91Ó°ÊÓ

1. The length of a soccer field, 120 m (three significant figures), to feet

\(\)\(\begin{array}{c}120\,\,m\,\, = \,\,120\,\,\left( {m\,\, \times \,\frac{{3.28084\,ft}}{m}} \right)\,\,\\ = \,\,393.701\,\,ft\,\,\\ = \,\,394\,\,ft\end{array}\)

Kilometers

\(\begin{array}{c}19,565\,\,ft\,\, = \,\,19,565\,\,\left( {ft\, \times \,\frac{{0.0003048\,km}}{{ft}}} \right)\,\,\\ = \,\,5.963412\,km\\ = 5.9634km\end{array}\)

c. The area of an 8.5 × 11-inch sheet of paper in cm²

\(\begin{array}{c}8.5\,\, \times \,\,11\, - inch\,\, = \,\,8.5\,inch\,\, \times \,11\,inch\,\,\\ = \,\,8.5\,\, \times \,11\,i{n^2}\,\,\\ = \,\,93.5\,\,i{n^2}\,\\\, = \,\,93.5\,\left( {i{n^2}\, \times \,\frac{{6.4516\,\,c{m^2}}}{{i{n^2}}}} \right)\,\,\\ = \,\,603.2246\\ = 6.0\, \times {10^2}\,c{m^2}\end{array}\)

(d) The displacement volume of an automobile engine, 161 in.3, to liters\(\begin{array}{c}161\,\,i{n^3}\,\, = \,\,161\,\,\left( {i{n^3}\, \times \,\,\frac{{16.387\,c{m^3}}}{{i{n^3}}}} \right)\,\,\\ = \,\,2638.32\,\,c{m^3}\\ = 2.64L\end{array}\)

(e) The estimated mass of the atmosphere, 5.6 × 1015 tons, to kilograms

\(\begin{array}{c}5.6\, \times \,{10^{15}}\,tons\,\, = \,\,5.6\, \times \,{10^{15}}\,\left( {tons\, \times \,\frac{{907.185\,\,kg}}{{tons}}} \right)\,\,\\ = \,\,5080.23\, \times \,{10^{15}}\,kg\,\,\\ = \,\,5.08023\,\, \times \,\,{10^{18}}\,kg\\ = 5.1 \times {10^{18}}\,kg\end{array}\)

(f) The mass of a bushel of rye, 32.0 lb, to kilograms

\(\begin{array}{c}32\,\,lb\,\, = \,\,32\,\,\left( {lb\, \times \,\frac{{0.453592\,\,kg}}{{lb}}} \right)\,\,\\ = \,\,14.515\,\,kg\\ = 14.5kg\end{array}\)

(g) The mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)

\(\begin{array}{c}5\,grain\, = \,5\,\, \times \,0.00229\,oz\,\\\, = \,\,0.01145\,\,oz\,\\\, = \,0.01145\,\,\left( {oz\, \times \,\frac{{28349.5\,\,mg}}{{oz}}} \right)\,\\\, = \,\,324mg\end{array}\)