Physics: Post your doubts here!

[goodluckayesha]

https://fbcdn-sphotos-h-a.akamaihd...._=1394792318_16dcac96b5bc8233430f3b7be709b4a5

thnx babii

 

[Grenade]

link plz

 

[Nibz]

Excuse Me. In your limits please.

There was absolutely no reason for you to have argued with someone publicly in a formal thread over such a petty thing.

My best guess is he is your tweedledum, so I am not into teen drama.

You should not have said this. And many other things.

manhus shakal ki bai sharam aurat!

Honestly, if there is anyone who is not in his/her limits, it's you. Using a language other than English in a proper thread, and that too an insulting one, is not allowed at all. So, you better watch what you are doing before telling others what to do and what not to.

And next time when you people find a post offensive, simply report it, and we will handle the rest. Do not act as backseat moderators.

 

[sma786]

There was absolutely no reason for you to have argued with someone publicly over such a petty thing.


You should not have said this. And many other things.


Honestly, if there is anyone who is not in his/her limits, it's you. Using a language other than English in a proper thread, and that too an insulting one, is not allowed at all.

And next time when you people find a post offensive, simply report it, and we will handle the rest. Do not act as backseat moderators.

I understand and I apologize to you for arguing in a formal thread

 

[sma786]

Oh could anybody explain Absolute Zero to me?

 

[Young Stunner]

These are the physics topical MCQ's for current electricity. Can someone plz give the answers???

http://www.mediafire.com/view/8x75fx0uun11t9o/9702_p1_current_electricity_all.pdf

 

[Musab100]

There was absolutely no reason for you to have argued with someone publicly in a formal thread over such a petty thing.


You should not have said this. And many other things.


Honestly, if there is anyone who is not in his/her limits, it's you. Using a language other than English in a proper thread, and that too an insulting one, is not allowed at all. So, you better watch what you are doing before telling others what to do and what not to.

And next time when you people find a post offensive, simply report it, and we will handle the rest. Do not act as backseat moderators.

well done Nibz the way u ended all this fight

 

[usama321]

can any1 help me with mcq 21
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf

There is more pressure near Y due to the height of the column. The water therefore is compressed and thus gains elastic potential energy.

As the rate of flow is constant, kinetic energy is out of the options.

 

[***amd***]

There was absolutely no reason for you to have argued with someone publicly in a formal thread over such a petty thing.


You should not have said this. And many other things.


Honestly, if there is anyone who is not in his/her limits, it's you. Using a language other than English in a proper thread, and that too an insulting one, is not allowed at all. So, you better watch what you are doing before telling others what to do and what not to.

And next time when you people find a post offensive, simply report it, and we will handle the rest. Do not act as backseat moderators.

I apologize for that. i am sorry sma786

 

[sma786]

I apologize for that. i am sorry sma786

Its fine

 

[danial 234]

can any1 solve this??

 

[hafizhah]

Can someone explain to me how to do may june 2004 paper 4 qn 3cii? :/

 

[usama321]

can any1 solve this??View attachment 37092

Moment due to child = 5 * 40g = 200g
As it is in equilibrium, the upward force at the spring would be
200g = 2 * F
F = 100g

100g = 10000 * compression
compression = 9.81 cm
C it is

 

[ahmed abdulla]

 

[usama321]

View attachment 37096

y = mx + c
m = 190 - 3840/ 100
m = -36.5
c = 3840

2300 = -36.5x + 3840
x = 42.2

 

[Muhammad TAimoor]

http://papers.xtremepapers.com/CIE/...and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Q3, b ii. Can anybody explain the steps?

 

[1357911]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_s05_qp_4.pdf
Q3, b ii. Can anybody explain the steps?

p = Nmc^2/3v
pv = Nmc^2/3
divide the numerator and the denominator by 2. If u divide and multiply by the same number, the ans is the same.
pv = 2N/3 (1/2 mc^2)
pv = Nrt and when N = 1 then pv = RT
so
RT = 2/3 N x KE (that 1/2 mc^2 is written as KE here)
3/2 (R/N) x T = KE
R/N = k (the Boltzman constant)
so
3/2 kT = KE

 

[Muhammad TAimoor]

p = Nmc^2/3v
pv = Nmc^2/3
divide the numerator and the denominator by 2. If u divide and multiply by the same number, the ans is the same.
pv = 2N/3 (1/2 mc^2)
pv = Nrt and when N = 1 then pv = RT
so
RT = 2/3 N x KE (that 1/2 mc^2 is written as KE here)
3/2 (R/N) x T = KE
R/N = k (the Boltzman constant)
so
3/2 kT = KE

Thanks but it's Q3 b ii part actually. You answered Q2

 

[1357911]

Thanks but it's Q3 b ii part actually. You answered Q2

My sincere Apologies.


I hope i am right?? Correct me If Im wrong.

 

[williamcwz]

Physics paper1 October November 2003 no.31 pls