A500.-kilogram roller coaster starts from rest at the top of an 80.0-meter hill. what is its speed at the bottom of this hill?

v = 39.6m/s

Explanation: Ignoring friction and all those things, we have that:

At the top of the hill, the roller coaster only has potential energy, and this potential energy is:

U = m*g*h where m is the mass, g is the gravity acceleration, and h is the height.

Now, when the roller coaster reaches the bottom of the hill, the potential energy is fully converted into kinetic energy, which can be written as:

K = (m/2)*v^2

Then we must solve the equation:

U = K

m*g*h = (m/2)*v^2

v^2 = g*h*2

v = √(2*g*h)

where g = 9.8m/s^2 and h = 80m

so v = (√(2*9.8*90))m/s = 39.6m/s

you can see that the mass of the roller coaster has not an impact on the velocity.

temp rises, and expansion

0.11 m/s and 0.28 m/s should be correct