Đáp án:
\(\begin{array}{l}
1.{R_1} = 0,16{R_2}\\
2.{S_2} = 6,{59.10^{ - 6}}{m^2}
\end{array}\)
Giải thích các bước giải:
Câu 1:
Ta có:
\(\begin{array}{l}
\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\rho \dfrac{l}{{{S_1}}}}}{{\rho \dfrac{l}{{{S_2}}}}} = \dfrac{{{S_2}}}{{{S_1}}} = \dfrac{{\pi \dfrac{{d_2^2}}{4}}}{{\pi \dfrac{{d_1^2}}{4}}} = \dfrac{{d_2^2}}{{d_1^2}} = \dfrac{{{2^2}}}{{{5^2}}} = 0,16\\
\Rightarrow {R_1} = 0,16{R_2}
\end{array}\)
Câu 2:
Ta có:
\(\begin{array}{l}
{R_1} = {R_2}\\
\Rightarrow {\rho _1}\dfrac{l}{{{S_1}}} = {\rho _2}\dfrac{l}{{{S_2}}}\\
\Rightarrow \dfrac{{1,{{7.10}^{ - 8}}}}{{{{4.10}^{ - 6}}}} = \dfrac{{2,{{8.10}^{ - 8}}}}{{{S_2}}}\\
\Rightarrow {S_2} = 6,{59.10^{ - 6}}{m^2}
\end{array}\)