When car is at the top of the hill its whole energy is stored in the form of gravitational potential energy
so when height of the car becomes half then its potential energy is given as
so final potential energy when car falls down by half of the height will become half of the initial potential energy
So it is U = 50 MJ after falling down
Now by energy conservation we can say that final potential energy + final Kinetic energy must be equal to the initial potential energy of the car
So here at half of the height kinetic energy of car = 100 - 50 = 50 MJ
so we can say at this point magnitude of potential energy and kinetic energy will be same
<em>A. the same as the potential energy at that point.</em>
I think it would be d because the spring or whatever was pushing until it reached the farthest it could then it would pull down but idrk
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(a) 
First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:
where
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 1.00 cm = 0.01 m is the displacement of the person
Solving for a,
And the average force on the person is given by
with m = 75.0 kg being the mass of the person. Substituting,
where the negative sign means the force is opposite to the direction of motion of the person.
b) 
In this case,
v = 0 is the final velocity
u = 20.0 m/s is the initial velocity
a is the acceleration
d = 15.00 cm = 0.15 m is the displacement of the person with the air bag
So the acceleration is
So the average force on the person is
