This is called average speed.Its really simple
Answer:
Part a)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
Explanation:
Part a)
As we know that frequency = 1 MHz
speed of electromagnetic wave is same as speed of light
So the wavelength is given as
![\lambda = \frac{c}{f}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7Bc%7D%7Bf%7D)
![\lambda = \frac{3\times 10^8}{1\times 10^6}](https://tex.z-dn.net/?f=%5Clambda%20%3D%20%5Cfrac%7B3%5Ctimes%2010%5E8%7D%7B1%5Ctimes%2010%5E6%7D)
![\lambda = 300 m](https://tex.z-dn.net/?f=%5Clambda%20%3D%20300%20m)
Part b)
As we know the relation between electric field and magnetic field
![E = Bc](https://tex.z-dn.net/?f=E%20%3D%20Bc)
![E = (9 \times 10^{-9})(3\times 10^8)](https://tex.z-dn.net/?f=E%20%3D%20%289%20%5Ctimes%2010%5E%7B-9%7D%29%283%5Ctimes%2010%5E8%29)
![E = 2.7 N/C](https://tex.z-dn.net/?f=E%20%3D%202.7%20N%2FC)
Part c)
Intensity of wave is given as
![I = \frac{1}{2}\epsilon_0E^2c](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cepsilon_0E%5E2c)
![I = \frac{1}{2}(8.85 \times 10^{-12})(2.7)^2(3\times 10^8)](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B1%7D%7B2%7D%288.85%20%5Ctimes%2010%5E%7B-12%7D%29%282.7%29%5E2%283%5Ctimes%2010%5E8%29)
![I = 9.68 \times 10^{-3} W/m^2](https://tex.z-dn.net/?f=I%20%3D%209.68%20%5Ctimes%2010%5E%7B-3%7D%20W%2Fm%5E2)
Pressure is defined as ratio of intensity and speed
![P = \frac{I}{c} = \frac{9.68\times 10^{-3}}{3\times 10^8}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7BI%7D%7Bc%7D%20%3D%20%5Cfrac%7B9.68%5Ctimes%2010%5E%7B-3%7D%7D%7B3%5Ctimes%2010%5E8%7D)
![P = 3.22 \times 10^{-11} N/m^2](https://tex.z-dn.net/?f=P%20%3D%203.22%20%5Ctimes%2010%5E%7B-11%7D%20N%2Fm%5E2)
Answer:
~~Now, you have left your question very open ended and didn't ask for any particular kind of answer so I'll do my best to get what you're looking for.~~
A physical change in a substance doesn't change what the substance is. It can possibly melt or freeze an object. I mean heat makes things expand while cooling makes them retract.... In chemical change where there is a chemical reaction, a new substance is formed and energy is either given off or absorbed.
The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as
![P = \frac{1}{2} \mu \omega^2 A^2 v](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Cmu%20%5Comega%5E2%20A%5E2%20v)
Here,
= Linear mass density of the string
Angular frequency of the wave on the string
A = Amplitude of the wave
v = Speed of the wave
At the same time each of this terms have its own definition, i.e,
Here T is the Period
For the linear mass density we have that
![\mu = \frac{m}{l}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7Bm%7D%7Bl%7D)
And the angular frequency can be written as
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
Replacing this terms and the first equation we have that
![P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7Bm%7D%7Bl%7D%29%282%5Cpi%20f%29%5E2%20A%5E2%28%5Csqrt%7B%5Cfrac%7BT%7D%7B%5Cmu%7D%7D%29)
![P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7Bm%7D%7Bl%7D%29%282%5Cpi%20f%29%5E2%20A%5E2%20%28%5Csqrt%7B%5Cfrac%7BT%7D%7Bm%2Fl%7D%7D%29)
![P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})](https://tex.z-dn.net/?f=P%20%3D%202%5Cpi%5E2%20f%5E2A%5E2%28%5Csqrt%7BT%28m%2Fl%29%7D%29)
PART A ) Replacing our values here we have that
![P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})](https://tex.z-dn.net/?f=P%20%3D%202%5Cpi%5E2%20%28105%29%5E2%281.8%2A10%5E%7B-3%7D%29%5E2%28%5Csqrt%7B%2829.0%29%282.95%2A10%5E%7B-3%7D%2F0.79%29%7D%29)
![P = 0.2320W](https://tex.z-dn.net/?f=P%20%3D%200.2320W)
PART B) The new amplitude A' that is half ot the wavelength of the wave is
![A' = \frac{1.8*10^{-3}}{2}](https://tex.z-dn.net/?f=A%27%20%3D%20%5Cfrac%7B1.8%2A10%5E%7B-3%7D%7D%7B2%7D)
![A' = 0.9*10^{-3}](https://tex.z-dn.net/?f=A%27%20%3D%200.9%2A10%5E%7B-3%7D)
Replacing at the equation of power we have that
![P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})](https://tex.z-dn.net/?f=P%20%3D%202%5Cpi%5E2%20%28105%29%5E2%280.9%2A10%5E%7B-3%7D%29%5E2%28%5Csqrt%7B%2829.0%29%282.95%2A10%5E%7B-3%7D%2F0.79%29%7D%29)
![P = 0.058W](https://tex.z-dn.net/?f=P%20%3D%200.058W)
That seems like a statement more than a question. Where's the question?