Answer:
Velocity of the electron at the centre of the ring, ![v=1.37\times10^7\ \rm m/s](https://tex.z-dn.net/?f=v%3D1.37%5Ctimes10%5E7%5C%20%5Crm%20m%2Fs)
Explanation:
<u>Given:</u>
- Linear charge density of the ring=
![0.1\ \rm \mu C/m](https://tex.z-dn.net/?f=0.1%5C%20%5Crm%20%5Cmu%20C%2Fm)
- Radius of the ring R=0.2 m
- Distance of point from the centre of the ring=x=0.2 m
Total charge of the ring
![Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C](https://tex.z-dn.net/?f=Q%3D0.1%5Ctimes2%5Cpi%20R%5C%5CQ%3D0.1%5Ctimes2%5Cpi%200.4%5C%5CQ%3D0.251%5C%20%5Crm%20%5Cmu%20C)
Potential due the ring at a distance x from the centre of the rings is given by
![V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\](https://tex.z-dn.net/?f=V%3D%5Cdfrac%7BkQ%7D%7B%5Csqrt%7B%28R%5E2%2Bx%5E2%29%7D%7D%5C%5C)
The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by
![\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V](https://tex.z-dn.net/?f=%5CDelta%20V%3D%5Cdfrac%7BkQ%7D%7BR%7D-%5Cdfrac%7BkQ%7D%7B%5Csqrt%7B%28R%5E2%2Bx%5E2%29%7D%7D%5C%5C%5CDelta%20V%3D%7B9%5Ctimes10%5E9%5Ctimes0.251%5Ctimes10%5E%7B-6%7D%7D%20%5Cleft%28%20%5Cdfrac%7B1%7D%7B0.4%7D-%5Cdfrac%7B1%7D%7B%5Csqrt%7B%280.4%5E2%2B0.2%5E2%29%7D%7D%20%5Cright%20%29%5C%5C%5CDelta%20V%3D5.12%5Ctimes10%5E2%5C%20%5Crm%20V)
Let
be the change in potential Energy given by
![\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J](https://tex.z-dn.net/?f=%5CDelta%20U%3De%5Ctimes%20%5CDelta%20V%5C%5C%5CDelta%20U%3D1.67%5Ctimes10%5E%7B-19%7D%5Ctimes5.12%5Ctimes10%5E%7B2%7D%5C%5C%5CDelta%20U%3D8.55%5Ctimes10%5E%7B-17%7D%5C%20%5Crm%20J)
Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron
![\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s](https://tex.z-dn.net/?f=%5CDelta%20U%3D%5Cdfrac%7Bmv%5E2%7D%7B2%7D%5C%5C8.55%5Ctimes10%5E%7B-17%7D%3D%5Cdfrac%7B9.1%5Ctimes10%5E%7B-31%7Dv%5E2%7D%7B2%7D%5C%5Cv%3D1.37%5Ctimes10%5E7%5C%20%5Crm%20m%2Fs)
So the electron will be moving with ![v=1.37\times10^7\ \rm m/s](https://tex.z-dn.net/?f=v%3D1.37%5Ctimes10%5E7%5C%20%5Crm%20m%2Fs)
By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.
Hope this helps
<span />
By striking another object that is free to move, the moving object can exert a force and cause the second object to shift its position. While the object is moving, it has the capacity for doing work. Energy means the ability to do work, so all moving things have energy by virtue of their motion.
Translation: Al golpear otro objeto que se puede mover libremente, el objeto en movimiento puede ejercer una fuerza y hacer que el segundo objeto cambie de posición. Mientras el objeto se mueve, tiene la capacidad de realizar un trabajo. Energía significa la capacidad de realizar un trabajo, por lo que todas las cosas en movimiento tienen energía en virtud de su movimiento.
Answer:
The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m
Energy approach has been used to sole the problem.
The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring
The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved
Explanation:
The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.
As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .
x = compression of the spring = 0.89
Gravity is the correct answer.