Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Answer:
Star A is closer than Star B
Explanation:
As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle
Here we can say

so we have

so here we have
angle subtended by Star A = 1 arc sec
angle subtended by star B = 0.75 arc sec
now we have
distance for star A is given as

distance of star B is given as

So star A is closer than star B
Answer:
16.8 lb is the force on the brake pad of one wheel.
Explanation:
Force applied on the piston = 
Area of the piston = 
Force applied on the brakes = 
Area of the brakes = 
Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.


16.8 lb is the force on the brake pad of one wheel.