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

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2