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

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    (Original post by HenryHein)
    Only have to if they ask you to. Of course it is good general practice to do so anyway.
    So if they don't ask they will except either?
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    (Original post by lukebiddell)
    Does anyone know how to go about answering this question? I got A but the answer is B.
    Field strength proportional to (1/r2).
    Say distance to M is x and distance to 4M y:
    At zero point:
    M/x2 = 4M/y2
    So 1/x2 = 4/y2
    So y2/x2 = 4
    so y/x = 2
    so y= 2x

    As y + x = d
    Subbing in y=2x:
    3x = d
    so x= d/3 which is B
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    (Original post by coqthepoliceman)
    does anyone have any good pdfs i could use to revise from
    See the first post I uploaded a revision guide I made, and there's other useful links.


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    Could someone explain why question 1 of this paper is C?
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    (Original post by michaelotty)
    So if they don't ask they will except either?
    Yes, although I always go by 3 sf unless it is implied you must quote to what is given in the question. An exception may be if you are asked to leave something in atomic mass units but binding energy is Unit 5 of course.
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    (Original post by HenryHein)
    Field strength proportional to (1/r2).
    Say distance to M is x and distance to 4M y:
    At zero point:
    M/x2 = 4M/y2
    So 1/x2 = 4/y2
    So y2/x2 = 4
    so y/x = 2
    so y= 2x

    As y + x = d
    Subbing in y=2x:
    3x = d
    so x= d/3 which is B
    Thanks so much for the reply, but I'm still a bit confused. Isn't y the distance from M? I know you're right, I just can't get my head around why it isn't
    M/y2 = 4M/x2
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    (Original post by CD223)
    See the first post I uploaded a revision guide I made, and there's other useful links.


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    thanks; by the way could you help me on how to go about answerig q13 and 14 of the june 2010 paper?
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    (Original post by HenryHein)
    Field strength proportional to (1/r2).
    Say distance to M is x and distance to 4M y:
    At zero point:
    M/x2 = 4M/y2
    So 1/x2 = 4/y2
    So y2/x2 = 4
    so y/x = 2
    so y= 2x

    As y + x = d
    Subbing in y=2x:
    3x = d
    so x= d/3 which is B
    do you know what situation will mean that M/x2 = - 4M/y2

    Sometimes I cant tell when the minus matters. It might only be for potential.
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    anyone know what the marks will be for a 6 marker on the motor effect?
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    Could someone please explain why you need AC for a cyclotron?

    Thanks
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    http://filestore.aqa.org.uk/subjects...1-QP-JAN13.PDF -

    could someone please explain number 9 and 10 please
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    (Original post by lukebiddell)
    Thanks so much for the reply, but I'm still a bit confused. Isn't y the distance from M? I know you're right, I just can't get my head around why it isn't
    M/y2 = 4M/x2
    Ah, that may just be because I messed up the units. I meant for the distance to M to be Y and the distance to 4M to be X
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    (Original post by Plasmapause)
    Could someone explain why question 1 of this paper is C?
    Ft=mv

    So v =Ft/m

    (0.5*10*40)/0.25=800

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    (Original post by Kennethm)
    do you know what situation will mean that M/x2 = - 4M/y2

    Sometimes I cant tell when the minus matters. It might only be for potential.
    If it says potential is zero and one of the charges on negative would you need to include the negative or not? SOmetimes they do sometimes not (although I think they do)
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    (Original post by SilenceOrNoise)
    Because the magnetic field lines are not passing through the charged particles at RIGHT ANGLES. Therefore, the magnetic force would not act where the particles would move unaffected.
    Thanks, got mixed up with electric fields
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    (Original post by RemainSilent)
    If it says potential is zero and one of the charges on negative would you need to include the negative or not? SOmetimes they do sometimes not (although I think they do)
    ive been trying to find out lol

    https://60abffc9b401b1c0936e01291c15...%20Physics.pdf

    Q13, the answer is A meaning you dont make it minus on the other side. Which makes me wonder do you do the same or the opposite if they were positive?
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    (Original post by Kennethm)
    do you know what situation will mean that M/x2 = - 4M/y2

    Sometimes I cant tell when the minus matters. It might only be for potential.
    Never would be the case as gravity is always attractive and hence the force exerted on a point mass by other masses will always have the same sign.

    In gravitational fields a minus is used for potential. The idea is that being in a field is like being in an "energy well" and energy must be done to be in a "neutral state" (i.e. out of the field) in the same way electron energy levels are given negative values.
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    Anyone got any ideas what the QWC will be on?
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    (Original post by Patar31)
    Anyone got any ideas what the QWC will be on?
    I think it would be on comparing electric filds magnetic fields and gravitational fields

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    (Original post by a.a.k)
    Ft=mv

    So v =Ft/m

    (0.5*10*40)/0.25=800

    Posted from TSR Mobile
    Thanks for the reply.
    Just wondering but where did the 0.5 come from? :confused:
 
 
 
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