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Gnesinka [82]
3 years ago
5

An AM radio transmitter broadcasts 63.2 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike

the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.5 km away
Physics
1 answer:
egoroff_w [7]3 years ago
6 0

Answer:

Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

Explanation:

Given;

power of radio transmitter, P = 63.2 kW = 63200 W

distance of transmission, r = 30.5 km

Intensity of the transmitted radio wave is calculated as follows;

I = \frac{P}{4\pi r^2}

where;

I is the intensity of the transmitted radio wave

Substitute the given values and calculate the intensity of the transmitted radio wave;

I = \frac{P}{4\pi r^2} = \frac{63200}{4\pi (30500)^2} = 5.406 *10^{-6} \ W/m^2

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

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Answer:

<h2>Case i) if \omega L > \frac{1}{\omega c}</h2><h2>So initially if the circuit is inductive in nature then its net impedance will decrease after this</h2><h2>Case ii) if \omega L < \frac{1}{\omega c}</h2><h2>So initially if the circuit is capacitive in nature then its net impedance will increase after this</h2>

Explanation:

As we know that the impedance of the circuit is given as

z = \sqrt{(\omega L - \frac{1}{\omega c})^2 + R^2}

when we join another identical capacitor in parallel with previous capacitor in the circuit then we will have for parallel combination

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So initially if the circuit is inductive in nature then its net impedance will decrease after this

Case ii) if \omega L < \frac{1}{\omega c}

So initially if the circuit is capacitive in nature then its net impedance will increase after this

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