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Vikki [24]
3 years ago
14

A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?

Physics
1 answer:
NeX [460]3 years ago
5 0

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

W =F*d

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

W = 294*5\\W = 1470 [J]\\

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%

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Explanation:

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2 years ago
a 1500 kg car accelerates uniformly from rest to 10.0 meters per secound in 3.0 secound .what is the work done on the car in thi
zubka84 [21]

Answer:

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Explanation:

Given,

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                     v = u + at

                     a = (v - u) / t

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                    a = (10 - 0) / 3

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Using the second equations of motion

                      s = ut + 1/2 at²

                         = 0 + 0.5 x 3.33 x 3²

                         = 15 m

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                         F = m x a

                            = 1500 Kg x 3.33 m/s²

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The work done by the car,

                          W = F x S

                               = 5000 N x 15 m

                               = 75,000 J

Hence, the work done on the car is, W = 75,000 J

The power delivered by the engine,

                            P = W / t

                                = 75,000 J / 3 s

                                 = 25,000 watts

The power delivered by the engine, P = 25,000 watts

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