L=1.0mS=2×10−7m2R=0.1Ωρ−?solution:R=ρ×LS⇒ρ=R×SL;ρ=0.1×2×10−71=0.2×10−7Ω×m.Answer: the electrical conductivityof the wire equals 0.2×10−7Ω×m.L=1.0m\\S=2\times10^{-7}m^2\\R=0.1\Omega\\\rho-?\\solution:\\R=\rho\times \frac{L}{S} \Rightarrow\rho=\frac{R\times S}{L};\\\rho=\frac{0.1\times2\times10^{-7} }{1}=0.2\times 10^{-7}\Omega\times m.\\Answer:\;the \;electrical\; conductivity\\ of \;the\; wire\;equals\;0.2\times 10^{-7}\Omega\times m.L=1.0mS=2×10−7m2R=0.1Ωρ−?solution:R=ρ×SL⇒ρ=LR×S;ρ=10.1×2×10−7=0.2×10−7Ω×m.Answer:theelectricalconductivityofthewireequals0.2×10−7Ω×m.
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