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Question

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?
12 years ago

Answers

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

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