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my pleasure (^_^)thank you soo much!
btw.. there's another thing, you got sometime?
yup
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my pleasure (^_^)thank you soo much!
btw.. there's another thing, you got sometime?
Yesis this how i should do it ?
Could someone just give me a quick explanation with tips on how I should find the area under a graph when I'm asked to do so? Do I count the squares? Then what? Thank you
thank you very much
1eV=1.6x10^-191eV=1.6x10^-19 J
Why in some calculations is 1.6x10^-13J used?:S
cn u post the ms as well?1 http://prntscr.com/75zebj
2- http://prntscr.com/75zdy9
can you do this question,
and also explain this a bit. I kind of have missed this topic altogether.
like if for side AB force, we use length of BC, why's that.
and if the coil was rotated and now is perpendicular, how will we calculate force on AB now.
mehria
1 http://prntscr.com/75zebj
2- http://prntscr.com/75zdy9
can you do this question,
and also explain this a bit. I kind of have missed this topic altogether.
like if for side AB force, we use length of BC, why's that.
and if the coil was rotated and now is perpendicular, how will we calculate force on AB now.
mehria
We know that alpha( the intensity reflection coefficient ) is also = Ir/I
It is correct, since E=hc/lambda
thank youApply 2as=v^2-u^2
a is -g, v is 0 and u is vsinα.
Insert these values to get C.
Nov 13/43from which paper is this?
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