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3 years ago

Find the resistance at 50°c of copper wire 2mm in diameter and 3m long

1 answer:

8 0

0.0179 ohms for copper.

0.0184 ohms for annealed copper

Ď = R (A/l) where

Ď = electrical resistivity

R = electrical resistance of a uniform specimen

A = cross sectional area

l = length

Solve for R by multiplying both sides by l/A

R = Ď(l/A)

The cross section of the wire is pi * 1^2 mm = 3.14159 square mm = 3.14159e-6 square meters.

The length is 3 meters. So l/A = 3/3.14159e-6 = 9.5493e5

Ď for copper is 1.68e-8 so 1.68e-8 * 9.5493e5 = 1.60e-2 ohms at 20 C

But copper has a temperature coefficient (Î±) of 0.00386 per degree C.

So the resistance value needs to be adjusted based upon how far from 20 C the temperature is.

50 - 20 = 30 C

So 0.00386 * 30 = 0.1158 meaning that the actual resistance at 50 C will be 11.58% higher.

So 1.1158 * 0.016 = 0.0179 ohms.

If you're using annealed copper, the values for Ď and the temperature coefficient change.

Ď = 1.72e-8

Î± = 0.00393

Doing the math, you get

1.72e-8 * 9.5493e5 * (1 + 30 * 0.00393) = 0.0184 ohms

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4 years ago

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snow_lady [41]

Answer:

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b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

$m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}$ ..........(1)

$m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s$

Substituting the above values into equation (1) and make V2 the subject of the formula:

$0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\$

$V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}$ ..................(2)

Using the law of conservation of kinetic energy:

$0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\$

$2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}$ .......(3)

Substitute equation (2) into equation (3)

$2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8}) ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2} }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s$

Substituting $V_{1}$ into equation(2)

$V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s$

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3 years ago

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Answer:

C. Replacing one gas by another under the same conditions, has no effect on pressure.

Explanation:

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