91影视

Using the power loss formula P=I虏R and given I = 50 A and R=0.15 惟/km: Power loss in Copper cable per km = \((50)^2 \times 0.15 = 375000 \mathrm{W}\) For a distance of 500 km, total power loss in Copper cable = \(375000 \times 500 = 1.875 \times 10^8 \mathrm{W}\) Similarly, the power loss in Aluminum cable per km = \(375000 \mathrm{W}\) For a distance of 500 km, total power loss in Aluminum cable = \(375000 \times 500 = 1.875 \times 10^8 \mathrm{W}\)

Using the formula R = 蟻L/A, solve for A, the cross-sectional area (A) of the cable. We are given R = 0.15 惟/km, and the resistivities 蟻 (Copper) = 1.72e-8 惟m and 蟻 (Aluminum) = 2.82e-8 惟m. For Copper: \(0.15 = (1.72\times 10^{-8})(1000)/A\) Solving for A, we get A = \(1.72\times 10^{-5} \mathrm{m}^2\) Now, we can find the diameter (d) using the formula \(A = \pi(d/2)^2\). Solving for d, we get d (Copper) = 0.00467 m or 4.67 mm. For Aluminum: \(0.15 = (2.82\times 10^{-8})(1000)/A\) Solving for A, we get A = \(2.82\times 10^{-5} \mathrm{m}^2\) Now, we can find the diameter (d) using the formula \(A = \pi(d/2)^2\). Solving for d, we get d (Aluminum) = 0.00599 m or 5.99 mm.

To find the mass per meter, we first find the volume (V) of a 1 m length cable by using the formula V = A*L. For Copper: Volume (V) = \(1.72\times 10^{-5} \mathrm{m}^2 * 1 \mathrm{m} = 1.72\times 10^{-5} \mathrm{m}^3\) Now using mass density, mass per meter = density * V; Mass per meter (Copper) = (8920 kg/m鲁) * (\(1.72\times 10^{-5} \mathrm{m}^3\)) = 0.1534 kg/m For Aluminum: Volume (V) = \(2.82\times 10^{-5} \mathrm{m}^2 * 1 \mathrm{m} = 2.82\times 10^{-5} \mathrm{m}^3\) Now using mass density, mass per meter = density * V; Mass per meter (Aluminum) = (2702 kg/m鲁) * (\(2.82\times 10^{-5} \mathrm{m}^3\)) = 0.0762 kg/m Summary: a) Total power loss in both Copper and Aluminum cable: \(1.875 \times 10^8 \mathrm{W}\) b) Necessary cable diameter: Copper = 4.67 mm, Aluminum = 5.99 mm. c) Mass per meter of the cable: Copper = 0.1534 kg/m, Aluminum = 0.0762 kg/m.