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FOR A liquid drop of radius r falling in air with terminal velocity, power delivered by gravitational force is propotional to

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Answer:

P = m g v

Step-by-step explanation:

Power is defined as the relationship between work and time

P = W / t

The work is defined by the scalar product of the force and the displacement, in the case of a drop falling the vertical force and the displacement has the same direction, therefore the angle between them is zero and the cosines of zero is one (cos 0 = 1)

W = F y

the gravitational force is proportional to the weight of the drop (F = mg)

W = mg y

we substitute

P = m g y / t

the terminal velocity of the drop is constant

P = m g v

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