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atroni [7]
3 years ago
13

An 800-kHz radio signal is detected at a point 4.5 km distant from a transmitter tower. The electric field amplitude of the sign

al at that point is 0.63 V/m. Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed. What is the magnetic field amplitude of the signal at that point
Physics
1 answer:
Hitman42 [59]3 years ago
3 0

Answer:

B_2=2.1nT

Explanation:

From the question we are told that:

Frequency F=800kHz

Distance d=4.5km

Electric field amplitude B_2=0.63V/m

Generally the equation for momentum is mathematically given by

 B=\frac{E}{C}

Therefore

 B_2=\frac{0.63}{3*10^8}

 B_2=0.21*10^{-8}

 B_2=2.1nT

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Answer:

\displaystyle w=3.478\ rad/sec

M=0.0182\ J

v=0.398\ m/s

Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

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\displaystyle w=\sqrt{\frac{9.8}{0.81}}

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(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

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Y=0.0081\ m

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\displaystyle \frac{mv^2}{2}=0.0182

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\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}

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