Ta có: \(\dfrac{T'}{T}=\dfrac{2\pi\sqrt{\dfrac{l'}{g}}}{2\pi\sqrt{\dfrac{l}{g}}}=\sqrt{\dfrac{l'}{l}}=\dfrac{1,2T}{T}=1,2\Rightarrow l'=1,44l\)

\(\Rightarrow\dfrac{l'-l}{l}100\%=\dfrac{1,44l-l}{l}100\%=44\%\)

Chọn D

Ta có: \(T=2\pi\sqrt{\dfrac{l}{g}}\Rightarrow l=\dfrac{T\sqrt{g}}{2\pi}\)

Theo đề: \(\left\{{}\begin{matrix}l_3=l_1+l_2\\l_4=l_1-l_2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}l_1=\dfrac{l_3+l_4}{2}\\l_2=\dfrac{l_3-l_4}{2}\end{matrix}\right.\)

Ta có: \(l_1=\dfrac{\sqrt{g}\left(T_3+T_4\right)}{4\pi}=0,8\)

\(l_2=\dfrac{g\left(T_3^2-T_4^2\right)}{8\pi^2}=0,64\)

Ta có: \(\left\{{}\begin{matrix}T=2\pi\sqrt{\dfrac{l}{g}}=\dfrac{t}{12}\\T'=2\pi\sqrt{\dfrac{l-16}{g}}=\dfrac{t}{20}\end{matrix}\right.\) 

\(\Rightarrow\sqrt{\dfrac{l-16}{l}}=\dfrac{12}{20}\)

\(\Rightarrow l=25cm\)

Chọn B

\(R_{tđ}=R_1+R_2=25+40=65\Omega\\ I=\dfrac{U}{R_{tđ}}=\dfrac{26}{65}=0,4A\\ R_1ntR_2\Rightarrow I=I_1=I_2=0,4A\\ U_1=R_1.I=25.0,4=10V\\ U_2=U-U_1=26-10=16V\)

 1l = 1dm³

⇒ 10l = 10dm³

1l = Dm³

^{vậy 10l = 10dm3}

\(4x^2-2x^3=0\\ \Rightarrow2x^2\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x^2=0\\2-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(a)R_{tđ}=R_1+\dfrac{R_2.R_3}{R_2+R_3}=15+\dfrac{30.30}{30+30}=30\Omega\\ b)I=\dfrac{U}{R_{tđ}}=\dfrac{12}{30}=0,4A\\ R_1ntR_{23}\\ \Rightarrow I=I_1=I_{23}=0,4A\\ U_1=R_1.I=15.0,4=6V\\ U_{23}=12-6=6V\\ R_2//R_3\\ \Rightarrow U_{23}=U_2=U_3=6V\\ I_2=\dfrac{U_2}{R_2}=\dfrac{6}{30}=0,2A\\ I_3=I_{23}-I_2=0,4-0,2=0,2A\)

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