Chapter 2, Problem 16E

Interpretation Introduction

(a)

Interpretation:

The given metric-metric conversion is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions.

Answer to Problem 16E

The given metric-metric conversion of $\text{5}\text{.00}\text{Mm}$ into $\text{m}$ is $\text{5}\text{.00}\times {\text{10}}^6\text{m}$.

Explanation of Solution

The relation between meter and megameter is given below as,

$1\text{Mm}={\text{10}}^6\text{m}$ …(1)

The conversion of $\text{5}\text{.00}\text{Mm}$ into $\text{m}$ can be written using equation (1) as,

$\begin{array}{rll}1\text{Mm}& =& {\text{10}}^6\text{m}\\ \text{5}\text{.00}\text{Mm}& =& \frac{{\text{10}}^6\text{m}}{\text{1}\text{Mm}}\times 5.00\text{Mm}\\ & =& \text{5}\text{.00}\times {\text{10}}^6\text{m}\end{array}$

Conclusion

The given metric-metric conversion of $\text{5}\text{.00}\text{Mm}$ into $\text{m}$ is $\text{5}\text{.00}\times {\text{10}}^6\text{m}$.

Interpretation Introduction

(b)

Interpretation:

The given metric-metric conversion is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions.

Answer to Problem 16E

The given metric-metric conversion of $\text{5}\text{.00}\text{mg}$ into $\text{g}$ is $\text{5}\text{.00}\times {\text{10}}^{-3}\text{g}$.

Explanation of Solution

The relation between gram and milligram is given below as,

$1\text{mg}={\text{10}}^{-3}\text{g}$ …(2)

The conversion of $\text{5}\text{.00}\text{mg}$ into $\text{g}$ can be written using equation (2) as,

$\begin{array}{rll}1\text{mg}& =& {\text{10}}^{-3}\text{g}\\ \text{5}\text{.00}\text{mg}& =& \frac{{\text{10}}^{-3}\text{g}}{1\text{mg}}\times 5.00\text{mg}\\ & =& \text{5}\text{.00}\times {\text{10}}^{-3}\text{g}\end{array}$

Conclusion

The given metric-metric conversion of $\text{5}\text{.00}\text{mg}$ into $\text{g}$ is $\text{5}\text{.00}\times {\text{10}}^{-3}\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The given metric-metric conversion is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions.

Answer to Problem 16E

The given metric-metric conversion of $\text{5}\text{.00}\text{mL}$ into $\text{L}$ is $\text{5}\text{.00}\times {\text{10}}^{-3}\text{L}$.

Explanation of Solution

The relation between liter and milliliter is given below as,

$1\text{mL}={\text{10}}^{-3}\text{L}$ …(3)

The conversion of $\text{5}\text{.00}\text{mL}$ into $\text{L}$ can be written using equation (3) as,

$\begin{array}{rll}1\text{mL}& =& {\text{10}}^{-3}\text{L}\\ \text{5}\text{.00}\text{mL}& =& \frac{{\text{10}}^{-3}\text{L}}{1\text{mL}}\times 5.00\text{mL}\\ & =& \text{5}\text{.00}\times {\text{10}}^{-3}\text{L}\end{array}$

Conclusion

The given metric-metric conversion of $\text{5}\text{.00}\text{mL}$ into $\text{L}$ is $\text{5}\text{.00}\times {\text{10}}^{-3}\text{L}$.

Interpretation Introduction

(d)

Interpretation:

The given metric-metric conversion is to be stated.

Concept introduction:

The metric system unit is the unit system which is accepted as the standard system of unit internationally. The nonmetric units are the common unit system which is not standard for the parameter to be calculated. The unit factors are calculated by taking the ratio of two equivalent measurement systems and these are used in unit conversions.

Answer to Problem 16E

The given metric-metric conversion of $\text{5}\text{.00}\text{ds}$ into $\text{s}$ is $0.50\text{s}$.

Explanation of Solution

The relation between second and decisecond is given below as,

$1\text{ds}={\text{10}}^{-1}\text{s}$ …(4)

The conversion of $\text{5}\text{.00}\text{ds}$ into $\text{s}$ can be written using equation (4) as,

$\begin{array}{rll}1\text{ds}& =& {\text{10}}^{-1}\text{s}\\ \text{5}\text{.00}\text{ds}& =& \frac{{\text{10}}^{-1}\text{s}}{1\text{ds}}\times 5.00\text{ds}\\ & =& 0.50\text{s}\end{array}$

Conclusion

The given metric-metric conversion of $\text{5}\text{.00}\text{ds}$ into $\text{s}$ is $0.50\text{s}$.