2020-09-02T11:03:22-04:00
an electric power station is connected to a factory by two wires 2.5 km apart of the potential difference between two terminals of the wire at the station is 240 v and at the factory 220 v and the factory uses a current intensity 80 a
1
2020-09-03T14:10:22-0400
Δ V = 240 − 220 = 20 V R = 20 80 = 0.25 Ω R l = 0.25 5000 = 5.0 ⋅ 1 0 − 5 Ω ⋅ m − 1 \Delta V=240-220=20\ V\\R=\frac{20}{80}=0.25\Omega
\\\frac{R}{l}=\frac{0.25}{5000}=5.0\cdot 10^{-5}\Omega\cdot m^{-1} Δ V = 240 − 220 = 20 V R = 80 20 = 0.25Ω l R = 5000 0.25 = 5.0 ⋅ 1 0 − 5 Ω ⋅ m − 1
A = ρ l R = 1.57 ⋅ 1 0 − 8 5.0 ⋅ 1 0 − 5 = 3.14 ⋅ 1 0 − 4 m 2 A=\frac{\rho l}{R}=\frac{1.57\cdot 10^{-8}}{5.0\cdot 10^{-5}}=3.14\cdot 10^{-4}\ m^2 A = R ρl = 5.0 ⋅ 1 0 − 5 1.57 ⋅ 1 0 − 8 = 3.14 ⋅ 1 0 − 4 m 2
r = 3.14 ⋅ 1 0 − 4 π = 0.01 m r=\frac{\sqrt{3.14\cdot 10^{-4}}}{\pi}=0.01\ m r = π 3.14 ⋅ 1 0 − 4 = 0.01 m
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