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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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At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
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Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

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Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

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The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

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