A tiny water droplet of radius . descends through air from a high building. Calculate its terminal velocity. Given that for air = × − −− and density of the water = -3

Correct question is;

A tiny water droplet of radius 0.010 cm descends through air from a high building .Calculate its terminal velocity . Given that η of air = 19 × 10^(-6) kg/m.s and density of water ρ = 1000kg/ms

1.146 m/s

Explanation:

We are given;

Radius; r = 0.010 cm = 0.01 × 10^(-2) m

η = 19 × 10^(-6) kg/m.s

ρ = 1000 kg/ms

The formula for the terminal velocity is given by;

V_t = 2r²ρg/9η

g is acceleration due to gravity = 9.8 m/s²

Thus;

V_t = (2 × (0.01 × 10^(-2))² × 1000 × 9.8)/(9 × 19 × 10^(-6))

V_t = 1.146 m/s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

read more on -

answer and explanation;