\(U_{C\left(max\right)}\Rightarrow I_{max}\) => trong mạch xảy ra cộng hưởng \(\Rightarrow Z_L=Z_C\)
\(\Rightarrow\omega L=\dfrac{1}{\omega C}\Rightarrow\omega=\dfrac{1}{\sqrt{LC}}\)
\(I_0=\omega.C.U_0=\dfrac{1}{\sqrt{LC}}.C.U_0=U_0.\sqrt{\dfrac{C}{L}}=5.\sqrt{\dfrac{2000.10^{-12}}{20.10^{-6}}}=0,05\left(A\right)\)
\(\Rightarrow P=\dfrac{I_0^2.R}{2}=\dfrac{0,05^2.2}{2}=2,5.10^{-3}\left(W\right)\)
Mới sáng sớm mà đã gặp bài khó ăn vậy rồi :v
a/ \(\omega=\gamma t=\left(0,5+0,1t\right)t=0,5t+0,1t^2\left(rad/s\right)\)
\(v=\omega.R=\left(0,5t+0,1t^2\right).2=t+0,2t^2\left(m/s\right)\)
b/ Gia tốc tiếp tuyến: \(a_t=\gamma.R=\left(0,5+0,1t\right).2=1+0,2t\left(m/s^2\right)\)
Gia tốc hướng tâm: \(a_{ht}=\omega^2.R=\left(0,5t+0,1t^2\right)^2.2=\left(0,25t^2+0,05t^3+0,01t^4\right).2=0,5t^2+0,1t^3+0,02t^4\left(m/s^2\right)\)
\(\Rightarrow a=\sqrt{a_t^2+a_{ht}^2}=\sqrt{\left(1+0,2t\right)^2+\left(0,5t^2+0,1t^3+0,02t^4\right)^2}\left(m/s^2\right)\)
c/ \(v=\omega R=8+0,2.64=20,8\left(m/s\right)\)
\(d\varphi=\omega.dt\Rightarrow\varphi=\int_0^8d\varphi=\int_0^8\gamma.dt\)
\(=\int_0^8\left(0,5+0,1t\right).dt=0,5.8+0,1.8.8=10,4\left(m/s\right)\)
\(v=\omega R=10,4.2=20,8\left(m/s\right)\)
\(\varphi=\omega_0t+\frac{1}{2}.\gamma t^2=\frac{1}{2}.\left(0,5+0,8\right).8^2=41,6\left(rad\right)\)
Bài 2:
\(Cauchy:A_1+A_2\ge2\sqrt{A_1A_2}\Leftrightarrow10\ge2\sqrt{A_1A_2}\Rightarrow A_1A_2\le25\)
Have: \(A_1A_2=\sqrt{x_1^2+\frac{v_1^2}{\omega^2}}.\sqrt{x_2^2+\frac{v_2^2}{\omega^2}}=\sqrt{\left(x_1^2+\frac{v_1^2}{\omega^2}\right)\left(x_2^2+\frac{v_2^2}{\omega^2}\right)}\)
\(Bunhiacopxki:\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)\ge\left(a_1b_1+a_2b_2\right)^2\)
\(\Rightarrow\left(x_1^2+\frac{v_1^2}{\omega^2}\right)\left(x_2^2+\frac{v_2^2}{\omega^2}\right)\ge\left(x_1.\frac{v_2}{\omega}+x_2.\frac{v_1}{\omega}\right)^2\)
\(\Rightarrow A_1A_2\ge\left(x_1.\frac{v_2}{\omega}+x_2\frac{v_1}{\omega}\right)\Leftrightarrow25\ge\left(\frac{x_1.v_2+x_2v_1}{\omega}\right)\)
\(\Leftrightarrow x_1v_2+x_2v_1\le25\omega\Leftrightarrow9\le25\omega\)
\(\Rightarrow\omega\ge\frac{9}{25}=0,36\left(rad/s\right)\)
This exercise is hardest :<
Bài 3:
Lại đạo hàm :<
Have: \(\left(\frac{x}{v}\right)'=\frac{x'v-v'x}{v^2}\)
Have also: \(\left\{{}\begin{matrix}v=x'\\v'=a=-\omega^2x\end{matrix}\right.\)
\(\Rightarrow\left(\frac{x}{v}\right)'=\frac{v^2+\omega^2x^2}{v^2}=1+\frac{x^2}{\frac{v^2}{\omega^2}}=1+\frac{x^2}{A^2-x^2}\)
Đạo hàm 2 vế theo thời gian biểu thức: \(\frac{x_1}{v_1}+\frac{x_2}{v_2}=\frac{x_3}{v_3}\) :
\(\left(1+\frac{x_1^2}{A_1^2-x_1^2}\right)+\left(1+\frac{x_2^2}{A_2^2-x_2^2}\right)=1+\frac{x_3^2}{A_3^2-x_3^2}\)
\(\Rightarrow1+\frac{x_1^2}{A_1^2-x_1^2}+\frac{x_2^2}{A_2^2-x_2^2}=\frac{x_3^2}{A_3^2-x_3^2}\Rightarrow\left|x_3\right|=3,4\left(cm\right)\)
Trọng lượng của thùng hàng là
\(P=10m=5000\) N
Độ cao mà nó được nâng lên là
\(h=\frac{A}{P}=7,5\) m
