1-30 할리데이 10판 솔루션 일반물리학

(a) t(mMAX) =? $$ \begin{aligned} \dot m &= \dxt{m} = \dt\(5.00t^{0.8}-3.00t + 20.00\) = 0 \\ &= \frac{4}{\sqrt[5]{t}}-3 = 0 \end{aligned} $$ $$ \begin{aligned} \\ t &= \frac{1024}{243}\ut{s} \\ &\approx 4.2139917695473251029\ut{s} \\ &\approx 4.21\ut{s} \end{aligned} $$
(b) mMAX =? $$ \begin{aligned} m(t_{MAX}) &= 5.00(t_{MAX})^{0.8}-3.00(t_{MAX}) + 20.00\ut{g} \\ &= 5.00\(\frac{1024}{243}\)^{0.8}-3.00\(\frac{1024}{243}\) + 20.00\ut{g} \\ &= \frac{1876}{81}\ut{g} \\ &\approx 23.160493827160493827\ut{g} \\ &\approx 23.2\ut{g} \end{aligned} $$
(c) $$ \begin{aligned} \dot m(t) &= \(\frac{4}{\sqrt[5]{t}}-3\)\ut{g/s} \\ \dot m(2.00)&= \(\frac{4}{\sqrt[5]{2.00}}-3\)\ut{g/s}\cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \frac{60\ut{s}}{1\ut{min}} \\ &= \frac{3}{50} \(2\cdot2^{\frac{4}{5}}-3\)\ut{kg/min} \\ &\approx 0.028932135191069793393\ut{kg/min} \\ &\approx 2.89\times 10^{-2}\ut{kg/min} \end{aligned} $$
(d) $$ \begin{aligned} \dot m(5.00)&= \(\frac{4}{\sqrt[5]{5.00}}-3\)\ut{g/s}\cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \frac{60\ut{s}}{1\ut{min}} \\ &= \frac{3}{50} \(\frac{4}{\sqrt[5]{5}}-3\)\ut{kg/min} \\ &\approx -0.0060528807173530724471\ut{kg/min} \\ &\approx -6.05\times 10^{-3}\ut{kg/min} \end{aligned} $$