Acceleration = change in velocity/time
= 40/5
=8m/s^2
Answer:
E=12.2V/m
Explanation:
To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.
The equation is given by,
![V=\frac{I}{nAq}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BI%7D%7BnAq%7D)
Where,
V= Drift Velocity
I= Flow of current
n= number of electrons
q = charge of electron
A = cross-section area.
For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is
![\frac{I}{q} = 1.2*10^{18}](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7Bq%7D%20%3D%201.2%2A10%5E%7B18%7D)
![A= 1.3*10^{-8}m^2](https://tex.z-dn.net/?f=A%3D%201.3%2A10%5E%7B-8%7Dm%5E2)
![n=6.3*10^{28} e/m^3](https://tex.z-dn.net/?f=n%3D6.3%2A10%5E%7B28%7D%20e%2Fm%5E3)
Mobility
We can find the drift velocity replacing,
![V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B1.2%2A10%5E%7B18%7D%7D%7B%281.3%2A10%5E%7B-8%7D%29%286.3%2A10%5E%7B28%7D%29%7D)
![V= 1.465*10^-3m/s](https://tex.z-dn.net/?f=V%3D%201.465%2A10%5E-3m%2Fs)
The electric field is given by,
![E= \frac{V}{\omicron{O}}](https://tex.z-dn.net/?f=E%3D%20%5Cfrac%7BV%7D%7B%5Comicron%7BO%7D%7D)
![E=\frac{1.465*10^-3}{1.2*10^{-4}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1.465%2A10%5E-3%7D%7B1.2%2A10%5E%7B-4%7D%7D)
![E=12.2V/m](https://tex.z-dn.net/?f=E%3D12.2V%2Fm)
The Coulomb force between two or more charged bodies is the force between them due to Coulomb's law. If the particles are both positively or negatively charged, the force is repulsive; if they are of opposite charge, it is attractive. ... Like the gravitational force, the Coulomb force is an inverse square law.
V = IR
By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.
So i believe the answer would be : The string will remain lit
hope this helps
Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.