Acube of mass m1 = 7.7 kg is sitting on top of another cube of the same size and mass m2 = 1.4 kg while they are both in free-fall. ignoring any air resistance, what is the magnitude of the normal force with which the bottom cube is acting on the top cube?

N=0

Explanation:

If we take the two cubes as one whole object, we can see that there is no floor to support them, therefore there is no force acting against the cubes when free falling.

If we look at each individual cube, they will hardly be in contact, the bottom cube has no additional support (does not lie on any surface) to go against the weight of the top cube, therefore it doesn't apply any force to the cube on top.

They will be both falling with the same acceleration at the same rate. So there will be no force acting between the two of them.

This is the exact same situation you have on a falling elevator, you will have an ilusion of there not being gravity, but this is because the elevator and you are falling at the same rate, so there is no normal force between the floor of the elevator and you.

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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.