11판/3. 벡터

3-44 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 25. 18:37
{r=(5.25i^+4.90j^+3.00k^)[m] \begin{cases} \vec r &= \(5.25\i+4.90\j+3.00\k\)\ut{m} \end{cases} (a)\ab{a} r=5.252+4.92+32=12024229[m]7.782833674183202[m]7.78[m] \begin{aligned} r &= \sqrt{5.25^2+4.9^2+3^2}\\ &=\frac{1}{20}\sqrt{24229}\ut{m}\\ &\approx 7.782833674183202\ut{m}\\ &\approx 7.78\ut{m}\\ \end{aligned} (b)\ab{b} l<r ?l\lt r ~? [Impossible]\title{Impossible} (c)\ab{c} l>r ?l\gt r ~? [Possible]\title{Possible} (d)\ab{d} l=r ?l = r ~? [Possible]\title{Possible} (e)\ab{e} r=(5.25i^+4.90j^+3.00k^)[m]\vec r = \(5.25\i+4.90\j+3.00\k\)\ut{m} (f)\ab{f} l1=5.252+(4.90+3)2=3598920l2=32+(5.25+4.90)2=4480920l3=4.902+(3+5.25)2=3682920 \begin{aligned} l_1 &= \sqrt{5.25^2+(4.90+3)^2}\\ &=\frac{\sqrt{35989}}{20}\\ l_2 &= \sqrt{3^2+(5.25+4.90)^2}\\ &=\frac{\sqrt{44809}}{20}\\ l_3 &= \sqrt{4.90^2+(3+5.25)^2}\\ &=\frac{\sqrt{36829}}{20}\\ \end{aligned} lmin=l1=35989209.48538349251099[m]9.49[m] \begin{aligned} l_{\min}&=l_1 \\ &= \frac{\sqrt{35989}}{20}\\ &\approx 9.48538349251099\ut{m}\\ &\approx 9.49\ut{m}\\ \end{aligned}