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

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    (Original post by CD223)
    I did. Looks like you didn't flop!


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    Yes by taking two radius away from 6!
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    (Original post by ubisoft)
    You just multiply the rad/s by the time, then change it into degrees
    i cant really remember the paper but didnt ask for the angle turned in 1 sec given the radius and tangential velocity
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    (Original post by chizz1889)
    What was the correct answer to why it doesn't reach 1.4m? I put there is air resistance resisting the vertical motion and also energy is lost in the circuit due to resistance
    internal resistance of circuit and in practice you can't always assume the mass will remain perfectly vertical - It may sway from left to right causing additional forces.
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    (Original post by ubisoft)
    For the capacitor graph, was it above or below their original curve?
    I thinks it was below...bigger resistance, less current, hence smaller area, but started and finished at same places cos same pd :teeth:
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    (Original post by DanielWall96)
    i cant really remember the paper but didnt ask for the angle turned in 1 sec given the radius and tangential velocity
    i can't remember too tbh only had 5 mins left when i started Q5 I wasn't even reading the whole question haha
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    Grade boundaries lower this year compared to last year? I've definitely dropped 8, 4 on the damping one, 3 on the electric fields and i got the scales question as 0, any predictions on UMS?
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    (Original post by CD223)
    Yeah that's what I did lol.


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    I use the term relatively
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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:
    AQA made a mistake on that question and put 300,000 instead of 300 so my guess is it would be discounted
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    (Original post by BanoffeeGuy)
    Amanzz! Hey, hmmm. I agree with the other guys answer I think, smaller reduction in velocity/deceleration is less for given drag force etc. How did you find the paper?
    Relatively okay, I just put up Kampas' email so I don't know. Overall I think I did alright. I need my A, don't want any more or less, just get me to Imperial already haha! You? PS I thought that explanation take place but Alex is never wrong
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    Does anyone remember an answer in the written paper being "38 revolutions"
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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:
    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)
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    Dropped 10 marks by my count after going over everything. Probably lose 3/4 more after wording etc but that should still be an A* should it not?
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    (Original post by Amanzz)
    Relatively okay, I just put up Kampas' email so I don't know. Overall I think I did alright. I need my A, don't want any more or less, just get me to Imperial already haha! You? PS I thought that explanation take place but Alex is never wrong
    aha me too, just need an A for imperial but I was concentrating so much on further maths A* and neglected physics...
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    (Original post by CD223)
    Yeah that's what I did lol.


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    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...
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    (Original post by NEWT0N)
    Does anyone remember an answer in the written paper being "38 revolutions"
    It was 2
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    (Original post by ubisoft)
    Yesss I need every mark i can lol, did you use BANwsinwt anywhere and get like 0.06 i think?
    I remember doing that
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    (Original post by NEWT0N)
    Does anyone remember an answer in the written paper being "38 revolutions"
    I only got 2.01 revolutions for the circular motion question.


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    Name:  image.jpg
Views: 198
Size:  505.4 KB 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!!
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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}).

    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...
    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
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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)
    Yes I put area under graph, it was for finding charge

    Charge=area under graph
 
 
 
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