Chapter 18

1. $\frac{180 s}{4 min} = \frac{180 s}{240 s} = \frac{3}{4}$
2. $\begin{array}{rcl} \frac{1.8 m}{20.0 mm} & = & \frac{x}{14.5 mm} \\ 20.0 x & = & 1.8 (14.5) \\ x & = & 1.3 m \end{array}$
3. $\begin{array}{rcl} L & = & k T \\ L & = & 2.7 cm for T = 150 ° C \\ 2.7 & = & k (150) \\ k & = & 0.018 cm / ° C \\ L & = & 0.018 T \end{array}$
4. Let $x = length$ of the image (in in.)

$\begin{array}{l} \frac{1.00 in .}{2.54 cm} = \frac{x}{7.24 cm} \\ x = \frac{7.24}{2.54} = 2.85 in . \end{array}$
5. $2 l + 2 w = 210.0$

$\begin{array}{rcl} l & = & 105.0 - w \\ \frac{l}{w} & = & \frac{7}{3} \\ \frac{105.0 - w}{w} & = & \frac{7}{3} \\ 315.0 - 3 w & = & 7 w \\ 10 w & = & 315.0 \\ w & = & 31.5 in . \\ l & = & 105.0 - 31.5 \\ = & 73.5 in . \end{array}$
6. $p = k d h$

Using values of water.

$\begin{array}{rcl} 1.96 & = & k (1000) (0.200) \\ k & = & 0.00980 kPa \cdot m^{2} / kg \end{array}$

For alcohol,

$\begin{array}{rcl} p & = & 0.00980 (800) (0.300) \\ = & 2.35 kPa \end{array}$
7. $\begin{array}{rcl} Let L_{1} & = & crushing load of first pillar \\ L_{2} & = & crushing load of second pillar \end{array}$

$\begin{array}{l} L_{2} = \frac{k r_{2}^{4}}{l_{2}^{2}} L_{1} = \frac{k {(2 r_{2})}^{4}}{{(3 l_{2})}^{2}} \\ \frac{L_{1}}{L_{2}} = \frac{\frac{k {(2 r_{2})}^{4}}{{(3 l_{2})}^{2}}}{\frac{k r_{2}^{4}}{l_{2}^{2}}} = \frac{16 k r_{2}^{4}}{9 l_{2}^{2}} \times \frac{l_{2}^{2}}{k r_{2}^{4}} = \frac{16}{9} \end{array}$

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Basic Technical Mathematics, 11th Edition by Allyn J. Washington, Richard Evans