Answer:
Explanation:
Given
speed of Electron ![u=2\times 10^7\ m/s](https://tex.z-dn.net/?f=u%3D2%5Ctimes%2010%5E7%5C%20m%2Fs)
final speed of Electron ![v=4\times 10^7\ m/s](https://tex.z-dn.net/?f=v%3D4%5Ctimes%2010%5E7%5C%20m%2Fs)
distance traveled ![d=1.2\ cm](https://tex.z-dn.net/?f=d%3D1.2%5C%20cm)
using equation of motion
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
where v=Final velocity
u=initial velocity
a=acceleration
s=displacement
![(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}](https://tex.z-dn.net/?f=%284%5Ctimes%2010%5E7%29%5E2-%282%5Ctimes%2010%5E7%29%5E2%3D2%5Ctimes%20a%5Ctimes%201.2%5Ctimes%2010%5E%7B-2%7D)
![a=5\times 10^{16}\ m/s^2](https://tex.z-dn.net/?f=a%3D5%5Ctimes%2010%5E%7B16%7D%5C%20m%2Fs%5E2)
acceleration is given by ![a=\frac{qE}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BqE%7D%7Bm%7D)
where q=charge of electron
m=mass of electron
E=electric Field strength
![5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B16%7D%3D%5Cfrac%7B1.6%5Ctimes%2010%5E%7B-19%7D%5Ccdot%20E%7D%7B9.1%5Ctimes%2010%5E%7B-31%7D%7D)
Answer:
Position A/Position E
, ![U = 0](https://tex.z-dn.net/?f=U%20%3D%200)
Position B/Position D
,
, for ![0 < x < 1](https://tex.z-dn.net/?f=0%20%3C%20x%20%3C%201)
Position C
, ![U = E](https://tex.z-dn.net/?f=U%20%3D%20E)
Explanation:
Let suppose that ball-Earth system represents a conservative system. By Principle of Energy Conservation, total energy (
) is the sum of gravitational potential energy (
) and translational kinetic energy (
), all measured in joules. In addition, gravitational potential energy is directly proportional to height (
) and translational kinetic energy is directly proportional to the square of velocity.
Besides, gravitational potential energy is increased at the expense of translational kinetric energy. Then, relative amounts at each position are described below:
Position A/Position E
, ![U = 0](https://tex.z-dn.net/?f=U%20%3D%200)
Position B/Position D
,
, for ![0 < x < 1](https://tex.z-dn.net/?f=0%20%3C%20x%20%3C%201)
Position C
, ![U = E](https://tex.z-dn.net/?f=U%20%3D%20E)
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.
![R= 6370*10^3 m](https://tex.z-dn.net/?f=R%3D%206370%2A10%5E3%20m)
![v = 239m/s](https://tex.z-dn.net/?f=v%20%3D%20239m%2Fs)
![a = 16.5m/s^2](https://tex.z-dn.net/?f=a%20%3D%2016.5m%2Fs%5E2)
The circumference of the earth would be
![\phi = 2\pi R](https://tex.z-dn.net/?f=%5Cphi%20%3D%202%5Cpi%20R)
Velocity is defined as,
![v = \frac{x}{t}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D)
![t = \frac{x}{v}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bx%7D%7Bv%7D)
Here
, then
![t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Cphi%7D%7Bv%7D%20%3D%20%5Cfrac%7B2%5Cpi%20%286370%2A10%5E3%29%7D%7B239%7D)
![t = 167463.97s](https://tex.z-dn.net/?f=t%20%3D%20167463.97s)
Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds