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ALL CIE PHYSICS DOUBTS HERE !!!
Dania:
Thanks :)
Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?
Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.
ruby92:
M/j 2010 physics p41
q2 part c(iii)
ashish:
--- Quote from: ruby92 on November 05, 2010, 12:58:01 pm ---M/j 2010 physics p41
q2 part c(iii)
--- End quote ---
I will give you a tip try it
Internal energy = PEs + KEs
since ideal gas does not have intermolecular forces of attraction PE =o
Internal energy= KEs
= number of gas molecules * KE of one atom
TJ-56:
Can sumone help me with diagrams in nov 2003 qp2, the first one is q3 (c) (i) just wanna make sure if its right and explain if you can pls,
the second one is q5 (c) why isn't it a straight line from origin, why does it have a curve as it says on the mark scheme?
thanks so much for any help in advance
Deadly_king:
--- Quote from: Dania on November 05, 2010, 11:24:14 am ---Thanks :)
Also, there's something I always struggle to grasp. There are questions where they give a diagram with a wave on a graph & tell you that there's another wave, giving you the phase angle of the latter one. I don't know if you get me. But what I'm trying to say is that how would I make the necessary calculations to enable me to draw the second wave? Like the time difference, etc. Would you happen to have a link that could help me?
Ref. October/November 2002, Paper 2, Q. (b) (i).
Thanks in advance.
--- End quote ---
Yeah, I understand what you mean. Once it was rather complicated for me too. ;)
You should know this formula => Phase angle = 2(pie)/lambda
This formula applies when you have a graph of x against wavelength(lambda). Since in this case the graph is x against time(t), we can change the formula to => Phase angle = 2(pie)/T, where T is the period of oscillation and is the time difference.
It has been said that both waves have the same waveform, which implies same wavelength and same period. The only difference is that one would lead the other.
Take 2(pie) as 360o since the phase angle has been given in degrees and not radians.
Hence 60o = 2(180)/3 ---> = 0.5s
Therefore the new wave will lag behind by 0.5s, i.e it will have its first maximum at 0.5s and the first minimum at 2s.
Hope you get it. :)
Am sorry but I don't have any specific sites for this topic, but here are some links which a member has been so kind to look for us. :D
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