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

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    (Original post by n0cturnal)
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    mom before = 8000x2.5=20,000kgms

    mom after = (8000+12000)x2.5=50,000kgms

    50k-20k=30k but thats not on the answer what am i doing wrong😢
    I think you're assuming that the velocity after the collision is the same as the initial velocity. I want you to find it yourself


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    (Original post by Disney0702)
    Hello everyone!

    I was wondering if anyone could help me with Q8 Jun '06 in the following link.

    My answer was D but the correct answer is C and I cannot see how.

    Can someone please explain?
    Sorry for bad writing

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    (Original post by n0cturnal)
    Attachment 420091
    mom before = 8000x2.5=20,000kgms

    mom after = (8000+12000)x2.5=50,000kgms

    50k-20k=30k but thats not on the answer what am i doing wrong😢
    Also you need to find the change in momentum of the 8000kg body only once you have the final velocity


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    (Original post by a.a.k)
    Sorry for bad writing

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    WOW! Thank you so much, I totally understand!

    I did it slightly different.

    I worked out Q to be 1.5 C.

    I then worked out current, I = 1.5 / 5 x 10-3
    I then worked out curre nt, I= 300 A

    Then I did P = VI
    Then I did P = 30,000 x 300
    Then I did P = 9,000,000 W

    I do not know why my method is wrong, do you?
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    (Original post by Disney0702)
    WOW! Thank you so much, I totally understand!

    I did it slightly different.

    I worked out Q to be 1.5 C.

    I then worked out current, I = 1.5 / 5 x 10-3
    I then worked out curre nt, I= 300 A

    Then I did P = VI
    Then I did P = 30,000 x 300
    Then I did P = 9,000,000 W

    I do not know why my method is wrong, do you?
    You can use this formula if current is constant (it should say in the question)

    In this case current is not constant so you cant use this method.

    Hope this helps

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    (Original post by Mehrdad jafari)
    I think you're assuming that the velocity after the collision is the same as the initial velocity. I want you to find it yourself


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    aw **** i did thanks man i know what i did wrong

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    how would i do this one? mg/2 would be equal to mv^2/r rearrange i get v=sqrt gr/2 would i then sub v in for omega (v/r)? i tried that but im getting sqrt gr^2/r confused af right now
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    (Original post by a.a.k)
    You can use this formula if current is constant (it should say in the question)

    In this case current is not constant so you cant use this method.

    Hope this helps

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    Yes that does help a lot, thank you very much.
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    Can someone help me with Jun '07 Q9 in the followinglink.

    It'll be much appreciated
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    (Original post by n0cturnal)
    aw **** i did thanks man i know what i did wrong

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    how would i do this one? mg/2 would be equal to mv^2/r rearrange i get v=sqrt gr/2 would i then sub v in for omega (v/r)? i tried that but im getting sqrt gr^2/r confused af right now
    Yeah, you could do that but that would lack confidence as you could equate frictional force to centripetal force in terms of angular velocity in the first place.
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    (Original post by Disney0702)
    Can someone help me with Jun '07 Q9 in the followinglink.

    It'll be much appreciated
    Check if that gets you the answer
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    (Original post by Disney0702)
    Can someone help me with Jun '07 Q9 in the followinglink.

    It'll be much appreciated
    Is the answer B
    Coz E is directly proportional to V^2

    I hate ratio questions

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    (Original post by Disney0702)
    Can someone help me with Jun '07 Q9 in the followinglink.

    It'll be much appreciated
    Oh, apologise for getting the wrong question. I thought it was q 8. Let me see if there is a remedy for it


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    (Original post by Mehrdad jafari)
    Yeah, you could do that but that would lack confidence as you could equate frictional force to centripetal force in terms of angular velocity in the first place.
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    oh i see! yeah equating it directly is the easier method! thanks man appreciated.

    in terms of the qwc questions what are everyones opinions on what theyll ask this year? they asked on the capacitor circuit last yr and comparison of fields the year before what more can they really ask?
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    (Original post by Disney0702)
    Can someone help me with Jun '07 Q9 in the followinglink.

    It'll be much appreciated
    To be honest i don't know why the ratio doesn't hold for energy
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    EDIT: for questions like this one it's better not to use the ratio as you are not given any voltage, but only by how much it's changed so the ratios are exact multiples and can easily be figured out thinking about it. If you were given inexact values if voltage then you will have to use ratio as you can't figure out by logic.

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    (Original post by Disney0702)
    Hello everyone!

    I was wondering if anyone could help me with Q8 Jun '06 in the following link.

    My answer was D but the correct answer is C and I cannot see how.

    Can someone please explain?
    

P = \dfrac{\Delta E}{\Delta t}

    

\therefore P = \dfrac{0.5 \times 50 \times 10^{-6} \times (30 \times 10^{3})^2}{5 \times 10^{-3}}

    

\therefore P = 4.5 \text{MW}


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    (Original post by Disney0702)
    WOW! Thank you so much, I totally understand!

    I did it slightly different.

    I worked out Q to be 1.5 C.

    I then worked out current, I = 1.5 / 5 x 10-3
    I then worked out curre nt, I= 300 A

    Then I did P = VI
    Then I did P = 30,000 x 300
    Then I did P = 9,000,000 W

    I do not know why my method is wrong, do you?
    Sorry - didn't realise someone had already answered!

    I can tell you why this doesn't work though - you've worked out the average current - the capacitor discharge current falls exponentially and therefore is not a constant value

    Edit: and I've just realised this was answered too - I'm not having a good day 😂


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    (Original post by getback339)
    Could someone help me with june 2013 question 1 part c? thanks a lot
    anyone?
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    (Original post by getback339)
    anyone?
    Here's a similar question from the multiple choice section one year:

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    The shorter pendulum will have completed one more oscillation than the longer pendulum when they are next in phase. Use this to find the number of oscillations, then multiply by the period

    In the case of June 2013, you could have calculated 20 oscillations of the shorter pendulum (1.9s period) or 19 for the longer one (2.0s period). In either case, multiplying the number of oscillations by the period will give 38 seconds before they are next in phase.




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    Name:  image.jpg
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Size:  502.8 KBIs this the best way to do these kind of questions? Jan 12, question 15 multiple choice? Thanks
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    Hi sorry to be a pain but can someone just read this and make sure it makes sense and possibly add to it.

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