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Ivanshal [37]
3 years ago
13

A series R-L-C circuit is connected to a 70.0 Hz ac source that has Vrms = 80.0 V . The circuit has a resistance of 80.0 Ω and a

n impedance of 102 Ω at this frequency. What average power is delivered to the circuit by the source?
Physics
1 answer:
Andreyy893 years ago
3 0

Answer:

62.7 VA

Explanation:

The apparent power (power supplied by the source) of an AC circuit is determined by

S = \frac{(Vrms)^2}{Z}

The impedance(Z) depends on the frequency, but in this case it is given to us that it is 102 ohms at this frequency.

So:

S = \frac{(80 V)^2}{102 \Omega} = 62.7 VA

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Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

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speed of sound in gas, C_g = 376 m/s

speed of sound in liquid, C_l = 1640 m/s

The of the sound wave is given by;

I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}

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P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535

3 0
3 years ago
A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that
FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

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