솔루션피아

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

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

$$\begin{cases} \phi &= 36.0\degree\\ d &= 0.900\ut{m}\\ v_0 &= 3.56\ut{m/s} \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$H = d\sin\phi$$ $$\begin{aligned} 2a\Delta y&=v_y^2-v_{0y}^2,\\ 2(-g)(d\sin\phi)&=(0)^2-(v_0\sin\theta_0)^2\\ \sin\theta_0&=\frac{\sqrt{2dg\sin\phi} }{v_0}\\ \end{aligned}$$ $$\begin{aligned} \theta_0&=\sin^{-1}\frac{\sqrt{2dg\s..

$$\begin{cases} H &= 2.160\ut{m}\\ v_0 &= 15.00\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ -H&=v_0t\sin\theta-\frac{1}{2}gt^2\\ -(2.160)&=(15.00)t\sin\theta-\frac{1}{2}gt^2\\ \end{aligned}$$ $$ \therefore t=\frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)$$ $$\be..

$$\begin{cases} d_{1.50} &= -25.0\ut{m}\\ \theta_{1.50} &= 60.0\degree\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $h_0=?$ $$\begin{aligned} d&=v_x t\\ v_x&=\frac{25}{1.5}\\ &= \frac{50}{3} \end{aligned}$$ $$\begin{aligned} \vec v_{1.5}&=(-v_{x1.5},-v_{y1.5})\\ &=(-v_x,-\sqrt{3}v_x)\\ &=\(-\frac{50}{3},-\frac{50}{\sqrt{3}}\)\\ \end{aligned}$$ ..

$$\begin{cases} t_0 : \text{Start}, 0\ut{sec}\\ \end{cases}$$ $$\begin{cases} v_0 &= 31\ut{m/s}\\ v_{2.5} &= 19\ut{m/s}\\ v_5 &= 31\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $\Delta x_{0\rarr\text{Ground}}=?$ $$\begin{aligned} \abs{\vec v}_{\min}&=\vec v_{\text{HighMax}},\\ t_{2.5}&=t_{\text{HighMax}}\\ \end{aligned}$$ $$\begin{align..

$$\begin{cases} y_0 &= 12\ut{m}\\ v_x &= 2.50\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1=0.900\ut{sec}\\ t_2 : \text{Finish}\\ \end{cases}$$ (a) $\Delta x_1=?$ $$\begin{aligned} \Delta x &= v_xt,\\ &=2.25\ut{m} \end{aligned}$$ (b) $y_1=?$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{..