A 1210 kg rollercoaster car is moving 8.31 m/s. Before entering
the station, it first rolls up a 2.26 m
hill. How fast is it moving then?

4.98 m/s

Explanation:

You solve these kinetic energy, potential energy problems by using the fact P.E.+ K.E. = a constant as long as friction is ignored.

PEi = 0 in this case

KEi = ½mVi² = PEf+KEf = mghf + ½mVf²

½1210*8.31² = 1210*9.8*2.26 + ½1210*Vf²

½1210*Vf² = ½1210*8.31² - 1210*9.8*2.26

Vf² = 8.31² - 2*9.8*2.26 = 4.98² so Vf = 4.98m/s

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.