11판/4. 2차원 운동과 3차원 운동

4-61 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 13. 22:00
{amax=23.0gg=9.80665[m/s2]v=0.00200cc=299792458[m/s] \begin{cases} a_{\max}&=23.0g\\ g&=9.80665\ut{m/s^2}\\ v&=0.00200c\\ c&=299792458\ut{m/s} \end{cases} (a)\ab{a} a=v2R,a=\frac{v^2}{R}, R=v2a=(0.00200c)223g=c25750000g1.5938699604448848×109[m]1.59×109[m] \begin{aligned} R&=\frac{v^2}{a}\\ &=\frac{(0.00200c)^2}{23g}\\ &=\frac{c^2}{5750000g}\\ &\approx 1.5938699604448848\times 10^9\ut{m}\\ &\approx 1.59\times 10^9\ut{m}\\ \end{aligned} (b)\ab{b} v=2πRT,v=\frac{2\pi R}{T}, Ans=T4=2πR4v=π2c25750000g10002c=πc23000g4175.630527795919[s]69.59384212993199[min]69.6[min] \begin{aligned} \Ans&=\frac{T}{4}\\ &=\frac{2\pi R}{4v}\\ &=\frac{\pi}{2}\cdot\frac{c^2}{5750000g}\cdot\frac{1000}{2c}\\ &=\frac{\pi c}{23000 g}\\ &\approx 4175.630527795919\ut{s}\\ &\approx 69.59384212993199\ut{min}\\ &\approx 69.6\ut{min}\\ \end{aligned}