Question

A particle of mass 2.0×10^2 kg is moving with uniform speed of 4.0 metre per second in circular path of radius 1.0 M find with the help of a vector diagram i) the change in velocity vector B minus vector a of the particle in moving from a to b ii) the change in the centripetal force vector FB - vector FA

Answer 1

Since the velocities r perpendicular

Delta v=vb-va

=under root(4×4+4×4)

=under root (16+16)

=under root (32)

= 5.7m/s

Answer 2

hy dear ...❣......✌✌✌✌