Giúp mik câu này với

Câu 6.1: C

Ta có: 

\(\begin{array}{l}
\omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{2} = \pi \left( {rad/s} \right)\\
 \Rightarrow \sqrt {\dfrac{g}{l}}  = \pi  \Rightarrow g = {\pi ^2}l = 9,86\left( {m/{s^2}} \right)
\end{array}\)

Câu 6.2: B

Ta có: 

\(\begin{array}{l}
\omega  = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{1} = 2\pi \left( {rad/s} \right)\\
 \Rightarrow \sqrt {\dfrac{g}{l}}  = 2\pi  \Rightarrow l = \dfrac{g}{{4{\pi ^2}}} = 0,248\left( m \right) = 24,8\left( {cm} \right)
\end{array}\)

Câu 6.3: A

Ta có: 

\(\begin{array}{l}
\omega  = \sqrt {\dfrac{g}{l}}  = \sqrt {\dfrac{{{\pi ^2}}}{1}}  = \pi \left( {rad/s} \right)\\
T = \dfrac{{2\pi }}{\omega } = \dfrac{{2\pi }}{\pi } = 2\left( s \right)
\end{array}\)

Câu 6.4: B

Ta có:

\(\begin{array}{l}
\dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{\omega _2}}}{{{\omega _1}}} = \dfrac{{\sqrt {\dfrac{g}{{{l_2}}}} }}{{\sqrt {\dfrac{g}{{{l_1}}}} }} = \sqrt {\dfrac{{{l_1}}}{{{l_2}}}}  \Rightarrow 2 = \sqrt {\dfrac{{{l_1}}}{{{l_2}}}} \\
 \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = 4 \Rightarrow {l_2} = \dfrac{{{l_1}}}{4}
\end{array}\)

Câu 7.1: B

Môi trường càng nhớt thì dao động tắt dần càng nhanh.