Question

3.
A particle is moving with a constant speed of 20 ms-'
in a circular path of radius 14 m. The magnitude of
average acceleration of the particle between two
diametrically opposite point is
​

Answer 1

Answer:

A particle is moving with a constant speed of 20 ms-'

in a circular path of radius 14 m. The magnitude of

average acceleration of the particle between two

diametrically opposite point is

​Answer = 200/11

Explanation:

Speed (v) = 20 m/s

Radius of the path = 14 m

The magnitude of average acceleration = $\frac{2(20*20)}{22*2}$

==> 800/44 ==> 200/11