Answer:
i) 0.7
ii) 1.39
iii) 0.6
Next time, when compiling a Physics question, ensure you put the unit of each measurement.
Explanation:
i) T = time of flight = 
where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);
Subsituting the values, we have: T = 
= 0.7
ii) distance travel = Range = R = 
where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);
Subsituting values, we have: R = 
= 1.39
iii) Maximum Height = H = 
where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);
Subsituting values, we have: 
= 0.6
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
I believe it is water that is the medium for beach erosion
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,
Here,
D is diameter of the eye
The angle that relates the distance between the lights and the distance to the lamp is given by,
For small angle, 
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle 
Therefore,
Therefore the distance is 5.367km.
