Chapter 2, Problem 22E

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 22E

The value of $\text{65}\text{in}\text{.}$ is equal to $\text{165}\text{.1}\text{cm}$.

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 $\text{65}\text{in}\text{.}$ into $\text{cm}$ is given as,

$\begin{array}{c}\text{65}\text{in}\text{.=}\frac{\text{2}\text{.54}\text{cm}}{\text{1}\text{in}\text{.}}\text{×65}\text{in}\\ \text{= 165}\text{.1}\text{cm}\end{array}$

Conclusion

The value of $\text{65}\text{in}\text{.}$ is equal to $\text{165}\text{.1}\text{cm}$.

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 an universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 22E

The value of $\text{65}\text{lb}$ is equal to $\text{29510}\text{g}$.

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 $\text{65}\text{lb}$ into $\text{g}$ is given as,

$\begin{array}{rll}\text{65}\text{lb}& =& \frac{\text{454}\text{g}}{\text{1}\text{lb}}\text{×65}\text{lb}\\ & =& \text{29510}\text{g}\end{array}$

Conclusion

The value of $\text{65}\text{lb}$ is equal to $\text{29510}\text{g}$.

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 an universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 22E

The value of $\text{65}\text{mL}$ is equal to $\text{0}\text{.0687}\text{qt}$.

Explanation of Solution

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

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

The conversion of $\text{65}\text{mL}$ into $\text{qt}$ is given as,

$\begin{array}{rll}\text{65}\text{mL}& =& \frac{\text{1}\text{qt}}{\text{946}\text{mL}}\text{×65}\text{mL}\\ & =& \text{0}\text{.0687}\text{qt}\end{array}$.

Conclusion

The value of $\text{65}\text{mL}$ is equal to $\text{0}\text{.0687}\text{qt}$.

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 an universal system of measurement has been resisted by several countries. In science, measurements are regularly recorded in metric units.

Answer to Problem 22E

The value of $\text{6}\text{.50}\times {\text{10}}^{-3}\text{s}$ is equal to $\text{6}\text{.50}\times {\text{10}}^{-3}\text{sec}$.

Explanation of Solution

The value of $\text{1}\sec =\text{1}\text{.00}\text{s}$.

The conversion factor can be written as $\frac{\text{1}\text{sec}}{1.00\text{s}}$.

The conversion of $\text{6}\text{.50}\times {\text{10}}^{-3}\text{s}$ into $\text{sec}$ is given as,

$\begin{array}{rll}\text{6}\text{.50}\times {\text{10}}^{-3}\text{s}& =& \frac{\text{1}\text{sec}}{\text{1}\text{.00}\text{s}}\times \text{6}\text{.50}\times {\text{10}}^{-3}\text{s}\\ & =& \text{6}\text{.50}\times {\text{10}}^{-3}\text{sec}\end{array}$

Conclusion

The value of $\text{6}\text{.50}\times {\text{10}}^{-3}\text{s}$ is equal to $\text{6}\text{.50}\times {\text{10}}^{-3}\text{sec}$.