Physics in senior two (Law of resistance) 1. For a section of iron wire with uniform thickness, the diameter of cross section is D and the resistance is R. after drawing it into a uniform wire with diameter of D / 10, what is its resistance? 2. Cut a 64 ohm uniform resistance wire into equal length n segments, and then connect it in parallel. If the resistance side is 1 ohm, what is n equal to How should I think about these two questions?

Physics in senior two (Law of resistance) 1. For a section of iron wire with uniform thickness, the diameter of cross section is D and the resistance is R. after drawing it into a uniform wire with diameter of D / 10, what is its resistance? 2. Cut a 64 ohm uniform resistance wire into equal length n segments, and then connect it in parallel. If the resistance side is 1 ohm, what is n equal to How should I think about these two questions?

When the wire is in the original state, s = π R ^ 2 = π (D / 2) ^ 2 = (π d ^ 2) / 4R = ρ L / [(π d ^ 2) / 4] = 4L ρ / (π d ^ 2) after the wire is elongated, s' = π [(D / 10) / 2] ^ 2 = π d ^ 2 / 400 should be that the total volume of the wire remains unchanged and the cross-sectional area changes

If a uniform resistance wire with a resistance of 64 Ω is cut into equal length n segments and connected in parallel, the resistance becomes 1 Ω, then n is equal to ()
A. 32B. 24C. 12D. 8

According to the resistance law r = ρ ls, after a uniform resistance wire with 64 Ω is cut into N equal length segments, the resistance of each segment is r = 64n. From the characteristics of parallel circuit, we can get 1R = 164n × n, and the solution is n = 8. Therefore, D is correct

The law of physical resistance in the second year of senior high school
An electrical appliance is powered by a power supply 50m away, the current on the line is 5a, and the voltage loss on the line is required to be no more than 2.4V. What is the minimum cross-sectional area of the copper conductor? The resistivity of copper is 1.7 * 10 to the minus 8th power. M. thank you!

Because the required voltage is not more than 2.4V and the current is 5a, the maximum resistance on the wire can be obtained from Ohm's law as 2.4g5 = 0.49ohm
According to the resistance formula of conductor (resistance = material characteristics multiplied by length divided by cross-sectional area), it can be obtained that:
Cross sectional area s = 1.7 * 10 minus 8 times 50 divided by 0.49 = 1.77 * 10 minus 6 square meters