Chapter 5, Problem 42A

Interpretation Introduction

Interpretation: The distance in kilometers, meters and feet to be determined.

Concept introduction: Moving from one unit to another through factors of conversion (established on equivalence statements between the units) is also called dimensional analysis.

Answer to Problem 42A

190 mile = 306km, $\text{190mile=31}{\text{.6×10}}^{\text{4}}\text{m}$ , $\text{19}0\text{mile}=\text{1}{\text{.0×10}}^{\text{6}}\text{ft}$ .

Explanation of Solution

Relationship between two different units is equivalent statement. Example, 1 in=2.54 cm. Conversion of 2 units is as follows:

  $\text{Unit1×conversion}\text{factor=Unit2}$

Conversion of 190 miles to km is as follows:

  $\text{190}\cancel{\text{mile}}\text{×}\frac{\text{1}\text{.60934km}}{\text{1}\cancel{\text{mile}}}\text{=305}\text{.7746km}$

Thus,190 mile = 306km.

Conversion of 190 miles to meter is as follows:

  $\begin{array}{c}\text{190}\cancel{\text{mile}}\text{×}\frac{\text{1609}\text{.34m}}{\text{1}\cancel{\text{mile}}}\text{=305774}\text{.6m}\\ \text{=3}{\text{.1×10}}^5\text{m}\end{array}$

Thus $\text{190mile=31}{\text{.6×10}}^{\text{4}}\text{m}$ .

Conversion of 190 miles to feet is as follows:

  $\begin{array}{c}\text{190}\cancel{\text{mile}}\text{×}\frac{\text{5280feet}}{\text{1}\cancel{\text{mile}}}\text{=1003200ft}\\ \text{=1}{\text{.0×10}}^{\text{6}}\text{ft}\end{array}$

Thus $\text{19}0\text{mile}=\text{1}{\text{.0×10}}^{\text{6}}\text{ft}$ .

Conclusion

Conversion factor is a ratio for switching from one unit to next. This is used to convert one unit to another.