11판/1. 측정

1-29 할리데이 11판 솔루션 일반물리학

짱세디럭스 2022. 10. 7. 23:51
{V=40[cm]×40[cm]×26[cm] \begin{cases} V&=40\ut{cm}\times40\ut{cm}\times26\ut{cm} \end{cases} (a)\ab{a} V=41600[cm3]=41.6[L]=41.6[L]16[bottle]11.356[L]=1664002839[bottle]58.61218738992603[bottle]=(220+16+2+17382839)[bottle]=2[nebuchadnezzar]+1[balthazar]+1[magnum]+17382839[bottle] \begin{aligned} V&=41600\ut{cm^3}\\ &=41.6\ut{L}\\ &=41.6\ut{L}\cdot\frac{16\ut{bottle}}{11.356\ut{L}}\\ &=\frac{166400}{2839}\ut{bottle} \approx58.61218738992603\ut{bottle}\\ &=\(2\cdot20+16+2+\frac{1738}{2839}\)\ut{bottle}\\ &=2\ut{nebuchadnezzar}+1\ut{balthazar}+1\ut{magnum}+\frac{1738}{2839}\ut{bottle} \end{aligned} (b)\ab{b} Ans=17382839[bottle]0.6121873899260303[bottle]0.61[bottle] \begin{aligned} \Ans&=\frac{1738}{2839}\ut{bottle}\\ &\approx0.6121873899260303\ut{bottle}\\ &\approx0.61\ut{bottle}\\ \end{aligned} (c)\ab{c} Ans=17382839[bottle]=17382839[bottle]11.356[L]16[bottle]=8692000[L]=0.4345[L]0.43[L] \begin{aligned} \Ans&=\frac{1738}{2839}\ut{bottle}\\ &=\frac{1738}{2839}\ut{bottle}\cdot\frac{11.356\ut{L}}{16\ut{bottle}}\\ &=\frac{869}{2000}\ut{L}\\ &=0.4345\ut{L}\\ &\approx0.43\ut{L} \end{aligned}