You push a piece of wood under the surface of a swimming pool. after it is completely submerged, you keep pushing it deeper and deeper. as you do this, what will happen to the buoyant force on it? will the force keep increasing, stay the same, or decrease? why?

Same

Explanation:

If we keep on pushing the wooden block the buoyant force on it will remain same as it depends upon the volume of object submerged in liquid and density of liquid .

Buoyant force is given by

where buoyant force

it is independent of depth of object from free surface.

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.

490 n

f=125n a=2.5m/s2