trial 1 !^' :X Xf, vf Xi=0.02mxf= 0.65m - N₁ = 0.093 Vf = 1.09m =0.03m XF = 0.66m Xi Vi = 0.16 Vf = 1.10m Z Σ Σ а Xi = 203m xf=262m son rA |úan'cསྨ!\ X₁ = 0.02m xf=0>]m Vi= 0.17m XF=1.14m X;= 0.02m x₁ =953m V₁ = 0.103 | V₁ = 1.04 1 Vi- ༠་ W KE

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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W net (J) and KE (J) for data table and 

compare and explain using % difference

trial
1
!^' :X
Xf, vf
Xi=0.02mxf= 0.65m
-
N₁ = 0.093 Vf = 1.09m
=0.03m XF = 0.66m
Xi
Vi = 0.16
Vf = 1.10m
Z
Σ
Σ
а
Xi = 203m xf=262m
son rA |úan'cསྨ!\
X₁ = 0.02m xf=0>]m
Vi= 0.17m XF=1.14m
X;= 0.02m x₁ =953m
V₁ = 0.103 | V₁ = 1.04 1
Vi- ༠་
W
KE
Transcribed Image Text:trial 1 !^' :X Xf, vf Xi=0.02mxf= 0.65m - N₁ = 0.093 Vf = 1.09m =0.03m XF = 0.66m Xi Vi = 0.16 Vf = 1.10m Z Σ Σ а Xi = 203m xf=262m son rA |úan'cསྨ!\ X₁ = 0.02m xf=0>]m Vi= 0.17m XF=1.14m X;= 0.02m x₁ =953m V₁ = 0.103 | V₁ = 1.04 1 Vi- ༠་ W KE
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