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ser-zykov [4K]
3 years ago
7

Please help and quick!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
andrew-mc [135]3 years ago
3 0

Answer:

yes i know

Explanation:

Measuring Tsunamis. ... It consists of an anchored, ocean-bottom pressure recorder and a companion buoy (called DART, for Deep-Ocean Assessment and Reporting of Tsunamis). The recorder, sitting at a depth of up to 5,000 meters, measures changes in pressure due to changes in water level.

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An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
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To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

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