솔루션피아

$$\begin{cases} \vec A = A_x\hat i\\ B=6.0\\ \vec A + \vec B = \vec r\\ \vec r = R_y\hat j\\ r=2A\\ \end{cases} $$ $$ \vec B = (6\cos\theta_B)\hat i + (6\sin\theta_B)\hat j$$ $$ \Ans = A = \abs{A_x}$$ $$ \begin{aligned} \vec r =& \vec A + \vec B\\ =& (A_x+6\cos\theta_B)\hat i + (6\sin\theta_B)\hat j \end{aligned} $$ $$\therefore \begin{cases} A_x+6\cos\theta_B=0\\ 6\sin\theta_B=2A_x \end{cases} ..

$$ b=\sqrt{3^2+4^2}=5$$ $$ \theta_a=\tan^{-1}\(\frac{-1}{1}\)=-\frac{\pi}{4}$$ $$ \begin{aligned} \therefore \vec r&:\(5,-\frac{\pi}{4}\)\\ &=\bra{5\cos(-\frac{\pi}{4})}\hat i+\bra{5\sin(-\frac{\pi}{4})}\hat j\\ &=\(\frac{5}{\sqrt2}\)\hat i + \(-\frac{5}{\sqrt2}\)\hat j\\ \end{aligned} $$

$$ \begin{aligned} \vec E:& (6.00 \ut{m}, + 0.900\ut{rad})\\ \vec F:& (5.00 \ut{m}, -75.0^\circ)\\ \vec G:& (4.00 \ut{m},+ 1.20\ut{rad})\\ \vec H:& (6.00 \ut{m}, -210^\circ)\\ \end{aligned} $$ $$ \begin{aligned} \vec E &= (6\cos0.9)\hat i + (6\sin0.9)\hat j\\ \vec F &= (5\cos-75^\circ)\hat i + (5\sin-75^\circ)\hat j\\ &=\(\frac{5}{2} \sqrt{2-\sqrt{3}}\)\hat i + \(-\frac{5}{2} \sqrt{2+\sqrt{3}}\)..

$$\begin{cases} \vec a = 30.0\hat i + 40.0\hat j\\ \vec b = b_x\hat i + (-70.0)\hat j\\ \vec c = (-20.0)\hat i + c_y\hat j\\ \vec d = (-80.0)\hat i + (-70.0)\hat j\\ \vec r = (-140)\hat i + (-20.0)\hat j\\ \end{cases} $$ (a)$b_x=?$ $$ 30.0+b_x+(-20.0)+(-80.0)=(-140)$$ $$ b_x=-70.0 $$ (b)$c_y=?$ $$ 40.0+(-70.0)+c_y+(-70.0)=(-20.0)$$ $$ c=80.0$$ (c)$r=?$ $$ \begin{aligned} r&=\sqrt{(-140)^2+(-20.0..

$$\begin{cases} \vec r = 4.8\hat j\\ \vec a:(7.8,50^\circ)\\ \vec a+\vec b=\vec r\\ \end{cases} $$ $$ a= (7.8\cos50^\circ)\hat i + (7.8\sin50^\circ)\hat j$$ $$ \begin{aligned} \vec b &= \vec r - \vec a\\ &= (-7.8\cos50^\circ)\hat i + (4.8-7.8\sin50^\circ)\hat j\\ \end{aligned} $$ (a)$b=?$ $$ \begin{aligned} b&=\sqrt{(-7.8\cos50^\circ)^2+(4.8-7.8\sin50^\circ)^2}\\ &=\frac{3}{5} \sqrt{233-208 \cos..

$$\begin{cases} \vec p_1 = \hat i + 2\hat j\\ \vec p_2 = -2\hat i + 1\hat j\\ \vec p_3 = -1\hat i + 2\hat j\\ \end{cases} $$ $$\Sigma\vec p = -2\hat i + 5\hat j$$ (a)$\abs{\Sigma\vec p}=?$ $$ \begin{aligned} \abs{\Sigma\vec p}&=\sqrt{(-2)^2+(5)^2}\\ &=\sqrt{29}\\ &\approx 5.39\ut{m}\\ \end{aligned} $$ (b)$\theta_{\Sigma\vec p}=?$ $$ \begin{aligned} \theta_{\Sigma\vec p}&=\tan^{-1}\(\frac{5}{-2}\..

$$\begin{cases} \vec A + \vec B= \vec C\\ \vec A:(12.0,40^\circ)\\ \vec C:(16.0,200^\circ)\\ \end{cases} $$ $$\vec A = (12\cos40^\circ)\hat i + (12\sin40^\circ)\hat j $$ $$\vec C = (16\cos200^\circ)\hat i + (16\sin200^\circ)\hat j$$ $$ \begin{aligned} \vec B &= \vec C - \vec A\\ &=(16\cos200^\circ-12\cos40^\circ)\hat i + (16\sin200^\circ-12\sin40^\circ)\hat j\\ \end{aligned} $$ (a)B=? $$ \begin{..

$$\begin{cases} \vec a:(50,30^\circ)\\ \vec b:(50,195^\circ)\\ \vec c:(50,315^\circ)\\ \end{cases} $$ $$ \begin{aligned} \vec a &= (50\cos30^\circ)\hat i + (50\sin30^\circ)\hat j\\ &=\(25\sqrt3\)\hat i + 25\hat j \end{aligned} $$ $$ \begin{aligned} \vec b &= (50\cos195^\circ)\hat i + (50\sin195^\circ)\hat j\\ &=\(-25 \sqrt{2+\sqrt{3}}\)\hat i + \(25\sqrt{2-\sqrt{3}}\)\hat j \end{aligned} $$ $$ \..

$$\begin{cases} \vec a=(3.0\ut{m})\hat i + (4.0\ut{m})\hat j\\ \vec b=(5 .0 \ut{m})\hat i + (-2.0 \ut{m})\hat j\\ \vec a+\vec b=\vec r \end{cases} $$ (a) $\vec r =?$ $$\vec r = \(8.0\ut{m}\)\hat i + \(2.0\ut{m}\)\hat j$$ (b)$\abs{\vec r}=?$ $$ \begin{aligned} r&=\sqrt{8^2+2^2}\\ &=2\sqrt{17}\ut{m}\\ &\approx 8.2\ut{m} \end{aligned} $$ (c)$\theta_r=?$ $$\theta_r=\tan^{-1}\(\frac{2}{8}\)=\frac{11}..

$$\begin{cases} \vec a:(10.0, 30^\circ)\\ \vec b:(10.0, 30+105^\circ)\\ \end{cases} $$ $$ \begin{aligned} \vec a &= (10\cos30^\circ)\hat i + (10\sin30^\circ)\hat j\\ &=5\sqrt{3}\hat i + 5\hat j\\ \end{aligned} $$ $$ \begin{aligned} \vec b &= (10\cos135^\circ)\hat i + (10\sin135^\circ)\hat j\\ &=-5\sqrt2\hat i + 5\sqrt2\hat j\\ \end{aligned} $$ $$\vec r = \(5\sqrt3-5\sqrt2\)\hat i + \(5+5\sqrt2\)..