Đáp án:
C.162Ω
Giải thích các bước giải:
Điện trở của cuộn thứ hai là:$\begin{array}{l}\left\{ \begin{array}{l}{R_1} = \rho \dfrac{{{l_1}}}{{{S_1}}} = \rho \dfrac{{{l_1}}}{{\dfrac{{{d_1}^2}}{4}\pi }} \Rightarrow \rho = \dfrac{{{R_1}.\dfrac{{{d_1}^2}}{4}\pi }}{{{l_1}}}\\{R_2} = \rho \dfrac{{{l_2}}}{{{S_2}}} = \rho \dfrac{{{l_2}}}{{\dfrac{{{d_2}^2}}{4}\pi }} \Rightarrow \rho = \dfrac{{{R_2}.\dfrac{{{d_2}^2}}{4}\pi }}{{{l_2}}}\end{array} \right.\\ \Leftrightarrow \rho = \dfrac{{{R_1}.\dfrac{{{d_1}^2}}{4}\pi }}{{{l_1}}} = \dfrac{{{R_2}.\dfrac{{{d_2}^2}}{4}\pi }}{{{l_2}}}\\ \Leftrightarrow \dfrac{{{R_1}{d_1}^2}}{{\dfrac{V}{{{S_1}}}}} = \dfrac{{{R_2}{d_2}^2}}{{\dfrac{V}{{{S_2}}}}} \Leftrightarrow {R_1}.{d_1}^2.\dfrac{{{d_1}^2}}{4}\pi = {R_2}{d_2}^2.\dfrac{{{d_2}^2}}{4}\pi \\ \Leftrightarrow 2.0,{6^4} = {R_2}.0,{2^4} \Rightarrow {R_2} = 162\Omega \end{array}$
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Giải các tam giác vuông Giúp mik với ạ!
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Mọi người giúp em làm phần 1 với ạ Phần 2 không làm cũng không sao ạ Ai làm được phần 2 em sẽ cho câu trả lời hay nhất
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