The amount of internal energy needed to
raise the temperature of 0.26 kg of water by
0.3�C is 326.5 J.
How fast must a 0.26 kg baseball travel
in order for its kinetic energy to equal this
internal energy?
Answer in units of m/s
Require that (1/2)MV^2 = 326.5 J
Then solve for V.
V = 50.1 m/s
To find the velocity of the baseball, we need to equate its kinetic energy to the internal energy needed to raise the temperature of water. The kinetic energy of an object can be calculated using the formula:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
In this case, the mass of the baseball (m) is given as 0.26 kg, and the internal energy needed (KE) is given as 326.5 J. We can rearrange the formula to solve for v:
v = √(2KE / m)
Plug in the given values:
v = √(2 * 326.5 J / 0.26 kg)
v ≈ √(1253.85 m^2/s^2 / 0.26 kg)
v ≈ √(4813.27 m^2/s^2 / kg)
v ≈ √(4813.27 m^2/s^2) ≈ 69.4 m/s
Therefore, the baseball must travel at a velocity of approximately 69.4 m/s for its kinetic energy to equal the internal energy needed to raise the temperature of water.