The amount of internal energy needed to raise the temperature of 0.25 kg of water by 0.2°c is 209.3 j. how fast must a 0.25 kg baseball travel in order for its kinetic energy to equal this internal energy?

The baseball must be 40.92 m/s fast

Explanation:

Kinetic Energy: This maybe defined as the energy of a body, due to motion.

The S.I unit of kinetic energy is Joules (J)

Kinetic Energy is represented as

Ek = 1/2mv² Equation 1

making v the subject of the equation,

v = √(2Ek/m) Equation 2

Ek = Kinetic Energy of the base ball, m = mass of the base ball, v = velocity of the base ball.

Given: m = 0.25 kg, Ek = 209.3 J,

Note: The Kinetic energy of the baseball is equal to the internal energy of water.

Substituting these values into equation 2

v = √(2×209.3/0.25)

v = √1674.4

v = 40.92 m/s

Thus the baseball must be 40.92 m/s fast

b

explanation:

bc its just b

1/0.14nm = 10^10/1.4 a lot