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yubakkk said:my last question to u all...
cud u plzzz post em.Mobeen said:i have the answers for june 01.
Mobeen said:i have the answers for june 01.
yubakkk said:ok fromnext time i will give
will u xplain this 2 me????
rabib2013 said:alright anyone with the solution to the problem i just posted? :S
examfearz said:rabib2013 said:alright anyone with the solution to the problem i just posted? :S
Is the answer D?
I thnk u r rite....though i hav a confusion...y cant we use the K.E equation in this??hassam said:well cn this be an explantion...for this one just wanna knw abt ma concept....energy supplied by moter is used to increas p.e of m1 and decrease p.e of m2.......so total energy suppplied m1gh-m2gh...for rate...we cn replace h wid v....so migv-m2gv
Zizu1992 said:Please somebody explain my question too!![/qu
suppose initial velocity is v...so E=1/2 mv^2
At the highest point only the horizontal comp acts which is vcos45...so the K.e at the top will be 1/2(vcos45)^2=1/2v^2(0.5)
As 1/2 mv^2 = E so it becomes 0.5E which is A
examfearz said:I = A^2.....so for Amp 8.0 the Intensity shud be 8^2 = 64....in the equation given this I is for I = 1\r^2
now for distance 2r..the eq gives us I = 1/(2r)^2 whch is equal to 1\4r^2 which means that the intensity decreases by 4 times
so for 2r I becomes 64\4 = 16...and Amplitude= root of Intensity so it becomes 4.... hope I helped....
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