University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 1 - Units, Physical Quantities, and Vectors - Problems - Exercises - Page 29: 1.43

Answer

(a) $-104 \space m^2$ (b) $-148 \space m^2$ (c) $40.6 \space m^2$

Work Step by Step

As we have calculated in the exercise 1.27: $$A_x = 0 , A_y = -8.00 \space m$$ $$B_x = 7.50 \space m, B_y = 13.0 \space m$$ $$C_x = -10.9 \space m, C_y = -5.07 \space m$$ (a) $$\vec A \cdot \vec B = A_xB_x + A_yB_y = (0)(7.50 \space m) + (-8.00 \space m)(13.0 \space m)$$ $$\vec A \cdot \vec B = 0 - 104 \space m^2 ´= -104 \space m^2$$ (b) $$\vec B \cdot \vec C = B_xC_x + B_yC_y = (7.50 \space m)(-10.9 \space m) + (13.0 \space m)(-5.07 \space m)$$ $$\vec B \cdot \vec C = -148 \space m^2$$ (c) $$\vec A \cdot \vec C = A_xC_x + A_yC_y = (0)(-10.9 \space m) + (-8.00 \space m)(-5.07 \space m)$$ $$\vec B \cdot \vec C = 40.6 \space m^2$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.