Asked by Rosie
A 0.025kg rubber stopper is swung in a horizontal circle by a 0.95 meter threat that makes 25 revolutions every 15 minutes. a) what is the stoppers speed? b) what is the acceleration? c) what is the value of the centripetal force?
Answers
Answered by
Rosie
sorry not minutes but 15 seconds
Answered by
Damon
T = 15/25 = .6 second
C = 2 pi R = 2 * 3.14 * .95 = 5.97 meter
a) v = C/T = 9.94 m/s
b) v^2/R = 9.94^2/.95 = 104 m/s^2
wow, 10 g
c) F = m a = .025*104 = 2.6 Newtons
C = 2 pi R = 2 * 3.14 * .95 = 5.97 meter
a) v = C/T = 9.94 m/s
b) v^2/R = 9.94^2/.95 = 104 m/s^2
wow, 10 g
c) F = m a = .025*104 = 2.6 Newtons
Answered by
Rosie
thank you for your help
Answered by
Damon
You are welcome.
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