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The terminal velocity of a 3 × 10^-5 raindrop is about 9 m/s. Assuming a drag force Fd = −bv, determine (a) the value of the constant b and (b) the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

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mickymike92

Based on the data provided, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s.

What is terminal velocity?

The terminal velocity of a body is the velocity at which the body falls at constant velocity through a fluid.

For the falling raindrop, let positive direction be downwards and negative direction upwards,

  • mass of the raindrop, m = 3×10-5 kg
  • velocity at time t, is v(t)
  • terminal velocity, v0 = 9 m/s
  • gravitational acceleration, g = 9.81 m/s²

The raindrop experiences a downward gravitational force mg, and an upward drag force -bv.

The total force at a time t is given as

  • F(t) = mg - bv(t)

a)

Terminal velocity is achieved then the total force is 0,

0 = mg - bv0

Therefore

b = mg/v0

Substitutingthe values:

b = (3 × 10^-5 × 9.8)/9

b = 3.27 × 10^-5 kg/s

b) Applying Newton's Second Law

F = ma

where

  • a = v/t
  • F = mg

Therefore,

mg = mv/t t = v/g

however, t is at 63% velocity

thus:

t = 0.63v/g

t = 0.63 × 9 /9.8

t = 0.58 s

Therefore, the value of the constant b is 3.27 × 10^-5 kg/s and the time required to reach 63% of terminal velocity is 0.58 s

Learn more about terminal velocity at: https://brainly.com/question/12903433

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