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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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To solve this problem, we know that:

1 Albert = 88 meters

1 A = 88 m

The first thing we have to do is to square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

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v_y is the final vertical velocity

u_y = 0 is the initial vertical velocity (zero because the pebble is launched horizontally)

a=g=-9.8 m/s^2 is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

v_x = 20.0 m/s

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In this case, the pebble is launched straight up, so its initial vertical velocity is

u_y = 20.0 m/s

So we can find the final vertical velocity using the same suvat equation as before:

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