Physics, asked by parekhrutu65, 11 months ago

A particle is moving on a circular path of radius 5 m
with uniform speed 10 ms from point A to B. The
magnitude of average velocity of particle in moving
from A to B is​

Answers

Answered by rakshaomshivam
0

Answer:

Centripetal accl. is equal to v^2/r

ac = 5^2/5 = 5m/s^2.

Tangential accl.(at) is equal to

Change in velocity/ time taken.

Change in velocity = 5-(-5) =10.

Time period =( 2*pi. R) /V = 2*pi sec.

Time taken in half revolution is pi.

at = 10/pi.

Explanation:

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