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QveST [7]
2 years ago
14

1. How much heat is required to raise the temperature of 57.8kg of silver from 17.0°C to

Physics
1 answer:
Y_Kistochka [10]2 years ago
7 0
Answer: unknown so the second one
I hope this helps you with your stuff
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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
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<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:

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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

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