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gtnhenbr [62]
2 years ago
8

A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 .

At the window, the electric field of the wave has an rms value 0.0400 V/m .
How much energy does this wave carry through the window during a 30.0-s commercial?
Express your answer with the appropriate units.
Physics
1 answer:
Darina [25.2K]2 years ago
3 0

Answer:

The  value is  E =  8.9 *10^{-5} \  J

Explanation:

From the question we are told that

   The  area is  A =  0.700 \  m^2

   The  root mean square value  is E_{rms} =  0.0400 \  V/m

   The time taken is t =  30.0 \  s

Generally the energy is mathematically represented as

     E =  c *  \sepsilon_o *  A  *  t  * E_{rms}^2

=>    E =  3.0*10^{8} *  8.85*10^{-12} *  0.700 *  30 * (0.04)^2

=>     E =  8.9 *10^{-5} \  J

     

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6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air
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From  the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

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A 50g ball is released from rest 1.0 above the bottom of thetrack
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Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
2 years ago
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