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Initial KE = 2 * final KE ( because the ball rebounds to half its original height)
So, 1/2*m*u^2 = 2*1/2*m*v^2
1/2*m*u^2 = m*v^2
Rearrange in the form v/u and you get 1/(root) 2
thanks a lot :')
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Initial KE = 2 * final KE ( because the ball rebounds to half its original height)
So, 1/2*m*u^2 = 2*1/2*m*v^2
1/2*m*u^2 = m*v^2
Rearrange in the form v/u and you get 1/(root) 2
After all the times you've swooped in and saved me from going insane, no problemthanks a lot :')
can u explain a bit? y did u multiply by 2 when it was half hieght? :SInitial KE = 2 * final KE ( because the ball rebounds to half its original height)
So, 1/2*m*u^2 = 2*1/2*m*v^2
1/2*m*u^2 = m*v^2
Rearrange in the form v/u and you get 1/(root) 2
thanks for ur help ^^haha easy. spending time and short tempering will not help. its the game of concept
see in 13th mcq. you should recall that torque of a couple is one of the force into perpendicular distance between the two same forces. means F into 1.20 = 900 into .20 , this will give you force 150 N i.e the very dear and sweet B option!
for the minimum force to be applied we need to get same torque below and on the spindle in order to create rotation effect so that the poor weight can be lifted
15th: height decreases by 2 in both containers. pressure reduces by 4 times altogether...no convincing answer available at the moment.
can u explain a bit? y did u multiply by 2 when it was half hieght? :S
for 35 use da formula (2n-1)*75 so for first higher note it will be (2*2)-1=3*75=225
lykwise for da nxt note
hvnt done da nxt yet
oh okay thanks a lotinitial momentum of the ball is 2mv before collision... if collision was perfectly elastic one, all the momentum would have been gained by the ball... and this would have made the change in the momentum equal to 4mv.. 2mv-(-2mv)... now it is given that collision is inelastic, it wil losse some of its momentum to the wall.. so now change in momentum will be less than 4mv but cant be less than the initial momentum... so the answer is C.. i think you cannot verify the answer by the calculations...
for q.31, find the charge dats 10x1 = 10C
KE = o.5*m*v^2 , so if the height is halved, so is the velocity and also the KE.can u explain a bit? y did u multiply by 2 when it was half hieght? :S
for 1 closed end we use (2n-2)c/4l ryt? but we wont use l here and just da upper part...wait what formula is this??
for 1 closed end we use (2n-2)c/4l ryt? but we wont use l here and just da upper part...
for q.31, find the charge dats 10x1 = 10C
now divide it by 1.6x10^-19 to find the no. of electrons
np just plz pray 4 a gud pprwasn't even taught this thanks ♥
for 1 closed end we use (2n-2)c/4l ryt? but we wont use l here and just da upper part...
this myt helpso c is frequency right?
and what about open end?
this myt help
i dun think its specifically mentioned in the syllabus!!thanks ♥
but the problem my wonderful teacher never taught us that
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