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

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    (Original post by CD223)
    That makes sense! But then again I guess we could say that discrepancy between estimates of the age of the universe is due to uncertainty in {H_0}?


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    Yeah the hubble constant value is different according to different sources. I think AQA have it at 65 haha
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    (Original post by JJBinn)
    Yeah the hubble constant value is different according to different sources. I think AQA have it at 65 haha
    Yeah they do. Is there an accuracy that they want it to be in when you convert it to s^{-1}?


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    Hi, could anyone doing astrophysics help me on chapter 2.3 Summary Question 4 I cant seem to get the right answer
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    (Original post by CD223)
    Yeah they do. Is there an accuracy that they want it to be in when you convert it to s^{-1}?


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    I think the answer is supposed to come out at about 15000billion years
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    (Original post by frankiejayx)
    Hi, could anyone doing astrophysics help me on chapter 2.3 Summary Question 4c I cant seem to get the right answer
    

P=\sigma A T^{4}

    

{T_X} = 4,000K

{T_Y} = 8,000K

{T_Z} = 20,000K

{M_X} = -2

{M_Y} = +4

{M_Z} = +10

    The surface area is proportional to the diameter squared, so:

    

\dfrac{{A_X}}{{A_Z}} = \left(\dfrac{{D_X}}{{D_Z}}\right  )^{2}

    As in the course notes, the equation for ratios can be re-written as:

    

\dfrac{{P_X}}{{P_Z}} =  \left(\dfrac{{D_X}}{{D_Z}}\right  )^{2} \left(\dfrac{{T_X}}{{T_Z}}\right  )^{4}

    The power output ratio is dependent on the absolute magnitude difference between the two stars.

    There is a difference of (100)^{\frac{1}{5}} between each magnitude due to the logarithmic scale.

    In which case, as there is a difference of 12 magnitudes between X and Z:

    

\dfrac{{P_X}}{{P_Z}} = (100)^{\frac{12}{5}}

    So for X and Z:

    

(100)^{\frac{12}{5}} = \left(\dfrac{{D_X}}{{D_Z}}\right  )^{2} \left(\dfrac{{4,000}}{{20,000}}\  right)^{4}

    From which the ratio:

    

\dfrac{{D_X}}{{D_Z}}

    can be found.

    If you do the same with the figures for Y and Z, you'll end up with answers of:

    

\dfrac{{D_X}}{{D_Z}} = 6300

    and

    

\dfrac{{D_Y}}{{D_Z}} = 100

    Long post (sorry), but hope it helps.

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    (Original post by JJBinn)
    I think the answer is supposed to come out at about 15000billion years
    Yeah although my teacher said estimates for {H_0} can vary from 50 to 100 depending on the source, and as the rate is accelerating the value is subject to change as time goes on, producing a graph of V against D with a curved gradient instead of a straight line haha.


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    Is the ratio of:

    

\dfrac{number\ of\ \gamma\ photons\ incident\ on\ a\ detector}{number\ of\ \gamma\ photons\ produced\ by\ a\ source}

    Always equal to:
    

\dfrac{surface\ area\ of\ detector}{4 \pi r^{2}}
    ?


    Also, for the equation:

    

I=\dfrac{k{I_0}}{x^2}

    Does I represent the activity and count rate? Because doesn't

    

I = \dfrac{Power}{Area}
    ?
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    (Original post by CD223)
    Is the ratio of:

    

\dfrac{number\ of\ \gamma\ photons\ incident\ on\ a\ detector}{number\ of\ \gamma\ photons\ produced\ by\ a\ source}

    Always equal to:
    

\dfrac{surface\ area\ of\ detector}{4 \pi r^{2}}
    ?


    Also, for the equation:

    

I=\dfrac{k{I_0}}{x^2}

    Does I represent the activity and count rate? Because doesn't

    

I = \dfrac{Power}{Area}
    ?
    Hello ☺

    I believe the answer to the first ratio you were asking about can be found in Q13b)i) in the following link.

    http://www.cyberphysics.co.uk/Q&A/KS...ivity/Q13.html

    I'm afraid I do not know the answer to the second ratio yet, I will keep trying 😊
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    (Original post by Disney0702)
    Hello ☺

    I believe the answer to the first ratio you were asking about can be found in Q13b)i) in the following link.

    http://www.cyberphysics.co.uk/Q&A/KS...ivity/Q13.html

    I'm afraid I do not know the answer to the second ratio yet, I will keep trying 😊
    Hello! Thank you for that - it really helped


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    For determining the radius of a nucleus, how is the following equation found? I know it's electron diffraction and that:

    

\sin\theta = \dfrac{0.61\lambda}{R}

    Can be rewritten as:

    

\sin\theta = \dfrac{1.22\lambda}{D}

    For the radius R, or diameter D of a nucleus, but how is this derived from the diffraction pattern?

    In the CGP guide it says: "This is where the first minimum appears":
    Is it related to the single slit diffraction pattern we did last year?

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    Anyone gathered together the grade boundaries for the past papers to see which are the most difficult?
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    (Original post by 0151)
    Anyone gathered together the grade boundaries for the past papers to see which are the most difficult?
    The grade boundaries don't necessarily indicate a harder paper. Last year's paper was manageable compared to previous years and the boundaries were ridiculously low.


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    (Original post by 0151)
    Anyone gathered together the grade boundaries for the past papers to see which are the most difficult?
    Had nothing to do while watching tv so, unit five grade boundaries (for the astrophysics option - I might do them for the others if I have spare time but no promises).

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    (Original post by Lau14)
    Had nothing to do while watching tv so, unit five grade boundaries (for the astrophysics option - I might do them for the others if I have spare time but no promises).

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    Oh cool thank you! May I nab these for the first post please?

    I wasn't aware they varied with each option lol.


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    (Original post by CD223)
    Oh cool thank you! May I nab these for the first post please?

    I wasn't aware they varied with each option lol.


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    Of course!

    They must do because the optional bits can change, but I can't say I've ever checked!
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    (Original post by Lau14)
    Of course!

    They must do because the optional bits can change, but I can't say I've ever checked!
    True - how's your revision going? do you have many papers left?


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    Is this thread only for astrophysicists?


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    (Original post by Mehrdad jafari)
    Is this thread only for astrophysicists?


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    Not at all! All options - there just seems to be a bit of a bias towards Astro as that's what the majority of people on this thread so far seem to be doing - although medical/turning points/applied options are welcome!


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    (Original post by CD223)
    Not at all! All options - there just seems to be a bit of a bias towards Astro as that's what the majority of people on this thread so far seem to be doing - although medical/turning points/applied options are welcome!


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    That's cool


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    (Original post by CD223)
    True - how's your revision going? do you have many papers left?


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    Still pretty much ignoring physics, I've got 3 PHYA5 and 2 PHYA4 papers left and I'd like to go through and try most of the six markers again at some point. You?
 
 
 
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