Question

Small, slowly moving spherical particles experience a drag force given by Stokes' law: Fd = 6πηrv where r is the radius of the particle, v is its speed, and η is the coefficient of viscosity of the fluid medium.(a) Estimate the terminal speed of a spherical pollution particle of radius 1.20 multiply.gif 10-5 m and density of 2182 kg/m3.cm/s(b) Assuming that the air is still and that η is 1.80 multiply.gif 10-5 N · s/m2, estimate the time it takes for such a particle to fall from a height of 100 m.

Answer 1

Answer:

Explanation:

At the time of a body achieving terminal velocity, the drag force becomes equal to the weight of the body less the buoyant force by the surrounding medium which can be represented by the following equation

$\frac{4\pi\times r^3(d-\rho)}{3} =6\pi\times n\times r\times v$

Where r is radius of the body , d is density of the material of the body σ is density of the medium and n is coefficient of viscosity of the medium and v is terminal velocity.

Simplifying

v = $\frac{2\times r^2(d-\rho)}{9\times n}$

Assuming the value of density of air as 1.225 kg/m³ and putting other given values in the formula we get

v = $[tex]\frac{2\times (1.2\times10^{-5})^2(2182-1.225)}{9\times 1.8\times10^{-5}}$[/tex]

v = 387 x 10⁻⁵ m/s

Terminal velocity = 387 x 10⁻⁵ m/s

Time taken to fall a distance of 100 m

= $\frac{100}{387\times10^{-5}}$

= 2.6 x 10⁴ s.