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

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    (Original post by danny9253)
    PLEASE SOMEBODY HELP ME ANSWER THIS QUESTION

    I get 237.9688u,87.72405u, 4.03468u respectively for the masses but these seem to be greater than on the mark scheme and i can't understand why??
    I came across the same question, I assumed that since it was the syllabus they were given the mass already, and they had to use that


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    (Original post by oonic0rn)
    http://filestore.aqa.org.uk/subjects...1-QP-JUN14.PDF can anyone please explain why you dont include the neutrons in 1B)ii. ?? thanks
    I think in our course, There is no information on binding energy of protons and nucleons separately, Binding energy per nucleon is simply the average binding energy of a proton and neutron in a certain nucleus.

    Neutrons that are not in a nucleus thus have no binding energy as binding energy is the energy released when a nucleus formed, which is determined by the mass defect. As there is no nucleus, no mass defect, hence no binding energy for thermal neutrons.
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    (Original post by johnh545)
    If your calculating the energy released using the change in mass, then yes neutrons are included in the calculation. But in this situation it was just about the binding energy of the nuclei. Seeing as a neutron has already split from the neutron it has no binding energy and isnt included.
    ohh my bad i misread the question thanks
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    (Original post by Klaxoii)
    I came across the same question, I assumed that since it was the syllabus they were given the mass already, and they had to use that


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    That was my thinking too since you have to be given the relative masses in order to work out mass defect. Thanks
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    Anyone doing astro know the reason behind the answer to 4(b)(iii)?

    http://www.tomred.org/uploads/7/7/8/...p_jan_2004.pdf

    http://www.tomred.org/uploads/7/7/8/...s_jan_2004.pdf
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    Explanation of how to convert from apparent magnitude to brightness? Its something to do with 2.5^x???
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    (Original post by slaven123)
    Explanation of how to convert from apparent magnitude to brightness? Its something to do with 2.5^x???
    Apparent magnitude to intensity you mean? What sort of question have you seen that relates the two?


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    Can anyone help me with June 2013 4(b)(iii) about gas
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    any idea of definitions we need to know for section A? loads came up last year and I didn't know any of them haha
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    (Original post by DannySmith420)
    Can anyone help me with June 2013 4(b)(iii) about gas
    You just have to use n=PV/RT again but with the new values, then multiply n by the molar mass given to you in the previous question.
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    (Original post by CD223)
    Apparent magnitude to intensity you mean? What sort of question have you seen that relates the two?


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    No brightness. Question 2b(ii) on june 2010.
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
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    (Original post by CD223)
    Apparent magnitude to intensity you mean? What sort of question have you seen that relates the two?


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    How many times brighter one star is than another is 2.5^(delta m)
    I asked before if we need to know this but nobody knows
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    Hey, would anyone be able to post how to derive the kinetic theory model as how they would want you write it in an answer? I just want to learn it so I can nail it if it comes up
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    (Original post by 000alex)
    How many times brighter one star is that another is 2.5^(delta m)
    I asked before if we need to know this but nobody knows
    Ahh okay. They've asked about it before so I guess you do need to know it
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    (Original post by slaven123)
    No brightness. Question 2b(ii) on june 2010.
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
    All you need to know is that each magnitude difference represents a change of 100^{\frac{1}{5}}.


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    (Original post by 000alex)
    How many times brighter one star is than another is 2.5^(delta m)
    I asked before if we need to know this but nobody knows
    It's not exactly 2.5 - it's the fifth root of 100. Might sound stupid but when multiplying by large numbers it can give you the incorrect answer easily.


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    (Original post by CD223)
    All you need to know is that each magnitude difference represents a change of 100^{\frac{1}{5}}.


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    A change of what though? Intensity, brightness...?
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    (Original post by slaven123)
    A change of what though? Intensity, brightness...?
    "Brightness". Although equally I've seen them use intensity just as much before.


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    (Original post by CD223)
    It's not exactly 2.5 - it's the fifth root of 100. Might sound stupid but when multiplying by large numbers it can give you the incorrect answer easily.


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    The markscheme only used 2.5 so idk
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    (Original post by 000alex)
    The markscheme only used 2.5 so idk
    I'd just be careful. The mark schemes are so inconsistent.


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