Book edition 9th Edition

Author(s) Raymond A. Serway, John W. Jewett

Pages 1624 pages

ISBN 9781133947271

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Short Answer

Find the scalar product of the vectors in Figure P7.10.

The scalar product of the two vectors is 5.33 J.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Magnitude of the first vector

A =32.8 N

Magnitude of the second vector

$B = 17.3 cm = 17.3 \cdot (1 cm \times \frac{1 m}{100 cm}) = 0.173 m$

Large angle made by the first vector and y axis = $118^{\circ}$

Large angle made by the second vector and x axis = $132^{\circ}$

Step 2: Dot product of two vectors

The dot product of two vectors $\vec{A}$ and $\vec{B}$ is

$\vec{A} \cdot \vec{B} = ABcosθ . . . . . . . (I)$

Step 3: Determining the dot product of the given vectors

Large angle between the positive y axis and the positive x axis = $270^{°}$

Thus, angle between the two vectors

$θ = 270^{\circ} - 118^{\circ} - 132^{\circ} = 20^{\circ}$

From equation (I), the dot product of the two vectors is

$\vec{A} \cdot \vec{B} = 32.8 N \times 0.173 m \times \cos 20^{\circ} = 5.33 \cdot (1 N \cdot m \times \frac{1 J}{1 N \cdot m}) = 5.33 J$

Thus the required dot product is 5.33 J.

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Physics For Scientists & Engineers

Answers without the blur.

Short Answer

Find the scalar product of the vectors in Figure P7.10.

Step by Step Solution

Step 1: Given data

Step 2: Dot product of two vectors

Step 3: Determining the dot product of the given vectors

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Physics For Scientists & Engineers

Answers without the blur.

Short Answer

Find the scalar product of the vectors in Figure P7.10.

Step by Step Solution

Step 1: Given data

Step 2: Dot product of two vectors

Step 3: Determining the dot product of the given vectors

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