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)

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

answer:

yes

explanation:

answer:

a

explanation: