The amount of work (W) done on an object is equal to the force (F) needed to lift the object (the object’s weight) multiplied by the distance (d) the object is lifted: W = F ✕ d.
Use the weight of the ball that you calculated in activity B to determine how much work
would be required to lift the ball 2 meters above the zero position:

Work done during lifting up body is 0

Explanation:

From the question it is given that

Distance traveled by body during lifting is 2 m

since work done by body is given by

work done W = F x s x cosθ

Where F= mg = weight of body

s= distance traveled = 2m

θ = angle between force and distance vector

here θ = 90° as body is lifting up

on substituting the respected values we get

⇒ W = mg x 2 x cos90° = 0   {∵ cos90°= 0 }

Thus, work done during lifting up body is 0

(75°f − 32) × 5/9 = 23.889°c

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.