11판/4. 2차원 운동과 3차원 운동

4-63 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 14. 15:57
{aR=6.00×1014[m/s2]R=18.0[cm] \begin{cases} a_R &= 6.00\times10^{14}\ut{m/s^2}\\ R &= 18.0\ut{cm}\\ \end{cases} (a)\ab{a} aR=v2R,a_R=\frac{v^2}{R}, v=RaR=63×106[m/s]1.039230484541326×107[m/s]1.04×107[m/s] \begin{aligned} v&=\sqrt{Ra_R}\\ &=6\sqrt3\times10^6\ut{m/s}\\ &\approx 1.039230484541326\times10^7\ut{m/s}\\ &\approx 1.04\times10^7\ut{m/s}\\ \end{aligned} (b)\ab{b} v=2πRT,v=\frac{2\pi R}{T}, T=2πRv=2πRRaR=3π5×107[s]1.088279618540531×107[s]1.09×107[s] \begin{aligned} T&=\frac{2\pi R}{v}\\ &= \frac{2\pi R}{\sqrt{Ra_R}}\\ &=\frac{\sqrt3 \pi}{5}\times10^{-7}\ut{s}\\ &\approx 1.088279618540531\times10^{-7}\ut{s}\\ &\approx 1.09\times10^{-7}\ut{s}\\ \end{aligned}