Question

Find the terminal velocity of a spherical bacterium (diameter 2.00 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/m}³.

A. 1.20 × 10⁻⁶ , {m/s}
B. 2.40 × 10⁻⁶ , {m/s}
C. 4.80 × 10⁻⁶ , {m/s}
D. 9.60 × 10⁻⁶ , {m/s}

Answer 1

The terminal velocity of a bacterium falling in water can be calculated using Stokes' law. The formula for terminal velocity is V = (2R^2g(p - p₀))/(9n), where R is the radius of the bacterium, g is the acceleration due to gravity, p is the density of the bacterium, p₀ is the density of the water, and n is the viscosity of the water.

Step-by-step explanation:

The terminal velocity of a bacterium falling in water can be calculated using Stokes' law. Stokes' law states that the terminal velocity is given by the equation V = (2R^2g(p - p₀))/(9n), where V is the terminal velocity, R is the radius of the bacterium, g is the acceleration due to gravity, p is the density of the bacterium, p₀ is the density of the water, and n is the viscosity of the water.

Given that the diameter of the bacterium is 2.00 µm, the radius R is 1.00 µm. The density of the bacterium p is 1.10 × 10³ kg/m³, and the density of the water p₀ is 1000 kg/m³. The acceleration due to gravity g is approximately 9.81 m/s². The viscosity of water at room temperature is approximately 1.00 x 10⁻³ kg/(m·s).

Substituting these values into the equation, we get V = (2(1.00 x 10⁻⁶ m)²(9.81 m/s²)(1.10 × 10³ kg/m³ - 1000 kg/m³))/(9(1.00 x 10⁻³ kg/(m·s))). Calculating this expression will give us the terminal velocity of the bacterium in water.