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
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
Hanging mass (m1): 50 g= 0.050 kg
Glider mass (m2): 445.4 g= 0.4454 kg
T= M1M2g/M1+M2
Table and compare and explain using % difference
thank you!

Transcribed Image Text: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)

Transcribed Image Text: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
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