Final answer:
By equating the drag force given by Stokes' Law with the gravitational force and ignoring the buoyant force, we use the formula v = (2gr²(ρb - ρf))/(9η) to calculate the terminal velocity of the bacterium in water. We use the given values for the density and viscosity of the bacterium and water, along with the acceleration due to gravity, to find the terminal velocity in meters per second.
Step-by-step explanation:
To find the terminal velocity of a spherical bacterium falling in water, one needs to apply Stokes' Law, which relates the drag force to the velocity of a spherical object moving through a viscous fluid. At terminal velocity, the drag force equals the gravitational force minus the buoyant force. However, we are asked to ignore the buoyant force, simplifying the equation to equating the drag force with the gravitational force alone. Using the given parameters, the drag force, Fd, according to Stokes' Law is Fd = 6πηrv, where η is the viscosity of the fluid, r is the radius of the sphere, and v is the terminal velocity.
The gravitational force, Fg, acting on the bacterium is given by Fg = Vρbg, where V is the volume of the bacterium, ρb is the density of the bacterium, and g is the acceleration due to gravity. At terminal velocity, Fd = Fg, so by equating the two forces and solving for v, we get v = (2gr²(ρb - ρf))/(9η), where ρf is the density of the fluid. Inputting the given values of r = 1.4 × 10-6 m, ρb = 1.45 × 103 kg/m³, ρf = 998 kg/m³, η = 1.002 × 10-3 kg/m*s, and g = 9.81 m/s² into the equation yields the terminal velocity v in meters per second.