Question: At time t = 0, a 3.0 kg particle with velocity v = (5.0 m/s) (6.0 m/s) is a t x = 3.0 m,

At time = 0, a 3.0 kg particle with velocity = (5.0 m/s)î – (6.0 m/s)ĵ is at x = 3.0 m,= 8.0 m. It is pulled by a 7.0 N force in the negative direction. About the origin, what are

(a) The particle's angular momentum,

(b) The torque acting on the particle, and

(c) The rate at which the angular momentum is changing?

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