Giải bài 4 trang 205 – Bài 42 - SGK môn Vật lý lớp 10 Nâng cao

$\begin{align} & {{p}_{1}}=8,{{0.10}^{4}}\text{ }\left( Pa \right) \\ & {{v}_{1}}=2\text{ }\left( m/s \right) \\ & {{S}_{2}}=\dfrac{{{S}_{1}}}{4}\Leftrightarrow \dfrac{{{S}_{1}}}{{{S}_{2}}}=4 \\ & {{v}_{2}}.{{S}_{2}}={{v}_{1}}.{{S}_{1}}\Rightarrow {{v}_{2}}=\dfrac{{{S}_{1}}}{{{S}_{2}}}{{v}_{1}}=4.2=8\text{ }\left( m/s \right) \\ & {{p}_{2}}={{p}_{1}}+\dfrac{S}{2}\left( v_{1}^{2}-v_{2}^{2} \right)=8,{{0.10}^{4}}+\dfrac{{{10}^{3}}}{2}\left( {{2}^{2}}-{{8}^{2}} \right)={{5.10}^{4}}\text{ }\left( Pa \right) \\ \end{align}$