3-14 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} R&=22.0\ut{cm}\\ \vec P_1&=0\\ \vec P_2&=\cfrac{2\pi R}{2}\i+2Rj \end{cases} $$ $$ \begin{aligned} \Delta \vec P &= \vec P_2\\ &= \pi R\i+2R\j\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} P&=\sqrt{(\pi R)^2+(2R)^2}\\ &=R\sqrt{\pi^2+4}\\ &=22\sqrt{\pi^2+4}\ut{cm}\\ &\approx 81.93221912121781\ut{cm}\\ &\approx 81.9\ut{cm} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_P&=\tan^{-1}\frac{2R}{\pi R}\\ &=\tan^{-1}\frac{2}{\pi}\\ &\approx 0.567\ut{rad} \end{aligned} $$