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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    (Original post by a.a.k)
    Oh sorry my bad

    Yeah u find the number if moles then divide by avagadro constant to get number of particles or in this case number of molecules

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    oh ok thanks, why cant you use the normal formula with N?
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    Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??
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    (Original post by imyimy)
    oh ok thanks, why cant you use the normal formula with N?
    I dont know tbh i dont want to give u wrong advice

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    (Original post by imyimy)
    oh ok thanks, why cant you use the normal formula with N?
    You got the right answer it is +21
    So you CAN use pV=NkT
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    Hi I am really stuck on this question... Any help would be appreciated
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    (Original post by sykik)
    Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??
    It should be 600*0.35
    Whats the answer btw.

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    (Original post by sykik)
    Hi ... I don't get why you do 600/0.35 instead of 600 * 0.35 ??
    Hmm, I think it's because of this:

    The Power output is 6000MW, but this is only 35% of the overall energy supplied by the fuel rods, so dividing by 0.35 gives you 100% of the energy supplied by the fuel rods,, I guess from then you can work out the decrease in mass (based on part a? idk..)
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    1.14x10^-2 kg
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    (Original post by Cosmocos)
    Hmm, I think it's because of this:

    The Power output is 6000MW, but this is only 35% of the overall energy supplied by the fuel rods, so dividing by 0.35 gives you 100% of the energy supplied by the fuel rods,, I guess from then you can work out the decrease in mass (based on part a? idk..)
    doesn't 600/0.35 give u the input power?
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    (Original post by sykik)
    1.14x10^-2 kg
    Close i got 1.41x10^-3 kg

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    Actually its not close enough

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    (Original post by sykik)
    doesn't 600/0.35 give u the input power?
    Isn't the power input what is supplied by the fuel rods? lol I;m not sure tbh
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    [QUOTE=a.a.k;56386187]Close i got 1.41x10^-3 kg

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    Here is MS Answer
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    This is what i got

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    (Original post by sykik)
    Hi I am really stuck on this question... Any help would be appreciated
    After probably doing it the longest way possible. I got a mass of 80623 kg...lol
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    600MW is only elextrical energy which is only 35% other 75% energy is wasted whixh you need to take in account so you need to fins 100% energy by 600x100/35

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    (Original post by Cosmocos)
    After probably doing it the longest way possible. I got a mass of 80623 kg...lol
    That might be mass of crude oil:yy::yy:

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    Hey guys will the formula for closest approach of a scattered particle be given in the exam??
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    (Original post by santiago_carmelo)
    Hey guys will the formula for closest approach of a scattered particle be given in the exam??
    You mean nuclear radius

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    Tomato tomato is it given? or do i have to memorise the thing
    (Original post by a.a.k)
    You mean nuclear radius

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