prob_051 - × 10 –3 s = 9.47 × 10 –5 min The number of...

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51. REASONING AND SOLUTION The sample makes one revolution in time T as given by T = 2 π r / v . The speed is v 2 = ra c = (5.00 × 10 –2 m)(6.25 × 10 3 )(9.80 m/s 2 ) so that v = 55.3 m/s The period is T = 2 (5.00 × 10 –2 m)/(55.3 m/s) = 5.68
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Unformatted text preview: × 10 –3 s = 9.47 × 10 –5 min The number of revolutions per minute = 1/ T = 10 600 rev/min . _______________________________________________________________________________________ ______...
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This note was uploaded on 01/29/2010 for the course CALC 01:640:135 taught by Professor Soffer during the Spring '08 term at Rutgers.

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