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geniusboy [140]
3 years ago
13

What is the equivalent resistance of the circuit?

Physics
1 answer:
raketka [301]3 years ago
8 0

Answer:

I feel it would be D. 120.00 I'm really not sure

Explanation:

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An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t
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A. The athlete isn’t doing any work because he doesn’t move the weight.

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So when he's holding the weight, he doesn't do any work.

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Homogous chromosomes separate during.....?
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Which of the following would decrease the resistance to the flow of an electric current through a body?
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3 0
3 years ago
Read 2 more answers
A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power  over one period. The instantaneous power can be found as:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}

In terms of RMS values:

V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI

7 0
3 years ago
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