\(P_Đ=\dfrac{U_Đ^2}{R_Đ}\Rightarrow R_Đ=\dfrac{U_Đ^2}{P_Đ}=\dfrac{12^2}{24}=48\left(\Omega\right)\)

\(R_{tđ}=R_Đ+R_2=48+12=60\left(\Omega\right)\)

\(P_m=\dfrac{U^2}{R_{tđ}}=\dfrac{30^2}{60}=15\left(W\right)\)

\(I_Đ=I_m=\dfrac{U_m}{R_{tđ}}=\dfrac{30}{60}=0,5\left(A\right)\)

\(I_{đm}=\dfrac{12}{24}=0,5\left(A\right)\Rightarrow I_Đ=I_{đm}\Rightarrow\) đèn sáng bình thường

Lực đẩy Ác si mét tác dụng lên vật:

\(F_A=P_{ngoài}-P_{trong}=2,2-1,9=0,3\left(N\right)\)

Thể tích của vật:

\(F_A=d.V\Rightarrow V=\dfrac{F_A}{d}=\dfrac{0,3}{10000}=3.10^{-5}\left(m^3\right)\)

\(n_{N_2}=\dfrac{280}{14.2}=10\left(mol\right)\)

\(V_{N_2}=10.22,4=224\left(l\right)\Rightarrow D\)

\(t_1=\dfrac{S_1}{v_1}=\dfrac{S}{2v_1}=\dfrac{S}{24}\left(h\right)\)

\(v_{tb}=\dfrac{S_1+S_2}{t_1+t_2}=\dfrac{S}{\dfrac{S}{24}+\dfrac{S}{2v_2}}=\dfrac{S}{S\left(\dfrac{1}{24}+\dfrac{1}{2v_2}\right)}=8\Rightarrow\dfrac{1}{24}+\dfrac{1}{2v_2}=\dfrac{1}{8}\Rightarrow\dfrac{1}{2v_2}=\dfrac{1}{12}\Rightarrow v_2=6\left(\dfrac{km}{h}\right)\Rightarrow A\)

\(400cm^2=0,04m^2\)

\(p=\dfrac{F}{S}=\dfrac{P}{S}=\dfrac{100}{0,04}=2500\left(Pa\right)\)

Câu 1:

a) SGK

b) \(Q_{tỏa}=A=I^2Rt=3^2.70.15.60=567000\left(J\right)\)

Câu 2:

c) \(R=\rho\dfrac{l}{S}\Rightarrow S=\dfrac{\rho.l}{R}=\dfrac{1,7.10^{-8}.20}{25}=1,36.10^{-8}\left(m^2\right)\)

Nhiệt lượng tỏa ra trong 1h:

\(Q_{tỏa}=A=P.t=1000.1.60.60=3600000\left(J\right)=1\left(kWh\right)\)

Điện năng tiêu thụ của bếp trong 1 tháng:

\(A=P.t=1000.30.2.60.60=216000000\left(J\right)=60\left(kWh\right)\)

Tiền điện phải trả: \(1500.60=90000\left(đ\right)\)

a) \(A=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)

\(minA=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\)

\(minB=3\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=3\end{matrix}\right.\)

c) \(C=2x-x^2=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)

\(maxC=1\Leftrightarrow x=1\)

\(x^2-2x+3=\left(x^2-2x+1\right)+2=\left(x-1\right)^2+2\ge2\forall x\in R\)

\(\Rightarrow2x\left(x-2\right)+3\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(2x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)