InterviewSolution
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InterviewSolution
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A particle moves in a circle of radius 5 cm with constant speed and time period `0.2pis`. The acceleration of the particle isA. `25m//s^(2)`B. `36m//s^(2)`C. `5m//s^(2)`D. `15m//s^(2)` |
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Answer» Correct Answer - c Given `r=5 cm =5xx10^(-2)m and T= 0.2 pis` we know that acceleration `a=r omega^(2)=(4pi^(2))/(T^(2))r` `=(4xxpi^(2)xx5xx10^(-2))/((0.2pi)^(2))=5 ms^(-2)` |
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