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The Physics PHYA2 thread! 5th June 2013 watch

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1. (Original post by Sukura)

Question 6di from June 2010!

I'd truly appreciate it
need to do 1/300 to get the answer in mm, and then convert to m by multiplying by 10 to the power of -3
2. Could anyone help me with the wave questions? especially the diffraction garting and the light internally reflecting! ASAP please thank you!
3. (Original post by iamsocoollol)
hi guys can some one please upload the MS AND QP jan 2013 unit 2 pleaseeeeeeeeeeeeeeeeeeeeeeeeeee eeeeeeee

my stupid teacher forgot to email
Why not m8? Here you go!
Attached Images
4. AQA-PHYA2-W-MS-JAN13.pdf (182.1 KB, 69 views)
5. AQA-PHYA2-QP-Jan13.pdf (670.0 KB, 73 views)
6. (Original post by Edorrans)
Could anyone help me with the wave questions? especially the diffraction garting and the light internally reflecting! ASAP please thank you!
Please give the question and the exam paper (year). Thanks.
7. Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?
8. (Original post by x-Sophie-x)
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?
The area under the curve (is the triangle which is given by a=1/2ab). Your a is the x-axis and b is the y-axis. Or something like that. Just replace the energy stored equation with the area and replace those delta L and K with the a and b. Is that ok?
9. (Original post by x-Sophie-x)
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?
Average force exerted on spring = 0 + f divided by 2 = f/2

Work done = force x distance

Workdone/energy stored = 1/2f x deltaL
10. (Original post by x-Sophie-x)
Does anyone know how to prove the energy stored E=1/2kdeltaL formula..?
The energy is area under the graph which is the triangle area if f vs delta k is a straight line.

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11. The area under the graph is a triangle for (straight line) graphs with extension and force on the axes, which act as base and height in area of traingle =1/2 x base x height
12. Which question?
13. (Original post by StalkeR47)
Please give the question and the exam paper (year). Thanks.
Jan 12 - Question 5 please
14. (Original post by Sukura)

Question 6di from June 2010!

I'd truly appreciate it
There are slits per , so there is slit every
15. (Original post by BenChard)
Attachment 223191
Attachment 223192

the ms says 1.4 my calc gives 0.6 -.-
Draw a tangent and find the gradient ?
16. (Original post by Edorrans)
Jan 12 - Question 5 please
This?
Attached Images

17. Done all the past papers, this question was the only one really that made no sense at all.

The question asked to state the phase relationship between x and y, and x and z, I calculate that X and Y are 135 degrees out of phase, and x and z are 315 degrees out of phase. However the mark scheme for the June 2012 paper states that the phase relationship between x and y is completely out of phase/180 degrees, and between x and z is 360/0/2pi radians completely in phase.

need to do 1/300 to get the answer in mm, and then convert to m by multiplying by 10 to the power of -3
Oh!! I forgot to convert... Thank you oh so much! Was wondering why I was 10x-3 out
19. (Original post by StalkeR47)
Which exam paper and which q?
jun10 q3d
20. (Original post by StalkeR47)
This?
Yeah thats great thank you! Could you explain what the difference would be between the white light and the red laser in that drawing?
Done all the past papers, this question was the only one really that made no sense at all.

The question asked to state the phase relationship between x and y, and x and z, I calculate that X and Y are 135 degrees out of phase, and x and z are 315 degrees out of phase. However the mark scheme for the June 2012 paper states that the phase relationship between x and y is completely out of phase/180 degrees, and between x and z is 360/0/2pi radians completely in phase.

I did the same thing as you :L

X and Z are in phase because they are 'doing the same thing at the same time'
X and Y are antiphase because they are doing different things.
This applies because the wave is a stationary wave :')

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