A sample of blood is placed in a centrifuge of radius 20.0 cm. The mass of a red blood cell is 3.0 10-16 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 10-11 N. At how many revolutions per second should the centrifuge be operated?
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To find the number of revolutions per second at which the centrifuge should be operated, we can use the concept of centripetal force.
The centripetal force acting on an object moving in a circular path is given by the formula:
F = (mv^2) / r
where:
F is the centripetal force,
m is the mass of the object,
v is the velocity of the object, and
r is the radius of the circular path.
In this case, the force acting on the red blood cell settling out of the plasma is given as 4.0 × 10^(-11) N, and the mass of the red blood cell is 3.0 × 10^(-16) kg.
We need to find the velocity at which the red blood cell settles out of the plasma. To do that, we rearrange the formula as follows:
v = √((F * r) / m)
Substituting the given values:
v = √((4.0 × 10^(-11) N * 0.20 m) / (3.0 × 10^(-16) kg))
v = √(1.2 × 10^(-11) m^2/s^2 / ( 3.0 × 10^(-16) kg))
Now, to convert meters squared per second squared (m^2/s^2) to meters per second (m/s), we take the square root of the numerator and denominator, which results in:
v = (1.0954 x 10^(3)) m/s
Since revolutions per second (rps) is required, we need to convert the linear velocity to angular velocity. Recall that angular velocity is defined as the angle covered in a given time period, which in this case is 1 revolution.
The formula to convert linear velocity to angular velocity is:
v = ω * r
where:
v is the linear velocity,
ω is the angular velocity, and
r is the radius of the circular path.
Rearranging the formula to solve for ω, we get:
ω = v / r
Substituting the values:
ω = (1.0954 x 10^3 m/s) / (0.20 m)
ω ≈ 5477 rps
Therefore, the centrifuge should be operated at approximately 5477 revolutions per second.