Asked by Nancy
A 0.500 kg particle moves in a circle of R=0.0150 m at constant speed. The time for 20 complete revolutions is 31.7 s. What is the period T of the motion? What is the speed of the particle? What is the centripetal acceleration of the particle? What is the centripetal force on the particle?
Answers
Answered by
bobpursley
This is pretty simple and straightforward, and I am wondering what your question is.
You are given period: T=31.7s/20
Speed = distance/time=2PI*R/T
Centripetal acceleration? speed^2/R
Frankly, your teacher is to easy.
You are given period: T=31.7s/20
Speed = distance/time=2PI*R/T
Centripetal acceleration? speed^2/R
Frankly, your teacher is to easy.
Answered by
Nancy
Sorry just re-learning after 30+ years.I had part of it answered however, the 4th and 5th question info was pertaining to questions 3 and 4 so I had to ask the question from the beginning. My answer was T = 1.585 s.
for speed I calc 0.942/1.585s=0.594...right?
my real problem was I thought Ac of particle...use Ac= v2/r? Right? And not sure on force...Thanks
for speed I calc 0.942/1.585s=0.594...right?
my real problem was I thought Ac of particle...use Ac= v2/r? Right? And not sure on force...Thanks
Answered by
bobpursley
force=mass*centripetal acceleration
Answered by
Nancy
Ok so F=MA thanks, I do appreciate especially since I am going to 35th reunion and went back to school. Much harder than I remembered. Again thanks
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