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Olenka [21]
2 years ago
14

The magnitude of the force associated with the gravitational field is constant and has a value F. A particle is launched from po

int B with an initial velocity and reaches point A having gained U0 joules of kinetic energy. A resistive force field is now set up such that it is directed opposite the gravitational field with a force of constant magnitude 12F. A particle is again launched from point B. How much kinetic energy will the particle gain as it moves from point B to point A
Physics
1 answer:
Goryan [66]2 years ago
4 0

Answer:

Energy gained by the second particle = 12Uo

Explanation:

Given Data;

Resistant force = 12F

Initial kinetic energy = Uo

Calculating the kinetic energy gained, we have;

u = f *r

where f= resistant force = 20F

r = initial kinetic energy = Uo

Therefore,

U = 12 * uo

  = 12 Uo

Therefore, energy gained by the second particle = 12Uo

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Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

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