A. -1/2
B. -1
C. 2
D. 1/2
B
Có:
\(\begin{array}{l}
T = 2\pi \sqrt {\frac{\ell }{g}} \Rightarrow T \sim \frac{1}{{\sqrt g }}\\
\frac{{{T_1}}}{{{T_0}}} = 5 \Rightarrow \sqrt {\frac{g}{{{g_1}}}} = 5 \Rightarrow {g_1} = \frac{g}{{25}}\\
\Rightarrow {g_1} = g - {g_{{\rm{l1}}}}\left( {\overrightarrow {{g_{{\rm{l1}}}}} \uparrow \downarrow \overrightarrow g } \right)\\
\Rightarrow {g_{{\rm{l1}}}} = \frac{{24g}}{{25}} = \frac{{\left| {{q_1}E} \right|}}{m};{q_1} < 0\\
\frac{{{T_2}}}{{{T_0}}} = \frac{5}{7} \Rightarrow \sqrt {\frac{g}{{{g_2}}}} = \frac{5}{7} \Rightarrow {g_2} = \frac{{49}}{{25}}g\\
\Rightarrow {g_2} = g + {g_{{\rm{l2}}}}\left( {\overrightarrow {{g_{{\rm{l2}}}}} \uparrow \uparrow \overrightarrow g } \right)\\
\Rightarrow {g_{{\rm{l2}}}} = \frac{{24g}}{{25}} = \frac{{\left| {{q_2}E} \right|}}{m};{q_2} > 0
\end{array}\)
mà trái dấu \( \Rightarrow \frac{{{q_1}}}{{{q_2}}} = - 1\)
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