Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 3 - Vectors in Physics - Problems and Conceptual Exercises - Page 76: 1

Answer

(a) By a factor of 2. (b) By a factor of one. That is, the angle remains the same.

Work Step by Step

The vector $A$ has components $A_{x}$ and $A_{y}$. So: A $=\sqrt {(A_{x})^{2}+(A_{y})^{2}}$ and $\theta=tan^{-1}(\frac{A_y}{A_x})$ If $A_{x}$ and $A_{y}$ are doubled: (a) A' $=\sqrt {(2A_{x})^{2}+(2A_{y})^{2}}=\sqrt {4(A_{x})^{2}+4(A_{y})^{2}}=2\sqrt {(A_{x})^{2}+(A_{y})^{2}}$. A' = 2A (b) $\theta~'=tan^{-1}(\frac{2A_y}{2A_x})=tan^{-1}(\frac{A_y}{A_x})=\theta$
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