솔루션피아

(풀이자주:(c),(d)의 경우 모든t에 대한 v를 지칭하는지 t=3일때를 지칭하는지 애매합니다. 하여, 두가지 모두를 풀었습니다.) $$\vec r = 3.00t\i - 4.00t^2\j + 2.00\k$$ (a)$\vec v(t)=?$ $$ \begin{aligned} \vec v(t)=& \dt\vec r\\ =&\dt\(3.00t\i - 4.00t^2\j + 2.00\k\)\\ =&\dt\(3.00t\)\i+\dt\(-4.00t^2\)\j+\dt\(2.00\)\k\\ =&3.00\i-8.00t\j \end{aligned} $$ (b)$\vec v(3.00)=?$ $$ \begin{aligned} \vec v(3.00)=&3.00\i-8.00(3.00)\j\\ =&3.00\i-24.0\j \end{..

$$\begin{cases} t_{0\to1}=40.0\ut{min}\\ t_{1\to2}=20.0\ut{min}\\ t_{2\to3}=50.0\ut{min}\\ \end{cases} $$ $$\begin{cases} \vec v_{0\to1}=60.0\ut{km/h}\i\\ \vec v_{1\to2}:(60.0\ut{km/h},90^\circ-50^\circ)\\ \vec v_{2\to3}=-60.0\ut{km/h}\i=-\vec v_{0\to1}\\ \end{cases} $$ $$ \begin{aligned} \vec r_{0\to1}=&v_{0\to1}t_{0\to1}\\ =&60.0\ut{km/h}\i\cdot40.0\ut{min}\cdot\frac{1\ut{h}}{60\ut{min}}\\ =&4..

$$\begin{cases} t_0=15\ut{min}\\ t_1=30\ut{min}\\ t_2=t_1+30\ut{min}\\ t_3=t_2+60\ut{min}\\ \end{cases} $$ (a)(b)$\Delta \vec r_{0\to1}=?$ $$ \begin{aligned} \Delta\vec r_{0\to1} =& \vec r_1-\vec r_0\\ =&(12,\pi)-(12,\frac{\pi}{2})\\ =&(-12\j)-(12\i)\\ =&-12\i-12\j\\ \end{aligned} $$ (a)$\abs{\Delta r_{0\to1}}=?$ $$ \begin{aligned} \abs{\Delta r_{0\to1}}=&\sqrt{(-12)^2+(-12)^2}\\ =&12\sqrt2\\ \a..

$$ \vec r_0 =?$$ $$\vec r_0 + \Delta\vec r = \vec r$$ $$ \begin{aligned} \vec r_0 =& \vec r - \Delta\vec r\\ =&(4.0\j — 5.0\k) - (2.0\i - 4.0\j + 8.0\k) \\ =&-2.0\i+8.0\j-13\k\\ \end{aligned} $$

$$\begin{cases} p_x = -5.0\ut{m}\\ p_y = 8.0\ut{m}\\ p_z = 0.0\ut{m}\\ \end{cases} $$ (a)$\vec p=?$ $$ \vec p = (-5.0\ut{m})\i+8.0\ut{m}\j$$ (b)$p=?$ $$ \begin{aligned} p =& \sqrt{(-5)^2+8^2}\\ =& \sqrt{89}\\ \approx&9.4\ut{m} \end{aligned} $$ (c)$\phi_{p|\i}=?$ $$ \begin{aligned} \phi_{\vec p|\i}=&\cos^{-1}\(\frac{\vec p \cdot \i}{\abs{p}\abs{\i}}\)\\ =&\cos^{-1}\(\frac{-5}{\sqrt{89}}\)\\ \appr..

(a)$r=?$ $$ \begin{aligned} r=&\sqrt{6^2+(-4)^2+3^2}\\ =&\sqrt{61}\ut{m}\\ \approx&7.81\ut{m}\\ \end{aligned} $$ (b)

$$\begin{cases} \vec F = q\vec v \times \vec B\\ q = 3\\ \vec v = 2.0\i + 4.0\j +6. 0\k\\ \vec F = 4.0\i - 20\j + 1 2\k\\ \vec B = B_x\i+B_y\j+B_z\k\\ B_x=B_y\\ \end{cases} $$ $$\vec F = q\vec v \times \vec B$$ $$ \begin{aligned} \vec F=&q(v_x\i+v_y\j+v_z\k)\times (B_x\i+B_y\j+B_z\k)\\ =&q\begin{vmatrix} \i & \j & \k\\ v_x&v_y&v_z\\ B_x&B_y&B_z\\ \end{vmatrix}\\ =&q\begin{vmatrix} v_y&v_z\\ B_y&..

$$\begin{cases} \vec a:(3.00\ut{m},0)\\ \vec b:(4.00\ut{m},30.0^\circ)\\ \vec c:(10.0\ut{m},\frac{\pi}{2}+30.0^\circ)\\ \end{cases} $$ (a)(b) $$ \vec a = 3\i,$$ (a)$a_x=?$ $$a_x = 3.00$$ (b)$a_y=?$ $$a_y = 0$$ (c)(d) $$ \begin{aligned} \vec b =& 4\cos30.0^\circ\i+4\sin30.0^\circ\j\\ =&2\sqrt3\i+2\j\\ \end{aligned} $$ (c)$b_x=?$ $$ b_x=2\sqrt3 \approx3.46$$ (d)$b_y=?$ $$ b_y=2.00 $$ (e)(f) $$ \be..

$$\begin{cases} \vec d_1 = (4.0 \ut{m})\i + (5.0 \ut{m})\j\\ \vec d_2= (-3.0 \ut{m})\i + (4.0 \ut{m})\j\\ \end{cases} $$ (a) $\vec d_1 \times \vec d_2=?$ $$ \begin{aligned} \vec d_1 \times \vec d_2=&\begin{vmatrix} \i & \j & \k\\ 4&5&0\\ -3&4&0\\ \end{vmatrix}\\ =&\begin{vmatrix} 5&0\\ 4&0\\ \end{vmatrix}\i+ \begin{vmatrix} 0&4\\ 0&-3\\ \end{vmatrix}\j+ \begin{vmatrix} 4&5\\ -3&4\\ \end{vmatrix}..

$$\begin{cases} \vec a =4.0\i+4.0\j-4.0\k\\ \vec b=3.0\i+2.0\j-4.0\k\\ \end{cases} $$ $$ \begin{aligned} \vec a \cdot \vec b=&a_xb_x+a_yb_y+a_zb_z\\ =&4.0\cdot3.0+4.0\cdot2.0+(-4.0)\cdot(-4.0)\\ =&36 \end{aligned} $$ $$a=\sqrt{4^2+4^2+(-4)^2}=4\sqrt3$$ $$b=\sqrt{3^2+2^2+(-4)^2}=\sqrt{29}$$ $$ \begin{aligned} \phi_{ab}=&\cos^{-1}\(\frac{\vec a \cdot \vec b}{\abs{a}\abs{b}}\)\\ =&\cos^{-1}\(\frac{..