A2 Physics | Post your doubts here

[angelgirl:)]

p41/42/43 ???

iam soorrrryyy...its 42...

 

[optimistic]

oops sorry..i thought it was the other question...upp i guess that n09 one...sorry for that..

ok no problem dude!

 

[histephenson007]

iam soorrrryyy...its 42...

no problem, the answer is in page 44

 

[optimistic]

NOV 2008 Q5 and MAY/JUNE 2008 Q11 part 2. PLEASE HELP!

 

[angelgirl:)]

qno

p41/42/43 ???

I'm guessing 41 & 42. cuz 43 doesn't have a 2 iii lol

So, why is internal energy directly proportional to the temperature?

u = k.e. + p.e.
but we assume that there are no intermolecular forces in ideal gases, hence p.e. = 0
u = k.e
but according to ideal gas equations, we can deduce that k.e. = (3/2) kT (where k is the Boltzman's constant)

So, u = (3/2)kT

Qno. 2biii...iam sorry once more ...

 

[histephenson007]

NOV 2008 Q5 and MAY/JUNE 2008 Q11 part 2. PLEASE HELP!

For O/N 2008 Q5,
Use energy conservation :
K1+P1 = K2+P2

P1 is zero, because they are far from each other
K2 is zero, because they just come into contact
P2 = Q1Q2 / 4(pi)(E) * d
(d is the diameter which equals to the sum of both the radiuses of the deutirium nuclei)

 

[angelgirl:)]

qno

o/n 42...Qno. 2biii...iam sorry once more ...

pls help...

 

[histephenson007]

qno

Qno. 2biii...iam sorry once more ...

Just use W = -P (ΔV) to get first column values

And add heat supplied to get the third column

 

[optimistic]

For O/N 2008 Q5,
Use energy conservation :
K1+P1 = K2+P2

P1 is zero, because they are far from each other
K2 is zero, because they just come into contact
P2 = Q1Q2 / 4(pi)(E) * d
(d is the diameter which equals to the sum of both the radiuses of the deutirium nuclei)

Okay thanks! What about May June 08?

 

[angelgirl:)]

Just use W = -P (ΔV) to get first column values

And add heat supplied to get the third column

i didnt get? cud u pls show me the calculation of each values...?

 

[Zenzenzen]

Can someone explain part (b) of question 9 for me?

http://xtremepapers.com/papers/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_43.pdf

Thanks!

 

[histephenson007]

Okay thanks! What about May June 08?

For 1) The amplitude is constant, because it is frequency modulation.

2) Definition of frequency modulation : carrier frequency varies with the displacement of the signal wave.

Maximum frequency = Carrier wave frequency + Maximum displacement * frequency deviation of carrier wave
Max freq = 500 + 1.5(20)
Max freq =530 Hz
Min freq = 500 - 1.5(20) = 470 Hz

The number of times the carrier wave frequency changes from the min frequency to max frequency depends on the frequency of the signal wave which is 8kHz.

 

[SkyPilotage]

For 1) The amplitude is constant, because it is frequency modulation.

2) Definition of frequency modulation : carrier frequency varies with the displacement of the signal wave.

Maximum frequency = Carrier wave frequency + Maximum displacement * frequency deviation of carrier wave
Max freq = 500 + 1.5(20)
Max freq =530 Hz
Min freq = 500 - 1.5(20) = 470 Hz

The number of times the carrier wave frequency changes from the min frequency to max frequency depends on the frequency of the signal wave which is 8kHz.

its better to write the answer as 8000 times rather than 8kHz. because hz is number of time per second, while the question is asking for the number of times only. Please avoid losing unnecessary marks !

 

[angelgirl:)]

i didnt get? cud u pls show me the calculation of each values...?

???help ppl...o/n 42 2010..Q2biii

 

[angelgirl:)]

i didnt get? cud u pls show me the calculation of each values...?

???help ppl...o/n 42 2010..Q2biii...
i need detail calculation since iam not geting the answer after applying the corect formula as well...

 

[abdullah12]

Do we have to memorize the frequencies in the Electromagnetic spectrum for A2?

 

[angelgirl:)]

Do we have to memorize the frequencies in the Electromagnetic spectrum for A2?

noooo i dont think so...

 

[histephenson007]

???help ppl...o/n 42 2010..Q2biii...
i need detail calculation since iam not geting the answer after applying the corect formula as well...

dV is delta Volume

P -> Q : P = 4x10^5 , dV = -6x10^-4
............. W = - p(dV) = - (4x10^5)(-6x10^-4) = 240 J (first column)
............. u = q + w = -600 + 240 = -360

Q -> R : +720J

R -> P : omg, how'd they get -840?
.............. by reverse calculation, 840 = (1/2)*P*(dV) which is the area of the triangle in the graph.

Any light bulbs shining now?? Anyone?!
Thank you for asking the question, i thought it was easy

 

[angelgirl:)]

dV is delta Volume

P -> Q : P = 4x10^5 , dV = -6x10^-4
............. W = - p(dV) = - (4x10^5)(-6x10^-4) = 240 J (first column)
............. u = q + w = -600 + 240 = -360

Q -> R : +720J

R -> P : omg, how'd they get -840?
.............. by reverse calculation, 840 = (1/2)*P*(dV) which is the area of the triangle in the graph.

Any light bulbs shining now?? Anyone?!
Thank you for asking the question, i thought it was easy

 

[angelgirl:)]

dV is delta Volume

P -> Q : P = 4x10^5 , dV = -6x10^-4
............. W = - p(dV) = - (4x10^5)(-6x10^-4) = 240 J (first column)
............. u = q + w = -600 + 240 = -360

Q -> R : +720J

R -> P : omg, how'd they get -840?
.............. by reverse calculation, 840 = (1/2)*P*(dV) which is the area of the triangle in the graph.

Any light bulbs shining now?? Anyone?!
Thank you for asking the question, i thought it was easy

look at the answer here....