V^2/R = 5.4*10^4 g = 5.29*10^5 m/s^2
R = 0.079 m
Solve for V
Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.4×104 times the acceleration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.9cm.
1 answer