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Rus_ich [418]
3 years ago
12

A series RCL circuit is at resonance and contains a variable resistor that is set to 206Ω. The power dissipated in the circuit i

s 1.30 W. Assuming that the voltage remains constant, how much power is dissipated when the variable resistor is set to 532Ω? Note: The ac current and voltage are rms values and power is an average value unless indicated otherwise.
Physics
1 answer:
mash [69]3 years ago
5 0

Answer:

Power dissipated in resistor 532 ohm is 0.503 watt

Explanation:

We have given in first case resistance R_1=206ohm

Power dissipated in this resistance is P_1=1.30watt

Power dissipated in the resistor is equal to P=\frac{v_{rms}}^2{R}

We have to find the power dissipated in the resistor is 1.30 watt

From the relation we can say that \frac{P_1}{P_2}=\frac{R_2}{R_1}

\frac{1.3}{P_2}=\frac{532}{206}

P_2=0.503watt

So power dissipated in resistor 532 ohm is 0.503 watt  

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The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

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The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

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-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

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-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

2 (-2.72 m) (9.80 m/s²) = 2<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ - (<em>v</em>₀² + (15.9 m/s) <em>v</em>₀ + 63.0 m²/s²)

-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

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