Suppose a gust of wind has carried a 44-μm-diameter dust particle to a height of 310 m. if the wind suddenly stops, how long will it take the dust particle to settle back to the ground? dust has a density of 2700 kg/m3, the viscosity of 25∘c air is 2.0×10−5 n⋅s/m2, and you can assume that the falling dust particle reaches terminal speed almost instantly.

t=39.76 min

Explanation:

Given that

D= 44 μm

R= 22 μm

h=310 m

We know that drag force

F = 6πμRV

V=Terminal velocity

R=Radius

μ=Dynamic velocity

At the terminal condition

m g = 6πμRV              -----------1

m=ρ x Volume

Now by putting the value in equation 1

V= 0.13 m/s

So time taken by particle

t= h/V

t = 310/0.13

t=2384.6 sec

t=39.76 min

no

explanation:

no i dont know how to solve this you should probably try to google the answer