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hahah no problemokay you're a genius! thanks so much
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hahah no problemokay you're a genius! thanks so much
Q30- d sin theta=n lambda
sin theta=1 as its normal to the grating. then d=1/N. N=300 per mm means 300000 per m. so 1/300000=(690*10^-9)*n
n will be around 4. then multiply by 2 and u get 9
you always multiply by 2 to find the maxima. i think it was because n is only for one side or something...i dont really remember the reason but just remember 2 do it everytimehow come you multiply it by 2?
you always multiply by 2 to find the maxima. i think it was because n is only for one side or something...i dont really remember the reason but just remember 2 do it everytime
no diameter gets halved but the radius gets decreasd by 1/4th then. Try it using a value for diameter and find the area. Then repeat using half value of diameter. Area will be 1/4thand for 27..if the diameters being halved shouldnt it be 1/2?
sorry can you do 36 as well..last question
you have to split it into 2 parts. 1st take the positive part. I=2. then find P. P=I^2R=2*2*100=400Wand for 27..if the diameters being halved shouldnt it be 1/2?
sorry can you do 36 as well..last question
ok for 15 take f=kvhttp://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w10_qp_12.pdf
Can any one help me with the Question No:7,15,22...
Please anyone
ok for 15 take f=kv
mg=kv
make v the subject
mg/k
and substitute in the formula .5mv^2
u will get the ans D
Damn dnt u knw the formula of kinetic energy .5 mv^2
link plsM/J/12 paper 12 questrion no 30. If anyone can help, I would be very obliged.
see the cro rises to a maximum at a position X, falls to a minimum and then rises once again to a maximum at aM/J/12 paper 12 questrion no 30. If anyone can help, I would be very obliged.
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