Force f = (-6.0 n)ihat + (2.0 n) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.
(a) what is the torque on the particle aboutthe origin?
1 n·m ihat+ 2 n·m jhat+ 3 n·m khat

(b) what is the angle between the directions of rand f? (if there is no torque, enter 0.)
4°

Explanation:

Given that.

Force acting on the particle,

Position of the particle,

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r,

Magnitude of F,

Using dot product formula,

Therefore, this is the required solution.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.