The absolute magnitude of the terminal velocity of the E. coli bacteria in water is 1.450726 × 10⁻⁶ m/s.
How to determine absolute magnitude?
The terminal velocity of an object falling through a fluid is determined by the balance between the gravitational force pulling it down and the drag force exerted by the fluid resisting its motion. The drag force for a sphere in laminar flow is given by Stokes' law:
Fd = 6πηrv
where:
Fd = drag force
η = fluid viscosity
r = radius of the sphere
v = terminal velocity
In this case, given the following values:
η = 0.8 × 10⁻³ N·s/m²
r = 0.4 μm = 0.4 × 10⁻⁶ m
m = 480 fg = 480 × 10⁻¹⁸ kg
Calculate the terminal velocity by solving the following equation for v:
mg = 6πηrv
where:
g = acceleration due to gravity (9.81 m/s²)
Substituting in the given values:
(480 × 10⁻¹⁸ kg)(9.81 m/s²)
= (6)(0.8 × 10⁻³ N·s/m²)(0.4 × 10⁻⁶ m)(v)
Solving for v:
v = 1.450726 × 10⁻⁶ m/s
Therefore, the absolute magnitude of the terminal velocity of the E. coli bacteria in water is 1.450726 × 10⁻⁶ m/s.
Complete question:
An e coli bacteria is suspended in water, which has a viscosity of η=0.8×10−3 N*s/m2 and a density of 1000 kg/m3. Assuming the water is absolutely still (no net velocity), the bacteria will fall downward under the influence of gravity. Assuming the bacteria has a mass of 480 femtograms (1 femtogram = 10−18 kg) and a diameter of 0.4 μm, what is the absolute magnitude of the terminal velocity of this bacteria in water? (Ignore buoyancy for the purposes of this problem.) 1.450726x10^-6 m/s is the answer I have gotten continuously and yet it says its incorrect.