\(d_v=\dfrac{P}{V}\Rightarrow V=\dfrac{P}{d_v}=\dfrac{10m}{d_v}=\dfrac{10.2}{78000}=\dfrac{1}{3900}\left(m^3\right)\)

\(F_A=d_{nc}.V=10000.\dfrac{1}{3900}\approx2,56\left(N\right)\)

\(R=\dfrac{U}{I}=\dfrac{220}{2}=110\left(\Omega\right)\)

\(R=\rho\dfrac{l}{S}\Rightarrow S=\dfrac{\rho.l}{R}=\dfrac{0,4.10^{-6}.5,5}{110}=2.10^{-8}\left(m^2\right)\)

\(P=I^2.R\) chứ k phải \(P=I.R\) ạ

Bài 1:

\(U_{max}=I_{max}.R_{max}=2,5.50=125\left(V\right)\)

Bài 2:

\(P=I^2.R\Rightarrow R=\dfrac{P}{I^2}=\dfrac{1100}{5^2}=44\left(\Omega\right)\)

\(R=\rho\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{\rho}=\dfrac{44.0,5.10^{-6}}{1,1.10^{-6}}=20\left(m\right)\)

\(250cm^2=0,025m^2\)

a) \(p=\dfrac{F}{S}\Rightarrow F=p.S=10000.0,025=250\left(N\right)\)

b) \(F=P=10m\Rightarrow m=\dfrac{F}{10}=\dfrac{250}{10}=25\left(kg\right)\)

\(40cm=0,4m\)

a) \(p_1=d.h_1=10000.1,5=15000\left(Pa\right)\)

b) \(p_2=d.h_2=10000.\left(1,5-0,4\right)=11000\left(Pa\right)\)

Gọi số tiền 3 lớp 7A,7B,7C góp lần lượt là a,b,c(nghìn đồng)(a,b,c∈N*)

Áp dụng t/c dtsbn:

\(\dfrac{a}{8}=\dfrac{b}{9}=\dfrac{c}{10}=\dfrac{c-b}{10-9}=\dfrac{150}{1}=150\)

\(\Rightarrow\left\{{}\begin{matrix}a=150.8=1200\\b=150.9=1350\\c=150.10=1500\end{matrix}\right.\)(nhận)

Vậy...

Gọi số HS khối 6 là x(HS)(x∈N*,\(320\le x\le400\))

Theo đề bài ta có: \(\left(x-5\right)BC\left(10,12,15\right)=\left\{60;120;180;240;360;480;...\right\}\)

\(\Rightarrow x\in\left\{65;125;185;245;365;485;....\right\}\)

Kết hợp ĐKXĐ => x=365

Vậy....

a) ĐKXĐ: \(x\ge-\dfrac{1}{2}\)

\(pt\Leftrightarrow\sqrt{\left(x+1\right)^2}=2x+1\Leftrightarrow\left|x+1\right|=2x+1\)

\(\Leftrightarrow x+1=2x+1\left(do.x\ge-\dfrac{1}{2}\right)\Leftrightarrow x=0\left(tm\right)\)

b) \(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=-2\\3x=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

Cân nặng: \(\left(7.109,2\right):14=54,6\left(kg\right)\)