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

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Original post by Jimmy20002012
Could anyone explain the twin paradox doing turning points?


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We don't need to know it just that time goes slower for the moving object


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How I did the question:

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Has anyone got any tips on energy (heat specifically) transfer questions? I always seen equations in the mark scheme along the lines of: Heat supplied by X = Heat gain by mateiral Y + heat dissipated by Z. But I struggle forming these equations.
Original post by Protoxylic
Are the neutrons bounded to anything? They are just neutrons so they have a binding energy of zero


cheers
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....?


distance is 1/angle in arc seconds. so 1/0.02 gives 500 parsecs, which you convert to meters.
Original post by johnh545
Anyone?


Basically what this means is that for every 400 emitted rays the detecter only detects 1 of them. For example if the source emitted 800, the detector would detect 2.
for the equation r = r0 A1/3 is r0 always 1.4 x 10-15 m or will they give you a value depending on the question in the exam?
(edited 8 years ago)
hey can anyone explain question 3b(i) on this paper...

http://filestore.aqa.org.uk/subjects/AQA-PHYA5-1-QP-JUN12.PDF


i don't understand why they are using area of a sphere and why the radius of that sphere is then 0.18
What is r0 in the nuclear equation? Book says 1.05 fm but wiki says 1.25fm...
What do we want the 6 marker to be tomorrow?

And does anyone have a link to a good set of Applied notes?
Original post by Disney0702
for the equation r = r0 A1/3 is r0 always 1.4 x 10-15 or will they give you a value depending on the question in the exam?


I just asked that a few seconds after you!

book says 1.05fm, do they give you a value in the exams?
68cccfb74c5459516b69fb8077e20b2a.png

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Why is the mass of hydrogen 1.00728?
3b7ffe8deed8b58e0ca303c54d1bb347.png14.1KB
Original post by ubisoft
I just asked that a few seconds after you!

book says 1.05fm, do they give you a value in the exams?


Original post by Disney0702
for the equation r = r0 A1/3 is r0 always 1.4 x 10-15 m or will they give you a value depending on the question in the exam?


I don't think we're expected to know a value, we'll be given one/given information to calculate it in the exam I think.
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Nit
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Hey can anyone explain why you multiply by 0.5 for question 4(a)(i)
http://filestore.aqa.org.uk/subjects/AQA-PHYA5-2C-QP-JUN12.PDF

Thanks
Original post by donutellme
68cccfb74c5459516b69fb8077e20b2a.png

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Why is the mass of hydrogen 1.00728?


That Hydrogen only contains 1 proton so it's just the mass of a proton.

If you remember the binding energy graph and look at where 1 nucleon is, it's approximately 0 so binding energy of the atom is approximately zero -> just the mass of a proton.
Original post by Lau14
I don't think we're expected to know a value, we'll be given one/given information to calculate it in the exam I think.


Ah okay thank you :smile:
Original post by betbi3etwerrd
That Hydrogen only contains 1 proton so it's just the mass of a proton.

If you remember the binding energy graph and look at where 1 nucleon is, it's approximately 0 so binding energy of the atom is approximately zero -> just the mass of a proton.


Yeah, but wouldn't that just be 1au?
Original post by johnh545
Anyone?


Theres more than 1 way to do this calculation however i think the easiest way to look at it is if the detector counts 0.62/s but this is only 1/400 of the photons emitted. So if the detector was on the source the count rate would be 248/s (0.62 x 400) this is the activity of the source however only a small part of it as the area of the detector is 1.5x10^-3 and the rays are emitted uniformly in all directions. The total activity of the source is how many emitted per second. As worked out in b(i) you know the ratio of the area of dectector to area of emitted rays. So use 248 ÷ this (3.684x10^-3) and you get 67000 Bq
How do you work out the corrected count rate?
Original post by donutellme
Yeah, but wouldn't that just be 1au?


In this case it's just one proton so you just look at the mass of a proton from a your formula book and it gives it in u.

Definition of au is 1/12 th of the mass of a carbon atom.

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