Đáp án:
câu 1
$\left\{ \begin{array}{l}
x = - 3\sqrt 3 \left( {cm} \right)\\
v = 12\pi \left( {cm/s} \right)\\
a = 831,38\left( {cm/{s^2}} \right)
\end{array} \right.$
câu 2
$\begin{array}{l}
a.{Z_L} = 200\left( \Omega \right)\\
{Z_C} = 100\left( \Omega \right)\\
b.{Z_{AN}} = 20\sqrt {29} \left( \Omega \right)\\
c.NB:{C_1},{R_1}\\
{C_1} = {\kern 1pt} \frac{{{{10}^{ - 4}}}}{\pi }\left( F \right)\\
{R_1} = 70\left( \Omega \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
x = 6\cos \left( {4\pi t + \frac{\pi }{6}} \right)\\
\Rightarrow v = - 24\pi \sin \left( {4\pi t + \frac{\pi }{6}} \right)\\
t = 0,25\\
\Rightarrow \left\{ \begin{array}{l}
x = 6\cos \left( {4\pi .0,25 + \frac{\pi }{6}} \right) = - 3\sqrt 3 \left( {cm} \right)\\
v = - 24\pi \sin \left( {4\pi .0,25 + \frac{\pi }{6}} \right) = 12\pi \left( {cm/s} \right)\\
a = - {\omega ^2}x = - {\left( {4\pi } \right)^2}.\left( { - 3\sqrt 3 } \right) = 831,38\left( {cm/{s^2}} \right)
\end{array} \right.\\
\end{array}$
CÂU 2
$\begin{array}{l}
a.{Z_L} = L\omega = \frac{2}{\pi }.100\pi = 200\left( \Omega \right)\\
{Z_C} = \frac{1}{{C\omega }} = \frac{1}{{\frac{{{{10}^{ - 4}}}}{\pi }.100\pi }} = 100\left( \Omega \right)\\
b.{Z_{AN}} = \sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} = \sqrt {{{40}^2} + {{\left( {200 - 100} \right)}^2}} = 20\sqrt {29} \left( \Omega \right)
\end{array}$
c.
mạch AN có tính cảm kháng mà $\cos \varphi = 1$
nên $NB:{C_1},{R_1}$
$\begin{array}{l}
\Rightarrow {Z_L} = {Z_C} + {Z_{{C_1}}}\\
\Rightarrow 200 = 100 + \frac{1}{{\omega {C_1}}}\\
\Rightarrow \frac{1}{{100\pi {C_1}}} = 100\\
\Rightarrow {C_1} = {\kern 1pt} \frac{{{{10}^{ - 4}}}}{\pi }\left( F \right)\\
{Z_{AB}} = \frac{{{U_{AB}}}}{I} = \frac{{220}}{{\sqrt 2 .\sqrt 2 }} = 110\\
{Z_{AB}} = {R_1} + R\\
\Rightarrow 110 = 40 + {R_1}\\
\Rightarrow {R_1} = 70\left( \Omega \right)
\end{array}$
