\(A=P.h=5000.10=50000J=50kJ\)

\(\dfrac{2002}{2003}< 1;\dfrac{14}{13}>1\\ \Rightarrow\dfrac{2002}{2003}< \dfrac{14}{13}\)

\(1km=1000m\\ 1\left(\dfrac{m}{s}\right)=3,6\left(\dfrac{km}{h}\right)\\ 13\left(\dfrac{km}{h}\right)=\dfrac{65}{18}\left(\dfrac{m}{s}\right)\\ 5,5\left(\dfrac{km}{h}\right)=\dfrac{55}{36}\left(\dfrac{m}{s}\right)\\ 4\left(\dfrac{m}{s}\right)=\dfrac{1}{9}\left(\dfrac{cm}{h}\right)\)

đề là  \(v_2=\dfrac{v_1}{3}\) :))??

Nửa qđ AB là \(s'=\dfrac{s_{AB}}{2}=1,8km\) 

Tgian đi nửa quãng đầu

\(t_1=\dfrac{s'}{v_1}=\dfrac{1,8}{v_1}\)

Tgian đi nửa quãng sau

\(t_2=\dfrac{s'}{v_2}=\dfrac{1,8}{\dfrac{v_1}{3}}=\dfrac{5,4}{v_1}\)

mà theo đề bài ta có

\(\dfrac{1,8}{v_1}+\dfrac{5,4}{v_1}=\dfrac{3}{10}\left(18p=\dfrac{3}{10}\right)\\ \Leftrightarrow1,8\left(\dfrac{1}{v_1}+\dfrac{3}{v_1}\right)=\dfrac{3}{10}\\ \Leftrightarrow\dfrac{4}{v_1}=\dfrac{1}{6}\\ \Rightarrow v_1=24\left(\dfrac{km}{h}\right)\\ \Rightarrow v_2=\dfrac{24}{3}=8\left(\dfrac{km}{h}\right)\)

Cho m₂ = 2l

Ta có phương trình cân bằng nhiệt 

\(Q_{toả}=Q_{thu}\\ \Leftrightarrow m_3c_{Al}\left(t_2-t_{cb}\right)=\left(m_1c_{Cu}+m_2c_{H_2O}\right)\left(t_{cb}-t_1\right)\\\Leftrightarrow 0,5.880\left(100-t_{cb}\right)=\left(0,3.380+2.4200\right)\left(t_{cb}-30\right)\\ \Leftrightarrow44000-440t_{cb}=8664t_{cb}-259920\\ \Leftrightarrow303920=9124t_{cb}\\ \Rightarrow t_{cb}=33,3^o\)