Chapter 1, Problem 39E

Interpretation Introduction

Interpretation:

The unit of gauge pressure in $\frac{\text{lb}}{{\text{in}}^{\text{2}}}$ needs to be converted into $\frac{\text{g }}{{\text{cm}}^{\text{2}}}$ and $\frac{\text{kg}}{{\text{m}}^{\text{2}}}$.

Concept introduction:

The conversion factor is the factor which is required to determine the results in different or desired scale by equivalizing the different units.

The following conversion factor are given:

$\begin{array}{l}\text{1lb = 453}\text{.6 g}\hfill \\ \text{1 In = 2}\text{.54 cm}\hfill \\ {\text{1 In}}^{\text{2}}\text{= }{\left(\text{2}\text{.54}\right)}^{\text{2}}{\text{cm}}^{\text{2}}\text{= 6}{\text{.45 cm}}^{\text{2}}\hfill \end{array}$