Answer:
z = 0.8 (approx)
Explanation:
given,
Amplitude of 1 GHz incident wave in air = 20 V/m
Water has,
μr = 1
at 1 GHz, r = 80 and σ = 1 S/m.
depth of water when amplitude is down to 1 μV/m
Intrinsic impedance of air = 120 π Ω
Intrinsic impedance of water = ![\dfrac{120\pi}{\epsilon_r}](https://tex.z-dn.net/?f=%5Cdfrac%7B120%5Cpi%7D%7B%5Cepsilon_r%7D)
Using equation to solve the problem
![E(z) = E_0 e^{-\alpha\ z}](https://tex.z-dn.net/?f=E%28z%29%20%3D%20E_0%20e%5E%7B-%5Calpha%5C%20z%7D)
E(z) is the amplitude under water at z depth
E_o is the amplitude of wave on the surface of water
z is the depth under water
![\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B%5Csigma%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7B%28120%5Cpi%29%5E2%7D%7B%5CEpsilon_r%7D%7D)
![\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B%5Cdfrac%7B%28120%5Cpi%29%5E2%7D%7B80%7D%7D)
![\alpha =21.07\ Np/m](https://tex.z-dn.net/?f=%5Calpha%20%3D21.07%5C%20Np%2Fm)
now ,
![1 \times 10^{-6} = 20 e^{-21.07\times z}](https://tex.z-dn.net/?f=1%20%5Ctimes%2010%5E%7B-6%7D%20%3D%2020%20e%5E%7B-21.07%5Ctimes%20z%7D)
![e^{21.07\times z}= 20\times 10^{6}](https://tex.z-dn.net/?f=e%5E%7B21.07%5Ctimes%20z%7D%3D%2020%5Ctimes%2010%5E%7B6%7D)
taking ln both side
21.07 x z = 16.81
z = 0.797
z = 0.8 (approx)
Answer:
Styrofoam would be the best insulator because it traps the air in small pockets, blocking the flow of heat energy.
Explanation:
I cant read it, i could most likely help if i could read it.
Answer:
421.83 m.
Explanation:
The following data were obtained from the question:
Height (h) = 396.9 m
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
First, we shall determine the time taken for the ball to get to the ground.
This can be calculated by doing the following:
t = √(2h/g)
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 396.9 m
Time (t) =.?
t = √(2h/g)
t = √(2 x 396.9 / 9.8)
t = √81
t = 9 secs.
Therefore, it took 9 secs fir the ball to get to the ground.
Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:
Time (t) = 9 secs.
Initial velocity (u) = 46.87 m/s
Horizontal distance (s) =...?
s = ut
s = 46.87 x 9
s = 421.83 m
Therefore, the horizontal distance travelled by the ball is 421.83 m
Answer:
increase.
Explanation:
According to the newton’s second law of motion force is expressed as product of mass and acceleration.
F = m a
If the force acting is constant, then.
m∝ ![\frac{1}{a}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba%7D)
That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.
As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.
Thus, it won't stay the same.
As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.
Thus, it won't decrease.
As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.
Thus, it will increase.