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

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    anyone doing medical option?
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    (Original post by Klaxoii)
    Came across this question, am I missing something because I used the equation given in the data sheet, which has (1-v^2/c^2)^-0.5, however they did not have the minus power?



    Attachment 430551Attachment 430553


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    That's very strange. Which paper is that?


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    (Original post by SuperMushroom)
    Ill have a look at this now, thanks
    No worries!
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    Not feeling this at all. I just hope my university doesn't question why I want to major in Physics when Physics ends up being my worst A Level :lol:
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    (Original post by Mehrdad jafari)
    That's very strange. Which paper is that?


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    I'm not sure of the specific paper, it's on the physicandmathstutor.com website, in the special relativity questions, I think it's old syllabus however


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    (Original post by Klaxoii)
    I'm not sure of the specific paper, it's on the physicandmathstutor.com website, in the special relativity questions, I think it's old syllabus however


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    What's the tutor is turning points 6 on your tabs can j just ask


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    (Original post by gcsestuff)
    What's the tutor is turning points 6 on your tabs can j just ask


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    http://www.antonine-education.co.uk/...g_points_6.htm


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    Anyone know how to do 4b on last year's Astrophysics paper?

    Basically says distance to stars measured using angle of parallax method to a precision of 0.002 arc seconds. Calculate the maximum distance measurable?

    Mark scheme shows use of d=1/p....?
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    (Original post by gcsestuff)
    What's the tutor is turning points 6 on your tabs can j just ask


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    Also, I found this whilst procrastinating on reddit, and it helped me understand slightly better what actually happens

    http://www.reddit.com/r/explainlikei...ou_age/c28z810


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    (Original post by EconFan_73)
    Anyone know how to do 4b on last year's Astrophysics paper?

    Basically says distance to stars measured using angle of parallax method to a precision of 0.002 arc seconds. Calculate the maximum distance measurable?

    Mark scheme shows use of d=1/p....?
    

d = \dfrac{1}{p}

    This equation relates the maximum distance, d, measurable (the precision) of a telescope to the parallax angle it is capable of resolving, p.

    Subbing in the numbers gives:

    

d = \dfrac{1}{0.002} = 500 pc

    As it asks for the distance in metres,

    

d = 500 \times 3.08 \times 10^{16}m

    

\therefore d = 2 \times 10^{19}m\ \text{(1 sf)}


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    (Original post by Exams12343)
    Can someone help explain Q 2b on the june 2011 paper please?
    Nuclear and thermal or astro?
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    Milikans outcome-

    Discovered the charges of loads of masses.

    Realised that all charges where n* 1.6*10^19

    That must mean that n was the number of a certain charge

    This was to be discovered ad the charge of the electron

    Am I right?


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    Guys I know this is a dumb question but what does count rate actually mean?
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    (Original post by Disney0702)
    Guys I know this is a dumb question but what does count rate actually mean?
    Not a dumb question!

    The number of sparks/detections per unit time for a particular detector placed at a certain distance away from a radioactive source.


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    Is it true that the binding energy released is shared amongst products in the fraction of their masses over the total mass?

    I read this which confused me:

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    I need help with this question.
    i dont really understand what it means by 1 in 400 detected on the facing surface of the detector.
    were told earlier in the question that the detector has area 1.5x10^-3 and the count rate is 0.62
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    (Original post by CD223)
    Is it true that the binding energy released is shared amongst products in the fraction of their masses over the total mass?

    I read this which confused me:

    Name:  ImageUploadedByStudent Room1434540073.308046.jpg
Views: 101
Size:  135.8 KB


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    energy released is equal to the total change in binding energy.
    I hadnt read anything about the way the energy is divided up, but we know momentum has to be conserved so the fission fragments will have the energy divided up so that the momentum remains constant
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    (Original post by johnh545)
    energy released is equal to the total change in binding energy.
    I hadnt read anything about the way the energy is divided up, but we know momentum has to be conserved so the fission fragments will have the energy divided up so that the momentum remains constant
    I see. So those with lower mass will have greater velocity and hence a greater share of the total energy, as it depends on v^2?


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    (Original post by CD223)
    Not a dumb question!

    The number of sparks/detections per unit time for a particular detector placed at a certain distance away from a radioactive source.


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    Thank you very much, I usually get confused about that
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    (Original post by Disney0702)
    Thank you very much, I usually get confused about that
    No worries


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