Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air

Question

Aliun [14]

2 years ago

7

Taking the density of air to be 1.29 kg/m3, what is the magnitude of the angular momentum (in kg · m2/s) of a cubic meter of air

moving with a wind speed of 73.0 mi/h in a hurricane? Assume the air is 51.2 km from the center of the hurricane "eye."

Physics

1 answer:

Svet_ta [14] · Answer 1 · 2023-03-28T08:15:55+03:00

The magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s

The rotating equivalent of linear momentum in physics is called angular momentum. Because it is a conserved quantity—the total angular momentum of a closed system stays constant—it is significant in physics. Both the direction and the amplitude of angular momentum are preserved.

Given the density of air to be 1.29 kg/m3 and a wind speed of 73.0 mi/h

We have to find the magnitude of the angular momentum

Let,

ρ = Density of air = 1.29 kg/m^3

v = Speed of wind = 73.0 mi/h = 0.032 km/s

M = angular momentum of air

Let the volume of air be 1 m^3

Mass = Volume x ρ = 1 x 1.29 = 1.29 kg

Momentum = M = mass x velocity

Momentum = 1.29 x 0.0032

Momentum = 4.128 x 10^(-3) kg·m^2/s

Hence the magnitude of the angular momentum of air will be 4.128 x 10^(-3) kg·m^2/s