C physics Archive october x Wiley PLUS C edugen.wileyplus.com/edugen/student/mai

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C physics Archive october x Wiley PLUS C edugen.wileyplus.com/edugen/student/mainfr.uni Gradebook ORION Home Read, Study & Practice Assignment Assignment Open Assignment ONEXTD FULL SCREEN PRINTER VERSION HBACK ASSIGNMENT RESOURCES Chapter 09, Problem 001 Homework 9 Fall 2015 M Chapter 09 Concept A 1.20 kg particle has the xy coordinates (-1.92 m, 0.965 m and a 2.86 kg particle has the xy coordinates (0.232 m, -0.0104 m). Both lie on a horizontal plane. At what (a) x and (b) 01 y coordinates must you place a 1.10 kg particle such that the center of mass of the three-particle system has the coordinates (-0.0612 m, -0.155 m)? M Chapter 09 Concept 06 09, Prob (a) Number 001 Units 09, Prob 002 b) Number Units 09, Prob 004 09, Prob 015 SHOW SOLUTION 09, Prob 102 09, Prob 109 Question Attempts: 0 of 6 used SAVE FOR LATER SUBMIT ANSWER Review Results by Study object Copyright 2000-2015 by John Wiley & Sons, Inc. or related companies. All rights reserved License Agreement l Privacy Policy I G 2000-2015 John Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley Sons Inc. Version 4.16.1.7 Ask me anything 0/25/2015

Explanation / Answer

let third particle is at (x,y)

then x coordinate of center of mass=(1.2*(-1.92)+2.86*0.232+1.1*x)/(1.2+2.86+1.1)

=(1.1*x-1.64048)/5.16

given that x coordinate of center of mass=-0.0612 m

hence (1.1*x-1.64048)/5.16=-0.0612

==>1.1*x-1.64048=-0.315792

==>x=(1.64048-0.315792)/1.1

==>x=1.20426 m

part b:

y coordinate of center of mass=(1.2*0.965+2.86*(-0.0104)+1.1*y)/(1.2+2.86+1.1)

=(1.1*y+1.128256)/5.16

given that y coordinate of center of mass=-0.155 m

hence (1.1*y+1.128256)/5.16=-0.155

==>1.1*y+1.128256=-0.7998

==>y=(-0.7998-1.128256)/1.1

==>y=-1.75277 m

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