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

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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/Pages/Physics_5_Options/Turning_points/TP_06/turning_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....?
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/explainlikeimfive/comments/j44uq/how_does_time_dilation_work_why_should_you_age/c28z810


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CD223
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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....?


[br]d=1p[br][br]d = \dfrac{1}{p}[br]

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:

[br]d=10.002=500pc[br][br]d = \dfrac{1}{0.002} = 500 pc[br]

As it asks for the distance in metres,

[br]d=500×3.08×1016m[br][br]d = 500 \times 3.08 \times 10^{16}m[br]

[br]∴d=2×1019m (1 sf)[br][br]\therefore d = 2 \times 10^{19}m\ \text{(1 sf)}[br]


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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?
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:

ImageUploadedByStudent Room1434540073.308046.jpg


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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
Picture3.png33.1KB
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:

ImageUploadedByStudent Room1434540073.308046.jpg


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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 :smile:
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Original post by Disney0702
Thank you very much, I usually get confused about that :smile:


No worries :smile:


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Original post by CD223
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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exactly :smile:
Could anyone explain what we need to know of the parallax angle?
Original post by CD223
*


Thanks CD223. I understand how to get the answer once you know that d=1/p, it's just I had never seen that before anywhere, neither in the textbook nor spec notes online. At least I'll know it now if it comes up tomorrow.

P.s Cheers for your notes in the op, was useful to have a quick read through as there's little bits in the Nelson Thornes book not covered in the CGP one :smile:
Original post by you-only-live-once
anyone doing medical option?


Yeah I am :smile: it's annoying, every past paper has some different facts which you needed to know, most of which aren't in my textbook or notes -.- just going through loads of old-spec medical papers at the moment...
http://filestore.aqa.org.uk/subjects/AQA-PHYA51-QP-JUN14.PDF

hey guys can anyone explain why the markscheme doesnt take the binding energy of neutrons into account in Q 1b (ii) ? confused :frown:
Original post by Star Light
Yeah I am :smile: it's annoying, every past paper has some different facts which you needed to know, most of which aren't in my textbook or notes -.- just going through loads of old-spec medical papers at the moment...


Yes i know! what do you think will be the 6 marker? :smile:

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