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

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Why bother with a post grad course - waste of time? 17-10-2016
1. That paper was hard af.

Was hoping for a lot more mechanics as ive completed M1,2,3 but almost nothing came up.

Hopefully the boundaries are really low
2. I failed. Were the 3 melted parts nodes or antinodes?

I chose D for lift and C for bungee...not sure if its right 😔
4. (Original post by rifat12345)
I failed. Were the 3 melted parts nodes or antinodes?
Antinodes bro thats why they melted due to the vibration
So the frequency was 2.5x10^9 Hz
5. (Original post by rifat12345)
I failed. Were the 3 melted parts nodes or antinodes?
Melted parts were antinodes.
6. (Original post by thao.lephuong)

I chose D for lift and C for bungee...not sure if its right 😔
I said D for bungee rope and B for the lift...
7. I don't understand the frequency chocolate question... did you have to measure the line? I had no idea you had to do that and was really confused. :/
8. (Original post by thao.lephuong)

I chose D for lift and C for bungee...not sure if its right 😔
Wasnt D the one that experienced a lot of plastic deformation?
It wouldnt be suitable for a lift cos it would just stretch

I put C for bungee tho, but apparently some people are saying D is also a valid answer
9. (Original post by thao.lephuong)

I chose D for lift and C for bungee...not sure if its right 😔
I chose those but the other way around
10. (Original post by NYB)
Melted parts were antinodes.
What did you get for the frequency of the chocolate bar radiation? 2.5x10^9?
11. I think so, 12cm
12. (Original post by Chickenslayer69)
I don't understand the frequency chocolate question... did you have to measure the line? I had no idea you had to do that and was really confused. :/
I measured the distance between the two antinodes, and then multiplied by 2 to get the wavelength...
Then used the wave speed equation to calculate frequency.
13. (Original post by rifat12345)
I think so, 12cm
I got 11.8 cm
14. (Original post by Chickenslayer69)
I don't understand the frequency chocolate question... did you have to measure the line? I had no idea you had to do that and was really confused. :/
I measured the length across all three antinodes, which resulted in a length of 11.8cm. That is 0.118m wavelength
Frequency = Speed of light in a vacuo (3x10^8)/Wavelength (0.118)
So f = 2.5x10^9
15. (Original post by HasanRaza1)
What did you get for the frequency of the chocolate bar radiation? 2.5x10^9?
I got that as well...
16. (Original post by NYB)
I got that as well...
Great did u get 25% as the percentage change?
17. (Original post by HasanRaza1)
Great did u get 25% as the percentage change?
No. I got 605% but I am not sure.
18. Just came home, all i have to do know is wait for the next physics paper. The test was pretty hard but i answered most of the we questions. One thing though, how do u calculate the stopping potential using energy? U are not given coulombs.. So is there some other formula i havent noticed?
19. I failed. Were the centres nodes?
20. (Original post by icecubeinferno)
Just came home, all i have to do know is wait for the next physics paper. The test was pretty hard but i answered most of the we questions. One thing though, how do u calculate the stopping potential using energy? U are not given coulombs.. So is there some other formula i havent noticed?
Divide the energy by the charge of an electron (1.6x10^-19) to get voltage
As voltage = energy/charge as in the formula sheet

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