A2 Physics | Post your doubts here

[optimistic]

Hey guys please help me solve this question of MAY/JUNE 2008 Q8 part a and b.... I have partially done part b by the formula e=mc2 but can't figure out how to convert the energy into M.E.V

 

[user]

Did you get it?
If you are the centre of rotation, how far are you from the centre? zero! so centripetal force is zero hence weight = normal force.

aoa wr wb!
umm i didnt get this...

 

[user]

Hey guys please help me solve this question of MAY/JUNE 2008 Q8 part a and b.... I have partially done part b by the formula e=mc2 but can't figure out how to convert the energy into M.E.V

1eV = 1.6 x 10^-19
1MeV = 1.6 x 10^-19 x 10^6 = 1.6 x 10^-13 J

divide the value in joules by this number...and you'll get it in MeV

P.S. I didn't check the question exactly

 

[histephenson007]

Hey guys please help me solve this question of MAY/JUNE 2008 Q8 part a and b.... I have partially done part b by the formula e=mc2 but can't figure out how to convert the energy into M.E.V

a) Ah, the monstrous question! I just asked my teacher few days ago.
This question is about the conservation of momentum.

In question, it says that the kinetic energies of positron and electron are negliblie initially. So, v should be zero (almost).
So, we know that initial momentum is zero.
Conservation of momentum says that initial momentum is equal to final momentum.
Therefore final momentum must also be zero.
So, each of these photons are moving with the same magnitude of momentum, but towards opposite directions.(making total momentum zero)
Since momentum is equal, and m of the photons is equal, p=mv, v should also be equal
therefore, kinetic energy is also equal.

 

[SkyPilotage]

aoa wr wb!
umm i didnt get this...

If you arre the poles, this means that you are on the centre of rotation. For example, if you rotate around your self, you are on the axis of rotation. What is your centripetal force? its zero because centripetal force is M x R ( from centre ) x omega square. Since R is zero, the force is zero.
When you are at the poles, there is a graviattional force downwards, since centripetal force is zero, normal force is equal to weight
When you are at the equator, there is ur weight towards the Earth and a normal force. Centripetal force is not zero so normal force is less than weight!

 

[user]

If you arre the poles, this means that you are on the centre of rotation. For example, if you rotate around your self, you are on the axis of rotation. What is your centripetal force? its zero because centripetal force is M x R ( from centre ) x omega square. Since R is zero, the force is zero.
When you are at the poles, there is a graviattional force downwards, since centripetal force is zero, normal force is equal to weight
When you are at the equator, there is ur weight towards the Earth and a normal force. Centripetal force is not zero so normal force is less than weight!

jazakAllah khairen... i think i get some idea now..

P.S. need help with applications i got not much time left...and i dont know what to do...any help?

 

[SkyPilotage]

jazakAllah khairen... i think i get some idea now..

P.S. need help with applications i got not much time left...and i dont know what to do...any help?

The application booklet is the best source there is. Because its prepared by examiners. Its only 43 pages. 10 pages per hour, you will be done withh all chapter of applications today!
Any problems you got please do post them or inbox me!

 

[optimistic]

a) Ah, the monstrous question! I just asked my teacher few days ago.
This question is about the conservation of momentum.

In question, it says that the kinetic energies of positron and electron are negliblie initially. So, v should be zero (almost).
So, we know that initial momentum is zero.
Conservation of momentum says that initial momentum is equal to final momentum.
Therefore final momentum must also be zero.
So, each of these photons are moving with the same magnitude of momentum, but towards opposite directions.(making total momentum zero)
Since momentum is equal, and m of the photons is equal, p=mv, v should also be equal
therefore, kinetic energy is also equal.

Yayyyyy!!! You made it so easy! Thanks!

 

[optimistic]

I am having a problem in finding the frequency in Q11(ii) MAY/JUNE 2008. Can anyone help?

 

[user]

The application booklet is the best source there is. Because its prepared by examiners. Its only 43 pages. 10 pages per hour, you will be done withh all chapter of applications today!
Any problems you got please do post them or inbox me!

will try that inshaAllah..
P.S. Please note, there's a mistake in that mri part...it uses hydrogen 'atom' when it should be hydrogen 'nuclei'. ms says reject atoms, i dont know why do they then mention so in the booklet :/

anyways make sure u know that

 

[arlery]

http://www.xtremepapers.com/papers/...and AS Level/Physics (9702)/9702_w07_qp_4.pdf
Q 1b (ii)

 

[smzimran]

will try that inshaAllah..
P.S. Please note, there's a mistake in that mri part...it uses hydrogen 'atom' when it should be hydrogen 'nuclei'. ms says reject atoms, i dont know why do they then mention so in the booklet :/

anyways make sure u know that

Applications portion is better explained in Chris Mee book, although mostly the matter is the same!

 

[user]

I am having a problem in finding the frequency in Q11(ii) MAY/JUNE 2008. Can anyone help?

Assalamoaalikum wr wb!
Check attached..this might help...this is what helped me actually ....

 

[user]

Applications portion is better explained in Chris Mee book, although mostly the matter is the same!

I know right....
It's like you just read that...cuz it has almost the same things, with additions where application booklet is not clear enough...plus the mistake i mentioned above isn't there in that book..while the whole para is exactly the same, it uses 'nuclei' and not 'atoms'

 

[SkyPilotage]

will try that inshaAllah..
P.S. Please note, there's a mistake in that mri part...it uses hydrogen 'atom' when it should be hydrogen 'nuclei'. ms says reject atoms, i dont know why do they then mention so in the booklet :/

anyways make sure u know that

yup! hydrogen nuclei or "protons"
Thank you for the concern, brother

 

[angelgirl:)]

o/n 2010 Q2iii...pls help...

 

[histephenson007]

I am having a problem in finding the frequency in Q11(ii) MAY/JUNE 2008. Can anyone help?

For these sort of questions, I basically have a few rules after practising with them for a while :

For frequency modulated waves,
Rule 1 - Maximum frequency = Carrier frequency + (amplitude of signal wave)*(frequency deviation)
As I come to think of it, it suits the definition of frequency modulation. Which states that the carrier wave frequency is varied according to the displacement of the signal wave.
Similarly, *** Min freq = Carrier freq - Xo*frequency deviation***

Rule 2 - Frequency of Max freq. ---> Min freq. ---> Max freq. depends on the frequency of the signal wave.

Just try using these rules.

 

[optimistic]

Assalamoaalikum wr wb!
Check attached..this might help...this is what helped me actually ....

thanks but I can't find the answer to my question. On which page is the explanation given?

 

[histephenson007]

o/n 2010 Q2iii...pls help...

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

 

[user]

thanks but I can't find the answer to my question. On which page is the explanation given?

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