Final answer:
The impulse on a 0.17-kg baseball that changes velocity from -38 m/s to 62 m/s when hit by a bat is calculated to be 17 N·s, a value that is not listed in the provided options.
Step-by-step explanation:
To determine the magnitude of the impulse exerted upon the baseball by the bat, we can use the concept of impulse which is equal to the change in momentum of the ball. The impulse can be found using the formula Impulse (J) = Δp = mf * vf - mi * vi, where mf is the final mass of the ball, vf is the final velocity of the ball, mi is the initial mass of the ball, and vi is the initial velocity of the ball. In this case, the ball's mass doesn't change, so mf and mi are both 0.17 kg, the initial velocity (vi) is -38 m/s (since it is thrown toward the batter and comes back in the opposite direction, we consider it as a negative value), and the final velocity (vf) is 62 m/s. Plugging these values into the equation, we get:
Impulse (J) = 0.17 kg * 62 m/s - 0.17 kg * (-38 m/s) = 10.54 kg * m/s + 6.46 kg * m/s = 17 kg * m/s
The magnitude of the impulse exerted on the baseball by the bat is therefore 17 N·s, which is not one of the provided options (A. 2.4 N·s B. 3.4 N·s C. 5.4 N·s D. 6.8 N·s), suggesting there may be an error in the options given in the question or in the interpretation of the question itself.