trial AKE NM 3 ५ 5 0.310151- m = yus. yg 0.27270 N.m 0.26923N.m 0.25897 Nm 0.27753 Nm Vei Ji=0.62 V+= 1,29 Vi=0,65 V+= 125 Vi=0.61m/s Vf= 1.23m 15 vid Vf=125765 Vi= 0.70 m/s. Vf 1.30m/s 3 Printed 1. AKE= ½ mv^2= ½ mv^2 Sample calculations (run 1 & 2): Run 1: AKE= x 0.4454 x (1.29^2 - 0.62 ^2) = ½ x 0.4454 x (1.6641-0.3844) = 0.2849 J Run 2: AKE= x 0.4454 x (1.25^2 - 0.65 ^2) = 0.2538 J Run 3: 0.2540 J Run 4: 0.1461 J
trial AKE NM 3 ५ 5 0.310151- m = yus. yg 0.27270 N.m 0.26923N.m 0.25897 Nm 0.27753 Nm Vei Ji=0.62 V+= 1,29 Vi=0,65 V+= 125 Vi=0.61m/s Vf= 1.23m 15 vid Vf=125765 Vi= 0.70 m/s. Vf 1.30m/s 3 Printed 1. AKE= ½ mv^2= ½ mv^2 Sample calculations (run 1 & 2): Run 1: AKE= x 0.4454 x (1.29^2 - 0.62 ^2) = ½ x 0.4454 x (1.6641-0.3844) = 0.2849 J Run 2: AKE= x 0.4454 x (1.25^2 - 0.65 ^2) = 0.2538 J Run 3: 0.2540 J Run 4: 0.1461 J
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter2: Vectors
Section: Chapter Questions
Problem 21CQ: What is wrong with the following expressions? How can you correct them? (a) C=AB , (b) C=AB , (c)...
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Question
1. calculate ΔKE for each run (5 runs)
2. Error: Calculate Average Error for ΔKE and work done
3. Explain the difference between Δ KE and work done
(some calculations done previously, can you please check my work as well?) thank you!
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