Question: answer At a particular instant, a 1.0 kg particle's position is r = (2.01 - 6.0j + 5.0k) m, its velocity is v = (-1.01

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At a particular instant, a 1.0 kg particle's position is r = (2.01 - 6.0j + 5.0k) m, its velocity is v = (-1.01 + 6.0j + 1.0k) m/s, and the force on it is F = (13.01 + 16.0j) N. (Express your answers in vector form.) (a) What is the angular momentum (in kg . m/s) of the particle about the origin? kg . m2/s (b) What is the torque (in N . m) on the particle about the origin? N . m (c) What is the time rate of change of the particle's angular momentum about the origin at this instant (in kg . m2/s2)? di kg . m2/s2 dt

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