Chapter 2, Problem 21PE

(a)

Interpretation Introduction

Interpretation:

The quantity $\text{42}\text{.2}\text{in}\text{.}$ has to be converted into centimeter.

Answer to Problem 21PE

The quantity $\text{42}\text{.2}\text{in}\text{.}$ in centimeter is $\text{107}\text{cm}\text{.}$

Explanation of Solution

The given quantity is $\text{42}\text{.2}\text{in}\mathrm{..}$

Inches is converted into centimeters using the conversion factor,

  $\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\right)$

The quantity $\text{42}\text{.2}\text{in}\text{.}$ is converted as,

  $\begin{array}{l}\text{cm=}\left(\text{42}\text{.2}\text{in}\text{.}\right)\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\right)\\ \text{cm=107}\text{cm}\end{array}$

The quantity $\text{42}\text{.2}\text{in}\text{.}$ in centimeter is $\text{107}\text{cm}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The quantity $\text{0}\text{.64}\text{mi}$ has to be converted into inches.

Answer to Problem 21PE

The quantity $\text{0}\text{.64}\text{mi}$ in inches is $\text{4}{\text{.1×10}}^{\text{4}}\text{in}\mathrm{..}$

Explanation of Solution

The given quantity is $\text{0}\text{.64}\text{mi}\text{.}$

Miles is converted into inches using the conversion factor,

  $\left(\frac{\text{5280}\text{ft}}{\text{1}\text{mi}}\right)\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)$

The quantity $\text{0}\text{.64}\text{mi}$ is converted as,

  $\begin{array}{l}\text{in}\text{.=}\left(\text{0}\text{.64}\text{mi}\right)\left(\frac{\text{5280}\text{ft}}{\text{1}\text{mi}}\right)\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)\\ \text{in}\text{.=4}{\text{.1×10}}^{\text{4}}\text{in}\text{.}\end{array}$

The quantity $\text{0}\text{.64}\text{mi}$ in inches is $\text{4}{\text{.1×10}}^{\text{4}}\text{in}\mathrm{..}$

(c)

Interpretation Introduction

Interpretation:

The quantity $\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}$ has to be converted into centimeter squares $\left({\text{cm}}^{\text{2}}\right)\text{.}$

Answer to Problem 21PE

The quantity $\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}$ in centimeter square is $\text{12}\text{.9}{\text{cm}}^{\text{2}}\text{.}$

Explanation of Solution

The given quantity is $\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}.$

Square inches is converted into centimeter square using the conversion factor,

  ${\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\right)}^{\text{2}}$

The quantity $\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}$ is converted as,

  $\begin{array}{l}{\text{cm}}^{\text{2}}\text{=}\left(\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}\right){\left(\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\right)}^{\text{2}}\\ {\text{cm}}^{\text{2}}\text{=12}\text{.9}{\text{cm}}^{\text{2}}\end{array}$

The quantity $\text{2}\text{.00}\text{in}{\text{.}}^{\text{2}}$ in centimeter square is $\text{12}\text{.9}{\text{cm}}^{\text{2}}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The quantity $\text{42}\text{.8}\text{kg}$ has to be converted into pounds.

Answer to Problem 21PE

The quantity $\text{42}\text{.8}\text{kg}$ in pounds is $\text{94}\text{.4}\text{lb}\text{.}$

Explanation of Solution

The given quantity is $\text{42}\text{.8}\text{kg}\text{.}$

Kilograms is converted into pounds using the conversion factor,

  $\left(\frac{\text{2}\text{.205}\text{lb}}{\text{1}\text{kg}}\right)$

The quantity $\text{42}\text{.8}\text{kg}$ is converted as,

  $\begin{array}{l}\text{lb=}\left(\text{42}\text{.8}\text{kg}\right)\left(\frac{\text{2}\text{.205}\text{lb}}{\text{1}\text{kg}}\right)\\ \text{lb=94}\text{.4}\text{lb}\end{array}$

The quantity $\text{42}\text{.8}\text{kg}$ in pounds is $\text{94}\text{.4}\text{lb}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The quantity $\text{3}\text{.5}\text{qt}$ has to be converted into milliliters.

Answer to Problem 21PE

The quantity $\text{3}\text{.5}\text{qt}$ in milliliters is $\text{3}{\text{.3×10}}^{\text{3}}\text{mL}\text{.}$

Explanation of Solution

The given quantity is $\text{3}\text{.5}\text{qt}\text{.}$

Quarts is converted into milliliters using the conversion factor,

  $\left(\frac{\text{946}\text{mL}}{\text{1qt}}\right)$

The quantity $\text{3}\text{.5}\text{qt}$ is converted as,

  $\begin{array}{l}\text{mL=}\left(\text{3}\text{.5}\text{qt}\right)\left(\frac{\text{946}\text{mL}}{\text{1qt}}\right)\\ \text{mL=3}{\text{.3×10}}^{\text{3}}\text{mL}\end{array}$

The quantity $\text{3}\text{.5}\text{qt}$ in milliliters is $\text{3}{\text{.3×10}}^{\text{3}}\text{mL}\text{.}$

(f)

Interpretation Introduction

Interpretation:

The quantity $\text{20}\text{.0}\text{L}$ has to be converted into gallons.

Answer to Problem 21PE

The quantity $\text{20}\text{.0}\text{L}$ in gallons is $\text{5}\text{.29}\text{gal}\text{.}$

Explanation of Solution

The given quantity is $\text{20}\text{.0}\text{L}\text{.}$

Liters is converted into gallons using the conversion factor,

  $\left(\frac{\text{1}\text{qt}}{\text{0}\text{.946}\text{L}}\right)\left(\frac{\text{1}\text{gal}}{\text{4}\text{qt}}\right)$

The quantity $\text{20}\text{.0}\text{L}$ is converted as,

  $\begin{array}{l}\text{gal=}\left(\text{20}\text{.0}\text{L}\right)\left(\frac{\text{1}\text{qt}}{\text{0}\text{.946}\text{L}}\right)\left(\frac{\text{1}\text{gal}}{\text{4}\text{qt}}\right)\\ \text{gal=5}\text{.29}\text{gal}\end{array}$

The quantity $\text{20}\text{.0}\text{L}$ in gallons is $\text{5}\text{.29}\text{gal}\text{.}$