The direction of the force experienced by the positive charge is upward.
We can use the right-hand rule to understand the direction of the Lorentz force acting on the charge: let's put the thumb in the same direction of the current in the wire (eastward), while the other fingers "wrap themselves" around the wire. These other fingers give the direction of the Lorentz force in every point of the space around the wire. Since the charge is located north of the wire, in that point the fingers are directed upward, so the positive charge experiences a force directed upward.
(if it was a negative charge, we should have taken the opposite direction)
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
(a) The force exerted by the electric field on the plastic sphere is equal to

where

is the charge of the sphere and E is the strength of the electric field. This force should balance the weight of the sphere:

where m is the mass of the sphere and g is the gravitational acceleration.
Since the two forces must be equal, we have:

and so we find the intensity of the electric field

(b) Now let's find the direction of the field. The electric force must balance the weight of the sphere, which is directed downward, so the electric force should be directed upward. Since the charge is negative, the force is opposite to the electric field direction, and so the direction of the electric field is downward.