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Question:

Physics resistance help?

A sheet of copper is 5 mm thick and hassurface dimensions of 6 cm × 21 cm .If the long edges are joined to form atube 21 cm in length, what is the resistance between the ends? The mass densityof copper is 8920 kg/m3and its resistivity is1.7 × 10?8 Ω · m .Answer in units of ΩWhat mass of copper is required to manufacture a 1200 m-long spool of copper cable witha total resistance of 5.6 Ω?Answer in units of kgI got the first part but can‘t get the second. The answer to the first question is 1.19E-5 Ω. Please help with the second!

Answer:

copper is used the position you want to diminish losses, ie, have a low resistance, as in homestead cord. tungsten is used in incandescent mild bulbs for a style of motives, because it has a larger resistance, so it receives hotter, and it resists breakdown on the extreme temperatures reached by utilising a mild bulb filament. in case you tried to apply copper in a filament, it may could be a lot thinner or longer, and could dissipate immediately, if it did not melt first. .
Resistance of a wire in Ω R ρL/A ρ is resistivity of the material in Ω-m L is length in meters A is cross-sectional area in m? resistivity Cu 17.2e-9 Ω-m Cross-sectional area is 6 cm x 0.5 cm 3 cm? or 3e-4 m? R (17.2e-9)(0.21) / (3e-4) 0.000012 ohms or 12 ?Ω 2) R ρL/A 5.6 A (17.2e-9)(1200) / 5.6 3.69e-6 m? cylinder V πr?h Cross-sectional area x length V (3.69e-6)(1200) 0.00442 m? density copper 8960 kg/m? 8960 kg/m? x 0.00442 m? 39.6 kg
copper is used the position you want to diminish losses, ie, have a low resistance, as in homestead cord. tungsten is used in incandescent mild bulbs for a style of motives, because it has a larger resistance, so it receives hotter, and it resists breakdown on the extreme temperatures reached by utilising a mild bulb filament. in case you tried to apply copper in a filament, it may could be a lot thinner or longer, and could dissipate immediately, if it did not melt first. .
Resistance of a wire in Ω R ρL/A ρ is resistivity of the material in Ω-m L is length in meters A is cross-sectional area in m? resistivity Cu 17.2e-9 Ω-m Cross-sectional area is 6 cm x 0.5 cm 3 cm? or 3e-4 m? R (17.2e-9)(0.21) / (3e-4) 0.000012 ohms or 12 ?Ω 2) R ρL/A 5.6 A (17.2e-9)(1200) / 5.6 3.69e-6 m? cylinder V πr?h Cross-sectional area x length V (3.69e-6)(1200) 0.00442 m? density copper 8960 kg/m? 8960 kg/m? x 0.00442 m? 39.6 kg

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