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Parul singh Grade: 12th pass
        a=15m/s^2 represent total acceleration of a particle moving in the clockwise direction in a circle of radius R=2.5m at a given instant of time the speed of particle is
one year ago

Answers : (2)

Piyush Kumar Behera
199 Points
										
see this acceleration is the centripetal acceleration as the particle is moving in the constant velocity .
so a=v2/r
v=\sqrt{ar}
so speed of the particle is \sqrt{2.5*15}
=6.1237243569579452454932101867647m/s
hope it helps please approve!!
one year ago
Shefali
11 Points
										The acceleration given in question is the resultant of tangential and centripetal acceleration.So first we will calculate the centripetal acceleration. Here,Centripetal acceleration =a cos 30°                                   =a √3/2=15*√3/2We know,                Centripetal acceleration=v*v/r                           15*√3/2=v*v/2.5                            15*√3/2*2.5=v*v                             31.875= v*v                             √31.875=v                            5.7=v
										
3 months ago
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