Calculate M1 (50 g or 0.050 kg) M2 (445.4 g or 0.4454 kg) M1 m2 (system) (495.4 g or 0.4954 kg) KE (J) Wnet (J) xf,vf Wnet (J) KE (J)/AKE (J) xf= 0.65 m 0.0295 vf= 1.09 m/s KEI= 0.2645 KEf= 0.0297 AKE=0.0295 2 xi= 0.03 m vi= 0.16 m/s xf= 0.66 m vf= 1.10 m/s 0.2638 KEI= 0.0057 KEf=0.2695 Trial 1 xi, vi xi=0.02 m vi= 0.09 m/s 3 + LO AKE= 0.2387 AKE=0.2638 xi= 2.03 vi= 0.18 xf= 0.62 m vf= 1.06 m/s 0.2431 KEI= 0.0.0072 KEf= 0.2503 AKE= 0.2431 xi= 0.02 vi=0.17 xf= 0.77 m vf=1.14 m/s 0.2830 KEI= 0.00647 KEf= 0.2895 AKE= 0.2830 xi= 0.02 vi= 0.10 xf=0.57 m 0.2387 KEI= 0.00223 vf= 1.04 m/s KEf= 0.2409

Calculate M1 (50 g or 0.050 kg) M2 (445.4 g or 0.4454 kg) M1 m2 (system) (495.4 g or 0.4954 kg) KE (J) Wnet (J) xf,vf Wnet (J) KE (J)/AKE (J) xf= 0.65 m 0.0295 vf= 1.09 m/s KEI= 0.2645 KEf= 0.0297 AKE=0.0295 2 xi= 0.03 m vi= 0.16 m/s xf= 0.66 m vf= 1.10 m/s 0.2638 KEI= 0.0057 KEf=0.2695 Trial 1 xi, vi xi=0.02 m vi= 0.09 m/s 3 + LO AKE= 0.2387 AKE=0.2638 xi= 2.03 vi= 0.18 xf= 0.62 m vf= 1.06 m/s 0.2431 KEI= 0.0.0072 KEf= 0.2503 AKE= 0.2431 xi= 0.02 vi=0.17 xf= 0.77 m vf=1.14 m/s 0.2830 KEI= 0.00647 KEf= 0.2895 AKE= 0.2830 xi= 0.02 vi= 0.10 xf=0.57 m 0.2387 KEI= 0.00223 vf= 1.04 m/s KEf= 0.2409

An Introduction to Physical Science

14th Edition

ISBN:9781305079137

Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Publisher:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar Torres

Chapter1: Measurement

Section: Chapter Questions

Problem 9MC

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