<u>Record of magnetic reversals recorded in molten rock at mid-ocean ridges:</u>
Often in ocean, at the junction of two plates, lava often comes out of the junction because of the movement of plates away from each other as a result, forming a divergent plate boundaries. As the lava comes up, the magnetic minerals come along with it.
During the molten state, they arrange themselves along the magnetic poles of the earth. Then they gets solidified. As the magnetic reversals take place in earth, these rock can’t change the orientation, but they retain the data about the magnetic orientation of earth corresponding to the date of their formation.
Using the conservation of momentumwhere momentum = mass x velocity
so the let x be the initial velocity of the ball B
(30 kg) (-10 m/s) + (10 kg) (x) = (30 kg) (30 m/s) + ( 10 kg) (-10 m/s)
-300 + 10 x = 900 - 10010x = 900 - 100 + 30010x = 1100x = 1100 / 10x = 110 m/s
The number of turns of wire is directly proportional to voltage.
The higher the number of turns of wire the higher the voltage
Since .
Power (output) of a generator = voltage x current. Therefore the higher the voltage the higher the output of the generator.
So, the higher the number of turns of wire the higher the output of the generator.
I believe the answer is transmission
To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
![f = 4.11 *10^{12} Hz](https://tex.z-dn.net/?f=f%20%3D%204.11%20%2A10%5E%7B12%7D%20Hz)
![A = 1.23 * 10^{-11}m](https://tex.z-dn.net/?f=A%20%3D%201.23%20%2A%2010%5E%7B-11%7Dm)
The angular velocity of a body can be described as a function of frequency as
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
![\omega = 2\pi 4.11 *10^{12}](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%204.11%20%2A10%5E%7B12%7D)
![\omega=2.582*10^{13} rad/s](https://tex.z-dn.net/?f=%5Comega%3D2.582%2A10%5E%7B13%7D%20rad%2Fs)
PART A) The expression for the maximum angular velocity is given by the amplitude so that
![V = A\omega](https://tex.z-dn.net/?f=V%20%3D%20A%5Comega)
![V =( 1.23 * 10^{-11})(2.582*10^{13})](https://tex.z-dn.net/?f=V%20%3D%28%201.23%20%2A%2010%5E%7B-11%7D%29%282.582%2A10%5E%7B13%7D%29)
![V = = 317.586m/s](https://tex.z-dn.net/?f=V%20%3D%20%20%3D%20317.586m%2Fs)
PART B) The maximum acceleration on your part would be given by the expression
![a = A \omega^2](https://tex.z-dn.net/?f=a%20%3D%20A%20%5Comega%5E2)
![a =( 1.23 * 10^{-11})(2.582*10^{13})^2](https://tex.z-dn.net/?f=a%20%3D%28%201.23%20%2A%2010%5E%7B-11%7D%29%282.582%2A10%5E%7B13%7D%29%5E2)
![a= 8.2*10^{15}m/s^2](https://tex.z-dn.net/?f=a%3D%208.2%2A10%5E%7B15%7Dm%2Fs%5E2)