Small, slowly moving spherical particles experience a drag force given by Stokes’ law: where r is the radius of the particle, 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 1.00 x 10-5 m and density of 2000 kg/m3. (b) Assuming that the air is still and that η is 1.80 x 0-5 N · s/m2, estimate the time it takes for such a particle to fall from a height of 100 m.

Repaeat post. Already answered.

(a) Well, well, look who's causing a stir in the fluid medium! Let's calculate that terminal speed, shall we? According to Stokes' law, the drag force (F) acting on the particle can be written as F = 6πηrv.

Now, we know that at terminal speed, the drag force is equal to the gravitational force acting on the particle. So we have F = mg, where m is the mass of the particle and g is the acceleration due to gravity.

The mass of the particle can be calculated using its density and volume. The volume of a sphere is given by V = (4/3)πr^3, and the mass (m) is density (ρ) times volume (V). Plug in the numbers to get the mass.

Then we set F = mg equal to 6πηrv and solve for v, the terminal speed of our spherical troublemaker.

(b) Now that we have the terminal speed (v), we can calculate the time it takes for the particle to fall from 100 m. Since we're assuming the air is still, we know that the initial velocity (u) is zero.

We can use one of the kinematic equations to find the time (t) it takes to fall a certain distance using the equation: v = u + at. Since our initial velocity is zero, the equation simplifies to v = at.

Plug in the numbers and solve for t, my friend! Remember, it's wise to check if any other forces, like buoyancy, come into play. But in this case, we are assuming the air is still, so we've got that covered.

I hope I didn't leave you in a spin with all these equations!

To estimate the terminal speed of the particle, we can use Stokes' law formula:

Fd = 6πηrv

Where Fd is the drag force, η is the coefficient of viscosity, r is the radius of the particle, and v is its speed.

(a) To estimate the terminal speed, we need to find the maximum speed at which the drag force equals the gravitational force acting on the particle. At terminal speed, the net force on the particle becomes zero.

The gravitational force acting on the particle is given by:

Fg = m * g

Where m is the mass of the particle and g is the acceleration due to gravity.

The mass of the particle can be calculated using:

m = ρ * V

Where ρ is the density of the particle and V is its volume.

The volume of a sphere can be calculated using:

V = (4/3) * π * r^3

Substituting the values provided:

ρ = 2000 kg/m^3
r = 1.00 x 10^-5 m

We can now calculate the mass and gravitational force acting on the particle:

m = ρ * V = 2000 kg/m^3 * (4/3) * π * (1.00 x 10^-5 m)^3
m ≈ 8.38 x 10^-9 kg

Fg = m * g ≈ 8.38 x 10^-9 kg * 9.8 m/s^2
Fg ≈ 8.20 x 10^-8 N

For the drag force to be equal to the gravitational force at terminal speed, we have:

Fd = Fg

Using Stokes' law, we can rearrange the equation to solve for v:

Fd = 6πηrv
Fg = 6πηrv

Solving for v:

v = Fg / (6πηr)

Substituting the given values:

η = 1.80 x 10^-5 N · s/m^2
r = 1.00 x 10^-5 m

v = (8.20 x 10^-8 N) / (6π * (1.80 x 10^-5 N·s/m^2) * (1.00 x 10^-5 m))
v ≈ 2.99 x 10^-3 m/s

Therefore, the estimated terminal speed of the spherical pollution particle is approximately 2.99 x 10^-3 m/s.

(b) To estimate the time it takes for the particle to fall from a height of 100 m, we can use the kinematic equation:

s = ut + (1/2)at^2

Since the particle starts from rest (u = 0 m/s) and the only force acting on it is gravity, the acceleration is equal to g.

Substituting the given values:

s = 100 m
u = 0 m/s
a = g = 9.8 m/s^2

Rearranging the equation to solve for time:

t = sqrt((2s) / a)

t = sqrt((2 * 100 m) / 9.8 m/s^2)
t ≈ 4.52 s

Therefore, it would take approximately 4.52 seconds for the particle to fall from a height of 100 m.