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An electric fan is turned off, and its angular velocity decreases uniformly from 460rev/min to 160rev/min in a time interval of length 3.60s . Find the number of revolutions made by the motor in the time interval of length 3.60s .
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\[\omega _{f}^{2} = \omega _{i}^{2} +2\alpha \theta \] find alpha from \[\omega _{f} = \omega _{i} + \alpha t\] where omega final = 23/3 rev.s /sec omega initial = 8/3 rev.s /sec and t = 3.6 sec
sub.s alpha, omega final and omega initial in the top eqn. and solve for theta...
i got the 2 answers out of 3 :D thanks :D going to the next question :F
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