Asked by mike
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
rotor turn?
rotor turn?
Answers
Answered by
Elena
N=(n₂²-n₁²)/2a=(186²-226²)/(-2•6.4)=
=1287.5 rev
=1287.5 rev
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.