Cân bằng nhiệt: \(Q_{thu}=Q_{toa}=2\cdot460\cdot\left(100-50\right)=46000\left(J\right)\)

Ta có: \(Q_{thu}=mc\left(t_2-t_1\right)\)

\(\Leftrightarrow46000=2\cdot4200\cdot\left(50-t_1\right)\)

\(\Leftrightarrow46000=420000-8400t\)

\(\Leftrightarrow t=44,5^0C\)

Ta có:

\(A=Ph=mgh=600\cdot10\cdot4,5=27000\left(J\right)\)

\(\Rightarrow P=\dfrac{A}{t}=\dfrac{27000}{12}=2250\) (W)

Ta có: \(\left\{{}\begin{matrix}Q_1=0,3\cdot880\cdot\left(100-20\right)=21120\left(J\right)\\Q_2=1,2\cdot4200\cdot\left(100-20\right)=403200\left(J\right)\end{matrix}\right.\)

\(\Rightarrow Q=Q_1+Q_2=21120+403200=424320\left(J\right)\)

a.

\(Q=mc\left(t_2-t_1\right)=5\cdot880\cdot\left(80-30\right)=220000\left(J\right)\)

b.

Ta có: \(Q=Q_1+Q_2\)

\(\Leftrightarrow220000=m\cdot4200\cdot\left(100-20\right)+0,3\cdot880\cdot\left(100-20\right)\)

\(\Leftrightarrow220000=336000m+21120\)

\(\Leftrightarrow m\approx0,6\left(kg\right)\)

Ta có: \(F_{ht}=mg+T\)

\(\Leftrightarrow T^2=\left(mg\right)^2+F_{ht}^2\left(Pytago\right)\)

\(\Leftrightarrow T^2=F_{ht}^2+\left(mg\right)^2=\left(\dfrac{mv^2}{R}\right)+\left(mg\right)^2\)

\(\Rightarrow m=\dfrac{T}{\sqrt{g^2+\dfrac{v^4}{R^2}}}\approx0,8\left(kg\right)\)

a.

Ta có: \(n_2< n_1\left(50< 1000\right)\)

Vậy có thể dùng máy biến thế trên để giảm hiệu điện thế.

b. 

Ta có: \(\dfrac{U_1}{U_2}=\dfrac{n_1}{n_2}\Rightarrow U_2=\dfrac{U_1n_2}{n_1}=\dfrac{220\cdot50}{1000}=11V\)

Ta có: \(\dfrac{v_1}{v_2}=\dfrac{r_1\omega_1}{r_2\omega_2}\)

Mà: \(\dfrac{\omega_1}{\omega_2}=\dfrac{T_1}{T_2}=12\Rightarrow\dfrac{r_1}{r_2}=\dfrac{12}{12}\)

\(\Rightarrow r_2=\dfrac{12}{12}\cdot10=10\left(cm\right)\)

_{Bạn coi lại đề thử ha}

Áp dụng quy tắc momen lực:

\(\dfrac{F_1}{F_2}=\dfrac{d_2}{d_1}\Leftrightarrow\dfrac{300}{100}=\dfrac{d_2}{1}\Leftrightarrow d_2=3\left(m\right)\)

Độ dịch chuyển:

\(\theta=2\pi\dfrac{t}{T}=2\pi\dfrac{5\cdot60}{60\cdot60}=\dfrac{1}{6}\pi\left(rad\right)\)

Quãng đường đi:

\(s=\theta r=\dfrac{1}{6}\pi\cdot0,12\approx0,06\left(m\right)\)

Ta có:

\(R'=Rcos60^0=6400\cdot cos60^0=3200\left(km\right)=3200000\left(m\right)\)

Tốc độ góc: \(\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{24\cdot60\cdot60}\approx7,3\cdot10^{-5}\left(\dfrac{m}{s}\right)\)

Tốc độ của A: \(v=\omega R'=7,3\cdot10^{-5}\cdot3200000\approx232,7\left(\dfrac{m}{s}\right)\)