The angular speed of the rotor in a centrifuge decreases from 226 to 186.0 rev/s with an angular deceleration of 6.4 rev/s2. During this period, how many revolutions does the

Question

rotor turn?

Answer 1

N=(n₂²-n₁²)/2a=(186²-226²)/(-2•6.4)=
=1287.5 rev