Chapter 2, Problem 18E

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 18E

The given metric-metric conversion of $\text{7}\text{.50}\text{km}$ into $\text{Gm}$ is $\text{7}\text{.50}\times {\text{10}}^{-6}\text{Gm}$.

Explanation of Solution

The relation between meter and gigameter is given below as,

$1\text{Gm}={\text{10}}^9\text{m}$ …(1)

The relation between meter and kilometer is given below as,

$1\text{km}={\text{10}}^3\text{m}$ …(2)

The conversion of $\text{7}\text{.50}\text{km}$ into $\text{Gm}$ can be written using equation (1) and (2) as,

$\begin{array}{rll}1\text{km}& =& {\text{10}}^3\text{m}\\ \text{7}\text{.50}\text{km}& =& \frac{{\text{10}}^3\text{m}}{\text{1}\text{km}}\times \text{7}\text{.50}\text{km}\times \frac{1\text{Gm}}{{\text{10}}^9\text{m}}\\ & =& \text{7}\text{.50}\times {\text{10}}^{-6}\text{Gm}\end{array}$

Conclusion

The given metric-metric conversion of $\text{7}\text{.50}\text{km}$ into $\text{Gm}$ is $\text{7}\text{.50}\times {\text{10}}^{-6}\text{Gm}$.

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 18E

The given metric-metric conversion of $\text{750}\text{Mg}$ into $\text{Tg}$ is $\text{750}\times {\text{10}}^{-6}\text{Tg}$.

Explanation of Solution

The relation between gram and megagram is given below as,

$1\text{Mg}={\text{10}}^6\text{g}$ … (3)

The relation between gram and teragram is given below as,

$1\text{Tg}={\text{10}}^{12}\text{g}$ …(4)

The conversion of $\text{750}\text{Mg}$ into $\text{Tg}$ can be written using equation (3) and (4) as,

$\begin{array}{rll}1\text{Mg}& =& {\text{10}}^6\text{g}\\ \text{750}\text{Mg}& =& \frac{{\text{10}}^6\text{g}}{1\text{Mg}}\times \text{750}\text{Mg}\times \frac{1\text{Tg}}{{\text{10}}^{12}\text{g}}\\ & =& \text{750}\times {\text{10}}^{-6}\text{Tg}\end{array}$

Conclusion

The given metric-metric conversion of $\text{750}\text{Mg}$ into $\text{Tg}$ is $\text{750}\times {\text{10}}^{-6}\text{Tg}$.

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 18E

The given metric-metric conversion of $\text{0}\text{.750}\text{pL}$ into $\mu \text{L}$ is $\text{0}\text{.750}\times {\text{10}}^{-6}\mu \text{L}$.

Explanation of Solution

The relation between liter and picoliter is given below as,

$1\text{pL}={\text{10}}^{-12}\text{L}$ … (5)

The relation between liter and microliter is given below as,

$1\mu \text{L}={\text{10}}^{-6}\text{L}$ … (6)

The conversion of $\text{0}\text{.750}\text{pL}$ into $\mu \text{L}$ can be written using equation (5) and (6) as,

$\begin{array}{rll}1\text{pL}& =& {\text{10}}^{-12}\text{L}\\ \text{0}\text{.750}\text{pL}& =& \frac{{\text{10}}^{-12}\text{L}}{1\text{pL}}\times \text{0}\text{.750}\text{pL}\times \frac{1\mu \text{L}}{{\text{10}}^{-6}\text{L}}\\ & =& \text{0}\text{.750}\times {\text{10}}^{-6}\mu \text{L}\end{array}$

Conclusion

The given metric-metric conversion of $\text{0}\text{.750}\text{pL}$ into $\mu \text{L}$ is $\text{0}\text{.750}\times {\text{10}}^{-6}\mu \text{L}$.

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 18E

The given metric-metric conversion of $\text{0}\text{.000}\text{750}\text{ms}$ into $\text{ns}$ is $750\text{ns}$.

Explanation of Solution

The relation between second and nanosecond is given below as,

$1\text{ns}={\text{10}}^{-9}\text{s}$ …(7)

The relation between second and millisecond is given below as,

$1\text{ms}={\text{10}}^{-3}\text{s}$ …(8)

The conversion of $\text{0}\text{.000}\text{750}\text{ms}$ into $\text{ns}$ can be written using equation (7) and (8) as,

$\begin{array}{rll}1\text{ms}& =& {\text{10}}^{-3}\text{s}\\ \text{0}\text{.000750}\text{ms}& =& \frac{{\text{10}}^{-3}\text{s}}{1\text{ms}}\times \text{0}\text{.000750}\text{ms}\times \frac{1\text{ns}}{{\text{10}}^{-9}\text{s}}\\ & =& 750\text{ns}\end{array}$

Conclusion

The given metric-metric conversion of $\text{0}\text{.000}\text{750}\text{ms}$ into $\text{ns}$ is $750\text{ns}$.