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I cant do these 2 questions on C4 paper A3, please help. watch

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    I have attatched the paper and the answers.

    Ok heres what I need help on.

    5B - I have done it down to the second from last line but I dont understand how it simplifies down in terms of pi

    6B - I have no idea what they are doing there. I tried making 2 simultaneuous [sp] equations but that doesnt seem to work.


    Thanks.
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  1. File Type: doc C4 Practice Paper A3.doc (61.0 KB, 163 views)
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    Answers:
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  2. File Type: doc Answers.doc (96.0 KB, 72 views)
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    anyone? lol
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    Can you tell me where you got that paper from, if there are more?
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    http://www.thestudentroom.co.uk/t124400.html

    they are here.
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    Thanks a lot man
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    5b. Just plug in the limits and note that: cos(pi)=-1 and sin(pi)=0.

    6b.
    When t=0, N=7*10^18:
    7 * 10^18 = A e^(0) = A

    When t=8, N=3*10^18:
    3 * 10^17 = 7 * 10^18 . e^(-8k)
    e^(8k) = 70/3
    8k = ln(7/3)
    k = (1/8) ln(7/3)
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    Its quite simple.
    you are integrating between π and 0. so swap the x's with π and you will get:
    Code:
    π*( [2πsin(π) + 2cos(π) - π²cos(π)] - [2cos(0)])
    π*[(0 -2 + π²) - (2)] 
    = π[π² -4]
    ill have a look at the other one...
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    Ok 6b.
    rearranging part a gives you dN = -kN dt
    divide by N so you get 1/N dN = -k dt
    Integrate each side with respect to what ever they are, so you get
    ln N = -kt + c
    do a bit of e^ to get
    N = e^(-kt +c) = Ae^-kt
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    For the final question I get 1.29*10^16, anyone else get the same?
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    thanks ppl!
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    (Original post by dvs)
    5b. Just plug in the limits and note that: cos(pi)=-1 and sin(pi)=0.

    6b.
    When t=0, N=7*10^18:
    7 * 10^18 = A e^(0) = A

    When t=8, N=3*10^18:
    3 * 10^17 = 7 * 10^18 . e^(-8k)
    e^(8k) = 70/3

    8k = ln(7/3)
    k = (1/8) ln(7/3)
    I STILL dont get it!!! how do u do the steps in bold?
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    by dividing by 10^17 on both sides you get 3=70e^-kt
    3/70=e^-kt
    ln 3 - ln 70 = -kt
    k = ln 70-ln 3/t
 
 
 
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