AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread]

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Hey guys, I have another question.

Screenshot_2015-06-07-16-40-07~01.png

Why is the answer C. I thought it'd be A as the resistivity increases as you go along meaning that P should come out first.

Reply 2901

a.a.k

Original post by Disney0702

Hey guys, I have another question.

Screenshot_2015-06-07-16-40-07~01.png

Why is the answer C. I thought it'd be A as the resistivity increases as you go along meaning that P should come out first.

Think about relating resiatance and current.

Current inveraly proportional to resistance so higher resistance less current is induces.
And
Rember farades law

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Reply 2902

Fvthoms

Original post by Disney0702

Hey guys, I have another question.

Screenshot_2015-06-07-16-40-07~01.png

Why is the answer C. I thought it'd be A as the resistivity increases as you go along meaning that P should come out first.

You're right in saying that resistivity increases as you go along, but the force that makes the cylinder take longer to pass through the tube is due to the induced current caused by the magnet passing through; as resistivity (and therefore resistance) increases, the induced current becomes smaller, meaning the magnetic field that the cylinder generates from the induced current is reduced.
Remember that Lenz's law states that the e.m.f. generated from the change in flux is always an opposing and restoring force.

p.s. @NEWT0N, Your explanations are much appreciated.

Reply 2903

Disney0702

Original post by Fvthoms

Oh my! Completely disregarded Lenz's Law!
Thank you so much!

Reply 2904

Disney0702

Original post by NEWT0N

The greater the resistance of the tube, the less emf is induced in it as the magnet falls, and therefore the smaller the magnetic field it sets up to oppose the magnet's motion. Therefore, as the rubber tube has the highest resistance it opposes the bar magnet the least (i.e. magnet will emerge from the tube in a shorter time), and so on.

Thank you so much, it really cleared things up.

Reply 2905

Disney0702

Original post by a.a.k

Think about relating resiatance and current.

Current inveraly proportional to resistance so higher resistance less current is induces.
And
Rember farades law

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Thank you very much.

And do you mean Lenz's Law?

Reply 2906

a.a.k

Original post by Disney0702

Thank you very much.

And do you mean Lenz's Law?

Yeah i realised that but both laws work together. Lenz law is addition to faradays law i think

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Reply 2907

fruity97

how many sig figs are we actually meant to give answers to? I remember in AS my teacher kept saying 2sf (unless stated otherwise), but after doing past papers it seems it is 3sf for A2?

Reply 2908

Me123456789

A model car moves in a circular path of radius 0.8 m at an angular speed of rad s–1
. What is its displacement from point P, 6 s after passing P?

A zero
B 1.6 m
C 0.4 πm
D 1.6 πm

Can someone help me with this question, the answer is B but I don't know how to get to it.

Any help?

Original post by IWantSomeMushu

Any help?

http://www.thestudentroom.co.uk/showthread.php?p=56698247#post56698247

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Reply 2911

IWantSomeMushu

Original post by CD223

http://www.thestudentroom.co.uk/showthread.php?p=56698247#post56698247

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Ah thanks, I didn't pay attention to the word ''resultant'' ha!

Reply 2912

Me123456789

Original post by NEWT0N

You didn't include the speed but I'll assume it's pi/2 because I remember solving this. So in 6s it goes round (pi/2)*6=3pi=2pi+pi radians, that is a complete circle followed by a semicircle. The complete circle makes the car return to its original position and the semicircle makes it go to a point antipodal to its current position, i.e. a displacement equal to the diameter of the circle, i.e. 0.8*2=1.6m.

Oh, yeah I didn't notice, it is pi/2
It makes sense now, thank you :h:

Reply 2913

Lau14

Original post by fruity97

how many sig figs are we actually meant to give answers to? I remember in AS my teacher kept saying 2sf (unless stated otherwise), but after doing past papers it seems it is 3sf for A2?

If it doesn't say to give your answer to an appropriate number of significant figures, then either two or three is expected, no more or less - although you won't be penalised for more (I was reading this somewhere on the aqa site the other day, possibly in the spec or an examiners report). If it does say to give them to a specific number of significant figures, it should match the piece of data with the least significant figures as newton explained above.

Reply 2914

CD223

Given Newton has joined the conversation, I'm disappointed at the lack of Faraday, Coulomb, Lenz...

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Reply 2915

Disney0702

Can someone please explain Jan '11 Q5(c) from Section B.

I do not understand the mark scheme or the examiner's report.

Reply 2916

a.a.k

Original post by CD223

Given Newton has joined the conversation, I'm disappointed at the lack of Faraday, Coulomb, Lenz...

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Dirac was here before

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Reply 2917

a.a.k

Original post by NEWT0N

Robert Hooke, bogeyman of science. He is much too short to be standing on the shoulders of giants like me, the exalted demigod.

How is your buddy Huygen. Freinds forever

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Reply 2918

Sonnyjimisgod

Original post by CD223

[br]\dfrac{8}{2} = \dfrac{x^2}{(60-x)^2}[br]

Where x is the distance to the 8nC charge.

[br]\therefore 4 = \dfrac{x^2}{(60-x)^2}[br]

[br]\therefore 2 = \dfrac{x}{(60-x)}[br]

[br]\therefore 2(60-x) = x[br]

[br]\therefore 120 - 2x = x[br]

[br]\therefore 120 = 3x[br]

[br]\therefore 40 = x[br]

Does that help? :smile:

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Hello, great working out here, but was just wondering how did you derive the expression. It looks like you've done Q/q=R/(d-R)^2.

What formula did you use, thanks!