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jolli1 [7]
3 years ago
6

A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​

Engineering
1 answer:
dexar [7]3 years ago
5 0

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

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Find the following of the given sinusoidal wave
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a) The amplitude of the <em>sinusoidal</em> wave is 15.

b) The period of the <em>sinusoidal</em> wave is 0.5236.

c) The <em>phase</em> shift of the <em>sinusoidal</em> wave is - 1.338π.

d) The <em>vertical</em> shift of the <em>sinusoidal</em> wave is - 5.

e) The equation of the <em>sinusoidal</em> wave is y = 15 · sin [(12 · x - 1.338π)] - 5.

<h3>What is the equation of the sinusoidal graph?</h3>

Herein we have the graph of a <em>sinusoidal</em> wave, whose constants can be found by using the following expressions:

Amplitude - Vertical distance between the maximum and minimum of the function:

a = [10 - (- 20)] / 2

a = 15

The amplitude of the <em>sinusoidal</em> wave is 15.

Period - Horizontal distance between two <em>consecutive</em> maxima or two <em>consecutive</em> minima.

T = 2 · (0.3809 - 0.1191)

T = 0.5236

The period of the <em>sinusoidal</em> wave is 0.5236.

And the <em>angular</em> frequency is calculated by the following expression:

b = 2π /T

b = 2π / 0.5236

b = 12

The <em>angular</em> frequency of the <em>sinusoidal</em> wave is 12.

Vertical shift - The average of the minimum and maximum of the function.

k = (10 - 20) / 2

k = - 5

The <em>vertical</em> shift of the <em>sinusoidal</em> wave is - 5.

Phase shift - <em>Horizontal</em> translation of the <em>sinusoidal</em> wave, which can be found by algebraic handling:

0.3809 = 15 · sin [(12 · 0.3809) + c] - 5

5.3809 = 15 · sin [(12 · 0.3809) + c]

0.359 = sin [(12 · 0.3809) + c]

0.117π = (12 · 0.3809) + c

0.117π = 4.5708 + c

c = - 1.338π

The <em>phase</em> shift of the <em>sinusoidal</em> wave is - 1.338π.

The equation of the <em>sinusoidal</em> wave is y = 15 · sin [(12 · x - 1.338π)] - 5.

To learn more on sinusoidal waves: brainly.com/question/13260086

#SPJ1

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2 years ago
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