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olga nikolaevna [1]

3 years ago

14

The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries

a current of 170 mA. How much power is dissipated in the wire?

1 answer:

tester [92]3 years ago

6 0

Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance = 2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W

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Read 2 more answers

To construct an oscillating LC system, you can choose from a 11 mH inductor, a 6.0 μF capacitor, and a 4.2 μF capacitor. What ar

Free_Kalibri [48]

Answer:

a. 475.14 Hz

b. 1959 Hz

c. 2341.53 Hz , 3053.34 Hz

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$f = \frac{1}{2\pi*\sqrt{C*L}}$

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$f = \frac{1}{2\pi*\sqrt{C*L}}f=\frac{1}{2\pi*\sqrt{10.2uF*11mH}}$

$f = 475.14 Hz$

b. second smallest use the capacitive 6 uF so:

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{6uF*11mH}}$

$f = 1959Hz$

c. second largest and largest oscillation first combination so:

Use 4.2 uF

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$f = 2341.53 Hz$

And finally largest oscillation cap in serie so:

$C=\frac{c_1*c_2}{c_1+c_2}=\frac{4.2uF*6.0uf}{4.2uf+6.0uF}$

$C=2.47 uF$

$f = \frac{1}{2\pi*\sqrt{C*L}}=f = \frac{1}{2\pi*\sqrt{2.47uF*11mH}}$

$f = 3053.34 Hz$

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3 years ago

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aleksklad [387]

Answer:

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3 years ago