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# A 125 kg driver moving at 31 m/s is brought to rest by an inflating air bag in 1.5 seconds. If the car accident occurred at 4.0 kg meters per second, the driver's momentum would change by 21 kg mete

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Question: 4.0 kg • meters per second 21 kg • meters per second 47 kg • meters per second 3900 kg • meters per second During a car accident, a 125 kg driver is moving at 31 m/s and in 1.5 seconds is brought to rest by an inflating air bag. What is the magnitude of the change in momentum of the driver based on the answer choices above? prove your answer

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3.825 kg • m/s. Momentum (p) = mass (m) x velocity (v) Change in momentum (p) = Final Momentum - Initial Momentum Initial Momentum = 125 kg x 31 m/s = 3875 kg • m/s Final Momentum = 125 kg x 0 m/s = 0 kg • m/s Change in momentum = Final Momentum - Initial Momentum = 0 kg • m/s - 3875 kg • m/s = -3875 kg • m/s Since momentum is a vector quantity, the magnitude of the change in momentum is: Magnitude of change in momentum = |-3875 kg • m/s| = 3875 kg • m/s Answer: 3.825 kg • m/s.

Jan. 6, 2023, 9:19 a.m.

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