10판/3. 벡터

3-21 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 7. 21:32

{a=30.0i^+40.0j^b=bxi^+(70.0)j^c=(20.0)i^+cyj^d=(80.0)i^+(70.0)j^r=(140)i^+(20.0)j^\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)bx=?b_x=? 30.0+bx+(20.0)+(80.0)=(140) 30.0+b_x+(-20.0)+(-80.0)=(-140) bx=70.0 b_x=-70.0
(b)cy=?c_y=? 40.0+(70.0)+cy+(70.0)=(20.0) 40.0+(-70.0)+c_y+(-70.0)=(-20.0) c=80.0 c=80.0
(c)r=?r=? r=(140)2+(20.0)2=10025.2[m] \begin{aligned} r&=\sqrt{(-140)^2+(-20.0)^2}\\ &=100\sqrt2\\ &\approx 5.2\ut{m}\\ \end{aligned}
(b)θb=?\theta_b=? θb=tan1(20.0140)3.00[rad] \begin{aligned} \theta_b&=\tan^{-1}\(\frac{-20.0}{-140}\)\\ &\approx -3.00\ut{rad} \end{aligned}