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34kurt
2 years ago
4

A 4.00 kg stone is dropped from a height of 145 m. What is the stone's potential and kinetic energy respectively when it is 50.0

m from the ground?
Physics
1 answer:
kompoz [17]2 years ago
3 0
<h3><u>Answer;</u></h3>

D - 1,960 J, 3,720 J

<h3><u>Explanation</u>;</h3>
  • The stone is initially at a height of 100 m from where it is dropped. The gravitational potential energy of an object with mass m at a height h from the ground is PE = mgh where g is the acceleration due to gravity.
  • At a height of 145 m from the ground, the potential energy of the 3 kg stone is 4×9.8×145 = 5,684 J, approximately; 5680 Joules.

At a height of 50 m from the ground, the potential energy will be;

= 4 × 9.8 × 50

<u>= 1960 Joules</u>

This means that some of the energy possessed by the stone at a height of 145 m was converted to kinetic energy.

Therefore; the energy that was converted to kinetic energy will be;

= 5,680 J - 1960 J

= 3,724 Joules

Approximately kinetic energy is 3,720 Joules

Therefore;

The Potential energy is 1960 Joules and Kinetic energy is 3,720 Joules

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Answer:

Part a) When collision is perfectly inelastic

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Part b) When collision is perfectly elastic

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Explanation:

Part a)

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now we know that in order to complete the circle we will have

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now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

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v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

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2 years ago
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A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of
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Explanation:

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