An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic friction? 0.16
0.40
0.08
not enough information

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ F (vertical) = n - mg = 0   →   n = mg = 1876 N

where n is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ F (horizontal) = 747 N - f = 0   →   f = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, µ, such that

f = µn   →   µ = f / n = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6