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konstantin123 [22]

3 years ago

15

Ignoring reflection at the air-water boundary, if the amplitude of a 10 GHz incident wave in air is 20 V/m at the water surface,

at what depth will it be down to 1 µV/m? Water is characterized by er = 81, µr = 1, and σ = 0.1 S/m. Can water described above, at 10 GHz, be described as a low-loss dielectric, good conductor, or "in-between"?

1 answer:

dimaraw [331]3 years ago

7 0

Answer:

0.80267 m

Explanation:

E(z) = Electric field = 1 µV/m

$E_0$ = 20 V/m

z = Depth

$\sigma$ = Conductivity = 0.1 S/m

$\epsilon_r$ = 81

$\mu$ = Impedance of free space = $120\pi\ \Omega$

Frequency is given by

$E(z)=E_0e^{-\alpha z}$

Parameter is given by

$\alpha=\dfrac{\sigma}{2}\sqrt{\dfrac{\mu}{\epsilon_r}}\\\Rightarrow \alpha=\dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{81}}\\\Rightarrow \alpha=20.94395\ N_p/m$

From the first equation

$1\times 10^{-6}=20e^{-20.94395z}\\\Rightarrow ln\dfrac{1\times 10^{-6}}{20}=-20.94395z\\\Rightarrow z=\dfrac{ln\dfrac{1\times 10^{-6}}{20}}{-20.94395}\\\Rightarrow z=0.80267\ m$

The depth is 0.80267 m

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a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

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$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

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c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

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