# Physics 6th June 2011 G482!!!!!!! past papersssWatch

#1
I'm going to try out a physics past paper ocr g482, as I havnt done any yet!!!
I'm a little lazy when it comes to revision

just wondering if anyone wanted to do it with me, and we could see what we get lol

I was thinking june 2010 on OCR website

anyone???
1
7 years ago
#2
If you can post some questions on here people will attempt to walk you through them
1
#3
well I dont get question 5e on the june 2010 paper

oh and also 6a, and 7c

anyone help plzzzzzzzzz, it'd be much appreciated
0
#4
oh and I also dont get 5ciii

0
7 years ago
#5
Good luck guys, it's a dirty exam. By far the module I have hated most over the past 2 years out of all my subjects.
0
7 years ago
#6
I can't help if you don't post the questions.
0
#7
6a) describe, in terms of vibrations, the difference between longitudinal and transverse waves. give an example of each wave.
0
7 years ago
#8
^ Transverse - the wave has vibrations at right angles to the energy. Eg light
Longitudinal - the wave has vibrations in the same direction as the energy. Eg vibrations in gases
3
#9
(Original post by Pythag)
^ Transverse - the wave has vibrations at right angles to the energy. Eg vibrations in gases
Longitudinal - the wave has vibrations in the same direction as the energy. Eg light

thanks alot mate
0
7 years ago
#10
Just if it helps anyone on here. I made an annotated formula sheet for this unit, which I, and those in my class have found very useful. It is great to help during revision, and then you should become familiar with all of the information on it, and only need to to occassionaly check things. It has all the equations, all the symbols used, and what they mean, (e.g. n=number of charge carries per meter cubed), and simple names for them which give their use. I hope it is able to help some of you, it has kindly been uploaded here:

Preview
2
7 years ago
#11
(Original post by Pythag)
^ Transverse - the wave has vibrations at right angles to the energy. Eg vibrations in gases
Longitudinal - the wave has vibrations in the same direction as the energy. Eg light
isn't light transverse?
0
7 years ago
#12
Yes, sorry must've got them mixed up.
0
7 years ago
#13
Can anyone help me with the a question on g482 jan 2010 past paper?

i dont get question 7(b) (ii)
0
#14
Can anyone help me with the a question on g482 jan 2010 past paper?

i dont get question 7(b) (ii)

just checked it out to see if i could help, but I guess not, Im finding it quite difficult too lol
I dont get how your meant to derive the current lol
0
7 years ago
#15
i know right! how does I = ne?
0
#16
i know right! how does I = ne?

yh exactly, and how the hell are you meant to get 0.2 from???

I need someone to tell me right now, or I wont be able to do anything else lol I hope a question like this doesnt come up.....(now it probably will)
0
7 years ago
#17
(Original post by taunt)
just checked it out to see if i could help, but I guess not, Im finding it quite difficult too lol
I dont get how your meant to derive the current lol
i know right! how does I = ne?

the method i did was as follows
3x1015 photons
20% of that is 6x14 electrons.

In the previous part it tells you 3x10^15 photons emitted by the laser per second, we're expected to know interactions with electrons are instantaneous (though its really about 10^-9 s )
So every second 3x10^15 photons emitted, 20% of these release electrons, so every second there's 6x10^14 electrons released, they reach the conducting layer very quickly since the hint they give you is that a a potential difference is applied which cause the electrons to move from one layer to the other.

I= Q/t
T=1s Q= n*electrons = 6x10^14 * 1.6x10^-19 =9.6x10^-5 A
0
#18
(Original post by sohail.s)
the method i did was as follows
3x1015 photons
20% of that is 6x14 electrons.

In the previous part it tells you 3x10^15 photons emitted by the laser per second, we're expected to know interactions with electrons are instantaneous (though its really about 10^-9 s )
So every second 3x10^15 photons emitted, 20% of these release electrons, so every second there's 6x10^14 electrons released, they reach the conducting layer very quickly since the hint they give you is that a a potential difference is applied which cause the electrons to move from one layer to the other.

I= Q/t
T=1s Q= n*electrons = 6x10^14 * 1.6x10^-19 =9.6x10^-5 A

Woah!!

thanks!!!

I've been sat here like a fat lump, trying to work it out...thanks so much!!!

btw are you going to do this paper in june too??
0
#19
(Original post by sohail.s)
the method i did was as follows
3x1015 photons
20% of that is 6x14 electrons.

In the previous part it tells you 3x10^15 photons emitted by the laser per second, we're expected to know interactions with electrons are instantaneous (though its really about 10^-9 s )
So every second 3x10^15 photons emitted, 20% of these release electrons, so every second there's 6x10^14 electrons released, they reach the conducting layer very quickly since the hint they give you is that a a potential difference is applied which cause the electrons to move from one layer to the other.

I= Q/t
T=1s Q= n*electrons = 6x10^14 * 1.6x10^-19 =9.6x10^-5 A

OMG!!! Im from halifax toooooo

and

got same GCSE's

wierd....................
2
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