1.- Let. R=A+B where A=15.9, A = 51.8 degrees, B = 18.5 +B= 254 degrees. You find Ra= Normal format and 3 SF. Ax= A cos QA 15.9 cos (51.8)= 9.8 Вх = В сод ок > 18.5 cos (254) = -5.1 Rx= Ax + Bxc = 18.7// 18.664 18.7

1.- Let. R=A+B where A=15.9, A = 51.8 degrees, B = 18.5 +B= 254 degrees. You find Ra= Normal format and 3 SF. Ax= A cos QA 15.9 cos (51.8)= 9.8 Вх = В сод ок > 18.5 cos (254) = -5.1 Rx= Ax + Bxc = 18.7// 18.664 18.7

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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Can you help me to solve this problem when I get the value of Bx= -5.1this is a negative value in this subject is addition of vectors, I am trying to get the value of RX , I don’t know if this value is negative how should I add it to get RX please explain step by step and how I should work in this cases

B
1. Let. R=A+B where A=15.9, A = 51.8 degrees, 13 = 18.5
B = 254 degrees. You find Ra=
Normal format
and 3 SF.
Ax= A COS QA
15.9 cos(51.8)= 9.8
Вх= В сод ор
18.5 cos (254) = -5.1
Rx= Ax + Bxc =
= 9.8 +
16.7/
18.664 = 18.7

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