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March 29, 2024, 10:12:49 am

Author Topic: VCE Physics Question Thread!  (Read 603393 times)  Share 

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Zealous

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Re: VCE Physics Question Thread!
« Reply #615 on: September 24, 2014, 05:15:51 pm »
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can i ask another one?

what is the purpose of slip rings in AC generator?
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.

I've got a quick Electric Power question.

How come the magnetic field lines diverge in this image? Can't they all be parallel? (VCAA 2008)



Current flows in an anti-clockwise direction, by the way.
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Makutar

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Re: VCE Physics Question Thread!
« Reply #616 on: September 24, 2014, 07:43:39 pm »
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The lines diverge because the magnetic field (B) gets smaller the further away from the loop you go. 

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Re: VCE Physics Question Thread!
« Reply #617 on: September 24, 2014, 08:20:26 pm »
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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?
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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #618 on: September 24, 2014, 09:34:30 pm »
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Just on that, wouldn't the magnetic field strength have no effect on the direction?

field strength is represented by field line density (closer lines equals stronger field) so if they stayed parallel, you'd be saying that the strength of field is the same no matter how far away from the loop you go

this isn't the case, and the only way for the strength to dissipate in a field-line drawing is if the lines spread out. the direction IS consistent with these lines and does spread out



something else that might help you is this; i assume you used a right hand rule to determine that an anticlockwise current (as viewed from the side the arrows point in this case) causes the field to point the way it does. this actually comes from the 'grip rule' for a single current-carrying wire, and happens to also work for a loop.

if you imagine gripping the wire anywhere around the loop, with your right thumb pointing in the direction of the conventional current (anticlockwise) you'll see that inside the loop, the current around the wire points in that direction, (this is the same everywhere in the loop).
the circular field around a wire wrapped in a loop gives the net result of having a field which is stronger inside the loop because all points around the loop contribute to that direction. but their contributions are only parallel in the plane of the loop, and everywhere else (including outside) they are weaker and not parallel.

As per this image, just showing the lines in one plane;
Spoiler

it's not very important to draw the lines that do loop back around and it's pretty hard to draw clearly on an exam, but they are there
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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #619 on: September 24, 2014, 09:47:36 pm »
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Can a light dependant resistor be used as a demodulator?
i know LED are used for the modulator

LDR's have too slow of a response rate to be at all effective in this application. you need a component that is fast enough to respond to the rapid changes in signal, but an LDR's response rate is never much faster than milliseconds, and is usually order 0.1s

For this reason, they're more useful as light sensors as in those circuits the response time is acceptable
In actual fact, sometimes they're used in audio compression for softening audio signals (utilising their delayed response)
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Brunette15

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Re: VCE Physics Question Thread!
« Reply #620 on: September 26, 2014, 05:32:45 pm »
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Can someone please provide a worked solution to this qu from the 2007 exam? The answer is 1.01m...
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kinslayer

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Re: VCE Physics Question Thread!
« Reply #621 on: September 26, 2014, 05:38:18 pm »
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The light begins from rest and then travels under constant acceleration of g m/s^2.

The equation of motion to use here is s = ut + 1/2gt^2 where u = 0 and t = 0.45s.

s = 1/2*(9.8 m/s^2)(0.45 s)^2 = 0.99 m

The tail-light was 0.99 m above the ground before it fell off.

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?  ???
« Last Edit: September 26, 2014, 05:43:18 pm by kinslayer »

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Re: VCE Physics Question Thread!
« Reply #622 on: September 27, 2014, 12:30:44 am »
+1
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?  ???

VCE Physics exams have g as 10m/s^2 on the formula sheet and in the study design, probably to save on calculations and make it more about the physics. The mechanics section of Specialist Maths uses 9.8m/s^2, and VCE physics questions aren't marked wrong if you use 9.8 instead of 10 (nor in fact any error in sigfigs unless you give like 7 decimal places or something silly) afaik.
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Brunette15

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Re: VCE Physics Question Thread!
« Reply #623 on: September 27, 2014, 01:45:34 pm »
+1
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 :/

The ans is B...
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Re: VCE Physics Question Thread!
« Reply #624 on: September 27, 2014, 02:37:56 pm »
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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 :/

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.
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Brunette15

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Re: VCE Physics Question Thread!
« Reply #625 on: September 27, 2014, 04:23:08 pm »
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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.

Ah yes...thankyou!  :D
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #626 on: October 03, 2014, 10:14:34 pm »
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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.

Not quite. The torque is a product of a force and a distance, but this distance isn't just the distance from the pivot point to the force. That only holds if the force acts perpendicular to the vector from the pivot point to the point of application of the force.

Think about a see-saw. Logically, it would move the most if you pushed straight down on the seesaw, not if you pushed diagonally down on it. The angle matters too.
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silverpixeli

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Re: VCE Physics Question Thread!
« Reply #627 on: October 03, 2014, 10:47:47 pm »
+1
I always thought of it as the distance between the force and the pivot and the perpendicular component of the force, since there's also a component parallel which causes no torque if your force is at an angle. I know taking the angle into account on the force or the distance is equivalent. Anyway, in this case, if the distance is 0, which it is here, it doesnt matter what the force is or what the angle of the force is, the torque will still be 0 if d is 0 :D
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Re: VCE Physics Question Thread!
« Reply #628 on: October 07, 2014, 11:00:44 am »
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Can someone please clarify this concept?
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?


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Re: VCE Physics Question Thread!
« Reply #629 on: October 07, 2014, 03:23:06 pm »
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Why is it that no work is done when the component is in a perpendicular direction to the objects movement?

thanks in advance :)