\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{20.30}{20+30}=12\left(\Omega\right)\)

\(U=U_1=U_2=5V\)

\(\left\{{}\begin{matrix}I=\dfrac{U}{R_{tđ}}=\dfrac{5}{12}\left(A\right)\\I_1=\dfrac{U_1}{R_1}=\dfrac{5}{20}=\dfrac{1}{4}\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{5}{30}=\dfrac{1}{6}\left(A\right)\end{matrix}\right.\)

\(v_{tb}=\dfrac{S_1+S_2}{t_1+t_2}\Rightarrow10=\dfrac{S_{tổng}}{\dfrac{S_{tổng}}{15}+\dfrac{S_{tổng}}{v_2}}=\dfrac{S_{tổng}}{S_{tổng}\left(\dfrac{1}{15}+\dfrac{1}{v_2}\right)}=\dfrac{1}{\dfrac{1}{15}+\dfrac{1}{v_2}}\)

\(\Rightarrow\dfrac{1}{v_2}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{1}{30}\Rightarrow v_2=30\left(\dfrac{km}{h}\right)\)

\(P=U.I\Rightarrow I=\dfrac{P}{U}=\dfrac{1000}{220}=\dfrac{50}{11}\left(A\right)\)

\(R=\dfrac{U}{I}=\dfrac{200}{\dfrac{50}{11}}=44\left(\Omega\right)\)

\(P=\dfrac{U^2}{R}=\dfrac{200^2}{44}=\dfrac{10000}{11}\left(W\right)\)

\(Q_{tỏa}=A=P.t=\dfrac{10000}{11}.4.60.60\approx13090909\left(J\right)\approx\dfrac{40}{11}\left(kWh\right)\)

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{4.12}{4+12}=3\left(\Omega\right)\)

\(U=U_1=U_2=I.R_{tđ}=2.3=6\left(V\right)\)

\(\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{6}{4}=\dfrac{3}{2}\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{6}{12}=\dfrac{1}{2}\left(A\right)\end{matrix}\right.\)

\(8cm=0,08m;5cm=0,05m\)

\(\left\{{}\begin{matrix}p_1=d.h_1=10000.0,08=800\left(Pa\right)\\p_2=d.h_2=10000.0,05=500\left(Pa\right)\end{matrix}\right.\)

a) \(R_{tđ}=R_1+R_2=15+25=40\left(\Omega\right)\)

b) \(I=I_1=I_2=\dfrac{U}{R_{tđ}}=\dfrac{9}{40}\left(A\right)\)

c) \(\left\{{}\begin{matrix}U_1=I_1.R_1=\dfrac{9}{40}.15=\dfrac{27}{8}\left(V\right)\\U_2=I_2.R_2=\dfrac{9}{40}.25=\dfrac{45}{8}\left(V\right)\end{matrix}\right.\)

Tham khảm trên mạng thì ghi thêm "tham khảo" vào nha bạn

Làm rồi nhé bạn

Quãng đường mà âm thanh di chuyển từ khi phát ra âm đến khi thu được âm phản xạ:

\(S'=2S=2.680=1360\left(m\right)\)

Thời gian tiếng âm của vật đó đến bức tường và phản xạ lại:

\(t=\dfrac{S'}{v}=\dfrac{1360}{340}=4\left(s\right)\)