<span>There is no special name for that. Physics is usually just concerned with "forces", and doesn't specify whether the force pushes or pulls. If you want to be more specific, you can just call it a "pulling force".
I hoped this was satisfying!:)</span>
Here we know that



now from kinematics we have

now from above all values we have



so final angular speed is -12.6 rad/s
Answer:
. always start on the north pole and terminate (end) on the South Pole
Explanation:
Speed = distance / time
3.4cm / 0.1s = 34 cm/sec