솔루션피아

$$ \begin{cases} \rho_A &= 10\ut{EA/U.S.~pint}\\ \rho_B &= 30\ut{EA/U.S.~pint}\\ V &= 1.0\ut{m}\times20\ut{cm}\times20\ut{cm}\\ 1\ut{U.S.~pint}&=0.4732\ut{L} \end{cases} $$ $$ \begin{aligned} V&=100\cdot20\cdot20\ut{cm^3}\\ &=40000\ut{cm^3} \end{aligned} $$ $$ \begin{aligned} \Ans &= N_B - N_A \\ &=\rho_B \cdot V - \rho_A \cdot V\\ &= V\(\rho_B - \rho_A\)\\ &=40000\ut{cm^3}\cdot20\ut{EA/U.S.~pin..

$$ \begin{cases} 1\ut{AU}&=1.50\times10^8\ut{km}\\ c&=3.0\times10^8\ut{m/s} \end{cases} $$ $$ \begin{aligned} c&=\cfrac{3.0\times10^8\ut{m}}{1\ut{sec}}\times\cfrac{1\ut{km}}{1000\ut{m}}\times\cfrac{1\ut{AU}}{1.50\times10^8\ut{km}}\times\cfrac{3600\ut{sec}}{1\ut{h}}\\ &=\cfrac{3.0\cdot3600}{1000\cdot1.50}\ut{AU/h}\\ &=\cfrac{36}{5}\ut{AU/h}\\ &=7.2\ut{AU/h} \end{aligned} $$

$$ \begin{cases} 1\ut{mol}&=6.02\times10^{23}\ut{EA}\\ \end{cases} $$ $$ \begin{aligned} N^3 &= 0.2\ut{mol}\\ N &= \sqrt[3]{0.2\times6.02\times10^{23}}\\ &\approx 4.93789854178\times10^7\\ &\approx 4.9\times10^7\ut{EA}\\ \end{aligned} $$ $$ \begin{aligned} \Ans&=4.9\times10^7\ut{cm}\\ &=4.9\times10^7\ut{cm}\times \cfrac{1\ut{m}}{100\ut{cm}}\times\cfrac{1\ut{km}}{1000\ut{m}}\\ &=4.9\times10^2\ut{..

$$ \begin{cases} 1\ut{diplacement~ton}&=7\ut{barrel~bluk}\\ 1\ut{freight~ton}&=8\ut{barrel~bluk}\\ 1\ut{register~ton}&=20\ut{barrel~bluk}\\ 1\ut{barrel~bluk}&=0.1415\ut{m^3}\\ 1\ut{m^3}&=28.378\ut{U.S~bushel} \end{cases} $$ $$A = 73\ut{diplacement~ton}$$ $$ \ab{a}$$ $$ \begin{aligned} \Ans &= 73\ut{freight~ton}-73\ut{diplacement~ton}\\ &=73\cdot\(1\ut{freight~ton}-1\ut{diplacement~ton}\)\\ &=73\..

$$ \begin{cases} m_H &= 1.0078\ut{u}\\ 1\ut{u} \times1\ut{mol}&= 1\ut{g}\\ N_A&=6.02214076\times10^{23} \ut{EA/mol}\\ M_H&=3.30\ut{kg}\\ \end{cases} $$ $$ \begin{aligned} \Ans&=\cfrac{3.30\ut{kg}}{1.0078\ut{u}}\times\cfrac{1\ut{u}\cdot1\ut{mol}}{1\ut{g}} \times\cfrac{1000\ut{g}}{1\ut{kg}} \times\cfrac{6.02214076\times10^{23}\ut{EA}}{1\ut{mol}}\\ &=\cfrac{3.30\cdot1000\cdot6.02214076\times10^{23}..

$$ \begin{cases} 1 \ut{tuffet} &= 2\ut{pecks}\\ &=0.50\ut{Imperial~bushel}\\ 1\ut{Imperial~bushel}&=36.3687\ut{L} \end{cases} $$ $$ 5.60\ut{tuffet}=?$$ $$\ab{a}$$ $$ \text{in} \ut{peck} ? $$ $$ \begin{aligned} 5.60\ut{tuffet}&=5.60\ut{tuffet}\times\frac{2\ut{pecks}}{1 \ut{tuffet}}\\ &=11.2\ut{pecks} \end{aligned} $$ $$\ab{b}$$ $$ \text{in} \ut{Imperial~bushel} ? $$ $$ \begin{aligned} 5.60\ut{tuf..

$$\ab{a}$$ $$ 1\ut{shake} = 10^{-8}\ut{s} $$ $$ \title{shake in second}$$ $$ \begin{aligned} 1\ut{s} &= 1\ut{s} \times \frac{1\ut{shake}}{10^{-8}\ut{s}}\\ &=10^8 \ut{shake} \end{aligned} $$ $$ \title{second in year}$$ $$ \begin{aligned} 1\ut{year} &= 1\ut{year}\times\frac{365\ut{day}}{1\ut{year}}\times\frac{24\ut{h}}{1\ut{day}}\times\frac{3600\ut{s}}{1\ut{h}}\\ &=365\times24\times3600\ut{s}\\ &=..

$$ \begin{cases} \Delta x_1 &= 23\ut{cm}\\ \Delta y_1 &= 19\ut{cm}\\ \Delta x_2 &= 28\ut{cm}\\ \Sigma \Delta y &= 9.50\ut{m}\\ \end{cases} $$ $$ \Sigma \Delta x_2 - \Sigma \Delta x_1=? $$ $$ \begin{aligned} N &= \frac{\Sigma \Delta y}{\Delta y_1} \\ &= \frac{9.50\ut{m}}{19\ut{cm}}\\ &= 50 \end{aligned} $$ $$ \begin{aligned} \Ans&=\Sigma \Delta x_2 - \Sigma \Delta x_1\\ &= N \cdot \Delta x_2 - N ..

$$ \begin{cases} 100 \ut{acre} &\le 1\ut{hide}\le120 \ut{acre}\\ 1\ut{acre} &= 4047\ut{m^2}\\ 1\ut{wapantake} &= 100\ut{hide}\\ 1\ut{barn} &= 1\times10 ^{-28}\ut{m^2}\\ \end{cases} $$ $$ 43\ut{wapentake}:8.0\ut{barn}=? $$ $$ \begin{aligned} &\min\(\frac{43\ut{wapantake}}{8.0\ut{barn}}\)\\ &= \frac{43\ut{wapantake}}{8.0\ut{barn}} \times\frac{100\ut{hide}}{1\ut{wapantake}}\times\frac{100\ut{acre}}..

$$\begin{cases} \text{StartPoint}=O&=(0,0)\ut{m}\\ \text{1stTower}=A&=(23,15)\ut{m}\\ \text{2ndTower}=B&=(23+\frac{23}{2},15)\ut{m}\\ \text{Goal}=C&=(R,0)\ut{m}\\ \end{cases}$$ $$\begin{cases} v_0 &= 26.5\ut{m/s}\\ \theta_0 &= 53\degree\\ \end{cases}$$ $$\begin{cases} t_0&=\text{Start Time}\\ t_1&=\text{Over 1stTower Time}\\ t_2&=\text{Over 2ndTower Time}\\ t_3&=\text{Goal Time}\\ \end{cases}$$ ..