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Year bhi batao dono qtns ke
1eV=(1.6*10^-19) JWkhan860 and can you please explain me what's this eV bcoz there are so many questions in new papers and I don't how to solve them as I haven't been taught this thing by my sir.
Its oct/nov 09 var is 11.Year bhi batao dono qtns ke
How can a voltage thing be equal to energy.1eV=(1.6*10^-19) J
Its a unit of energy
As the collision Is elastic.....both masses wont stick nd move in oppoaite directions after collisions
Its a new addition to syllabus I reckonH
How can a voltage thing be equal to energy.
there's no article regarding that in Cambridge endosed book.
Can you plx solve and post the written solution here on the thread?As the collision Is elastic.....both masses wont stick nd move in oppoaite directions after collisions
Furthermore the relative speed of approach shuld be equal to relative speed of seperation whch is satisfied by opt A
thus ans is A
I'm sorry I didn't get itIts a new addition to syllabus I reckon
Its the energy an electron gains when accelarated by 1 V of pd
Im srry but I hav only dis muchI'm sorry I didn't get it
If you have any article regarding that then plx lemme know bout that.
How did you make it as 5/3 and 1/3 . And why do we revert the signs while doing momentum calculations?
OkhayIm srry but I hav only dis much
As far as I hav dn qtns related to eV...its just abt conversion of energy into eV
nd for tht I hage told u the value
Send me a qtn related to tht nd Ill tell u how to deal wid itOkhay
I have seen that value in the question paper as well but the point is idk how to use and cz of this I'm losing straight forwards 3 marks for all these calculations.
5/3 nd 1/3 are their speeds after collision as given In opt AHow did you make it as 5/3 and1/3 . And why do we revert the signs while doing momentum calculations?
And ty
Here we used Sin90 cux we hav to find all orders md multiplied It by 2, to fimd orders on either side of zero line
I'll send you tomorrow okhay?Send me a qtn related to tht nd Ill tell u how to deal wid it
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