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Spherical proteins of an unknown radius have a density of 5.90 x 10^3 kg/m^3 and are suspended in blood which has a viscosity 3.5 times that of water. the mixture is placed in a centrifuge of radius 11.2 cm which spins at 9000 rpm. if the proteins have a terminal velocity of 1.60 x10^-6 m/s, what is their radius?

User Zuim
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Final answer:

To find the radius of a spherical protein with a known terminal velocity, density, and suspension medium's viscosity, we use Stokes' law which equates the drag force to gravitational force and buoyant force, and solve for the radius.

Step-by-step explanation:

To calculate the radius of a spherical protein that is falling at a terminal speed in a liquid, we use the balance of forces as described by Stokes' Law. Since the drag force (Fs) is given by 6πηRv, where η is the coefficient of viscosity, R is the sphere's radius, and v is the terminal velocity, we can solve for R once we have the viscosity (taken to be 3.5 times that of water) and the given terminal velocity (1.60 x 10-6 m/s).

Here's the equation from Stokes' law that fits our scenario:

V = 2R2g(Ps - P1) / 9η,

where V is the terminal velocity, Ps is the density of the sphere, P1 is the density of the fluid, g is the acceleration due to gravity (9.80 m/s2), and R is the radius.

Assuming that η (viscosity of blood) is 3.5 times that of water (about 0.001 kg/m·s), we can rearrange the equation to solve for R:

R = √((9ηV) / (2g(Ps - P1)))

Substituting the known values, including the given density of the spherical protein (Ps = 5.90 x 103 kg/m3) and assuming the density of blood (about 1060 kg/m3), we can compute the radius.

User Jonathan King
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