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 145.0 mm from the central spin axis, and if the machine hits 46500 rev/min, what would be the centripetal force exerted on a giant amoeba of mass 1.10E-8 kg at the bottom of the tube?
why is there no answer >_<
To find the centripetal force exerted on the giant amoeba at the bottom of the test tube, we need to use the equation for centripetal force.
The formula for centripetal force is:
F = (m * v^2) / r
Where:
F is the centripetal force
m is the mass of the object (1.10E-8 kg)
v is the linear velocity of the object in meters per second
r is the radius of the circular path in meters
First, we need to convert the given speed from revolutions per minute (rev/min) to meters per second (m/s).
1 revolution = 2π radians
1 minute = 60 seconds
So, to convert rev/min to m/s:
v = (46500 rev/min) * (2π radians/revolution) * (1 min / 60 s)
Next, we need to convert the given bottom distance from millimeters (mm) to meters (m).
r = 145.0 mm * (1 m/1000 mm)
Now, we have all the values needed to calculate the centripetal force.
F = (1.10E-8 kg) * (v^2) / r
Substitute the values for v and r into the equation and calculate the answer.