Charge of electron = 1.6×10−¹⁹
(1.6×10−¹⁹)(1×10²) (2e)
= 3.2×10−¹⁷ J
Answer:
a)
, b) ![v \approx 7745.967\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%5Capprox%207745.967%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Explanation:
a) The potential energy is:
![U_{e} = Q \cdot \Delta V](https://tex.z-dn.net/?f=U_%7Be%7D%20%3D%20Q%20%5Ccdot%20%5CDelta%20V)
![U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)](https://tex.z-dn.net/?f=U_%7Be%7D%20%3D%20%2830%5C%2CC%29%5Ccdot%20%281.0%5Ctimes%2010%5E%7B9%7D%5C%2CV%29)
![U_{e} = 3 \times 10^{10}\,J](https://tex.z-dn.net/?f=U_%7Be%7D%20%3D%203%20%5Ctimes%2010%5E%7B10%7D%5C%2CJ)
b) Maximum final speed:
![U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }](https://tex.z-dn.net/?f=U_%7Be%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20m%20%5Ccdot%20v%5E%7B2%7D%5C%5Cv%20%3D%20%5Csqrt%7B%5Cfrac%7B2%5Ccdot%20U_%7Be%7D%7D%7Bm%7D%20%7D)
The final speed is:
![v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cfrac%7B2%5Ccdot%20%283%20%5Ctimes%2010%5E%7B10%7D%5C%2CJ%29%7D%7B1000%5C%2Ckg%7D%20%7D)
![v \approx 7745.967\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%5Capprox%207745.967%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
h = 9.57 seconds
Explanation:
It is given that,
Initial speed of Kalea, u = 13.7 m/s
At maximum height, v = 0
Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :
![v=u-gt](https://tex.z-dn.net/?f=v%3Du-gt)
![u=gt](https://tex.z-dn.net/?f=u%3Dgt)
![t=\dfrac{u}{g}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bu%7D%7Bg%7D)
![t=\dfrac{13.7}{9.8}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B13.7%7D%7B9.8%7D)
t = 1.39 s
Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :
![h=ut+\dfrac{1}{2}at^2](https://tex.z-dn.net/?f=h%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2)
Here, a = -g
![h=ut-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=h%3Dut-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
![h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2](https://tex.z-dn.net/?f=h%3D13.7%5Ctimes%201.39-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%209.8%5Ctimes%20%281.39%29%5E2)
h = 9.57 meters
So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.
What are the options u can't say that and expect me to know wat u talkin bout
High energy waves have Gamma rays