104k views
0 votes
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
d

User Aswath
by
8.1k points

1 Answer

2 votes

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.

User LBA
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.