Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 3 - Vectors and Coordinate Systems - Exercises and Problems: 20

Answer

(a) $E = 3.6$ $F = 2.8$ (b) $E+F = 4.1$ (c) $-E-2F = 6.1$

Work Step by Step

$E = 2\hat{i} + 3\hat{j}$ $F = 2\hat{i} -2 \hat{j}$ (a) We can find the magnitude of the vector E. $E = \sqrt{E_x^2+E_y^2}$ $E = \sqrt{(2)^2+(3)^2}$ $E = 3.6$ We can find the magnitude of the vector F. $F = \sqrt{F_x^2+F_y^2}$ $F = \sqrt{(2)^2+(-2)^2}$ $F = 2.8$ (b) $E+F = 4\hat{i} + 1\hat{j}$ We can find the magnitude of the vector E+F. $E+F = \sqrt{(4)^2+(1)^2}$ $E+F = 4.1$ (c) $-E-2F = -(2\hat{i} + 3\hat{j})-2~(2\hat{i} -2 \hat{j})$ $-E-2F = -6\hat{i} + 1\hat{j}$ We can find the magnitude of the vector -E-2F. $-E-2F = \sqrt{(-6)^2+(1)^2}$ $-E-2F = 6.1$
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