Question: At time t = 0 t = 0 t = 0 , a 4 . 2 1 kg particle has position vector r = (

At time t=0t =0t=0, a 4.21 kg particle has position vector r=(7.00m)i^(1.00m)j^\vec{r}=(7.00\,\text{m})\,\hat{i}-(1.00\,\text{m})\,\hat{j}r=(7.00m)i^(1.00m)j^ relative to the origin. Its velocity is given by v=(9.98t2m/s)i^\vec{v}=(-9.98\, t^2\,\text{m/s})\,\hat{i}v=(9.98t2m/s)i^ for ttt in seconds. About the origin with t=15.6st =15.6\,\text{s}t=15.6s, what are (a) the particle's angular momentum and (b) the torque acting on the particle, both in unit-vector notation?

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