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# AQA Physics AS exam 24th MAY and 9th JUNE 2016 (Thread)

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1. (Original post by Kraixo)
so is the distance 0.12 cm or 0.24 cm
0.118cm is one complete wavelength
2. (Original post by Kraixo)
so is the distance 0.12 cm or 0.24 cm
Looking at that time wavelength is 8cm then, I thought is sed the melting was just at the centre
3. (Original post by HasanRaza1)
Nah, since there were three antinodes so its one complete wave they showed?
I thought it looks like something like in this diagram? Could you check if its alright?
Attached Images

4. (Original post by musslih)
I thought it looks like something like in this diagram? Could you check if its alright?
Ah i see what youve done
Antinodes are places of maximum displacement, so there is an antinode at the max positive AND max negative displacement
You have missed out the max negative displacment
5. (Original post by musslih)
I thought it looks like something like in this diagram? Could you check if its alright?
the microwave is a max and min amplitude, it doesn't go from rest to max amplitude, it goes from rest, max amplitude, rest, then min amplitude (by min amplitude i mean the max negative amplitude)
6. for the last two questions i got

Anti Neutron -> Anti Proton + Position(e+) + Electron Neutrino(Ve)
and Weak interaction
7. markscheme pls
8. (Original post by HasanRaza1)
Ah i see what youve done
Antinodes are places of maximum displacement, so there is an antinode at the max positive AND max negative displacement
You have missed out the max negative displacment
OH! now i know! alright thanks! I just thought the antinodes in this case was the maximum positive displacement, but is there an explanation for why the maximum negative displacement also melted the chocolate?
9. https://youtu.be/3HZIrJtA6jc In case anyone wanted it i made one.
10. Someone make a unofficial markscheme please
11. (Original post by musslih)
OH! now i know! alright thanks! I just thought the antinodes in this case was the maximum positive displacement, but is there an explanation for why the maximum negative displacement also melted the chocolate?
No worries
Antinodes are, by definition, points of maximum displacement on the wave
It doesnt matter if its positive or negative, the particles move/vibrate the same distance so it will warm up regardless
12. (Original post by 83457)
what did you guys put for percentage change of diameter? i put 3%, but i think it was 20%
its 0.8%, i put numbers in it and stuff, took me bloody 5 minutes :P
13. (Original post by 123chem)
its 0.8%, i put numbers in it and stuff, took me bloody 5 minutes :P
I've got the same answer! Yes it took a long time going through the equation again, then finding the % change
14. Did people use newton's first law or third? Used first one here
15. (Original post by 123chem)
Did people use newton's first law or third? Used first one here

I used the third law as the engine exerts a force on the air hence the air produces an equal opposite force in the direction of motion.
16. (Original post by BiopredatorZz)
for the last two questions i got

Anti Neutron -> Anti Proton + Position(e+) + Electron Neutrino(Ve)
and Weak interaction
Yep what I got
17. (Original post by chrisdale98)
Oh god no i did that too
I am pretty sure that as it is deceleration it is 2.71. That means it was accelerating at -2.71.

If you said the deceleration was -2.71, that would mean it is actually accelerating and so it would be speeding up not coming to a halt.

So 2.71 would be correct do not worry
18. (Original post by Akashi)
I used the third law as the engine exerts a force on the air hence the air produces an equal opposite force in the direction of motion.
Yep I put that and I also used the second law and said the rate of change of momentum was equal to the force exerted.
19. (Original post by Akashi)
I used the third law as the engine exerts a force on the air hence the air produces an equal opposite force in the direction of motion.
I used second and third just to be safe!

Engine exerts a force on air so air exerts a force on engine.
Air is accelerated as it passes through the engine so can also use f=ma
20. (Original post by aipomaniac)
I used second and third just to be safe!

Engine exerts a force on air so air exerts a force on engine.
Air is accelerated as it passes through the engine so can also use f=ma
How many marks was that question

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