1-20 할리데이 11판 솔루션 일반물리학

(풀이자주:문제의 유효숫자가 애매하여 다양하게 표현했습니다.) $$\ab{a}$$ $$ \begin{aligned} \Ans=&230\ut{ft^2/gal}\\ =&230\ut{ft^2/gal}\times\(\frac{12\ut{in}}{1\ut{ft}}\cdot\frac{2.54\ut{cm}}{1\ut{in}}\cdot\frac{1\ut{m}}{100\ut{cm}}\)^2\\ &\times \frac{1\ut{gal}}{231\ut{in^3}}\cdot\(\frac{1\ut{in}}{2.54\ut{cm}}\)^3\times\frac{1000\ut{cm^3}}{1\ut{L}}\\ =&\frac{230\cdot\br{12\cdot2.54}^2\cdot1000}{100^2\cdot231\cdot2.54^3}\ut{m^2/L}\\ =&\frac{55200}{9779}\ut{m^2/L}\\ \approx&5.644748951835566\ut{m^2/L}\\ \approx&5.64\ut{m^2/L}\\ \approx&5.6\ut{m^2/L}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans=&\frac{55200}{9779}\ut{m^2/L}\\ =&\frac{55200\ut{m^2}}{9779\ut{L}}\times\frac{1000\ut{L}}{1\ut{m^3}}\\ =&\frac{55200000}{9779}\ut{/m}\\ \approx&5644.748951835566\ut{/m}\\ \approx&5.644748951835566\times10^3\ut{/m}\\ \approx&5.64\times10^3\ut{/m}\\ \approx&5.6\times10^3\ut{/m}\\ \approx&5.6\ut{/mm}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \Ans=&\frac{1}{230}\ut{gal/ft^2}\\ =&\frac{9779}{55200000}\ut{m}\\ \approx&0.0001771557971014493\ut{m}\\ \approx&1.77\times10^{-4}\ut{m}\\ \approx&177\ut{\mu m}\\ \end{aligned} $$ $$\ab{d}$$ 1제곱피트를 칠하는데 1/230갤런의 페인트가 필요하다. 페인트를 칠했을때의 두께는 177μm이다.