Chapter 1, Problem 1.38QP

(a)

Interpretation Introduction

Interpretation:

$1.42lightyears$ has to be converted into miles.

Concept introduction:

Light year: The distance light can travel in one year it is approximately equal to 10 trillion kilometres or 6 trillion miles.

$\begin{array}{l}1mile=1609meter\\ speedoflight=3\times {10}^{8}m/s\end{array}$

Explanation of Solution

$1.42lightyears$ can be converted into miles as follows,

$1.42\cancel{year}\times \frac{365\cancel{days}}{1\cancel{year}}\times \frac{24\cancel{h}}{1\cancel{day}}\times \frac{3600\cancel{s}}{1\cancel{h}}\times \frac{3.00\times {10}^8\cancel{m}}{1\cancel{s}}\times \frac{1mile}{1609\cancel{m}}=8.35\times 10^{12}mile$

Hence, $1.42lightyears$ can be converted into miles is $8.35\times 10^{12}mile.$

(b)

Interpretation Introduction

Interpretation:

$32.4yd$ has to be converted into centimeters.

Concept introduction:

Yard is a length measurement.

It is equal to 3 feet.

$\begin{array}{cc}1feet& 12inches\\ 1yard& 3feet\\ 1inch& 2.54cm\\ 1mile& 5280feet\end{array}$

Explanation of Solution

$32.4yd$ can be converted into centimeters as follows,

$32.4\cancel{yard}\times \frac{36\cancel{inches}}{1\cancel{yard}}\times \frac{2.54cm}{1\cancel{inches}}=2.96\times 10^{3}cm$

Hence, $32.4yd$ can be converted into centimeters is $2.96\times 10^{3}cm.$

(c)

Interpretation Introduction

Interpretation:

$\text{3}\text{.0}\text{×}{\text{10}}^{\text{10}}\text{cm/s}$ has to be converted into $\text{ft/s}$.

Concept introduction:

$\begin{array}{cc}1inch& 2.54cm\\ 1feet& 12inches\end{array}$

Explanation of Solution

$\text{3}\text{.0}\text{×}{\text{10}}^{\text{10}}\text{cm/s}$ can be converted into $\text{ft/s}$.

$\frac{\text{3}\text{.0}\text{×}{\text{10}}^{\text{10}}\cancel{\text{cm}}}{\text{1s}}\text{×}\frac{\text{1}\cancel{\text{inch}}}{\text{2}\text{.54}\cancel{\text{cm}}}\text{×}\frac{\text{1}\text{feet}}{\text{12}\text{i}\cancel{\text{nches}}}\text{=}9.8\times 10^{8}feet/s$

Hence, $\text{3}\text{.0}\text{×}{\text{10}}^{\text{10}}\text{cm/s}$ can be converted into $\text{ft/s}$ is $9.8\times 10^{8}feet/s.$

(d)

Interpretation Introduction

Interpretation:

${47.4}^{0}F$ has to be converted into degree Celsius.

Concept introduction:

${}^{0}C={(}^{0}F-32)\times \frac{5^{0}C}{9^{0}F}$

Explanation of Solution

${47.4}^{0}F$ has to be converted into degree Celsius

Use,

${}^{0}C={(}^{0}F-32)\times \frac{5^{0}C}{9^{0}F}$

${}^{0}C=({47.4}^{0}F-32)\times \frac{5^{0}C}{9^{0}F}={8.5}^{0}C$

${47.4}^{0}F$ is equal to ${8.5}^{0}C$

(e)

Interpretation Introduction

Interpretation:

$-{273.15}^{0}C$ has to be converted into degree Fahrenheit.

Concept introduction:

${}^{0}F=\frac{9^{0}C}{5^{0}F}\times {(0}^{}C)+({32}^{0}F)$

Explanation of Solution

$-{273.15}^{0}C$ has to be converted into degree Fahrenheit.

Use,

${}^{0}F=\frac{9^{0}C}{5^{0}F}\times {(}^{0}C)+({32}^{0}F)$

${}^{0}F=\frac{9^{0}C}{5^{0}F}\times (-{273.15}^{0}C)+({32}^{0}F)=-{459.67}^{0}F$

$-{273.15}^{0}C$ is equal to $-{459.67}^{0}F$

(f)

Interpretation Introduction

Interpretation:

$71.2c{m}^3$ to cubic meter has to be converted.

Concept Introduction:

The procedure that is used to convert between units in solving problems is called dimensional analysis.

By definition, $1c{m}^3=1\times {10}^{-6}{m}^3$

Explanation of Solution

Let’s consider the conversion of centimeter to cubic meter,

$71.2c{m}^3=?m^3$

By definition, $1c{m}^3=1\times {10}^{-6}{m}^3$

So $71.2c{m}^3=71.2\times {10}^{-6}{m}^3=71.2\times {10}^{-5}{m}^3$

Thus $71.2c{m}^3$ is equal to $71.2\times {10}^{-5}{m}^3$

(f)

Interpretation Introduction

Interpretation:

$7.2m^3$ to liters has to be converted.

Concept Introduction:

The procedure that is used to convert between units in solving problems is called dimensional analysis.

By definition, $1m^3=1\times {10}^{3}Litre$

Explanation of Solution

Let’s consider the conversion of meter cubic to litre,

$7.2m^3=?L$

By definition, $1m^3=1\times {10}^{3}Litre$

So $7.2m^3=7.2\times {10}^{3}Litre=7200L$

Thus $7.2m^3$ is equal to $7200L$