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svetoff [14.1K]
1 year ago
15

What's the real power of a purely resistive circuit in which the RMS voltage measures 80 volts and the RMS current measures 5 am

ps?
A. 85 watts

B. 120 watts

C. 240 watts

D. 400 watts
Physics
1 answer:
Nat2105 [25]1 year ago
7 0

Answer:

D.400 watt

Explanation:

  • B/c <em>Pave</em><em>=</em><em>Irms</em><em>Vrms</em>
  • <em>Pave</em><em>=</em><em>(</em><em>5</em><em>)</em><em>(</em><em>8</em><em>0</em><em>)</em>
  • <em>Pave</em><em>=</em><em>4</em><em>0</em><em>0</em><em> </em><em>watt</em>

<em>where</em><em> </em><em>Pave</em><em> </em><em>=</em><em> </em><em>average</em><em> </em><em>power</em><em> </em>

<em>Irms</em><em>=</em><em>RMS</em><em> </em><em>current</em><em> </em><em>&</em>

<em>Vrms</em><em>=</em><em>RMS</em><em> </em><em>voltage</em>

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The answer is A. Hope this helps you with your work.
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An object is transported to three different planets in our solar system. Which of the following is true about that object?
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<u>Answer;</u>

<em>D.  The object’s weight changes, but its mass stays the same.</em>

<u>Explanation;</u>

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You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g
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Answer:

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(b) L =  182.56 m

(c) L =  114.28 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

<u>L =  128.75 m</u>

<u></u>

(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

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<u>L =  182.56 m</u>

<u></u>

(c)

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ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

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Therefore,

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L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

<u>L =  114.28 m</u>

<u></u>

(d)

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Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

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(e)

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<u>Cost of Gold Wire = $ 307040</u>

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