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 31: 1.83

Answer

The magnitude of $B$ is 28.0 m.

Work Step by Step

We can find the angle $\phi$ between the two vectors. $\phi = (90.0^{\circ}-39.0^{\circ}) + 28.0^{\circ} = 79.0^{\circ}$ We know that the scalar product $\vec{A} \cdot \vec{B} = AB~cos(\phi)$, where $\phi$ is the angle between the two vectors. $\vec{A} \cdot \vec{B} = AB~cos(\phi) = 48.0~m^2$ $B = \frac{48.0~m^2}{A~cos(\phi)} = \frac{48.0~m^2}{(9.00~m)~cos(79.0^{\circ})}$ $B = 28.0~m$ The magnitude of $B$ is 28.0 m.
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