Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where 
Difference in potential between the points is
![kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}](https://tex.z-dn.net/?f=kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7B0.2%5Ctext%7B%20m%7D%7D%20-%5Cleft%28%20-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D%20%3D%20%5Cdfrac%7BkQ%7D%7B0.2%5Ctext%7B%20m%7D%7D%20%3D%20%5Cdfrac%7B9%5Ctimes10%5E9%5Ctext%7B%20F%2Fm%7D%5Ctimes3%5Ctimes10%5E%7B-6%7D%5Ctext%7B%20C%7D%7D%7B0.2%5Ctext%7B%20m%7D%7D)
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
![270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]](https://tex.z-dn.net/?f=270%5Ctimes10%5E3%20%3D%20kQ%5Cleft%5B-%5Cdfrac%7B1%7D%7Bx%7D-%5Cleft%28-%5Cdfrac%7B1%7D%7B0.1%5Ctext%7B%20m%7D%7D%5Cright%29%5Cright%5D)
The charge chould be moved to infinity
The nucleus that completes the equation given above is 218/85 At, which is option D.
<h3>What is a nuclear fission?</h3>
Nuclear fission is a nuclear reaction in which a large nucleus splits into smaller ones with the simultaneous release of energy.
According to this question, the following nuclear fission reaction is given: 222/87Fr = 4/2He+ ?
To find the missing nucleus, we subtract 4 from 222 to get 218, therefore, the nucleus that completes the equation given above is 218/85 At.
Learn more about nuclear fission at: brainly.com/question/913303
#SPJ1
Answer:
Zero Acceleration.
Explanation:
If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s.
Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
= 
+ 
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
= 
+ 
= 
= 
⇒ f = 
= 0.37
The focal length of the eye is 0.37 cm.
