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8090 [49]
3 years ago
5

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area 0.320 m2. at

the window, the electric field of the wave has rms value 0.0195 v/m. how much energy does this wave carry through the window during a 26.0-s commercial?
Physics
1 answer:
AlexFokin [52]3 years ago
5 0
Remember, half of the energy in an EM wave is in the E field, the rest is in the B field. Thus, multiply E field energy by 2.
To calculate the energy of the wave you must then use the following equation: W = A*t*c*2*(1/2*E^2*Eo). Where, A = Area, t = time, c = speed of light (which is a constant), E = Electric field, E0 = vacuum permittivity (8.85*10^-12 Nm^2/C^2). Substituting W =(0.320)*(26)*(3*10^8)*(2)*((1/2)*(1.95*10^-2)^2*(8.854*10^-12)) = 8.40*10^-6 J
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Answer:

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b ) charge transported in the circuit is   q = i * t  

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d ) work done through the resistor is w = P * t

                                                            = ( V)^2 / R * t

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3 years ago
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Sergio039 [100]

Answer:

because of Rayleigh scattering

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I believe the answer is C.

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3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim
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Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

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a) distance between deer and car

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   d₁ = v x t

   d₁ = 20 x 0.5 = 10 m

   distance travel after you apply brake

   using equation of motion

   v² = u² + 2 a s

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D = d₁ + d₂

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b) now, maximum speed car.

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distance left between them

   d₂ = 38 - d₁

   d₂ = 38 - 0.5 V

   distance travel after you apply brake

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now, solving the quadratic equation

  x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

  V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}

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rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

 

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