<u>what's</u><u> </u><u>your </u><u>question </u><u>O_</u><u>-</u><u> </u><u>?</u><u>?</u>
<u>you </u><u>have </u><u>given</u><u> </u><u>the</u><u> </u><u>es</u><u>s</u><u>ay,</u><u> </u><u>but </u><u>wts </u><u>the </u><u>topic </u><u>?</u><u> </u><u>lol</u>
The induced electromotive force in any closed circuit is equal to the negative of the time rate of change of the magnetic flux enclosed by the circuit. That is the textbook definition that is widely accepted.
It depends most on the temperature of the gas.
Answer:
18.7842493212 W
Explanation:
T = Tension = 1871 N
= Linear density = 3.9 g/m
y = Amplitude = 3.1 mm
= Angular frequency = 1203 rad/s
Average rate of energy transfer is given by
![P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7BT%5Cmu%7D%5Comega%5E2y%5E2%5C%5C%5CRightarrow%20P%3D%5Cdfrac%7B1%7D%7B2%7D%5Csqrt%7B1871%5Ctimes%203.9%5Ctimes%2010%5E%7B-3%7D%7D%5Ctimes%201203%5E2%5Ctimes%20%283.1%5Ctimes%2010%5E%7B-3%7D%29%5E2%5C%5C%5CRightarrow%20P%3D18.7842493212%5C%20W)
The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W
The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.
By definition the gravitational potential energy is given by,
![PE=\frac{GMm}{r}](https://tex.z-dn.net/?f=PE%3D%5Cfrac%7BGMm%7D%7Br%7D)
Where,
m = Mass of Moon
G = Gravitational Universal Constant
M = Mass of Ocean
r = Radius
First we calculate the mass through the ratio given by density.
![m = \rho V](https://tex.z-dn.net/?f=m%20%3D%20%5Crho%20V)
![m = (1030Kg/m^3)(7*10^8m^3)](https://tex.z-dn.net/?f=m%20%3D%20%281030Kg%2Fm%5E3%29%287%2A10%5E8m%5E3%29)
![m = 7.210*10^{11}Kg](https://tex.z-dn.net/?f=m%20%3D%207.210%2A10%5E%7B11%7DKg)
PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon
Now we define the radius at the most distant point
![r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m](https://tex.z-dn.net/?f=r_1%20%3D%203.84%2A10%5E8%20%2B%206.4%2A10%5E6%20%3D%203.904%2A10%5E8m)
Then the potential energy at this point would be,
![PE_1 = \frac{GMm}{r_1}](https://tex.z-dn.net/?f=PE_1%20%3D%20%5Cfrac%7BGMm%7D%7Br_1%7D)
![PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}](https://tex.z-dn.net/?f=PE_1%20%3D%20%5Cfrac%7B%286.61%2A10%5E%7B-11%7D%29%2A%287.21%2A10%5E%7B11%7D%29%2A%287.35%2A10%5E%7B22%7D%29%7D%7B3.904%2A10%5E8%7D)
![PE_1 = 9.05*10^{15}J](https://tex.z-dn.net/?f=PE_1%20%3D%209.05%2A10%5E%7B15%7DJ)
PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.
At the nearest point we perform the same as the previous process, we calculate the radius
![r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m](https://tex.z-dn.net/?f=r_2%20%3D%203.84%2A10%5E8-6.4%2A10%5E6%20-%203.776%2A10%5E8m)
The we calculate the Potential gravitational energy,
![PE_2 = \frac{GMm}{r_2}](https://tex.z-dn.net/?f=PE_2%20%3D%20%5Cfrac%7BGMm%7D%7Br_2%7D)
![PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}](https://tex.z-dn.net/?f=PE_2%20%3D%20%5Cfrac%7B%286.61%2A10%5E%7B-11%7D%29%2A%287.21%2A10%5E%7B11%7D%29%2A%287.35%2A10%5E%7B22%7D%29%7D%7B3.776%2A10%5E8%7D)
![PE_2 = 9.361*10^{15}J](https://tex.z-dn.net/?f=PE_2%20%3D%209.361%2A10%5E%7B15%7DJ)