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nata0808 [166]
3 years ago
10

A 8.1 kg object initially at rest is pushed down a 15.0 m tall hill. What is the speed of the object at the bottom of the hill?

Physics
1 answer:
topjm [15]3 years ago
8 0

Answer:

The velocity of the object at the bottom is, v = 17.15 m/s

Explanation:

Given data,

The initial velocity of the object, u = 0

The height of the hill, h = 15 m

Let 'S' be the distance of the slope of the hill and 'Ф' be the slope of the hill formed with the ground.

The acceleration due to gravity component along the slope is given by,

                                      a = g Sin Ф

The distance of the slope since height 'h' of the hill is given,

                                       s = h / Sin Ф

Using the III equation of motion,

                                      v² = 2 as                    (∵ u = 0)

                                      v² = 2 x g Sin Ф x h / Sin Ф

                                           = 2 gh

Therefore,

                                      <em> v = √(2gh)</em>

Substituting the given values,

                                       v = √(2x9.8x15)

                                          = 17.15 m/s

Hence, the velocity of the object at the bottom is, v = 17.15 m/s

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Explanation:

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A carpenter is driving a 15.0-g steel nail into a board. His 1.00-kg hammer is moving at 8.50 m/s when it strikes the nail. Half
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Answer: The increase in temperature of the nail after the three blows is 8.0636 Kelvins. The correct option is (d).

Explanation:

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54.1875 J=0.015 kg\times 448 J/kg K\times \Delta T

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The increase in temperature of the nail after the three blows is 8.1  Kelvins.Hence, correct option is (d).

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