Chapter 2, Problem 13E

Interpretation Introduction

(a)

Interpretation:

Two unit factors for the given metric relationships are 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.

Answer to Problem 13E

Two unit factors for the given metric relationships are $\frac{1\text{km}}{1\times {10}^3\text{m}}$ and $\frac{1\times {10}^3\text{m}}{1\text{km}}$.

Explanation of Solution

The kilo $\left(\text{k}\right)$ stands for a factor of ${10}^3$. $\text{"m"}$ is the symbol used to denote meter.

The relation showing unit equation between meter $\left(\text{m}\right)$ and kilometers $\left(\text{km}\right)$ is given below as,

$1\text{km}=1\times {10}^3\text{m}$

The unit factors are calculated by taking the ratio of two equivalent measurement systems. Thus, the unit factor for meter $\left(\text{m}\right)$ and kilometers $\left(\text{km}\right)$ can be written as,

$\frac{1\text{km}}{1\times {10}^3\text{m}}$

Another unit factor can be written as,

$\frac{1\times {10}^3\text{m}}{1\text{km}}$

Conclusion

Two unit factors for the given metric relationships are $\frac{1\text{km}}{1\times {10}^3\text{m}}$ and $\frac{1\times {10}^3\text{m}}{1\text{km}}$.

Interpretation Introduction

(b)

Interpretation:

Two unit factors for the given metric relationships are 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.

Answer to Problem 13E

Two unit factors for the given metric relationships are $\frac{1\text{Mg}}{1\times {10}^6\text{g}}$ and $\frac{1\times {10}^6\text{g}}{1\text{Mg}}$.

Explanation of Solution

The mega $\left(\text{M}\right)$ stands for a factor of ${10}^6$. $\text{"g"}$ is the symbol used to denote grams.

The relation showing unit equation between grams $\left(\text{g}\right)$ and mega grams $\left(\text{Mg}\right)$ is given below as,

$1\text{Mg}=1\times {10}^6\text{g}$

The unit factors are calculated by taking the ratio of two equivalent measurement systems. Thus, the unit factor for grams $\left(\text{g}\right)$ and mega grams $\left(\text{Mg}\right)$ can be written as,

$\frac{1\text{Mg}}{1\times {10}^6\text{g}}$

Another unit factor can be written as,

$\frac{1\times {10}^6\text{g}}{1\text{Mg}}$

Conclusion

Two unit factors for the given metric relationships are $\frac{1\text{Mg}}{1\times {10}^6\text{g}}$ and $\frac{1\times {10}^6\text{g}}{1\text{Mg}}$.

Interpretation Introduction

(c)

Interpretation:

Two unit factors for the given metric relationships are 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.

Answer to Problem 13E

Two unit factors for the given metric relationships are $\frac{1\text{mL}}{1\times {10}^{-3}\text{L}}$ and $\frac{1\times {10}^{-3}\text{L}}{1\text{mL}}$.

Explanation of Solution

The milli $\left(\text{m}\right)$ stands for a factor of ${10}^{-3}$. $\text{"L"}$ is the symbol used to denote liters.

The relation showing unit equation between liters $\left(\text{L}\right)$ and milliliters $\left(\text{mL}\right)$ is given below as,

$1\text{mL}=1\times {10}^{-3}\text{L}$

The unit factors are calculated by taking the ratio of two equivalent measurement systems. Thus, the unit factor for liters $\left(\text{L}\right)$ and milliliters $\left(\text{mL}\right)$ can be written as,

$\frac{1\text{mL}}{1\times {10}^{-3}\text{L}}$

Another unit factor can be written as,

$\frac{1\times {10}^{-3}\text{L}}{1\text{mL}}$

Conclusion

Two unit factors for the given metric relationships are $\frac{1\text{mL}}{1\times {10}^{-3}\text{L}}$ and $\frac{1\times {10}^{-3}\text{L}}{1\text{mL}}$.

Interpretation Introduction

(d)

Interpretation:

Two unit factors for the given metric relationships are 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.

Answer to Problem 13E

Two unit factors for the given metric relationships are $\frac{1\mu \text{s}}{1\times {10}^{-6}\text{s}}$ and $\frac{1\times {10}^{-6}\text{s}}{1\mu \text{s}}$.

Explanation of Solution

The micro $\left(\mu \right)$ stands for a factor of ${10}^{-6}$. $\text{"s"}$ is the symbol used to denote seconds.

The relation showing unit equation between seconds $\left(\text{s}\right)$ and micro seconds $\left(\mu \text{s}\right)$ is given below as,

$1\mu \text{s}=1\times {10}^{-6}\text{s}$

The unit factors are calculated by taking the ratio of two equivalent measurement systems. Thus, the unit factor for seconds $\left(\text{s}\right)$ and micro seconds $\left(\mu \text{s}\right)$ can be written as,

$\frac{1\mu \text{s}}{1\times {10}^{-6}\text{s}}$

Another unit factor can be written as,

$\frac{1\times {10}^{-6}\text{s}}{1\mu \text{s}}$

Conclusion

Two unit factors for the given metric relationships are $\frac{1\mu \text{s}}{1\times {10}^{-6}\text{s}}$ and $\frac{1\times {10}^{-6}\text{s}}{1\mu \text{s}}$.