1) what is the potential energy at the end of the loop? 2)what is the kinetic energy at the end of the loop(based upon photogate time?

release the car and record the time from the photogate at the bottom of the hill and answer the top questions:
i attached an image that shows the recorded time this is a class work but i’m trying to get better at it

The particles of the gas inside the ball heats up, bouncing off each other and spreading farther apart.hope my answer has you : )

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge,    then the following gives us the potential difference,      the formula  may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get    using q = 8.00  × 10⁻¹⁹ c and  δv = 2.00 v, we get    the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.