Complete Question
Find the terminal velocity of a spherical bacterium (diameter [tex]2.00 \mu m[/tex] ) 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 [tex]1.10 *10^{3} \ kg/m^3[/tex]
Answer:
The value is [tex]v_t = 2.38*10^{-6} \ m/s[/tex]
Explanation:
From the question we are told that
The diameter of the spherical bacterium is [tex]d = 2.0\mu m = 2.0 *10^{-6} \ m[/tex]
The density of the bacterium is [tex]\rho =1.10 *10^{3} \ kg/m^3[/tex]
Generally the radius is mathematically represented as
[tex]r = \frac{d}{2}[/tex]
=> [tex]r = \frac{2.0 *10^{-6}}{2}[/tex]
=> [tex]r = 1.0*10^{-6}[/tex]
Generally the volume is mathematically represented as
[tex]V = \frac{4}{3} * \pi * r^2[/tex]
=>[tex]V = \frac{4}{3} * 3.142 * (1.0*10^{-6})^3[/tex]
=>[tex]V = 4.189 *10^{-18} \ m^3[/tex]
Generally the mass is mathematically represented as
[tex]m = V * \rho[/tex]
=> [tex]m = 4.189 *10^{-12} * 1.10 *10^{3}[/tex]
=> [tex]m = 4.608 *10^{-15} \ kg[/tex]
Generally the terminal velocity is mathematically represented as
[tex]v_t = \frac{m * g}{ 6\pi * r * \eta}[/tex]
Here [tex]\eta[/tex] is the viscosity of water with value [tex]\eta = 1.005 *10^{-3} \ kg/m\cdot s[/tex]
So
[tex]v_t = \frac{4.608 *10^{-15} * 9.8}{ 6 * 3.142 *1.0*10^{-6} * 1.005 *10^{-3}}[/tex]
=> [tex]v_t = 2.38*10^{-6} \ m/s[/tex]
