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31 tháng 10 2021
Câu 6:
\(U=I.R=10.1,5=15\left(V\right)\)
Câu 7:
\(P=U.I\Rightarrow U=\dfrac{P}{I}=\dfrac{10}{0,5}=20\left(V\right)\)
Câu 8:
\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{10.30}{10+30}=7,5\left(\Omega\right)\)
31 tháng 10 2021
Câu 3:
c: Xét ΔAHB vuông tại H có HE là đường cao
nên \(AE\cdot AB=AH^2\left(1\right)\)
Xét ΔAHC vuông tại H có HF là đường cao
nên \(AF\cdot AC=AH^2\left(2\right)\)
Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot AC\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
25 tháng 12 2021
\(Cau.23:\\ N=\left(A_1+T_1+G_1+X_1\right).2=\left(100+200=300+400\right).2=2000\left(Nu\right)\\ L=\dfrac{N}{2}.3,4=\dfrac{2000}{2}.3,4=3400\left(A^o\right)\\ Chon.C\)
19 tháng 2 2023
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Bài 1:
b. \(U=IR=I\left(R1+R2\right)=0,4\left(15+2\right)=6,8\left(V\right)\)
c. \(I=I1=I2=\dfrac{U'}{R}=\dfrac{60}{15+2}=\dfrac{60}{17}\simeq3,5\left(A\right)\left(R1ntR2\right)\)
Bài 2:
\(5400kJ=1500\left(Wh\right)\)
a. \(A=Pt\Rightarrow P=\dfrac{A}{t}=\dfrac{1500}{1}=1500\left(W\right)\)
b. \(P=UI\Rightarrow I=\dfrac{P}{U}=\dfrac{1500}{220}=\dfrac{75}{11}\simeq6,82\left(A\right)\)
Bài 3:
a. \(R=p\dfrac{l}{S}=1,1.10^{-6}\dfrac{3}{0,05.10^{-6}}=66\left(\Omega\right)\)
b. \(P=UI=U\left(\dfrac{U}{R}\right)=220.\left(\dfrac{220}{66}\right)=733,33\left(W\right)\)
c. \(A=Pt=733,33.\left(\dfrac{30}{60}\right)=366,665\left(Wh\right)=0,366665\left(kWh\right)=1319994\left(J\right)\)
Bài 4:
a. \(S=\pi\dfrac{d^2}{4}=\pi\dfrac{1^2}{4}=0,785\left(mm^2\right)\)
\(\Rightarrow R=p\dfrac{l}{S}=5,5.10^{-8}\dfrac{10}{0,785.10^{-6}}=\dfrac{110}{157}\simeq0,7\left(\Omega\right)\)
b. \(A=Pt=UIt=U\left(\dfrac{U}{R}\right)t=70\left(\dfrac{70}{0,7}\right).\dfrac{1}{3}=2333,33\left(Wh\right)=2,33333\left(kWh\right)\simeq8400000\left(J\right)\)