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The bases are loaded, and a 0.53 kg baseball is flying towards you at [tex]97.3 \, \text{m/s}[/tex]. What impulse did you apply to the ball?

A) [tex]0.53 \, \text{kg m/s}[/tex]

B) [tex]-0.53 \, \text{kg m/s}[/tex]

C) [tex]50.0 \, \text{kg m/s}[/tex]

D) [tex]-50.0 \, \text{kg m/s}[/tex]

Answer :

Final answer:

The correct answer to the question is Option D, which is -50.0 kg m/s. It represents the closest estimated impulse applied to the baseball to bring it to a stop from its initial velocity of 97.3 m/s, using the mass of the baseball which is 0.53 kg.

Explanation:

The question is asking about the impulse applied by the player to the baseball that was originally moving at 97.3 m/s towards the player. Impulse can be calculated using the formula Impulse = Δp = m Δv, where Δp is the change in momentum, m is the mass of the object, and Δv is the change in velocity.

Assuming the player brought the ball to a stop, the change in velocity (Δv) would be -97.3 m/s (since the ball was originally moving at +97.3 m/s and came to rest, the velocity change is the negative of the initial velocity). With a mass (m) of 0.53 kg, the impulse applied to the ball would be:

Impulse = 0.53 kg · (-97.3 m/s) = -51.569 kg·m/s.

Because answers offered in the question don't include this exact value, the closest answer is -50.0 kg m/s (Option D), which represents the magnitude and direction (negative, indicating opposite to the ball's original motion) of the impulse.

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