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Brums [2.3K]
3 years ago
15

A solid cylinder is released from the top of an inclined plane of height 0.81 m. From what height, in meters, on the incline sho

uld a solid sphere of the same mass and radius be released to have the same speed as the cylinder at the bottom of the hill?
Physics
1 answer:
Jlenok [28]3 years ago
3 0

Answer:

same 0.81m

Explanation:

in this problem if we assume there no resistance of any sort. and we apply the energy conservation

change in Potential energy = change in kinetic energy

mgh = 0.5mv^2

gh = 0.5v^2

the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.

so at the bottom

put h = 0.81m

9.81 * 0.81 * 2 = v^2

v=3.99 m/s

both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size

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Explanation:

In the given situation two forces are working. These are:

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It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

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         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

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Since, the electron is travelling downwards it means that it looses the potential energy.

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