a = 200438 g's
=200438 * 9.81 m/s^2
You have got to be kidding
r = 0.69 meters
a = omega^2 r
solve for omega
which is in radians/second
divide by 2 pi to get revolutions per second
multiply by 60 to get revolutions per minute
How fast (in rpm) must a centrifuge rotate if a particle 69-cm from the axis of rotation is to experience an acceleration of 200438 g's?
(can you show me how you get the answer) be thorough
2 answers
centrifuge means centripetal force
v^2 / r = 200438 g ... v in m/s
rpm = v * 60 / (2 π r)
v^2 / r = 200438 g ... v in m/s
rpm = v * 60 / (2 π r)