Find the terminal velocity (in m/s) of a spherical bacterium (diameter 1.81 µm) falling in water. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.10 ✕ 103 kg/m3. (Assume the viscosity of water is 1.002 ✕ 10−3 kg/(m · s).)

The terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

Explanation:

The terminal velocity of the bacterium can be calculated using the following equation:

   (1)

Where:

F: is drag force equal to the weight

η: is the viscosity = 1.002x10⁻³ kg/(m*s)

r: is the radium of the bacterium = d/2 = 1.81 μm/2 = 0.905 μm

v: is the terminal velocity

Since that F = mg and by solving equation (1) for v we have:

We can find the mass as follows:

Where:

ρ: is the density of the bacterium = 1.10x10³ kg/m³

V: is the volume of the spherical bacterium

Now, the terminal velocity of the bacterium is:

Therefore, the terminal velocity of a spherical bacterium falling in the water is 1.96x10⁻⁶ m/s.

I hope it helps you!

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation:

i dont know but

explanation:

whoever your teacher is, deserves a raise for using pokemon as an example