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Q10P

Expert-verifiedFound in: Page 150

Book edition
9th Edition

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

Pages
1624 pages

ISBN
9781133947271

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

The scalar product of the two vectors is 5.33 J.

Magnitude of the first vector

A =32.8 N

Magnitude of the second vector

$B=17.3\mathrm{cm}\phantom{\rule{0ex}{0ex}}=17.3\xb7\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right)\phantom{\rule{0ex}{0ex}}=0.173\mathrm{m}$

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

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

**The dot product of two vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ is**

**$\overrightarrow{\mathbf{A}}{\mathbf{\xb7}}\overrightarrow{\mathbf{B}}{\mathbf{=}}{\mathbf{ABcos\theta}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\mathbf{.}}{\left(I\right)}$**

Large angle between the positive y axis and the positive x axis = ${270}^{\xb0}$

Thus, angle between the two vectors

$\theta ={270}^{\circ}-{118}^{\circ}-{132}^{\circ}\phantom{\rule{0ex}{0ex}}={20}^{\circ}$

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

$\overrightarrow{A}\xb7\overrightarrow{B}=32.8\mathrm{N}\times 0.173\mathrm{m}\times \mathrm{cos}{20}^{\circ}\phantom{\rule{0ex}{0ex}}=5.33\xb7\left(1\mathrm{N}\xb7\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}\xb7\mathrm{m}}\right)\phantom{\rule{0ex}{0ex}}=5.33\mathrm{J}$

Thus the required dot product is 5.33 J.

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