Final answer:
The terminal velocity of a spherical bacterium falling in water can be found using Stokes' law. At terminal velocity, the drag force equals the weight of the bacterium. The acceleration, drag force, and buoyant force can also be calculated using equations from Newton's second law, Stokes' law, and Archimedes' principle. The correct option is a) Terminal velocity
Step-by-step explanation:
To find the terminal velocity of a spherical bacterium falling in water, we can use Stokes' law. Stokes' law states that the drag force experienced by a spherical particle moving through a viscous fluid is given by Fs = 6πηrv. In this equation, Fs is the drag force, η is the viscosity of the fluid, r is the radius of the particle, and v is the velocity of the particle.
Since the bacterium is falling at its terminal velocity, the drag force is equal to its weight. We can write this as Fs = mg, where m is the mass of the bacterium and g is the acceleration due to gravity.
Equating these two expressions for the drag force, we get 6πηrv = mg. From this equation, we can solve for the terminal velocity v: v = mg / (6πηr).
To calculate the other quantities, we need additional information. The acceleration of the bacterium can be found using Newton's second law: F = ma, where F is the net force on the bacterium. At terminal velocity, the net force is zero, so we have 0 = -mg + FD + FB, where FD is the drag force and FB is the buoyant force. From this equation, we can solve for the acceleration a: a = g - (FD + FB) / m.
The drag force can be calculated using Stokes' law, as mentioned earlier.
The buoyant force can be found using Archimedes' principle: FB = ρfVf * g, where ρf is the density of the fluid and Vf is the volume of the fluid displaced by the bacterium.
The volume of a spherical bacterium can be calculated using the formula V = (4/3)πr³.
Finally, the buoyant force can be expressed as FB = (4/3)πρf * r³ * g, where ρf is the density of the fluid. The correct option is a) Terminal velocity