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

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    (Original post by JJBinn)
    Thanks again, these are perfect
    No worries!

    Can you think of any other uses of these radiation types?


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    (Original post by frankiejayx)
    Anyone know any other disadvantages of a CCD other than expensive and need to be kept cool?
    It's difficult to think of other disadvantages I must say - advantages are a lot easier.

    Perhaps the sensitivity is the only main disadvantage. Has it come up as a past paper question before?


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    (Original post by CD223)
    No worries!

    Can you think of any other uses of these radiation types?


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    Radiation is used to power pacemakers I think. Can't remember the detail haha, not much use
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    (Original post by JJBinn)
    Radiation is used to power pacemakers I think. Can't remember the detail haha, not much use
    Oh I see! I'm sure it'll be an application of radiation they give us rather than us mentioning it specifically.


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    (Original post by CD223)
    Oh I see! I'm sure it'll be an application of radiation they give us rather than us mentioning it specifically.


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    Probably, although it could come at the end of a question on radiation after a few calculations involving a certain type of radiation: State and explain a use of *beta/gamma/alpha* radiation for 2 or 3 marks.
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    (Original post by JJBinn)
    Probably, although it could come at the end of a question on radiation after a few calculations involving a certain type of radiation: State and explain a use of *beta/gamma/alpha* radiation for 2 or 3 marks.
    I suppose. Those three would be my safe uses. Not too sure on others! I couldn't name 3 for each type.


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    Astrophysics: June 2011, Q1b (ii)
    I understand how the angular resolution and angular separation was calculated using:

    \theta = \frac{\lambda}{D}

    and

    \theta = \frac{4.8 \times 10^{3}}{1.4 \times 10^{9}}

    But...
    Why is the "minimum" angular resolution the smallest angle in this instance?

    I've always seen the maximum angular resolution being quoted as the smallest numerical value. So in this case, I thought that when commenting on the discrepancy between the two angles, you could say that:

    "The angle calculated in part ii was the minimum angular resolution, so it may resolve two objects at a smaller angle, making the claim valid".

    Is there a definitive time when the terms "minimum" and "maximum" resolution are used? Because I've been caught out too many times!


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    How has everyone been doing on past papers? Ive given up on 2012 and 2013 because they were sooo difficult, like stuff I didn't even know we had to know (if that makes sense, lol). On the first 2 I got an A however so mixed feelings
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    (Original post by Jed-Singh)
    How has everyone been doing on past papers? Ive given up on 2012 and 2013 because they were sooo difficult, like stuff I didn't even know we had to know (if that makes sense, lol). On the first 2 I got an A however so mixed feelings
    Have you done June 2014?

    I've done June 2010, 2011 and 2014 for PHYA5. Got 2012 and 2013 to go. What made them hard?


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    For those doing Astro:

    Is d=\dfrac{1}{p}

    for parallax distances to stars an equation we have to remember off by heart?

    It came up in June 2014 in my mock today so I was just unsure as I couldn't find it on the formula sheet.


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    (Original post by CD223)
    For those doing Astro:

    Is d=\dfrac{1}{p} for parallax distances to stars an equation we have to remember off by heart?

    It came up in June 2014 in my mock today so I was just unsure as I couldn't find it on the formula sheet.


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    I wasn't aware of this equation, is p parallex distance? If so what does this even mean?

    I thought you work out distances to an object by right angled triangles between the earth, sun, and object in question.
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    (Original post by JJBinn)
    I wasn't aware of this equation, is p parallex distance? If so what does this even mean?

    I thought you work out distances to an object by right angled triangles between the earth, sun, and object in question.
    That's essentially is it. The equation drops out of the right angled triangle geometry.

    d is the distance to the object in parsecs, p is the parallax angle in arcseconds

    Where:
    1 arcsecond = {(\dfrac{1 degree}{3600})}


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    (Original post by CD223)
    That's essentially is it. The equation drops out of the right angled triangle geometry.

    d is the distance to the object in parsecs, p is the parallax angle in arcseconds

    Where:
    1 arcsecond = {(\dfrac{1 degree}{3600})}


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    That's a relief, thought there was something I'd missed my teacher today asked us to show that 1/H is the age of the universe. I didn't know we needed to know that until today.
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    (Original post by CD223)
    Have you done June 2014?

    I've done June 2010, 2011 and 2014 for PHYA5. Got 2012 and 2013 to go. What made them hard?


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    Hi, yes, I got a B on June 2014. But 12/13 had like really difficultly worded questions and I found it difficult to give myself marks. At least I know now where to improve
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    (Original post by Jed-Singh)
    Hi, yes, I got a B on June 2014. But 12/13 had like really difficultly worded questions and I found it difficult to give myself marks. At least I know now where to improve
    That's true! I get my mark back tomorrow 😁


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    (Original post by JJBinn)
    That's a relief, thought there was something I'd missed my teacher today asked us to show that 1/H is the age of the universe. I didn't know we needed to know that until today.
    Oh right! Is that involving converting H from kms^{-1}Mpc^{-1} into s^{-1}?


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    (Original post by CD223)
    Oh right! Is that involving converting H from kms^{-1}Mpc^{-1} into s^{-1}?


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    Yeah, you rearrange hubbles law to give 1/H = distance/velocity, which is time. Then you convert 1/h into s^-1 yeah.
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    (Original post by CD223)
    That's true! I get my mark back tomorrow


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    Best of luck, I find the easy questions really really easy, but the hard questions are just rock solid impossible. For me theres no inbetween lol
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    (Original post by JJBinn)
    Yeah, you rearrange hubbles law to give 1/H = distance/velocity, which is time. Then you convert 1/h into s^-1 yeah.
    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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    (Original post by Jed-Singh)
    Best of luck, I find the easy questions really really easy, but the hard questions are just rock solid impossible. For me theres no inbetween lol
    I found the Astro paper easier than the core PHYA5 paper... That six marker was solid. The Astro one was much easier haha.


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