솔루션피아

(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..

$$ \begin{aligned} \\ \Ans &= 28.9\ut{picul} = ? \\&= 28.9\ut{picul} \cdot \frac{100\ut{gin}}{1\ut{picul}} \cdot \frac{16\ut{tahil}}{1\ut{gin}}\cdot \frac{10\ut{chee}}{1\ut{tahil}}\cdot \frac{10\ut{hoon}}{1\ut{chee}}\cdot \frac{0.3779\ut{g}}{1\ut{hoon}}\cdot \frac{1\ut{kg}}{1000\ut{g}} \\&= \frac{1092131}{625}\ut{kg} \\&= 1747.4096\ut{kg} \\&\approx 1.75 \times 10^3\ut{kg} = 1750\ut{kg} \end{ali..

$$ \begin{cases} c &= 299792458\ut{m/s} \\ 1\ut{MeV} &= 1.60218 \times 10^{-13}\ut{J} \\ 1\ut{u} &= 1.66054 \times 10^{-27}\ut{kg} \end{cases} $$ $$ \begin{aligned} \\ \Ans &= E = ? \\&= E = mc^2 \\&= 1\ut{u} \cdot (299792458\ut{m/s})^2 \\&= 1\ut{u} \cdot (299792458\ut{m/s})^2 \cdot \frac{1.66054 \times 10^{-27}\ut{kg}}{1\ut{u}}\cdot \frac{1\ut{MeV}}{1.60218 \times 10^{-13}\ut{J}} \\&= \frac{186..

물체=상자 로 가정함. $$ \begin{aligned} & \begin{cases} m_B &= 2.500\ut{kg} \\ m_{G1} &= 21.15\ut{g} \\ m_{G2} &= 21.16\ut{g} \end{cases}\\ \\ \Ans &= m_B - \Sigma m_G \\ &= m_B - (m_{G1}+m_{G2}) \\ &= 2.500\ut{kg} - (21.15\ut{g}+21.16\ut{g}) \\ &= 2.500\ut{kg} - (21.15\ut{g}+21.16\ut{g})\cdot \frac{1\ut{kg}}{1000\ut{g}} \\ &= 2.45769\ut{kg} \\ &\approx 2.458\ut{kg} \end{aligned} $$

$$ \begin{aligned} \Ans &= \rho = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \\ &= \frac{5.324\ut{g}}{2.5\ut{cm^3}} \cdot \frac{1\ut{kg}}{1000\ut{g}}\cdot \left(\frac{100\ut{cm}}{1\ut{m}}\right)^3 \\ &= \frac{10648}{5}\ut{kg/m^3} \\ &= 2129.6\ut{kg/m^3} \\ &\approx 2.1\times 10^3\ut{kg/m^3} =2.1\ut{tonne/m^3} \\ &= 2.1\ut{kg/L}=2.1\ut{g/cm^3} \end{aligne..

(a) $$ \begin{cases} R &= 1.90\ut{\AAA} \\ N_0 &= 6.023 \times 10^{23}\ut{EA/mol} \\ m_{\mathrm{Na}} &= 22.9898\ut{g/mol} \end{cases} $$ $$ \begin{aligned} \Ans &= \rho_{\mathrm{Na}} = ? \\ &= \frac{\mathrm{Mass}}{\mathrm{Volume}} \\ &= \frac{m_{\mathrm{Na}}}{N_0} \cdot \frac{3}{4\pi R^3} \\ &= \frac{22.9898\ut{g/mol}}{6.023 \times 10^{23}\ut{EA/mol}} \cdot \frac{3}{4\pi (1.90\ut{\AAA})^3} \\ &=..

$$ \begin{cases} R &= 60\ut{\mu m} \\ \rho_{\text{SiO}_2} &= 2600\ut{kg/m^3} \\ L &= 1.00\ut{m} \end{cases} $$ $$ \begin{aligned} \\ n\cdot S_{\text{particle}} &= S_{\text{cube}} \\ n &= \frac{S_{\text{cube}}}{S_{\text{particle}}} \end{aligned} $$ $$ \begin{aligned} \\ \Ans &= \Sigma m = nm =n\rho V_{\text{particle}} \\ &= \frac{S_{\text{cube}}}{S_{\text{particle}}}\rho V_{\text{particle}} \\ &=..

$$ \begin{cases} S &= 6.00\ut{m^2} \\ m &= 324\ut{kg} \\ \rho &= 2.70 \times 10^3 \ut{kg/m^3} \end{cases} $$ d=? $$ \begin{aligned} \\ \Ans &= d = \frac{Volume}{Area} \\ &= \frac{m}{\rho \cdot S} \\ &= \frac{324\ut{kg}}{2.70 \times 10^3\ut{kg/m^3} \cdot 6.00\ut{m^2}} \\ &= \frac{1}{50}\ut{m} \\ &= 0.02\ut{m} = 2.00\ut{cm} \end{aligned} $$

$$ \begin{cases} \rho &= 19.32\ut{g/cm^3} \\ m &= 29.34\ut{g} \end{cases} $$ (a) S=? $$ h = 1.000\ut{\mu m} $$ $$\begin{aligned} \\ \Ans &= S = \frac{\text{Volume}}{\text{Hight}} \\ &= \frac{m}{\rho}\cdot\frac{1}{h} \\ &= \frac{29.34\ut{g}}{19.32\ut{g/cm^3}}\cdot \frac{1}{1.000\ut{\mu m}} \\ &= \frac{489}{322}\ut{cm^3}\cdot \frac{1}{1.000\ut{\mu m}}\cdot \frac{10^6\ut{\mu m}}{1\ut{m}}\cdot \left..

$$ \begin{aligned} \Ans &= \rho = \frac{\text{Mass}}{\text{Volume}} \\ &= \frac{9.05\ut{g}}{3.5\ut{cm^3}} \\ &= \frac{181}{70}\ut{g/cm^3} \\ &\approx 2.5857142857142857143\ut{g/cm^3} \\ &\approx 2.6\ut{g/cm^3} \end{aligned} $$