Chapter 2, Problem 23E

Interpretation Introduction

(a)

Interpretation:

The following metric-English conversions are to be performed.

Concept introduction:

Metric system is a measurement system which has units of meter, liter and gram as the units for measuring length, volume and mass. Adoption of the metric system as a universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 23E

The value of $\text{72}\text{in}\text{.}$ is equal to $\text{1}\text{.8288}\text{m}$.

Explanation of Solution

The value of $\text{1}\text{in}\text{.}=\text{2}\text{.54}\text{cm}$.

The conversion factor can be written as $\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}$.

The conversion of $\text{72}\text{in}\text{.}$ into $\text{cm}$ is given as,

$\begin{array}{c}\text{72}\text{in}\text{.=}\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\text{×72}\text{in}\\ \text{= 182}\text{.88}\text{cm}\end{array}$

The value of $\text{1}\text{m}=\text{100}\text{cm}$.

The conversion factor can be written as $\frac{\text{1}\text{m}}{\text{100}\text{cm}}$.

The conversion of $\text{182}\text{.88}\text{cm}$ into $\text{m}$ is given as,

$\begin{array}{c}\text{182}\text{.88}\text{cm=}\frac{\text{1}\text{m}}{\text{100}\text{cm}}\text{×182}\text{.88}\text{cm}\\ \text{= 1}\text{.8288}\text{m}\end{array}$

Conclusion

The value of $\text{72}\text{in}\text{.}$ is equal to $\text{1}\text{.8288}\text{m}$.

Interpretation Introduction

(b)

Interpretation:

The following metric-English conversions are to be performed.

Concept introduction:

Metric system is a measurement system which has units of meter, liter and gram as the units for measuring length, volume and mass. Adoption of the metric system as a universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 23E

The value of $\text{72}\text{lb}$ is equal to $\text{32}\text{.688}\text{kg}$.

Explanation of Solution

The value of $\text{1}\text{lb}=\text{454}\text{g}$.

The conversion factor can be written as $\frac{\text{454}\text{g}}{\text{1}\text{lb}}$.

The conversion of $\text{72}\text{lb}$ into $\text{g}$ is given as,

$\begin{array}{rll}\text{72}\text{lb}& =& \frac{\text{454}\text{g}}{\text{1}\text{lb}}\text{×72}\text{lb}\\ & =& \text{32688}\text{g}\end{array}$

The value of $\text{1}\text{kg}=\text{1000}\text{g}$.

The conversion factor can be written as $\frac{\text{1}\text{kg}}{\text{1000}\text{g}}$.

The conversion of $\text{32688}\text{g}$ into $\text{kg}$ is given as,

$\begin{array}{rll}\text{32688}\text{g}& =& \frac{\text{1}\text{kg}}{\text{1000}\text{g}}\text{×32688}\text{g}\\ & =& \text{32}\text{.688}\text{kg}\end{array}$

Conclusion

The value of $\text{72}\text{lb}$ is equal to $\text{32}\text{.688}\text{kg}$.

Interpretation Introduction

(c)

Interpretation:

The following metric-English conversions are to be performed.

Concept introduction:

Metric system is a measurement system which has units of meter, liter and gram as the units for measuring length, volume and mass. Adoption of the metric system as a universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 23E

The value of $\text{0}\text{.500}\text{qt}$ is equal to $\text{0}\text{.473}\text{L}$.

Explanation of Solution

The value of $\text{1}\text{qt}=\text{946}\text{mL}$.

The conversion factor can be written as $\frac{946\text{mL}}{\text{1}\text{qt}}$.

The conversion of $\text{0}\text{.500}\text{qt}$ into $\text{mL}$ is given as,

$\begin{array}{rll}\text{0}\text{.500}\text{qt}& =& \frac{\text{946}\text{mL}}{\text{1}\text{qt}}\text{×0}\text{.500}\text{qt}\\ & =& \text{473}\text{mL}\end{array}$.

The value of $\text{1}\text{L}=\text{1000}\text{mL}$.

The conversion factor can be written as $\frac{1\text{L}}{\text{1000}\text{mL}}$.

The conversion of $\text{473}\text{mL}$ into $\text{L}$ is given as,

$\begin{array}{rll}\text{473}\text{mL}& =& \frac{1\text{L}}{\text{1000}\text{mL}}\text{×473}\text{mL}\\ & =& 0.\text{473}\text{L}\end{array}$.

Conclusion

The value of $\text{0}\text{.500}\text{qt}$ is equal to $\text{0}\text{.473}\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The following metric-English conversions are to be performed.

Concept introduction:

Metric system is a measurement system which has units of meter, liter and gram as the units for measuring length, volume and mass. Adoption of the metric system as a universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 23E

The value of $\text{4}\text{.55}\times {\text{10}}^{-4}\min$ is equal to $\text{0}\text{.273}\text{ds}$.

Explanation of Solution

The value of $\text{1}\min =\text{60}\text{s}$.

The conversion factor can be written as $\frac{60\text{s}}{1\min }$.

The conversion of $\text{4}\text{.55}\times {\text{10}}^{-4}\min$ into $\text{s}$ is given as,

$\begin{array}{rll}\text{4}\text{.55}\times {\text{10}}^{-4}\min & =& \frac{60\text{s}}{1\min }\text{×4}\text{.55}\times {\text{10}}^{-4}\min \\ & =& 0.0273\text{s}\end{array}$

The value of $\text{1}\text{s}=\text{10}\text{ds}$.

The conversion factor can be written as $\frac{\text{10}\text{ds}}{\text{1}\text{s}}$.

The conversion of $0.0273\text{s}$ into $\text{ds}$ is given as,

$\begin{array}{rll}\text{0}\text{.0273}\text{s}& =& \frac{\text{10}\text{ds}}{\text{1}\text{s}}\text{×0}\text{.0273}\text{s}\\ & =& \text{0}\text{.273}\text{ds}\end{array}$.

Conclusion

The value of $\text{4}\text{.55}\times {\text{10}}^{-4}\min$ is equal to $\text{0}\text{.273}\text{ds}$.