11판/1. 측정

1-22 할리데이 11판 솔루션 일반물리학

짱세디럭스 2022. 10. 4. 04:48
{1[AU]=92.9×106[mi]1[ly]=186000[mi]1[sec]1[year] \begin{cases} 1\ut{AU}=&92.9\times10^6\ut{mi}\\ 1\ut{ly}=&\cfrac{186000\ut{mi}}{1\ut{sec}}\cdot1\ut{year} \end{cases} (a)\ab{a} L=rθL=r\theta 1[AU]=1[pc]1[]1[°]3600[]π[rad]180[°]=π648000[pc]4.84813681109536×106[pc]4.85×106[pc] \begin{aligned} 1\ut{AU}=&1\ut{pc}\cdot1\ut{''}\cdot\frac{1\ut{\degree}}{3600\ut{''}}\cdot\frac{\pi\ut{rad}}{180\ut{\degree}}\\ =&\frac{\pi}{648000}\ut{pc}\\ \approx&4.84813681109536\times10^{-6}\ut{pc}\\ \approx&4.85\times10^{-6}\ut{pc}\\ \end{aligned} (b)\ab{b} Ans=1[AU]=1[AU]×92.9×106[mi]1[AU]×1[ly]1[sec]186000[mi]1[year]×1[h]3600[sec]×1[day]24[h]×1[year]365.25[day]=92.9106186000360024365.25[ly]=92958697136[ly]0.000015827007300662847[ly]1.58×105[ly] \begin{aligned} \Ans=&1\ut{AU}\\ =&1\ut{AU}\times\frac{92.9\times10^6\ut{mi}}{1\ut{AU}}\times\frac{1\ut{ly}\cdot1\ut{sec}}{186000\ut{mi}\cdot1\ut{year}}\\ &\times\frac{1\ut{h}}{3600\ut{sec}}\times\frac{1\ut{day}}{24\ut{h}}\times\frac{1\ut{year}}{365.25\ut{day}}\\ =&\frac{92.9\cdot10^6}{186000\cdot3600\cdot24\cdot365.25}\ut{ly}\\ =&\frac{929}{58697136}\ut{ly}\\ \approx&0.000015827007300662847\ut{ly}\\ \approx&1.58\times10^{-5}\ut{ly} \end{aligned}