is the horse running/moving? if so then kinetic i believe
The electric field generated by a point charge is given by:

where

is the Coulomb's constant
Q is the charge
r is the distance from the charge
We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.
Let's calculate first the electric field generated by the positive charge at that point:

where the positive sign means its direction is away from the charge.
while the electric field generated by the negative charge is:

where the negative sign means its direction is toward the charge.
If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
Answer
given,
mass of the piano = 170 kg
angle of the inclination = 20°
moves with constant velocity hence acceleration = 0 m/s²
neglecting friction
so, force required to pull the piano
F = m g sin θ
F = 170 × 9.81 × sin 20°
F = 570.39 N
so, force required by the man to push the piano is F = 570.39 N
Answer:
(A) Distance will be equal to 1.75 km
(B) Displacement will be equal to 1.114 km
Explanation:
We have given circumference of the circular track = 3.5 km
Circumference is given by 
r = 0.557 km
(a) It is given that car travels from southernmost point to the northernmost point.
For this car have to travel the distance equal to semi perimeter of the circular track
So distance will be equal to 
(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track
So displacement will be equal to d = 2×0.557 = 1.114 m
Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.
Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).
At the start:
ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J