A175 g ball containing 4.20×10^8 excess electrons is dropped into a 120 m vertical shaft. at the bottom of the shaft, the ball suddenly enters a uniform horizontal 0.200 t magnetic field directed from east to west. part a: if air resistance is negligibly small, find the magnitude of the force that this magnetic field exerts on the ball just as it enters the field.
part b: if air resistance is negligibly small, find the direction of the force that this magnetic field exerts on the ball just as it enters the field. (north, east, south, west)

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

Because they're in different places for this the temperature varies.

Hey did u get the answer?