The upper half of an inclined plane of inclination theta is perfectly
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For first half acceleration = g sin theta Therefore, velocity after travelling ball distance v^(2) = 2(g sin theta)l For second half, acceleration = g(sin theta - mu(k) cos theta) so 0^(2) = v^(2) = 2g (sin phi - mu(k) cos phi)l Solving (i) and (ii) we get mu(k) = 2 tan theta
SOLVED: The upper half of an inclined plane with inclination alpha is perfectly smooth, while the lower half is rough. A body starts from rest at the top of the incline and
The upper half of an inclined plane of inclination is perfectly smooth while lower half is rough. A
The upper half of an inclined plane of inclination andtheta; is perfectly smooth while lower half is rough. A blockstarting from rest at the top of the plane will again come to
The upper half of an inclined plane of inclination theta is perfectly smooth while the lower half rough. A block starting from rest the of the plane will again come to rest
SOLVED: 'The upper half of an inclined plane of inclination 0 is perfectly smooth while lower half is rough: A block starting from rest at the top of the plane will again
The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. - Sarthaks eConnect
If A body is released from the top of an inclined plane of inclination theta. It reaches the bottom with velocity V. If keeping the length same the angle of the inclination
The upper half of an inclined plane of inclination is perfectly smooth while the lower half rough. A block starting from rest the of the plane will again come to rest the
The upper half of an inclined plane of inclination theta is perfectly smooth while lower has rough. A block starting from rest the of the plane will again come to rest the
The upper half of an inclined plane with inclination θ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest
