Asked by Jon
A 0.50-kg ball is attached to a string of 0.50 m and swung in a horizontal circle with a velocity of 1.0 m/s. Find the centripetal force of the ball.
ac=1.0^2/0.5
=2
2(.50)
= 1.0 N
ac=1.0^2/0.5
=2
2(.50)
= 1.0 N
Answers
Answered by
drwls
Correct.
a = V^2/R = 2 m/s^2 and
F = Ma = 1.0 N
a = V^2/R = 2 m/s^2 and
F = Ma = 1.0 N
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.