Chapter 2, Problem 17E

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 17E

The given metric-metric conversion of $\text{6}\text{.50}\text{Tm}$ into $\text{Mm}$ is $\text{6}\text{.50}\times {\text{10}}^6\text{Mm}$.

Explanation of Solution

The relation between meter and megameter is given below as,

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

The relation between meter and terameter is given below as,

$1\text{Tm}={\text{10}}^{12}\text{m}$ …(2)

The conversion of $\text{6}\text{.50}\text{Tm}$ into $\text{Mm}$ can be written using equation (1) and (2) as,

$\begin{array}{rll}1\text{Tm}& =& {\text{10}}^{12}\text{m}\\ \text{6}\text{.50}\text{Tm}& =& \frac{{\text{10}}^{12}\text{m}}{\text{1}\text{Tm}}\times \text{6}\text{.50}\text{Tm}\times \frac{1\text{Mm}}{{\text{10}}^6\text{m}}\\ & =& \text{6}\text{.50}\times {\text{10}}^6\text{Mm}\end{array}$

Conclusion

The given metric-metric conversion of $\text{6}\text{.50}\text{Tm}$ into $\text{Mm}$ is $\text{6}\text{.50}\times {\text{10}}^6\text{Mm}$.

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 17E

The given metric-metric conversion of $\text{650}\text{Gg}$ into $\text{kg}$ is $\text{650}\times {\text{10}}^6\text{kg}$.

Explanation of Solution

The relation between gram and kilogram is given below as,

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

The relation between gram and gigagram is given below as,

$1\text{Gg}={\text{10}}^9\text{g}$ …(4)

The conversion of $\text{650}\text{Gg}$ into $\text{kg}$ can be written using equation (3) and (4) as,

$\begin{array}{rll}1\text{Gg}& =& {\text{10}}^9\text{g}\\ \text{650}\text{Gg}& =& \frac{{\text{10}}^9\text{g}}{1\text{Gg}}\times \text{650}\text{Gg}\times \frac{1\text{kg}}{{\text{10}}^3\text{g}}\\ & =& \text{650}\times {\text{10}}^6\text{kg}\end{array}$

Conclusion

The given metric-metric conversion of $\text{650}\text{Gg}$ into $\text{kg}$ is $\text{650}\times {\text{10}}^6\text{kg}$.

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 17E

The given metric-metric conversion of $\text{0}\text{.650}\text{cL}$ into $\text{dL}$ is $\text{0}\text{.065}\text{dL}$.

Explanation of Solution

The relation between liter and centiliter is given below as,

$1\text{cL}={\text{10}}^{-2}\text{L}$ …(5)

The relation between liter and deciliter is given below as,

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

The conversion of $\text{0}\text{.650}\text{cL}$ into $\text{dL}$ can be written using equation (5) and (6) as,

$\begin{array}{rll}1\text{cL}& =& {\text{10}}^{-2}\text{L}\\ \text{0}\text{.650}\text{cL}& =& \frac{{\text{10}}^{-2}\text{L}}{1\text{cL}}\times \text{0}\text{.650}\text{cL}\times \frac{1\text{dL}}{{\text{10}}^{-1}\text{L}}\\ & =& \text{0}\text{.065}\text{dL}\end{array}$

Conclusion

The given metric-metric conversion of $\text{0}\text{.650}\text{cL}$ into $\text{dL}$ is $\text{0}\text{.065}\text{dL}$.

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 17E

The given metric-metric conversion of $\text{0}\text{.000}\text{650}\text{ns}$ into $\text{ps}$ is $0.650\text{ps}$.

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 picosecond is given below as,

$1\text{ps}={\text{10}}^{-12}\text{s}$ …(8)

The conversion of $\text{0}\text{.000}\text{650}\text{ns}$ into $\text{ps}$ can be written using equation (7) and (8) as,

$\begin{array}{rll}1\text{ns}& =& {\text{10}}^{-9}\text{s}\\ \text{0}\text{.000650}\text{ns}& =& \frac{{\text{10}}^{-9}\text{s}}{1\text{ns}}\times \text{0}\text{.000650}\text{ns}\times \frac{1\text{ps}}{{\text{10}}^{-12}\text{s}}\\ & =& 0.650\text{ps}\end{array}$

Conclusion

The given metric-metric conversion of $\text{0}\text{.000}\text{650}\text{ns}$ into $\text{ps}$ is $0.650\text{ps}$.