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garik1379 [7]
3 years ago
15

Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to

the clothing; a current from the 12 V battery of the motorcycle passes through the wires, warming them. One recipe for a vest calls for 10 m of 0.25-mm-diameter copper wire. How much power will this vest provide to warm the wearer?
Physics
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

P=42.075W

Explanation:

The power provided by a resistor (wire in this case) is given by:

P=\frac{V^2}{R}.

The resistance of a wire is given by:

R=\frac{\rho L}{A}

Where for the resistivity the one of the copper should be used: \rho=1.68\times10^{-8}\Omega m.

The area A is that of a circle, which written in terms of its diameter is:

A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}

Putting all together:

P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}

Which for our values is:

P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W

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From here, you can use the dimensional analysis technique. Also, you should know that 1 kilowatt is equal to 1000 watts, and 24 hours is equal to 86,400 seconds. Then,

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4 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
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Answer:

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Explanation:

Given that,

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T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

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T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

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T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

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agasfer [191]

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