given it in the question. I'd probably say "The AC current through the primary coil creates a rapidly changing flux through the secondary coil. This results in an EMF induced in the secondary coil, as dictated by Faraday's Law. The magnitude of the EMF induced is directly proportional to the amount of turns in the secondary coil." or something like that. But there's nothing wrong with your definition, just saying it in different ways.
When a coil is placed in an external magnetic field and connected with slip rings, does the coil oscillate 90 degrees back and forth? Give or take an extra bit because of the momentum of the coil.
And when it's connected with slip rings to a DC power source, does it just align itself with the external field?Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
I'm not sure if you're talking about generators or motors. If you're talking about AC power generation, the coil will still rotate in the same direction (not going back and forth), but the current induced will create a sinosidual curve (sine curve) because the change in flux is positive and negative at different parts in the rotation. The momentum of the coil in rotation will completely depend on what you've got rotating the coil to generate the power.
Connecting a coil to a DC power source through slip rings is just like connecting it directly to the circuit. Once current goes through it, it will rotate to a position of maximum flux and get stuck at that position.
How can we conclude that the path differences are the same?You can make that conclusion because the position Y is in the exact same spot with both patterns. The light from both slits has still travelled the exact same distance to reach point Y, so the difference in the path they take will still be the same.
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks :)
Could someone explain slip rings in reference to AC generators for me please? I'm having trouble understanding textbook definitions, thanks :)
Also compared to the commutator, slip-rings remain stationary and do not rotate with the coil. The only place they are seen as far as the physics course is concerned is on the AC generator. One problem with slip rings though is because of the metal to metal contact between the coil and the rings they are very prone to wear and also cannot handle vibration too well as the coil can temporarily lose contact with the slip rings and cause sparks. This is why in many motor vehicles an alternator is used which rotates the magnet rather than the coil and hence eliminates the need for slip rings.How is the current directed in the coil, though? Like a split-ring commutator reverses the current every 180degrees but slip rings maintain an AC current within the coil? How does this keep it rotating? Maybe I should have been more specific as this is really what I want to know. That gave me heaps of useful information though - so thank you!
Went a bit off track but hope that covered all bases ;D
For the exam, are you not allowed to write anywhere in the border? Or is it only on the side closest to the centre?
Would somebody be able to explain modulation and demodulation? Thanks :)
Modulation is the process of encoding the information into the carrier wave, and in VCE physics we talk about intensity/amplitude modulation which means we encode the changes in signal into changes in amplitude of the high-frequency carrier. (there's also frequency modulation where changes are mapped onto changes in frequency).
So if we were using a light beam, would modulation be that we encode the changes in the signal into changes in the light brightness?
Is it necessary to put a negative in front of voltage gain if the amplifier is inverting? Do you lose marks if you don't?
How should we deal with the negative sign in Faraday's law when computing the average emf induced?
Quite clearly we have emf = -N(change in flux/change in time), which is likely to come out as a negative value - how should we 'convert' this to a positive average emf induced, in terms of mathematically sound working?
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
Or am I missing something?
Is there solutions to the sample exam VCAA provided last year? http://www.vcaa.vic.edu.au/Documents/exams/physics/physics-specs-samp-w.pdf
Or am I missing something?
It's a bit late but oh well. How do I tackle these questions? I found the acceleration and then consider about the forces in question.The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?
However I always get 'em wrong :(
The net force on Box A is equal to the 140N driving force minus the frictional force of Box A.
Frictional force is 0.2 x ( 10 x 9.8 ) = 19.6N. Therefore the net force on Box A is 120.4N to the right, which is now exerted on to Box B.
Net force on box B = 120.4 - (0.2 x (9.8 x 7)) = 106.68 to the right, which is exerted on to Box C.
i.e. the Force of Box B on Box C is 106.68N to the right, is this correct?
And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?
Why is that?
In part c I considered the forces acting on box C. I found net force from F=ma and then subtracted friction force and since thats the 'driving' force on box C then Force of B on C must be equal to it?
However I always get 'em wrong :(
And one more sorry - For Q7 they times the torque supplied by the pole by 1.5. I thought you times it by 3 since its 3m from the pivot point?
Why is that?
What are some devices that can act as modulators and demodulators, and what can act as a carrier wave?Modulators: you're mainly looking for transducers (electro-optic) which convert electrical signals to some physical signal, where their output can be adjusted based on an input signal. So things like an LED (can't think of any others right now).
EDIT: Also, how do you go about finding the rebound Force of a wall that is holding a cantilever as well as the angle of said force?Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.
Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.
(http://questions.transtutors.com/Transtutors001/Images/Transtutors001_e5ae4d3b-0bf7-4ba8-9555-a895c99d2e5b.PNG)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.
Hi,
Could someone please explain how standing waves support the existence of discrete energy levels in the atom?
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
ans says k =33?
how would i use their formula f=kx?
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.
hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2
thank you
Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.Gg man gg
Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m
I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.
Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.
The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.
In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.
(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.
(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.
The car will need to apply a force to accelerate the object at 2ms^{-2} as well as overcoming friction so the required force is:
Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.
(http://i1282.photobucket.com/albums/a531/Ovazealous/diagram_zps1564eb19.png)
There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force. We also know the car by itself will also accelerate at 2ms^{-2}. Therefore the NET force acting on just the car is F=ma=1400x2=2800.
Now we can setup an equation of forces:
F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.
You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s^{2}. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.
So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.
Hope this helps!
I know this question is meant to be quite easy but I can't seem to get my head around it:
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s/s, and keeps accelerating at this rate until it has passed the bike.
How far does the police car travel before it overtakes the motorbike?
Thanks!
That's the thing with physics, it appears easy on the surface but is actually a bit challenging underneath.
Is the answer 612.5m? Here's how I worked it out:
d (motorbike) = 35t
d (car) = (0)t + 1/2(4)t^2
Equate the two formulas, and t = 0 or t = 17.5
sub in 17.5 into the initial motorbike formula, and you get 612.5m.
Did they first assume light was the particle model or the wave model? who?
Thank you
Hi guys can some one help me with this question.
So a 2.5kg mass is rotated in a conical pendulum where the length of the string is 0.68 metres and thr angle between the string and the vertical is 35 degrees.
Find
a. The tension in the string
b. The speed of the mass
If the pendulum is now spun faster so that it's period is now 1.2 seconds find
A. The tension in the string
b. The angel the string makes with the vertical
How do I express direction in terms of degrees. eg. 53.1 degrees counterclockwise from the ground (is there a more "scientific" way to express this?)53.1 degrees from the horizontal is probably best. I definitely wouldn't use bearings.
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.
What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
A person pushes a lawnmower of mass 20kg at a constant speed with a force of 100N directed along the handle, which is at 35 degrees to the horizontal.
What force must the person exert on the lawnmower in order to give it a velocity of 2.0 m/s in 2.5 seconds?
Lawnmover moving at constant speed therefore acc. to Newton's first law forces are balanced therefore frictional forces=100 cos(35)=81.9 N.
assimung it accelerates from rest:
u=0, v=2, t=2.5, a=?
v=u+at
2=0+2.5a
a=0.8 ms^-2
Net force acting on lawnmower=ma=16 N=driving force (in the direction of motion)-frictional forces=F cos(35)-81.9
F cos(35)=97.9 N
F=120 N
Person must exert a force of 120 N.
I keep getting this question wrong.
A bicycle accelerates from rest, covering 16m in 4 s. The total mass of the bicycle and its rider is 90 kg. What is its average acceleration?
Thanks!
u=0, t=4, s=16, a=?
s=ut+1/2at^2
16=0+1/2*a*16=8a
a=2 ms^-2
Just out of curiosity, do you have to use the motion equations?
Why couldnt you do something along the lines of:
V = 16/4 = 4m/s
Acceleration = DeltaV/Time
= 4-0/4
=1m/s^2
Why doesnt that work?
A softball of mass 250 g is thrown with an initial velocity of 16 m s–1 at an angle θ to the horizontal. When the ball reaches its maximum height, its kinetic energy is 16 J.
a What is the maximum height achieved by the ball from its point of release?
b Calculate the initial vertical velocity of the ball.
c What is the value of θ?
d Whatisthespeedoftheballafter1.0s?
e What is the displacement of the ball after 1.0 s?
f How long after the ball is thrown will it return to
the ground?
g Calculate the horizontal distance that the ball will
travel during its flight.
Need help with part b and part e specifically.
Thanks.
Need help with this question:
A 0.50kg puck rests on a level air table and is connected by a light thread passing through a hole in the table to support a hanging mass of 3.0 kg. A stable orbit is achieved when the puck is sent into a circular path of radius 0.15 metres around the hole.
(A) neglecting friction at the edge of the hole, calculate the period of revolution of the puck in its orbit.
(B) Suppose that the speed of the puck in its orbit is now doubled, while the radius remains fixed. what central mass will be needed to achieve a stable orbit.
I tried the first one but I'm pretty sure i was wrong ahaha
Our teacher has forced us to do the second type of question to death though so I'll give it a crack
A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
First of all I would convert to m/s
Car=22.2m/s
Cop=27.8m/s
It takes the cop 15 seconds to accelerate to 27.8m/s, during which time the car has travelled 333m
In the first 10 seconds the cop travels
In the following 5 seconds the cop travels bringing him to a total of 291.5m
The cop is travelling 5.5m/s faster than the car, and has to catch up 41.5m, which will take him 7.5 seconds, bringing the time up to 22.5 seconds
Usually it isnt that messy and you have a nice 20m/s and 30m/s values to work with (classic VCE Physics)
Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to the bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
I hope this is right ahaha
Yeah my bad, I ahould have drawn a graph but I was too lazy
I'm gonna have a crack at the first one but no promises
All good dude. Found my error with some help. The distance covered in the first 15 seconds travelled i calculated was right, when i was solving for time i forgot it should be (for the cop car) t-15 and not just t. Cause the distance added to the first 15 seconds is the time after 15 seconds (hence the t-15 bit. All good. Thanks buddy.(http://i.imgur.com/aUQgBHx.jpg)
Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).
1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?
2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.
3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.
(http://i.imgur.com/aUQgBHx.jpg)
Gave this one a go, have no idea if it's right but it's feasible
You can't just add initial speed in; remember that energy is proportional to speed squared, so you have to calculate kinetic energy initially, gravitational potential energy at the end and subtract and calculateI get it now, thanks so much! I asked the class genius on facebook how to do that, I didn't even doubt his answer for one second ahaha I can't believe he was wrong
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.Almost this exact same question got answered a few days ago if you scroll back, different values but same concepts
a) How far does the police car travel before it overtakes the motorbike?
b) At what time does the police car overtake the motorbike?
A speeding motorbike travels past a stationary police car at a speed of 35 m/s. The police car starts accelerating immediately at 4 m/s^2, and keeps accelerating at this rate until it has passed the bike.
a) How far does the police car travel before it overtakes the motorbike?
b) At what time does the police car overtake the motorbike?
I am having trouble working this one out as I can not seem to find time taken. I had read the previous question before asking.
In momentum questions which require the use of the formula m1u1+m2u2=m1v1+m2v2, do we assign directions?
E.g. Question
A cart with mass 2kg moving at 2m/s in westerly direction collides with cart moving at 3m/s in easterly direction. Upon collision, carts connect and move of at 1m/s in westerly direction
What's the mass of second cart?
Would we go;
(2)(2)-(3)m2=(2+m2)(1)?
As far as im aware the general rule of thumb is, if its a vector, assign a direction. Which should be first nature now.
That's not even the rule of thumb - if you don't assign a direction, the maths breaks, lol.
Hahha, gez cant get away from you. But yeah, thats what i was implying.
Hi,
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
The time of contact between the ball and the floor during the bounce was 0.050 s.
a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.
thanks
This question relates to impulse.
Since I = Δp = F(average) Δt
and I = Δp = m Δv
Always assign a positive direction! Let Up be positive.
Therefore F(average) Δt = m Δv
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)
Therefore I (floor on ball) = 1.44 N s in the upwards direction.
c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.
Hence I (ball on floor) = - 1.44 N s
Therefore I (ball on floor) = 1.44 N s in the downwards direction.
I think this is all correct! Hope it helps!
This question relates to impulse.
Since I = Δp = F(average) Δt
and I = Δp = m Δv
Always assign a positive direction! Let Up be positive.
Therefore F(average) Δt = m Δv
a) Sub in values: Δt = 0.050 s, m = 0.080 kg, Δv = Final - Initial = (8.0) - (- 10.0) = 18.0 m/s
F(average) x 0.050 = 0.080 x 18.0
Therefore F (average) = 1.44 / 0.050 = 28.8 N
Therefore average net force is 28.8 N in the upwards direction.
b) I (floor on ball) = Δp (ball) = p (final) - p (initial) = m ( v (final) - v (initial) ) = 0.080 x (8.0 - (-10.0) )
= 0.080 x 18.0 = 1.44 N s (standard unit for impulse: Newton second)
Therefore I (floor on ball) = 1.44 N s in the upwards direction.
c) I (floor on ball) and I (ball on floor) are a Newton's 3rd law action reaction pair.
Hence I (ball on floor) = - 1.44 N s
Therefore I (ball on floor) = 1.44 N s in the downwards direction.
I think this is all correct! Hope it helps!
Hi,
Can someone please fully explain the following question for me? I'm confused about the way gravity and the normal force are involved for these sort of questions:
A rubber ball of mass 80 g bounces vertically on a concrete floor. The ball strikes the floor at 10 m/s and rebounds at 8.0 m/s
The time of contact between the ball and the floor during the bounce was 0.050 s.
a) Calculate the average net force acting on the ball during its contact with the floor.
b) Calculate the average force that the floor exerts on the ball.
c) Calculate the average force that the ball exerts on the floor.
thanks
Thanks for the reply but I'm still slightly confused.
If the question asks for the 'average force' how can the answer be in N s? Should it not just be N?
Part a) you gave an answer in Newtons, but in part b) and c) you gave an answer in N s, even though it was asking for force? Can you please explain this to me because it's really got me confused.
Also
it might be worth mentioning that this question was taken from the god-awful Heinemann textbook - below is their version of the fully-worked solution for parts b) and c).
b) The forces acting are gravity Fg downwards and normal reaction force FN upwards.
Fg = mg = 0.78 N
totalF = Fg + FN
This gives FN = 29 N up.
c) As described by Newton’s third law, this is equal and opposite to the force that the floor exerts on the ball, so F = 29 N down.
Rendering me totally confused
thanks and sorry for my ignorance in advance
Im confused now, haha. It says average, when i think of average i think of how the value changes over a period of time (im thinking calculus) and as a result thought that it would how the force changes over that time interval of impact.
Additionally, when your looking at a object that the only two forces acting on it are the weight and normal reaction forces and then you form the equation
Fw = -Fn cause you have to take into account the different directions in which they act.
And therefore the Sum of forces = Fw + Fn.
Why above is it (Sum of forces) expressed as F (av) (floor on ball) - W (weight force) = F (av) (net)
Can somebody explain just to clarify for me.
Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?
In units 1 and 2 the topics in my school were:
- Nuclear Radiation
- Electricity
- Light
- Motion
- 2 detailed studies
In units 3 and 4 all schools cover:
- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)
I hope this helps!
hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks :)
The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.
E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.
Now, using the constant acceleration formula, v = u + at:
30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s^{2} ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.
F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.
Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.
Nope, that's for chem.
In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.
Of course, scientific notation can be used when very large numbers are present
To show what I'm saying:
Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)
Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate
Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth
g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg
Is work done force x displacement or force x distance?
Is work done force x displacement or force x distance?
Can someone explain this to me.
Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.
Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff
So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.
Or have i totally just screwed up with the theory?
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?
A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?
Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?Spoiler(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009024_645555852239499_729218626_n.jpg?oh=bc99983a1c2136046e262bc25a2760b3&oe=54E32AD8&__gda__=1424148677_19c6a042e53bcc13e1049ddbd6225092)
Need help with this question, the 'worked' solutions are useless.
I also don't understand the whole dilation of gravity thing
Thanks
(http://i.imgur.com/gl96p42.jpg)
I found 7a, but don't know what to do for 7b or 8.Spoiler(https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11009026_648111405317277_42696173_n.jpg?oh=f18f9743dda9710cc67b8e13a3037e4e&oe=54E8B477&__gda__=1424530282_c625470941f4d77c7d98f714267bf350)
Can anyone confirm answers to questions on the 2014 vcaa exam for Q2 about the spring?Yep, I did it last year and got it right (acc. to statement of marks)
Yep, I did it last year and got it right (acc. to statement of marks)Thank you :)
a) k=F/x=mg/x pretty simple show that
b) 80 cm
c) Kinetic energy not included
d) 2 ms^-1
For questions such as the following:
A person pushes a 14.5KG mower at a constant speed with a force of 80N directed along the handle, which is at an angle of 45 degrees to the horizontal
(A) Find the normal force
Do we need to go; .
9.8x14.5+sin(45)x80?
Normal force always acts perpendicular to the direction of motion. In this case, if its traveling along flat ground, the net force vertically is 0 as the mower isnt moving up or down, but rather left and right. Thus, using f=mg, f=14.5x10 (take gravity as 10m/s^2) gives us 145N (force of gravity acting downwards). Thus the reaction (normal) force is 145N in the opposite direction (acting upwards). In this case, the ground is pushing the mower with a force of 145N. Check the answers because i might of interpreted the question wrong (next time post a photo of the actual whole question with diagram if there is, makes it a bit easier).Thanks Cosec.
Thanks Cosec.
That's what I thought, however I believe the answer for the question adds in the verticle component of the force being exerted on the mower for the verticle force as it's close to 200
The average speed for the uphill part of the journey is 4ms^1 while the average speed for travelling the same distance for downhill is 8ms^1. What is the average speed of the rider for whole journey. You may ignore time taken to turn around. Theres an algebraic method, in letting distances equal right, but from there i get confused
Let the distance be x. Work out the time travelled going up and down hill in terms of x, find the total distance and total time to work out the average speed.I tried to find time as 4t=x and 8t=x however i get t=0?
Can someone help me with question 5. This is from the detailed study materials and structures, Young's Modulus.A strain percentage of 0.075% is equal to a strain of 0.00075. Also, note that while I don't convert the cm measurements to meters, the output is also in cm.
Thank you :) you were correct haha. Also what detailed study do most schools do? My teacher says everyone does materials and structures because it is the easiest.I think the examiners report tells you
I think the examiners report tells you
But yeah definately structures, it's so straightforward.
Could someone please explain to me how the equation for resistance total for parallel circuits proves the validity of Kirchoff's Law?
When finding the kinetic energy of a projectile in motion do we use the horizontal component, verticle component or the resultant vector component?
Hi guys,
So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.
Please help,
Thanks.
Hi guys,
So we did this prac in physics for circular motion. The aim was to derive the formula F=mv^2/r. The apparatus we used contained paper clip, tube, fishing line, rubber stoper and slotted masses. So what we did with my partner was measure the time and record it for the rubber stopper to make 10 revolutions and then keep radius constant change mass and calculate the velocity. Then, we repeated the same thing instead we changed the radius and kept the mass constant.
And now we have to use graphs to show that F=mv^2/r. We know that Fc=ma=F by the hanging masses but we dont know what graphs to draw in order to prove the formula.
Please help,
Thanks.
Graph F against v^2. You should get a straight line of slope m/r
We did a prac about friction on inclined planes. We got the length of hypotenuse (1m), 3 angles (0,12.7 and 24.8 degrees) and 3 times (0.81,1.07 and 1.17) respectively.
A 600g trolley was pulled by a 400g weight.
My calculation for 0 degrees, was friction = 0.95N.
I used constant acceleration formula to calculate acceleration to be 3.05.
Then
Fnet = Fapplied - Fopposing
3.05 = 4 - friction.
Friction = .95N
I have had trouble finding the friction when the plane was inclined. Do I have to somehow factor in the force down the plane due to the component of weight down? Could someone help please?
If you can see the attachment those are the results that I have. And the graph that Excel produces. Is that right? If yes then what?
I found a question that dealt with an IR emitter and an IR phototransistor receiver. The question asked if the voltage drop across the receiver would increase or decrease when the IR emitter was on. The answer stated that it would decrease, but didn't give an explanation, could someone please explain why the voltage drop would decrease when the IR began hitting the receiver?
Thanks
A geosynchronous satellite is one with a period of 24h positioned exactly above the equator. It appears motionless viewed from the surface of Earth. Explain why it must be in an orbit above the equator.1. I guess it's unlikely VCAA will test this
Two questions. 1: How important is knowing this theory. 2: Could someone explain in layman's terms what is going on.
Thanks.
Hoping someone can clarify this:
If I drop an object why does it stop when it hits the ground? I'm not challenging whether or not it should, but If I was asked this question in a test I doubt I could answer it satisfactorily.
I know that every action creates and equal and opposite reaction, but I don't understand how the size of the normal force can change during the contact time, and how equal and opposite reactions can cause a decelleration as a result of that.
Not sure if I've expressed my self properly here, but I hope someone follows my train of thought...
Thanks
Not sure where to post this but, I'm undertaking physics 1/2 a year earlier this year and want to build a solid foundation for 3/4. I'm aware of that electricity and motion overlap in to 3/4 but I'm not sure what else I can do to prepare myself for 3/4.
I'm looking to get at least 40, or so. What else can I do to prepare better? What test percentages should I be aiming to get? What should I do right now since I have more time?
It might be worth nothing that my confidence for physics is kind of low...I struggled with nuclear physics and will be lucky to get at least 80% on my outcome test when we get our results back.
16KWh means "sixteen kilowatt-hours" or "sixteen thousand watt-hours". What precisely is a 'watt-hour'?
Consider that the watt (W) is a unit of power. One formula for power is Power = Energy/Time. Specifically, 1 Watt = 1 Joule/second. This unit is useful when we want to talk about the energy output of something (such as a lightbulb, you may have noticed things like "60W" or "40W" on lightbulb cartons). Now that we know what a Watt (pardon the pun) is, let's think about what a "watt-hour" is. A 'watt-hour' is the amount of energy that a 1W object would produce in an hour. In other words, this would be (1 Joule/sec) * (3600sec) = 3600 Joules. In short, a "Watt-hour" is a unit of energy, and it is equivalent to 3600J. If you like, you can think of this as the result of multiplying the units 'watt' and 'hour' together.
Therefore, 16KWh = 16000Watt-hours = 16000*3600J = 57600000J = 5.76*10^7 J .
When finding tension in rope or rod of a object going in circular motion, why do some questions consider the weight force in finding T when others don't? And what's the difference between vertical and horizontal motion?
This image below might help:Thanks zealous, what happens in between the points of the top diagram?
(http://www.ic.sunysb.edu/Class/phy141md/lib/exe/fetch.php?media=phy141:lectures:ballonstring.png)
When we're looking at an object in circular motion we've got to consider all the forces acting on it - there's going to be gravity always acting downwards and a tension force which needs to apply an inwards force to the object so that it has a centripetal acceleration. At the bottom of the circle, the tension in the rope has to be able to overcome the force of gravity on the object, and then have the required extra force for the centripetal acceleration, that's why we have at the bottom. At the top of the circle, gravity is working in the direction required for centripetal acceleration, so the tension is decreased as dictated by . If you're looking at the tension when the object is on the edges, we don't need to consider the weight force as it is acting perpendicular to the tension in the rope so there's no component of the weight force working against the tension.
Horizontal circular motion questions usually look like this:
(http://www.a-levelmathstutor.com/images/kinetics/kin-conpend.jpg)
The main difference is that the forces just work in different directions, and there are sometimes angles involved, but you can resolve them using simple trigonometry.
Thanks zealous, what happens in between the points of the top diagram?
apply Newton's second law to circular motion in a vertical plane; consider forces at the highest and lowest positions only
From the study designThank you kel9901!
so don't worry about it
How do we distinguish between v and u, and which to put in a equation in projectile motion?The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity if a height or time is given from this point.
In general I'm struggling to know which equations to apply to different questions :^(
Can somebody explain why when connected in series, a resistor and a thermistor, with the output voltage being measured across the fixed resistor which is placed second after the thermistor, that when the temp increases, so does the resistance of the thermistor and thus the output voltage increases. Im kind of confused with it all.
The simplest thing i do to figure out which equation I should use is to write out all the values for u,v,x,t,a that have been given in the question and also find the one that i need to solve for. Match the equation on the information given. If you aren't already given a diagram, try to draw one and try to draw the values on the diagram to get a sense of where each part applies. You are always given the velocity at the max height and the acceleration (vertical component) so if you split the parabola that makes up the projectile motion into 2, you will always have a final velocity if you use the first half. Then depending on what the questions asks, you can solve from there. You can always solve for initial velocity if a height or time is given from this point.Thanks!
Sorry if i am not too clear or anything. It's my first time actually trying to help someone on here. Usually i just get help haha. I can try to explain it a bit more if you need me to or if no one else tries to help.
Hi, can someone help me with question 3????
if an object of mass m is moving uniformly (constant speed, v) in a circle of radius r:
1. state the equation that relates the centripetal force, F to m,v,r
F= mv^2/r
2. state the equation that relates the period for 1 revolution (T) to v,r
T= (2 pi r )/v
3. from the two previous equations, write an equation relating T to m&r?
Hi guys, can someone help me with this question:
The mass of Earth is 6.0*10^24 kg and the amass of the moon is 7.4*10^22 kg. The radius of the earth is 6.4*10^6 metres and the radius of the moon is 1.7*10^6 metres. The orbital radius of the moon around the Earth taken from the centre of the Moon and the centre of the Earth is 3.8*10^8 metres.
1) Calculate the gravitational force acting between the Earth and the moon
2) Calculate to two decimal places how long it takes for the moon to orbit the Earth in days
3) Calculate the orbital speed of the moon around the earth in km/hr
For electricity: does VIt equal energy or work?
Aren't they the same thing in this context?
I was a little confused about the difference between energy and work
Question about diodes.
There is a battery of 12v, a resistor of 150 ohms and a forward biased diode in circuit. The I-V characteristic of the diode says that the diode needs 0.6 v for current to flow.
Therefore, there is 11.4 volts across the resistor.
When working out the current flowing through the diode, why do we use 11.4/150 instead of 12/150? Since they're in series, shouldn't the current be constant across both the diode and the resistor? Why do we take away the 0.6v before we apply ohms law to find the current?
Yes I understood.
However, why isn't the 0.6 v included? Isn't the current Vtotal/Rtotal?
So what you're saying is that the diode literally takes 0.6v away from the circuit for current to flow through it? But why isn't that voltage counted in the calculation of the total current?
Vtot/Rtot only works if every circuit component you have in your circuit is ohmic, or V/I is constant for every device you have.
Ohm's law isn't V=IR. Ohm's law is actually an assertion that V/I is a constant and devices that satisfy this assertion are ohmic, like normal resistors.
As you can quite clearly, diodes are NOT ohmic. Hence you can't just use normal voltage divider techniques to solve this question. You have to consider the voltage drop across each component separately.
This is what I don't like about VCE physics. Lots of concepts are not taught very well IMO.
Nope. Not at all. The detailed study consists of 13 multiple choice questions, each worth 2 marks. It's either get 0 marks or 2 marks, and you answer on a multiple choice sheet and shade bubbles, so there's no way the computer will pick out working out.Okay, thanks for the clarification :)
However, older study design exams from maybe around 2000-2006 used to have sections for a detailed study where the answers were short answer, so working out made a difference back then.
Hey can someone help me with this question:
a bullet of mass 'm' grams is fired with a velocity 'v' m/s into a wooden block of mass 'M' kg at rest on a friction less table of height 'H'. The bullet is embedded in the block which slides off the table landing at a distance of 'D'.
a) What is the velocity of the block immediately after the bullet is embedded in it?
b) How much time expires before the block lands on the floor?
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.Anyone?
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys :)
Well i have started Electricity now and I've actually forgot a lot from last year about basic circuit stuff and i need some help with some questions from the image below.
For question 1, before the switch is closed is the circuit working at all? Or does closing the switch turn it on? What exactly is the switch doing here? Why does the current increase when it is closed? I thought the current was constant in series.
For question 2, how do R4 and R5 have more current flowing through them than R2 and R3? In parallel current is I(total) = I1 + 12.... isnt it? If the total current at a point before it hits the resistors is the same for both of them, then how can R4 and R5 and a greater current since I(total) = I1 + 12... Do not the first two have the greatest and equal current going through them?
Most likely my basic understanding of this is just all wrong, but hopefully if someone can help explain this i can get back on track.
Thanks guys :)
1) Yes the circuit will work fine without the switch closed. There is still wire connecting all the elements together. When the switch is open, current is forced to flow from A, to B then to C and encounter all of the resistance. So when you close the switch, you provide another path for current to flow, the current can flow from A to B, but then can split off which effectively decreases the resistance as more current can flow through the circuit.I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?
2) Similar reasoning to the previous question. Resistors in parallel will have a lower overall resistance than resisters in series. This is because by putting element in parallel, we provide more pathways for current to flow in which actually decreases the resistance overall. So, let's just imagine the resistor value for all the resistors in question 2 was 4 ohms and the voltage source was 12V.
For circuit 1, the current flowing will be V/R which is 12/4=3A.
For circuit 2, we can imagine R2 and R3 as one resistor of 8 ohms, so the current will be 12/8 = 1.5A. So 1.5A of current will flow through R2 and R3.
For circuit 3, since the elements are in parallel, they will have the same voltage drop or potential of 12V. Using V=IR, I=V/R which for R4 would be 12/4=3A.
You can see that the current through each resistor in the third circuit is double the current in the second circuit because they are in parallel.
I've run into some other questions where there is a switch like in question 1 of the above question. Because it isn't actually in parallel with another resistor does the current still split between them or does it all go through the switch circuit and turn off the light at C? Is it because there is no resistance in the other path and it is easier to pass through it?For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.
For example, for 1c. I don't think Globe C lights up because when the switch is closed it creates a short circuit. I'm not sure if this is the sort of question you are referring to.Yeah this is what I'm talking about. Why does that happen for?
Yeah this is what I'm talking about. Why does that happen for?
EDIT: Another question. Is this a correct way of drawing the circuit in a more simplified way?
EDIT 2: Wait I'm pretty sure it's wrong because the 10 and the 30 should be in series with the 5 right? Not parallel. And the 5 and the second 30 should actually be in parallel i now think. So would the 3rd picture be correct now?
Yeah picture number 2 is wrong, number 3 looks right!Everything makes much more sense now. Thanks so much!
You could 'clean' it a bit by drawing the 5 resistor going vertically, and then joining it with the 30-branch. Then it will look more like the circuits you are used to. You can't see the numbers but this is the shape I'm talking about:circuit(http://i.imgur.com/2EQNIu8.jpg)
I'm not sure if it's a technical term but I call this 'linearising' a circuit. The easiest way to do it is to start at the battery and trace yourself around the circuit, adding components in series and then branching into parallel when the circuit splits. When the splits come back together, you close those parallel bits off. This is really hard to explain by text haha.
As for the short circuit question, yeah no current will flow through a resistor if there's an alternate path with zero resistance.
here's my way of justifying it: consider a circuit with just a resistor in parallel with nothing (one of the branches is just an empty branch with only wire), and maybe another resistor somewhere else in series. The effective resistance is 0 because of the resistor-less branch, meaning the voltage across the component is also 0. Having no voltage across the resistor-less branch isnt a problem, current will still flow. But if there is zero volts across the resistor in the other branch, no current is going to be pushed through it.
When they say initial velocity does this always mean that time=0secs?Yeah, when t=0, v=u
When is it like or not like this?
When they say initial velocity does this always mean that time=0secs?
When is it like or not like this?
Depends on the context. Usually, yeah, initial velocity refers to velocity at time t=0. But t=0 is kinda arbitrary, right? Otherwise t=0 might refer to the start of the universe or whatever absolute zero point you want to define. The idea is, we define a zero point that works best for us.
In many kinds of motion, we set t=0 to the start of that motion of interest, so 'initial velocity' relates directly to that starting instant.
If we're ever talking about a change in velocity, initial velocity will refer to the start of the change.
Sometimes, in projectile motion questions, we'll deal with different values of u, (different 'initial velocities') at different points in the question. For example if you want to find the time taken to fall from the peak of the projectile path to the landing point, you'd consider the start of this motion to be the peak of the path, and hence vertical u=0. In another part of the question, you might take u to be the launch speed because that's where you're measuring from (this is far more common).
It would just increase linearly because it has no friction
The question hasn't actually said the acceleration is positiveLzxnl 1 odeaa 0
How would you draw a velocity time graph for the following situation. A particle starts with a positive velocity and undergoes constant acceleration until the end of the time period?
I'm interested to know what should I be doing in units 1/2 if I'm aiming for a 40SS for 3/4? I've been putting a significant amount of effort in to electricity and will be doing the some for motion but is there anything else?
Pretty sure only motion and electricity carry forward to 3/4, sounds like you're setting yourself up with some solid foundations for these topics in year 12. Place emphasis on developing really solid problem solving skills! You can of course work on them in year 12 as well, but you should ideally be able to tackle any practice question you see without any trouble.
I started doing Checkpoints Physics 1/2's electric circuit chapter, I'm getting most of the questions wrong or don't understand the terminology...killed my self-esteem.
I have a couple questions about diodes:
Firstly, I don't really understand the purpose of diodes. What are they used for exactly and where? How does forward and reverse biased work? Like why does it have almost no resistance and then infinite resistance? I don't understand. Why do they have a switch on voltage? Before the switch on voltage is reached, my book says that the diode does not conduct. Does this mean, if it were in series for example with another resistor, the circuit would never actually turn on/be complete? Is this phase exactly the same as what happens in reverse bias?
Secondly, in class we went over how they work, a lot of the chemistry behind them, but I really didn't understand most of it. Do we need to know this at all for any reason? Would someone be able to explain how they work anyway because I still want to understand it. I looked at a couple videos online but i still don't really understand.
Thanks :)
^Holy crap, thanks so much for all that information. That toll-gate analogy actually helps a lot :)
Can someone help me with understanding questions? It seems like a broad question but I think that I understand the theory and concepts behind electronics but I can't seem to interpret the questions and understand what they want me to do.Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.
It ranges from the different terminology used in questions and just sometimes just being confused..
Keep doing the questions. If you get them wrong or don't understand what it was asking specifically, work it out by looking at the answers, asking somebody or even posting here. You said that you understand the theories and concepts, so even if you don't understand a certain question, once you have it explained you should try to relate it back to what you know.Cheers for the help, I agree I probably do have gaps in my knowledge without me knowing. I'll definitely post questions I have trouble with in the future.
Try to find patterns in questions. There is only a certain number of ways they can ask you things. Maybe it involves graphs of ohmic devices. Gather a whole heap of questions and attempt them. Find patterns in them and maybe even methods or steps of what to look for first when solving them.
Maybe you do actually have some gaps in your knowledge and you don't actually know what you think you should know. If this is the case, broaden your knowledge from more than just the textbook. Look at videos online about the concepts. Maybe they will fill in the unfamiliar terminology your textbook is using or will just give you a different perspective of the topic.
I think you should start by just posting one of the questions you are having trouble with and getting help from us with it.
A scientist has a 120g sample of the radioisotope polonium-218. The first three steps in the decay series of polonium-218 are an alpha emission followed by two beta particle emissions. These decays have half-lives of 3, 27 and 19 mins respectively. Calculate how many polonium-218 remains after 15 mins.
Realise this is year 11 physics. Should totally know but don't despite doing 3+4 this year. T.T
Please help?
When it comes to photonics, what theory is a must in knowing? Like modulation, demodulation, how about attenuation? Particularly from the Jacaranda text book.
Also How do I do sketch modulation graphs? :-[
We have our electricity SAC coming up in 2 weeks, we're allowed to bring a double sided cheat sheet. I've just finished mine but I have a lot of spare space - a whole page worth. I'm wondering if I should use that space to write questions I had trouble on or to use it for additional information about the less important things such as electric shocks and resistivity of your skin, I know it's going to be on the MCQ as well.If you know what is coming up on the test for the most part and you know what you will struggle with, definitely try to fill it up with that. Specific examples of questions you struggled with. I think simple sentence definitions could suffice for the less important information or even just try to rote learn it.
This is for unit 1/2 if you guys don't mind, thanks.
Was just wondering, is it possible to find the elastic limit of a material without a stress-strain graph?
Thanks :)
If a resistor or even another diode is placed in series in front of a reversed biased diode (the current flows through the resistor before it hits the reverse biased diode), why doesn't current still flow through. For example why isn't any power dissipated? The current has to go through the resistors and such to get to the reverse biased diode?
What happens to the current in the diode when its in reverse biased. Does it send it back the other way? Does it just store it? Or does it get used 100% as heat energy?
For diode labeling conventions. Are we supposed to always change the polarity of the circuit or can we just draw the diode facing the other way? Does it matter at all? Because sometimes I just don't look at which way the circuit is and assume a diode is in forward bias when it is really in reverse.
What exactly determines the switch on voltage of a diode. I know the material but what are the differences in material. Is a diode technically still a resistor? Are there other non-ohmic resistor type things as well?
http://imgur.com/wPXnz5RWell first you can find the total resistance which should be
In this circuit, what is voltage at X? I got the answer but I'm still confused on the method of getting the answer since I just guessed. We found the total resistance in the question above.
I'm not sure if it's possible without SOME knowledge of the behaviour (a graph, a table of stress and strain values, a force/extension graph), like there's certainly no VCE level formula for calculating the elastic limit of a material based on other quantities. There may be some formula that involves a lot of other properties of the material (but I doubt it).
Good question. Current in series has to be the same everywhere. Think of a road with a fixed amount of lanes and lots and lots of cars. The rate of the slowest moving cars determines the rate of all cars because otherwise fast moving cars would catch up and crash into slower cars. This analogy has plenty of holes but the basic idea is that current cant be high in some sections and low in others if those sections are in series.
So, even when current wants to leave the battery, the electrons can't physically go anywhere because there's a blockage all the way along the wires at the diode. (a reverse biased diode is just like a hole in the circuit - no current flows around then either)
Thus, no current flows through the wires, or the resistor, and that's why P=VI=V*0=0.
Same explanation as first question, there is no current in this circuit. Nothing happens with it. If a small amount of current were to flow (all real reverse bias diodes let a tiny leakage current through, i think), it would just go round the circuit like normal - the diode acting like a very high-resistance resistor.
Conventional current flows out of the bigger terminal of the battery (the positive one) and if that direction around the circuit points the same way that the triangle points, the diode is forward biased.
Electrons actually flow the other way in a real circuit, but the convention still holds.
Always watch out for trick questions 'what is the current in this circuit' when they sneakily put it in reverse mode.
If you're asked to draw a circuit with the diode in the other mode, it's probably easiest to reverse the diode direction rather then the battery direction: it's a lot easier to tell the difference for the diode. An assessor might not see that you changed the battery.
This is to do with the magical chemistry that makes diodes function so I'll let someone else try to answer it.
BUT for the other bit, a diode can be thought of as a resistor but it has no well defined resistance because it is non ohmic (resistance is not the same for different voltages/currents, V-I graph not linear, etc)
You can calculate R=V/I for any given situation but your result is meaningless for the same diode in a different circuit. It's merely the 'effective resistance' of the diode, if you replaced it with a resistor of that same resistance you would get the same behaviour for that circuit.
Other non ohmic devices studied in VCE include temperature dependant resistors, light dependent resistors, and the other types of diodes (LED's and photodiodes)
Well first you can find the total resistance which should beThanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?
Then the total current should be
The current though 12 ohm resistor should be since votlage is constant in parallel.
That leaves 2 amps to be found in the series part of the circuit and since current is constant in series they both have the 2 amps through them.
That leaves the voltage across the 2 ohm resistor to be
The voltage in the 4 ohm resistor is
You can verify this because voltage is constant in parallel and the 2 series resistors voltage add up to 12 (4 + 8 )
I'm pretty sure this is right. I'm notorious for forgetting at least something which stuffs up all the other calculations. It might be a long working but i hoped it highlighted everything for you and answered the question :)
Thanks for the fast reply. Additionally, could I find the current of the first arm which would be 2A then substitute to find the voltage at X? I did it like that, would that also be correct in terms of method?If you mean find the total voltage of the series part of the circuit, then you would get a different answer to the voltage at X. You would get 6*2 = 12 V, which is the total voltage across the series arm. From there you could go and find the voltage across the 2 individual resistors, which is exactly the same as doing what i showed anyway. Otherwise as long as you get the correct value for current and you use it to find the actual voltage across X and not the entire series arm, then yes. The method would be correct.
Thank you for clearing everything up for me again. I really appreciate it :) For the part about the current flowing into a reverse biased diode: I know the answer stays the same, but does that mean that for the instant when it is turned on, current will flow up past everything to the diode? Also when you turn off the power source to a circuit, do the electrons that were currently in the circuit all leave the wire/circuit back into the power supply or can they get 'trapped' in there? Thanks again!
Thanks again Silverpixeli! But alas i have more questions again. This time about voltage amplifiers. I went over it in class today and we watched a few videos and i even watched the VTextbook video on it just then, but I'm still confused.
Firstly, i don't really understand how a transistor works in a circuit. From what i understood, it has a voltage input that goes through the collector and then one that comes through the base? If that's correct, are they actually 2 different power sources or do they come from the same power supply. If so, then how do you regulate how much voltage goes through it from the base input? From a picture i saw in class today, the base is apparently usually connected to a capacitor? I don't really know what a capacitor is either, but doesn't that store energy/voltage? Is that connected back to the same power supply of the circuit if that's the case. How do you regulate how much voltage goes through that anyway?
Next, from what i gathered, when no input voltage comes through the base, the resistance is infinite since it is basically 2 diodes facing each other and one of them will be in reverse bias? When there is a voltage input from the base that goes above the switch on voltage it will work and the resistance will basically be 0. How does the 'switch on voltage' allow the diode to work if it is approaching it in reverse bias? You would need like 50V to break it but it is still only about 0.7 V according to my book.
Even if that all makes sense to me, i still don't really know what it does. It amplifies the voltage but how? Is it because you are basically adding 2 voltages from 2 power sources together. Almost like as if you were adding ordinates of 2 sine graphs? I hardly understand what all the graphs are about either, but if someone can enlighten me on how all the above works, maybe i will be able to figure it out for myself.
Thanks :)
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?
Hi guys, just wondering, is it possible to make a black (light emitting) LED/Torch? For example, if you were in a brightly lit room (such as an art gallery) with lots of big blank white walls and you were to shine this torch, it would make a dark patch..?
I'm having a bit of trouble with the attached question. I have looked at the worked solutions for question 3 and i understand the steps but i don't understand how they thought the steps up and how they got to them. I find that a lot of these ratio questions stump me, especially when they were used in Gravity and Satellites. I can kind of see how it wants me to answer it but i can never actually go through with it and get the correct answer. Can someone show me how they went through the steps to solve this? Thanks :)
Another question. There is a circuit with 3 identical resistors of 100 Ohms that are connected with one in series and then the other 2 in parallel and the max power from any one resistor is 25W.
The question was to find the max voltage the can be applied. I used and had already found the total resistance and went for the 3 resistors, but my answer seems to be wrong.
Why are you adding ohms and watts together? I assume typo?Yes, that was a typo. My original workings were and solve for V such that . The total resistance is 150 ohms and if each resistor can have a max output of 25 W, then the total power output could be 75 W, so . Is the total power across the parallel resistors not 50W? Because i'm pretty sure my answer is wrong because of my watts working out. If the total voltage is the same and the resistors are identical, the the current is exactly split across them, which would then give 12.5W + 12.5W each, changing the equation to , but that also doesn't give the correct answer. The question was what is the total allowed voltage that can pass across them. I might still be interpreting it wrongly though. Should i try to find the voltage across the parallel and series separately and then add them together?
Okay so just because the maximum power of one resistor is 25 Watts doesnt mean it's operating at that power. That just means that as you increase the voltage applied, you have to be careful not to take ANY of them over the max. In particular, With more resistance and twice the current, the resistor OUTSIDe of the parallel section will probably have the most power at any given time. So you should solve for the total voltage in the circuit when all you know is that the series resistor has 25W, 100Ohms.
Yes, that was a typo. My original workings were and solve for V such that . The total resistance is 150 ohms and if each resistor can have a max output of 25 W, then the total power output could be 75 W, so . Is the total power across the parallel resistors not 50W? Because i'm pretty sure my answer is wrong because of my watts working out. If the total voltage is the same and the resistors are identical, the the current is exactly split across them, which would then give 12.5W + 12.5W each, changing the equation to , but that also doesn't give the correct answer. The question was what is the total allowed voltage that can pass across them. I might still be interpreting it wrongly though. Should i try to find the voltage across the parallel and series separately and then add them together?
What does it mean to have a negative voltage? Is it only when you have a photodiode or something in reverse bias and hence a negative current?
Firstly, what is a voltage?Legend, thanks for clearing that up
A voltage is a potential difference between two points. Well that doesn't help, does it?
The electric potential is defined as the electric potential energy per unit charge. So you know how things like to go from high to low potential energy? For a positive charge, this means go from high to low potential but for a negative charge, it means it'll want to go from low to high potential.
So a negative potential just means you've measured a positive potential in the other way. This has implications for devices that are direction-specific, like diodes.
Hey guys. I have a couple questions today about Voltage RMS.
I have a physics SAC tomorrow on the first half of the Electronics and Photonics Area of Study and my teacher said today that there would be a question about Voltage RMS on it, even though we haven't actually come across it in the book yet. He explained it to us and basically told us that the question was just reading the peak AC current and finding the RMS Voltage, which is easy enough. I understand the formula and how to use it for the question we will get, but i still don't understand how they get it. In the Heinemann textbook it comes up in the further electronics AOS and quickly in Electric Power AOS, but I still wasn't able to understand it.
So basically my questions are:
Where/how do they derive it?
What is it? I gathered it was like the average voltage provided by an AC signal. I also read that it is the same the DC voltage for something? What is the relationship between them?
How does the current RMS work? Where do you ever find the peak current? Are there usually graphs of these or is it something you can solve after finding the voltage RMS?
Also i have some questions about AC and DC signals that i don't think i ever really understood, and because we are finally using them a lot more now i think my knowledge isn't really matching up with what i'm learning. And some other random questions if anybody can be bothered answering them
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
How do you actually make an alternating current? Are magnets used to do this or something else?
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.
Thanks! :)
i'll just answer the first part.Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)
Try to think about it this way.
In AC, both voltage and current have a sinusoidal shape. When multiplied together to give power, it will be a graph that looks something like (sin(x))^2, ie something like this http://imgur.com/lMxj3Hf. It's pretty visually obvious that the 'average' power is half the 'peak' (top) power.
Hence, P(RMS)=P(peak)/2
V(RMS)I(RMS)=V(peak)I(peak)/2=(V(peak)/sqrt(2))(I(peak)/sqrt(2)
V(RMS)=V(peak)/sqrt(2), and I(RMS)=I(peak)/sqrt(2).
The RMS voltage of an AC signal represents the DC voltage that would provide the same average power.
current is pretty much treated the same way a voltage.
Okay, the explanation makes a lot more sense, but I'm a bit confused as to where the root2 came from though. Maybe it's just the formatting that's confusing me. Can you explain those steps? Other wise thanks, it all makes a lot more sense now :)
Can someone please help me with the electricity question attached?For the first one, all we need to do is go find the voltage across the 6000 ohm resistor done by V = IR. Voltage is constant in parallel and there are no other components in the circuit, so that equals the EMF of the battery.
Thanks :)
What actually happens in an AC supplied circuit. If it was connected to a light, does that constantly turn off and on?
No, the frequency (how quickly the current goes up and down) is so high that humans don't notice it; it appears to stay at constant intensity
What does a resistor actually do to the electrons in a circuit? Does it slow them down so less can pass through or..?
Pretty much slows them down, yeah. The resistor pretty much makes the electrons give up some energy (potential) to get past it.
In a normal DC circuit for example, do the same amount of electrons keep running the circuit or do they just get replaced each time they finish it? Also, how do they get the electrons in the first place?
The same electrons run the circuit, and are 'recharged' or given back the potential they lost when going through the resistors when they reach the battery. The electrons come from the wires/resistors, as they are usually metal and hence have free moving electrons.
What are some different uses for AC and DC current? Like which one is usually used in our lights at home etc..?
I think pretty much everything uses AC. The main exception I can think of, aside from lab experiments (lol), is trains, which use DC power.
How do you actually make an alternating current? Are magnets used to do this or something else?
You'll learn this in unit 4, but yeah, magnets.
Some of these might have simple answers that i probably already know. I feel like I've been trying to think to deep into some of these things and I keep confusing myself and throwing what i already know out the door. The first part is what i need help with now anyway, so if anybody can help with anything at all that would be much appreciated.
Thanks! :)
Thanks!
What's the best way to revise for a mid-year exam? Practice exams only?
Yolo, whats revision for practice exams?! Haha,our teacher gave us a USB with all the VCAA exams back to 1997, ask your library I think they will have a CD with all the past exams
Yeah, thats what ill be doing! Smashing dem out. If i can find them. Do you know any sources that has a bunch of them?
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?
Thanks!
Can someone help explain photodiodes to me. I'm still kind of confused how they work and especially how the I-V characteristic graph works. I'm not really interpreting it properly. If they are placed in reverse bias, how do they even work? I think i'm a bit confused about the "dark current" and how it is negative. Also, what is the photoconductive mode and the photovoltaic mode?
Thanks!
EDIT: Whoops i had another question as well (now attached). I don't exactly understand how they got that equation for part b, the number of photons part. Is this something we should know as well. Because i have never really come across anything like it where we introduce something into an equation like that.
quick and dirty answer because i gotta run, but basically you place the photodiode in reverse bias and when light lands on it it lets a little trickle of current through (negative w.r.t. photodiode, but since the diode is reversed this is positive w.r.t. circuit as in, it's in the same direction as the voltage drop across the photodiode and the way conventional current would flow if there wasnt a diode there blocking it). usefully, this means we can get current that depends on light. the dark current is just the trickle that leaks through the diode when there's no light falling on it.
The graphs for these are most interesting in the negative voltage region (whichever of photovoltaic/photoconducti