Question

A particle of mass m = 1kg moves in a circle of radius R = 2m with uniform speed v = 37 m/s. The
magnitude of impulse given by centripetal force to the particle in one second is
(1) 21 NS
(2) 3 1 NS
(3) 2.3 1 Ns
(4) 3.21 NS​

Answer 1

Given that,

Mass of a particle, m = 1 kg

Radius of circle, R = 2 m

Speed of particle, v = 37 m/s

To find,

The magnitude of impulse given by centripetal force to the particle in one second is.

Solution,

The centripetal force of a particle is given by :

$F=\dfrac{mv^2}{R}\\\\F=\dfrac{1\times (37)^2}{2}\\\\F=684.5\ N$

The magnitude of impulse given by centripetal force to the particle is :

$J=F\times t\\\\J=684.5\times 1\\\\J=684.5\ N-s$

So, the magnitude of impulse given by centripetal force to the particle is 684.5 N-s.

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