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
Answer:
Explanation:
Given parameters:
Weight of object = 49N
Force applied = 12N
Unknown:
Acceleration of object = ?
Solution:
The acceleration of the object is found by dividing the force by the weight;
Acceleration =
= 0.25m/s²
Answer:
A. Speed
Explanation:
A vector quantity is a quantity which has both magnitude and direction. Here in the given options, speed is a scalar quantity but not the vector quantity.
Answer:
The atmospheric pressure is
.
Explanation:
Given that,
Atmospheric pressure
drop height h'= 27.1 mm
Density of mercury 
We need to calculate the height
Using formula of pressure

Put the value into the formula



We need to calculate the new height




We need to calculate the atmospheric pressure
Using formula of atmospheric pressure

Put the value into the formula


Hence, The atmospheric pressure is
.
Answer:
Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.
Potential energy loss:
M g H = 21.7*9.81*3.5 = 745.1 J
Kinetic energy at bottom of slide:
= (1/2) M v^2 = 52.5 J