\(\Rightarrow P = \dfrac{A}{t} = \dfrac{F.h}{t} = \dfrac{50000.800}{2.60} = 333334W\)

\(A=F.s= 10500. 850= 8925000(J) \)

\(P=\dfrac{A}{t}= \dfrac{8925000}{90} = 99166,67W ≈99,167(kW) \)

\(1.C ( 8m^2 = 80000 cm^2) \)

\(2.A (10m^2 = 1000 dm^2)\)

\(3. 32km = 3 200 000 cm \)

Quãng đường đó dài :  \(3 200 000 : 80000 = 40(cm)\)

\(2R+O_2 \rightarrow2RO \)

\(m_{O_2} = 6-3,6= 2,4(g) \)

\(n_{O_2} = \dfrac{2,4}{32}= 00,075 (mol) \)

\(Theo PT : n_R=2n_{O_2} = 0,5(mol) \)

\(\Rightarrow M_R=\dfrac{3,6}{0,15} = 24(g/mol) \)

\(\rightarrow R:Mg ( Margie )\)

\(1 ) 2M+O_2\rightarrow 2MO n_M=n_{MO}\Leftrightarrow \dfrac{16}{M_M}=\dfrac{20}{m_M+16} \Rightarrow m_m = 64(g/mol) \rightarrow M : Cu \)

\(2) 2R+3Cl_2\rightarrow 2RCl_3 n_R=nn_{RCl_3}\Leftrightarrow \dfrac{16,2}{M_R}=\dfrac{80,1}{M_R+35,5.3}\Rightarrow M_R = 27(g/mol)\rightarrow R:Al \)

Giải : \(|2x-1|=x \Leftrightarrow \begin{cases} 2x-1=x\\ 2x-1=-x \end{cases} \begin{cases} x=1\\ x=\dfrac{1}{3} \end{cases} \)

\(\text{=> }y.\left(\dfrac{11}{12}+\dfrac{3}{4}\right)=\dfrac{10}{11}=>y.\dfrac{5}{3}=\dfrac{10}{11}=>y=\dfrac{10}{11}:\dfrac{5}{3}=\dfrac{6}{11}\)

Công thức : 

\(\dfrac{-3}{11}+\dfrac{11}{8}-\dfrac{3}{8}+\dfrac{8}{11}=\left(-\dfrac{3}{11}+\dfrac{8}{11}\right)+\left(\dfrac{11}{8}-\dfrac{3}{8}\right)=\dfrac{5}{11}+\dfrac{8}{8}=\dfrac{16}{11}\)