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Physics: Post your doubts here!

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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 (y)(y)(y)
 
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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
 
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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
 
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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
 
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Using conservation of momentum, the velocity of the 1kg trolley would be
2*2 = 1* x
x = 4 m/s

now total kinetic energy of the two trolleys would be equal to the total energy released from the spring

1/2 * 2 * 4 + 1/2 * 1 * 16
= 12 J

For the 20th one, the strain energy would be the area under the graph...
By just taking a triangle, we get
1/2 * .002 * 100 = .1
However, we will also consider the upwards bend of the curve.. Therefore, the answer has to be more than .1... Now, it can't be D because D is too large. D would actually equal to the area of the whole square if we don't multiply our number with 1/2... Therefore, it has to be C.
 
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