(796) Problem 5: Consider a spherical bacterium (radius 1.85 ) falling in water

Question

(796) Problem 5: Consider a spherical bacterium (radius 1.85 ) falling in water at 20° C. Randomized Variables 1.85 d=1.05+103 kum3 Find the terminal velocity of the spherical bacterium in ms 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 1.05.10 kgm. The viscosity of water at 20 Cis 1.005.10kgms Grade Summary Deductions -0% Potential 100 Submission mepts rem.aining 4 5 6 acos per attempt det "Degrees , Radians Submit Hint I give up int Geduction per hunt Hnt reuning Feedback: S desuction per fendback

Explanation / Answer

For a spherical object falling in a medium, Drag force is given by:

Fs = 6*pi*r*n*Vt

Vt = terminal Velocity = Fs/(6*pi*r*n)

Fs = Force on Spherical Bacterium = m*g

m = mass of bacterium = Volume*density

r = radius of bacterium = 1.85*10^-6 m

Volume = 4*pi*r^3/3 = 4*pi*(1.85*10^-6)^3/3 = 2.65*10^-17 m^3

density = 1.05*10^3 kg/m^3

m = (2.65*10^-17 m^3)*(1.05*10^3 kg/m^3) = 2.78*10^-14 kg

g = gravitational acceleration = 9.81 m/sec^2

n = Viscosity = 1.005*10^-3 kg/m.sec

So using these values:

Vt = 2.78*10^-14*9.81/(6*pi*1.85*10^-6*1.005*10^-3)

Vt = 7.78*10^-6 m/sec

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