A test tube in a centrifuge is pivoted so that it swings out horizontally as the machine builds up speed. If the bottom of the tube is 155.0 mm from the central spin axis, and if the machine hits 51500 rev/min, what would be the centripetal force exerted on a giant amoeba of mass 8.00E-9 kg at the bottom of the tube?
r = 0.155 meter
m = 8 *10^-9 kg
Ac = m v^2/r
but v = 2 pi r * 5.15*10^4 meters / 60seconds = 836 meters/second
so m v^2 / r is
8*10^-9 (6.99* 10^5 ) / 0.155 = 0.0361 N
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To solve this problem, we can use the formula for centripetal force:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass
v is the velocity
r is the radius
First, we need to calculate the velocity of the giant amoeba. To do this, we can convert the given 51500 rev/min to radians/second.
1 revolution = 2π radians
Therefore, 51500 rev/min = (51500 rev/min) * (2π radians/1 revolution) * (1 min/60 s)
Now, we have the angular velocity in radians/second. Multiply this by the distance from the central spin axis to get the velocity of the amoeba.
Now, we substitute the values into the centripetal force formula:
F = (m * v^2) / r
Plugging in the given values will give us the centripetal force exerted on the giant amoeba.