솔루션피아

$$ \begin{cases} m_H &= 1.0078\ut{u}\\ 1\ut{u} \times1\ut{mol}&= 1\ut{g}\\ N_A&=6.02214076\times10^{23} \ut{EA/mol}\\ M_H&=3.30\ut{kg}\\ \end{cases} $$ $$ \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}}..