A large box of mass 4kg slides across a frictionless surface with a velocity of 12m/s to the right. It collides with a smaller box of mass 2kg that is stationary. The boxes stick together. What is the velocity of the two combined masses after the collision.

Explanation:

Let's call the large box "1" and the smaller box "2". The equation for this works so that momentum is conserved, as the Law states.

That says that the momentum of the larger box plus the momentum of the smaller box before the collision has to equal the momentum of the combined masses since they stick together. Filling in:

[(4(12) + 2(0)] = [(4 + 2)v] and

48 + 0 = 6v so

v = 8 m/s to the right

b.a device that inputs heat into a cool object

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.