Asked by Tyra J
A 250.0 kg roller coaster car has 20,000 J of potential energy at the top of the hill. Neglecting frictional losses, what is the velocity of the car at the bottom of thee hill?
A.) 9.8 m/s
B). 12.6 m/s
C). 14.1 m/s
D). 8.9 m/s
A.) 9.8 m/s
B). 12.6 m/s
C). 14.1 m/s
D). 8.9 m/s
Answers
Answered by
Quidditch
Assuming the car has no kinetic energy at the top...
energy at the bottom = energy at the top
energy at the bottom = kinetic energy
kinetic energy = (1/2) * m * (v^2)
So,
(1/2)*m*(v^2)=20,000J
Solve for v
energy at the bottom = energy at the top
energy at the bottom = kinetic energy
kinetic energy = (1/2) * m * (v^2)
So,
(1/2)*m*(v^2)=20,000J
Solve for v
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.