VCE Physics Question Thread!

[jssantucci]

@SocialRhubarb this is what my text book says in the chapter about transformers...

When there is no load connected to the secondary coil no current flows in the secondary and so all the flux is being generated by the primary coil. However, this flux will induce a voltage in the primary coil that will oppose the change of flux. That is, as the current increases, this ‘back EMF’ will tend to reduce the current. In a good transformer, this process is very effective and very little current will flow in the primary if none is flowing in the secondary.

Not really sure how this works but it seems to be the answer to the question. Could someone clarify the above idea about 'back EMF' for me?

Thanks

[SocialRhubarb]

Hmm. I'm familiar with the concept of back EMF but I haven't seen it being induced in a solenoid by its own flux.

Basically, in a simple DC motor with one loop, when you run a current through the loop, it runs through the magnetic field and generates a force, and your motor starts turning. But when your motor is turning, its flux is constantly changing. We know that a change in flux through a coil induces a voltage in that coil, so the changing flux actually induces an EMF which counteracts the voltage we applied to make the motor spin. This EMF we call a 'back EMF'.

[Guest]

I originally posted this question in my unit 1&2 Physics thread but people suggest that it should be of more relevance here on this thread.
Question:
1) Sound travels with a speed of 343m/s in air. Find the speed of sound in steel where K=200 GPa and p= 7,870 kg/m^3, and in water where E=2,200 GPa and p=1000 kg/m^3. Compare to the speed in air.

Normally I can solve these questions by simply plugging the values into the equations but in this case the value of K is provided for a solid object and the value of E is given for the liquid. Shouldn't it be the other way around? Would some conversion be necessary? How would you "compare to the speed in air"?

The answer states that : the speed is 5,040 m/s in steel and 4.7*10^4 m/s in water. Could someone please explain how one might reach this conclusion?

Thanks, in advance

[Robert123]

What would be the direction of the magnetic force and how do we work it out?

You missed an important piece of information there, the direction of the current (read 4 lines down from that picture in 4a & 4b).
From there, you use the right hand slap rule, fingers pointing to the South Pole and thumb down for 4a and up for 4b.
Clear?

[lolipopper]

i dont get the question from Hinemann Physics 12 Chapter 11 Exercise 11.2, question 4c.

the question is attached.

Lots of love. ahha.

sorry *i didnt notice the questions thread*

edit: damn sorry guys i just understood it. tell me if you want me to explain it.

Moderator edit [2/cos(c)]: Merged double post

[Nato]

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

[lolipopper]

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?

[lzxnl]

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?

Actually your interpretation is slightly off as well. Impulse is just change in momentum. For a given collision, their initial and final velocities are going to be the same, so regardless of the time taken for the collision, the change in momentum is the same. The time, however, affects the force and acceleration that will be felt by the object as mentioned.

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.

[Alwin]

More time does not mean less impulse. It does, however, meant less force. Force and impulse are not the same quantity.

Since you're always soo nit-picky , ill add to this too: same impulse over the longer period of time means smaller average force (usually meaning smaller maximum force too). sorry, just had to go there


From a few posts ago:

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

Since it's rather a lot to explain, I found a link that may interest you:

http://sound.westhost.com/xfmr.htm

Also, just something interesting from the Heinamann textbook:

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.
p369 of the Heinemann textbook

Eddy currents are present in the core of transformers too.

I quite like how it includes topics outside the 3/4 course (I first came across this stuff in Singapore coursework so not sure what level it is in australia) but you might find it interesting too!

[jssantucci]

http://www.physics.usyd.edu.au/~khachan/PTF/Transformer%20explanation.pdf
 
for anyone interested, this link explains the answer to my question quite nicely.

[Homer]

i know the top of the beam experiences compression and the bottom tension, but what about the centre? is it a little of both?

[Phy124]

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.

These diagrams indicate the bending stress (RHS) and strain (LHS) patterns a beam which is symmetrical about its neutral axis experiences.



As you can see when the beam is in its elastic state the bending stresses have a linear relationship with distance from the neutral axis. However, when the beam is plastically deforming the bending stress is constant throughout.

*Note: As far as I know for VCE Physics purposes you only need to know about the elastic state for bending stresses.

[sin0001]

I know that the force produced by the presence of current in a magnetic field is a result of the interactions of the conductor & magnet's magnetic fields, but can someone explain why this force is at MAX. value when the external magnetic field is perpendicular to the conductor?

Also, is 'Forces on moving charges' part of the course?

[lzxnl]

Because VCE physics is inadequate and doesn't give you the full picture. Period.

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.

[sin0001]

Because VCE physics is inadequate and doesn't give you the full picture. Period.

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.
Didn't the formula for cross product contain a 'cos', where did the 'sin' come from? O.o
Thanks, sorta makes sense