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

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    (Original post by CD223)
    The answer is A. Have you read the MS slightly wrong? Essentially the electron is parallel to the field lines so no magnetic force is experienced. Could you double check? Because I'm 95% sure it's A haha.


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    You could either be right ir wrong so 50% sure is safe to say:awesome:

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    (Original post by Disney0702)
    Uh this is slightly embarassing but yes you're right.
    The answer is indeed A.

    I looked at the wrong part of the mark scheme.

    Im clearly having a bad day!

    Sorry to have wasted your time
    No worries

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    (Original post by Disney0702)
    Uh this is slightly embarassing but yes you're right.
    The answer is indeed A.

    I looked at the wrong part of the mark scheme.

    Im clearly having a bad day!

    Sorry to have wasted your time
    No worries! Happens to us all.


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    (Original post by Fvthoms)
    Came across these questions over the weekend, anyone want to have a go at them? (Mehrdad and CCD, if these have come up already, can you let me know the page?) The paper is PHYA4 January 2013Attachment 421391Attachment 421393Attachment 421395The mark scheme for 2 a) (ii). for this question, don't need numerical advice, just a description of the thought process and the method taken to calculate the answer.Attachment 421397
    12) g\propto M/r^2 and \rho\propto M/r^3

    \therefore g_e=6g_m\rightarrow M_e/r_e^2=6(M_m/r_m^2)\rightarrow M_e/M_m=6(r_e/r_m)^2

    and

    \rho_m=0.6\rho_e\rightarrow M_m/r_m^3=0.6(M_e/r_e^3)\rightarrow M_e/M_m=(1/0.6)(r_e/r_m)^3.

    Equating the M_e/M_m in both equations gives:

    (1/0.6)(r_e/r_m)^3=6(r_e/r_m)^2\rightarrow r_e/r_m=6\times 0.6=3.6

    \therefore answer is \mathbf{B}.
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    Hey guys, I have another question.

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    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.
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    (Original post by Disney0702)
    Hey guys, I have another question.

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    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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    (Original post by Fvthoms)
    Came across these questions over the weekend, anyone want to have a go at them? (Mehrdad and CCD, if these have come up already, can you let me know the page?) The paper is PHYA4 January 2013Attachment 421391Attachment 421393Attachment 421395The mark scheme for 2 a) (ii). for this question, don't need numerical advice, just a description of the thought process and the method taken to calculate the answer.Attachment 421397
    \text{V}=Q/(4\pi \varepsilon_0 r).

    Potential at P due to one of the charges is \text{V}=(+0.8\times 10^{-9})/(4\pi\times 8.85\times 10^{-12}\times 40\times 10^{-3})=179.83...V

    Potential addition is algebraic so potential due to both charges is just double this: 2\times 179.83...=359.67...=360V\ (2\ \text{s.f})

    \therefore answer is \mathbf{C}
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    (Original post by Disney0702)
    Hey guys, I have another question.

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    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.
    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.
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    (Original post by Disney0702)
    Hey guys, I have another question.

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    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.
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    (Original post by Fvthoms)
    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.
    Oh my! Completely disregarded Lenz's Law!
    Thank you so much!
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    (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.
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    (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?
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    (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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    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?
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    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.
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    Any help?
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    (Original post by IWantSomeMushu)
    Any help?
    http://www.thestudentroom.co.uk/show...7#post56698247


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    (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?
    This is the rule we were taught:
    Look at the data given in the question and pick out the data that is given to the smallest number of sig figs. Give your answer to this number of sig figs. If it isn't clear from the question even after trying to do this, then give to 3 s.f.

    Example: A fixed mass of ideal gas is trapped in a container of volume 0.2\text{m}^3. The pressure of the gas is 1.20\times 10^5\text{Pa} and its temperature is 200K. How many moles of the gas are there? Give your answer to an appropriate number of sig figs.

    Answer: We look at the data and see that all of them are given to 3 sig figs (including the 200K, because if they wanted to give it to only 1 sig fig they would write 2\times 10^2\text{K}, and if they wanted to give it to 2 sig fig they would write 2.0\times 10^2\text{K}) except for the volume of the gas which is given to 1 sig fig. So we will give our answer to 1 sig fig.

    pV=nRT\rightarrow n=pV/RT=(1.2\times 10^5\times 0.2)/(8.31\times 200)=14.44...=10 \text{mol}\ (1\ \text{s.f}).

    (Note: The reason you won't usually round answers this much, i.e. 14.44... --> 10, is because they don't usually give data to 1 sig fig like 0.2m^3.)
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    (Original post by 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.
    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.
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    Ah thanks, I didn't pay attention to the word ''resultant'' ha!
 
 
 
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