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2 years ago

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A 14.0 gauge copper wire of diameter 1.628 mmcarries a current of 14.0 mA. Part A: What is the potential difference across a 2.1

5 mlength of the wire?
Part B: What would the potential difference in part A be if the wire were silver instead of copper, but all else was the same?

1 answer:

FrozenT [24]2 years ago

7 0

Answer:

for copper potential difference is 2.4871 × $10^{-4}$ V

for silver potential difference is 2.1256 × $10^{-4}$ V

Explanation:

given data

diameter D = 1.628 mm = 0.001628 m

current = 14.0 mA = 0.0140 A

length L = 2.15 m

to find out

potential difference

solution

we consider here

resistivity of silver = 1.47 × $10^{-8}$ ohm-m

resistivity of copper = 1.72 × $10^{-8}$ ohm-m

so we apply here resistance formula that is

Resistance = ρ × L / A ...............1

so here area A = πr² = π(0.001628 /2)² = 2.0816 × $10^{-6}$ m²

for copper Resistance = ρ × L / A

Resistance = 1.72 × $10^{-8}$ × 2.15 / 2.0816 × $10^{-6}$

Resistance = 1.7765 × $10^{-2}$ ohm

for silver Resistance = ρ × L / A

Resistance = 1.47 × $10^{-8}$ × 2.15 / 2.0816 × $10^{-6}$

Resistance = 1.5183 × $10^{-2}$ ohm

so

potential difference is calculated as

for copper potential difference = current × resistance

potential difference = 0.0140 ×1.7765 × $10^{-2}$

potential difference = 2.4871 × $10^{-4}$ V

for silver potential difference = current × resistance

potential difference = 0.0140 × 1.5183 × $10^{-2}$

potential difference = 2.1256 × $10^{-4}$ V

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Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

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cross section area of point 3 = 1.6 x 10^{2} m

g

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$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

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$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

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