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

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    (Original post by chizz1889)
    It was 2
    it was indeed
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    (Original post by CD223)
    I only got 2.01 revolutions for the circular motion question.


    Posted from TSR Mobile
    I got two as well, that was the correct answer
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    do you realise how much better this would be if we just had the damn MC paper here!!!!!
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    (Original post by NEWT0N)
    The calculation was easy and more accurate. \Delta V = \frac{30\times 10^{-9}}{4\pi\varepsilon_0}(\frac{1}{  0.2}-\frac{1}{0.5}).
    Then multiply by 60*10^-9 C.

    Also does the sign of the work done matter? I was told to always state it as positive, but sometimes work is being given and sometimes taken away so I dunno...
    I found my method easier. The graph went down in an easy interval of 750V. The value of the charge could then be multiplied by this to give the work done.


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    I'm soooo glad my tutor was there to tell me that I was right in thinking AQA had made a mistake here too!!! It threw me for hours last night so I do apologise for my silly questions on here!! Max flux linkage IS when coil is perpendicular just as I thought! So someone else on here got it wrong too... Very glad that's settled in my head now though phew!!Name:  image.jpg
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    (Original post by ubisoft)
    hmm I drew it above but people are saying aqa made a typo??

    Did you write area under the graph for current I think (or charge cant remember)
    For calculating charge I said to draw a tangent at t=0 and then the calculate the area under the tangent. Not sure though
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    (Original post by chizz1889)
    It was 2
    Another whoopsie then

    do you remember the working?
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    (Original post by Kennethm)
    do you realise how much better this would be if we just had the damn MC paper here!!!!!
    Wish id stayed at college as my tutor does the answers straight away and I've stored all mine! But I'll have to wait now! Grr
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    (Original post by ajcoo)
    I thinks it was below...bigger resistance, less current, hence smaller area, but started and finished at same places cos same pd :teeth:
    Unsure about this because wouldn't the area under the graph have to be the same as that represents charge? Remembering that the pd and capacitance were the same
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    (Original post by JAW-97)
    For calculating charge I said to draw a tangent at t=0 and then the calculate the area under the tangent. Not sure though
    I just put area under the graph, I think the tangent stuff is too much for 1 mark
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    (Original post by chizz1889)
    That gives you change in potential so you have to multiply by 60*10^-9 and no as work is done by the particle which is what it asked
    Yes I was just showing how to find potential difference (I edited in the multiplying by 60^10^-9 part though)

    And good, I gave a positive answer
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    What did everyone put for the damping question? Did the mass increase damping, decrease damping or have no effect?
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    (Original post by JAW-97)
    For calculating charge I said to draw a tangent at t=0 and then the calculate the area under the tangent. Not sure though
    I just said area under 😳
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    (Original post by CD223)
    I only got 2.01 revolutions for the circular motion question.


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    Actually I might be getting mixed up with some question from a past paper :s

    But I don't remember the working out
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    (Original post by NEWT0N)
    Another whoopsie then

    do you remember the working?
    I don't know the numbers but I worked out the tangential speed, found the distance travelled and then divided it by the circumfrence
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    (Original post by NEWT0N)
    Another whoopsie then

    do you remember the working?
    You use F=mw^2r
    You know F, m and r so you calculate the new w
    then then do 2pi/w = T = 0.5 = 2 per second
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    (Original post by Sbarron)
    I just said area under 😳
    Agreed
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    (Original post by jgarnham)
    What did everyone put for the damping question? Did the mass increase damping, decrease damping or have no effect?
    No effect
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    For the voltage you cant use the rotating coil equation right because we didn't know the angular velocity?
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    (Original post by NEWT0N)
    Yes I was just showing how to find potential difference (I edited in the multiplying by 60^10^-9 part though)

    And good, I gave a positive answer
    dude at first I put a negative and at the last second I was like 'yolo' and made it positive hahahaha
 
 
 
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