HSC Physics Question Thread

[Neutron]

You'll be fine Neutron!! Just relax mate, you'll be sweet 

First question, leave it with me, nothing strikes me immediately.

Next, the answer is definitely D. Orbital decay causes the velocity of the satellite to decrease, thus causing its orbital radius to drop. However, the drop in orbital radius causes an increase in orbital velocity (smaller orbit, needs to move faster to stay in orbit), so it actually ends up travelling faster than it did before the drop. The net effect? Orbital decay causes satellites to gradually spiral inwards towards earth, travelling faster and faster as it goes.

Next, there is definitely changing flux in the answer. However, the changing flux on opposite sides of the loop cancel each other out. One side moves down, the other moves up,, the net effect is no change in magnetic field strength and thus no induced current. Every change in flux on one side is opposite in sign and equal in magnitude to the change in flux directly opposite. Everything cancels, no induced current. The answer is therefore B. Every other example, there is some change in magnetic field strength in some part of the loop. The diagram is pretty bad, the magnet is at the dead centre of the loop, which is why this occurs. If the magnet was somewhere closer to one side of the loop the outcome would be different.

Question 11, I agree with you, I would say the answer is C not D (any takers? Anything I missed?)

Question 7, you would normally need to calculate the velocity of the earth's surface at the equator. We can do this by considering the distance travelled (one circumference, the length of the equator) divided by the time taken (24 hours). However, the question doesn't really need it because it's the same number everywhere. If we launch from Woomera, the surface of the earth isn't moving as quickly as the equator.

To demonstrate this, pick up a ball of any kind. Put black dot on the very edge of the ball (where the equator would be), and then put one lower down. If you spin it, the black dot on the equator rotates faster. Why? Because it has a greater distance to travel in the same amount of time. The 'circumference' circumnavigated from Woomera is lower. Thus, the velocity is lower. We won't get as much benefit there. Therefore, the answer must be D.

omg you guys are actual legends, but just with the satellite one, if the surface velocity is lower in Woomera, how is that an advantage? And since all satellites have to have their centre of orbit pass through the centre of the Earth (this is going to be really dumb) why does that satellite go west to east? Like I know normally satellites go that way because that's the way the Earth rotates and therefore rotational velocity helps blah blah but if it starts off in Australia, doesn't it have to go up and around the Earth? So like South to North in order to sustain its orbit?If that does happen, and it just has an orbital inclination, won't it still be travelling at the same speed as a geostationary orbit? Or are you allowed to just have a tiny orbit around the bottom of the Earth that's not passing through the centre?? Thank you so much

[jamonwindeyer]

omg you guys are actual legends, but just with the satellite one, if the surface velocity is lower in Woomera, how is that an advantage? And since all satellites have to have their centre of orbit pass through the centre of the Earth (this is going to be really dumb) why does that satellite go west to east? Like I know normally satellites go that way because that's the way the Earth rotates and therefore rotational velocity helps blah blah but if it starts off in Australia, doesn't it have to go up and around the Earth? So like South to North in order to sustain its orbit?If that does happen, and it just has an orbital inclination, won't it still be travelling at the same speed as a geostationary orbit? Or are you allowed to just have a tiny orbit around the bottom of the Earth that's not passing through the centre?? Thank you so much

So launching from Woomera is less advantageous than launching from the equator, but it is still advantageous. You still get a boost, just not as much  and you are correct, we can't just have a random orbit that doesn't have the centre of the earth at its focus. However, the idea would be launching the satellite from Woomera, then adjusting once beyond the atmosphere to place the satellite in a proper orbit, probably just by bending the trajectory downwards until it is wrapping diagonally (picture a line with gradient -1). The point being, the orbit can be established later, we still like the speed boost 

[Aliceyyy98]

Sure!

Okay, so let's just assume we are working with electric field deflection (you can apply the same idea to magnetic). The idea is that by controlling the electric field strength and direction, we can change the path of the electron beam. In this case, this would be done by applying different voltages to the deflection plates 

Let's say we can apply a voltage between 100V and -100V to the plates (same max magnitude, just opposite polarity) to a pair of plates controlling vertical deflection. 0V would mean the electron beam is unaffected, and travels straight. If we apply 100V, the beam might (for example) be deflected such that it hits the very top of the screen. 99V means it hits JUST below that. 98V, just below that again. All the way to zero where we are back in the centre. Then, -1V, just below the centre. And the process continues, all the way to -100V where the beam hits the bottom of the screen.

If we do this for both horizontal and vertical deflection, we can hit ANY point on our two dimensional screen. The top right corner might be 100V applied to vertical plates, 100V applied to horizontal plates. A point in the bottom right would be -100V vertical, but still 100V horizontal. We adjust the voltages of each pair to move up/down, or left/right.

We can do creative things with this ideas (again, just assume electrical is used for everything to keep things numerically simple). For example, in an Oscilloscope, the horizontal plates have a time varying voltage. We start by applying -100V, and the beam is at the very left. We gradually increase this to 0, and then to 100, to cause the beam to sweep across the screen. Then, when we hit 100V, we immediately reset to -100V and start the sweep again (for prospective electrical engineers, this signal would be called a sawtooth wave).

So we have an electron beam sweeping across the middle of the screen (maybe once a second). If we then connect the vertical deflection plates to some source we want to measure (for example, an electrode connected to the chest to measure heartbeat), that means the electron beams vertical position is dictated by what we are measuring. Thus, we then have your typical cardiograph, with the heartbeat visible as it fluctuates over time! The beam sweeps across, and jumps up and down as it sweeps in correspondence with the heartbeat.

Television screens are more complex, but basically, both the vertical and horizontal deflection plates are time varying. The periods are set up such that the beam does a sweep across the top row of the screen, then the next row in the reverse direction, then the next row, etc etc. Remember Donkey Kong? How you'd go all the way to the right, then jump, then all the way to the left, then jump, etc? That's what happens here (roughly speaking). It's a little tougher to picture, but let's say we start in the top left corner:

Horizontal = 100V, Vertical = 100V

Then we want the 'phosphor dot' to its right:

Horizontal = 99V, Vertical = 100V

We keep sweeping to the right until we get to the top right corner:

Horizontal = -100V, Vertical = 100V

Then we shift down:

Horizontal = -100V, Vertical = 99V

Then we begin sweeping left:

Horizontal = -99V, Vertical = 99V

And the process repeats in this fashion 

Thanks heaps! Just one more question sorry my trials is tmr what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!

[jamonwindeyer]

Thanks heaps! Just one more question sorry my trials is tmr what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!

I think that it would not have too much effect! What have you read? 

[Spencerr]

Thanks heaps! Just one more question sorry my trials is tmr what happens when you place a glass block in between emitter and receiver for hertz radio wave experiment? Different places say different things!

Hey there, so what the Hertz radio wave experiment did is that it (serendipitously) discovered the photoelectric effect (Which was the emission of photoelectrons from the surface of the electrodes when exposed to UV light). Hertz decided to test this and so he placed a glass block between an emitter and receiver. He observed that the maximum spark length in the receiving loop DECREASED and when the glass block was removed, it was INCREASED. This suggested that the glass block in fact blocked or absorbed the UV whilst allowing the radio wave to pass through. He then did a similar experiment with quartz but found that quartz did nothing as it allowed the UV rays to pass.

[MysteryMarker]

Hey Guys

Been stuck on this question for quite some time now. I've tried GPE and KE formulas and i seem to be either getting A or C, but the answer is B.

Cheers Guys.

[conic curve]

As a result of being hit from behind by a toy truck, a 500g toycar, initially at rest, rolls 12.0m across a floor that applies a constant retarding force of 1.2N to it. The car stops 2.0s after being hit. If the truck was in contact with the car for 0.12s calculate the impulse given to the car.

[Aliceyyy98]

Hey there, so what the Hertz radio wave experiment did is that it (serendipitously) discovered the photoelectric effect (Which was the emission of photoelectrons from the surface of the electrodes when exposed to UV light). Hertz decided to test this and so he placed a glass block between an emitter and receiver. He observed that the maximum spark length in the receiving loop DECREASED and when the glass block was removed, it was INCREASED. This suggested that the glass block in fact blocked or absorbed the UV whilst allowing the radio wave to pass through. He then did a similar experiment with quartz but found that quartz did nothing as it allowed the UV rays to pass.

Thank you!
________________

Hi guys,

Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??

Thankyou!

Moderator action: In these situations, please try to merge your posts

[RuiAce]

Hi guys,

Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??

Thankyou!

A loudspeaker has one of those E-shaped magnets, where the two outer prongs are of the same pole and the inner prong is of a different pole. I.e.

----N      ----S
----S  or ----N
----N      ----S

A paper diaphragm is connected to the ends of the magnets. The diaphragm is essentially what will provide the sound waves.

An image from Physics in Focus is attached.

How are the sound waves generated? The coil is connected to the rest of the circuit. Now, the electrical signals are sent such that the nature of the current is AC, not DC. Note that from the diagram, because of how the coils are wrapped one end shows current going into the page, whereas the other end shows current going out of the page. (Do you see why? You have to analyse how the current is travelling in a loop going away from you.)

The motor effect states that a current carrying conductor experiences a force when placed in an external magnetic field. The right-hand push (aka palm) rule is demonstrated in the diagram. Overall, the wire will be pushed off the magnet.

But recall that this is AC current, not DC. When the direction of the current reverses, the right-hand push rule now predicts that the wire will be pulled back in again! So what should happen is that this wire is moving in and away from the magnet recursively.

But there's a problem - the wire is too tightly wounded. This is the point - because the wire is so tightly wounded, all it will do is vibrate on the magnet. Because we now have these vibrations, that's how the sound gets produced!

As for the nature of the sound:
If the frequency of the electrical signals increase, the pitch goes up.
If the amplitude of the electrical signals increase, the volume goes up.

(I'll let someone else do the galvanometer. Lost experience to word it properly.)

[jamonwindeyer]

Hey Guys

Been stuck on this question for quite some time now. I've tried GPE and KE formulas and i seem to be either getting A or C, but the answer is B.

Cheers Guys.

Hey! Let me have a go, that's the approach I'd take!

First, let's consider the loss of potential energy between the two positions (difference of 1 metre). In this case we approximate with the following formula:



Now let's see how much kinetic energy it has by the end of that drop!



We have a difference in energy here of 0.26 Joules, and this is what I'd say would be the energy converted to other forms in the bulb. So I actually get D.

Hmm, it's late, I'll have another look in the morning, any ideas people? 

[jakesilove]

Hey! Let me have a go, that's the approach I'd take!

First, let's consider the loss of potential energy between the two positions (difference of 1 metre). In this case we approximate with the following formula:



Now let's see how much kinetic energy it has by the end of that drop!



We have a difference in energy here of 0.26 Joules, and this is what I'd say would be the energy converted to other forms in the bulb. So I actually get D.

Hmm, it's late, I'll have another look in the morning, any ideas people? 

I continuously got that answer, no matter what method I used (even tried a projectile motion version!) but figured that it was late, so I was getting something wrong. If Jamon got the same answer, I'd say that D is correct!

[jamonwindeyer]

I continuously got that answer, no matter what method I used (even tried a projectile motion version!) but figured that it was late, so I was getting something wrong. If Jamon got the same answer, I'd say that D is correct!

Really? Awesome, I just did it again and I keep getting D as well. B seems dubious to me because it is more than the kinetic energy of the magnet at the end, it seems a little out of the ballpark

[Spencerr]

Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??
Moderator action: In these situations, please try to merge your posts

For the Galvanometer.

There are loops of coils wrapped around a soft iron core (soft iron core is used to concentrate the B field). The coils are connected to an external circuit which provides an input of DC current.
The soft iron core is placed inside a radial magnetic field (which ensures that the Magnetic field strength is always at a maximum no matter how much the coil turns).
Now remember in DC motors, that if you input current into a coil, that is in a B-field, the coil will experience a force (Via the Motor effect) and this force will produce Torque (T = FD) which causes the coil to turn.
Same situation here, when current passes through the coils wrapped around the soft iron core, the coils will rotate in one direction, however it is counterbalanced by a torsional spring which produces a retarding force on the rotational motion. This counterbalancing spring is the reason why the reading at 0 when current is 0.
Since F = B I L sine theta, sine theta equals 1 as there is a radial magnetic field, L is the same as length of the coil does not change, and B is the same as the magnetic field strength does not change. So the only thing Force is dependent and thus Torque is dependent on is I which is current.
A pointer is attached to the centre of the coil, so as the coil rotates it also moves linearly. A calibrated scale is used in conjunction with the pointer to provide measurements/readings of the magnitude of the current.

P.S. I'm not sure how to insert diagrams but a quick google search will show you all the different components of the galvanometer.
P.S. I also got D as well for the other question.
EDITED: Added a schematic diagram of a galvanometer

[RuiAce]

P.S. I'm not sure how to insert diagrams but a quick google search will show you all the different components of the galvanometer.

You either upload it to imgur or something and copy/paste the [img] code, or you upload it where you write the post - attachments are uploaded below the box you type your message

[jamonwindeyer]

Thank you!
________________

Hi guys,

Can someone explain how the galvanometer and speakers work in regards to motor effect in simple words??

Thankyou!

I'll take Rui's tag in

Okay, so a galvanometer basically consists of a coil inside a radial magnetic field. What this means is that, while regularly the torque produced by a current carrying coil varies based on its angle with the field, the torque produced here is constant. This is because, if the field is radial, it is always parallel to an equal amount of the field.

Now, the current we are measuring with a galvanometer is fed through this coil. This causes it to rotate, just like a motor, due to the Motor Effect. However, the coil is kept in place with a spring, that provides an opposing torque. Basically, it keeps it from spinning too far. The spring can stretch, but it doesn't let it spin.

The result is that the coil will rotate ever so slightly, until the opposing torque provided by the string equals the torque due to the motor effect. The coil is attached to the needle on your galvanometer, and so the slight rotation causes the needle to move up the scale to read 1mA, or whatever. If you have more current, you have more deflection, and so the needle will move higher up the scale 

Of course, there are digital ammeters now which do fulfil this function with transistors and such, but this covers the basic principles you'll need to answer HSC questions! 

Edit: Combine my response with Spencer's more mathematical one for a complete picture