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AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread] watch

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    (Original post by ahlam H)
    Did anyone get 2100J and 2500J for the last 2 questions on nuclear physics
    What were those questions?


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    (Original post by ahlam H)
    Did anyone get 2100J and 2500J for the last 2 questions on nuclear physics
    Yes that's right


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    (Original post by nnblccz)
    Swear I got 1.8x10^11, which is basically the specific charge of an electron


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    Same, I got 1.81x10^11. You just re-arrange r=mv/Be solve for e/m
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    (Original post by the_chosen_one97)
    Same, I got 1.81x10^11. You just re-arrange r=mv/Be solve for e/m
    Or you can use the 2V/B^2r^2 because they gave all data and answer was 1.9*10^11


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    (Original post by BigDaddy97)
    Yes that's right


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    i think i got them the other way round? wasnt specific heat capacity 2120? and the energy was 2490?
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    (Original post by the_chosen_one97)
    Same, I got 1.81x10^11. You just re-arrange r=mv/Be solve for e/m
    Erm you didn't have v - the velocity - you had the potential through which they are accelerated


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    (Original post by BigDaddy97)
    Or you can use the 2V/B^2r^2 because they gave all data and answer was 1.9*10^11
    Yep , I got this

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    Yep I got this
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    (Original post by Ac19951)
    1.22x10^8 AU for the last question in Astrophysics anyone? I think i forgot to square root my answer.

    Also what did you guys write for the first part of the CCD question? Regarding structure.
    it was 2.4x10^14 AU
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    Surprisingly not too bad. Think I've made a few mistakes but that was a nice paper regardless
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    128MeV???????
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    (Original post by aelahi23)
    Found the exam to easy quite. I did turning points for optional module

    although one question threw me off, the 4 mark question in turning points that asked to find specific charge I got 2.1 * 10^8 what did other people get?

    since F=Bev and F= Mv^2/r
    I did equated them to each other then rearranged to get
    e/m = v/Br

    "v" mean velocity
    "V" means PD

    To find v (velocity) I did v=V/Br

    so subbed in the v=V/Br into e/m = v/Br to get rid of velocity from the equation to get e/m= V/Br/Br
    which simplified to e/m= V/ B^2 * r^2
    Is that correct?

    Section A was very easy so I don't have any queries about that
    I used the equation eV = 1/2 mv^2
    Gave me a velocity of something like 23989839. I then used this in the equation e/m = v/Br and it gave the right answer which is 1.8x10^11.
    I think I did it the way you did to begin with but thought it obviously can't be that so changed it.. you should get a mark for the Unit provided you got that right ! Which to be fair is given on the formula sheet lmao
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    (Original post by Sid1234)
    it was 2.4x10^14 AU
    Me too!! Thank god
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    (Original post by Mango Milkshake)
    128MeV???????
    YES !
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    I wrote about Rutherford alpha scattering, wbu?
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    (Original post by Mango Milkshake)
    128MeV???????
    127.24 i got o.O
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    (Original post by cowie)
    What were those questions?


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    It was about the specfic capacity and latent for ice and water where the ice was raised from -25 to 0 degrees celcius
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    (Original post by txnilxnur)
    I used the equation eV = 1/2 mv^2
    Gave me a velocity of something like 23989839. I then used this in the equation e/m = v/Br and it gave the right answer which is 1.8x10^11.
    I think I did it the way you did to begin with but thought it obviously can't be that so changed it.. you should get a mark for the Unit provided you got that right ! Which to be fair is given on the formula sheet lmao
    Both methods will get the marks chill


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    (Original post by Sid1234)
    it was 2.4x10^14 AU
    i got this as well
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    anyone can post the solution for turning point ?
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    (Original post by txnilxnur)
    I used the equation eV = 1/2 mv^2
    Gave me a velocity of something like 23989839. I then used this in the equation e/m = v/Br and it gave the right answer which is 1.8x10^11.
    I think I did it the way you did to begin with but thought it obviously can't be that so changed it.. you should get a mark for the Unit provided you got that right ! Which to be fair is given on the formula sheet lmao
    No you can't do it that way - you're using the electron charge and mass from the formula booklet which they wouldn't have had at the time! The question made it clear: use THIS data


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