Aparticle is acted on by 2 torques about the origin:
has a magnitude of 2.0nm and is directed inthe positive direction
of the x axis, and
has a magnitude of 4.0nm and is directed inthe negative direction
of the y axis. in unit-vectornotation, find
where
is the angular momentum of the particleabout the origin.

We will start with the understanding that the first derivative of the angular momentum is equivalent to the torque, therefore,

Vectorially the given value can be expressed as,

This will allow us to find the magnitude of the total torque by the trigonometric properties of the vectors, for which:

Finally the angle will be given by the relation of the tangent,

Direction of 63.4° clockwise from x-direction

more than, less

less than, more

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer