A particle is emitted from a smoke stack with diameter of 0.05 mm. In order to determine how far downstream it travels it is important to find its terminal downward velocity. If it has a density of 1200 kg/m3, its terminal downward velocity (cm) is: (assume the drag coefficient is 24/Re and the volume of a sphere is 4/3 pi R3)

Answer: downward velocity = 6.9*10^-4 cm/s

Explanation: Given that the

Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 * 10^-5 m

Where radius r = 2.5 * 10^-5 m

Density = 1200 kg/m^3

Area of a sphere = 4πr^2

A = 4 * π * (2.5 * 10^-5) ^2

A = 7.8 * 10^-9 m^2

Volume V = 4/3πr^3

V = 4/3 * π * (2.5 * 10^-5) ^3

V = 6.5 * 10^-14 m^3

Since density = mass / volume

Make mass the subject of formula

Mass = density * volume

Mass = 1200 * 6.5 * 10^-14

Mass M = 7.9 * 10^-11 kg

Using the formula

V = sqrt (2Mg / pCA)

Where

g = 9.81 m/s^2

M = mass = 7.9 * 10^-11 kg

p = density = 1200 kg/m3

C = drag coefficient = 24

A = area = 7.8 * 10^-9m^2

V = terminal velocity

Substitute all the parameters into the formula

V = sqrt[ (2 * 7.9*10^-11 * 9.8) / (1200 * 24 * 7.8*10^-9) ]

V = sqrt[ 1.54 * 10^-9/2.25*10-4]

V = 6.9*10^-6 m/s

V = 6.9 * 10^-4 cm/s