An official major league baseball has a mass of 0.14 kg. A pitcher throws a 40 m/s fastball which is hit by the batter straight back up the middle at a speed of 46 m/s. a) What is the change in momentum of the ball during the collision with the bat?
b) If this collision occurs during a time of 0.012 seconds, what is the average force exerted by the bat on the ball?

(a) The change in momentum is 12.04 kg-m/s

(b) The force exerted by the bat is 1003.33 N

Explanation:

Given that,

The mass of a ball, m = 0.14 kg

Initial speed of the ball, u = 40 m/s

Final speed of the ball, v = -46 m/s

(a) The change in momentum of the ball during the collision with the bat is given by :

(b) Time for collision, t = 0.012 s

Now the force can be calculated as follows :

Hence, this is the required solution.

less than one you can't create energy

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.