Answer:
just search up a ven-diagram and then try to draw it or trace it then use it for ur question
Explanation:
C) Radiation that comes from Earth...... Hope it helps, Have a nice day :)
Yes, eg., when 2 bodies move in opposite directions
, the relative velocity of each is greater than the individual velocity of either
Answer:
![E_r(6)=4.35614\ MPa](https://tex.z-dn.net/?f=E_r%286%29%3D4.35614%5C%20MPa)
Explanation:
= Strain = 0.49
= 3.1 MPa
At t = Time = 32 s
= 0.41 MPa
= Time-independent constant
Stress relation with time
![\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)](https://tex.z-dn.net/?f=%5Csigma%3D%5Csigma%20_0exp%5Cleft%28-%5Cfrac%7Bt%7D%7B%5Ctau%7D%5Cright%29)
at t = 32 s
![0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s](https://tex.z-dn.net/?f=0.41%3D3.1exp%5Cleft%28-%5Cfrac%7B32%7D%7B%5Ctau%7D%5Cright%29%5C%5C%5CRightarrow%20exp%5Cleft%28-%5Cfrac%7B32%7D%7B%5Ctau%7D%5Cright%29%3D%5Cfrac%7B0.41%7D%7B3%7D%5C%5C%5CRightarrow%20-%5Cfrac%7B32%7D%7B%5Ctau%7D%3Dln%5Cfrac%7B0.41%7D%7B3%7D%5C%5C%5CRightarrow%20%5Ctau%3D-%5Cfrac%7B32%7D%7Bln%5Cfrac%7B0.41%7D%7B3%7D%7D%5C%5C%5CRightarrow%20%5Ctau%3D16.0787%5C%20s)
The time independent constant is 16.0787 s
![E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}](https://tex.z-dn.net/?f=E_%7Br%7D%28t%29%3D%5Cfrac%7B%5Csigma%28t%29%7D%7B%5Cepsilon_0%7D)
At t = 6
![\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}](https://tex.z-dn.net/?f=%5C%5C%5CRightarrow%20E_%7Br%7D%286%29%3D%5Cfrac%7B%5Csigma%286%29%7D%7B%5Cepsilon_0%7D)
From the first equation
![\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451](https://tex.z-dn.net/?f=%5Csigma%28t%29%3D%5Csigma%20_0exp%5Cleft%28-%5Cfrac%7Bt%7D%7B%5Ctau%7D%5Cright%29%5C%5C%5CRightarrow%20%5Csigma%286%29%3D3.1exp%5Cleft%28-%5Cfrac%7B6%7D%7B16.0787%7D%5Cright%29%5C%5C%5CRightarrow%20%5Csigma%286%29%3D2.13451)
![E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa](https://tex.z-dn.net/?f=E_r%286%29%3D%5Cfrac%7B2.13451%7D%7B0.49%7D%5C%5C%5CRightarrow%20E_r%286%29%3D4.35614%5C%20MPa)
![E_r(6)=4.35614\ MPa](https://tex.z-dn.net/?f=E_r%286%29%3D4.35614%5C%20MPa)
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)