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
<em>Steel: 11.0 – 12.5</em>
<em>T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶ ̶T̶e̶t̶s̶u̶t̶e̶t̶s̶u̶</em>
Thanks,
<em>Deku ❤</em>
A. Using a combination lens made up of lenses, each of which has a different index of refraction. Is the correct answer.
Answer:
0.099C
Explanation:
First, we need to get the common potential voltage using the formula
Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then
Therefore
Charge, Q is given by CV hence for the first capacitor charge will be 
Here, 
Ok i apologise for the messy working but I'll try and explain my attempt at logic
Also note i ignore any air resistance for this.
First i wrote the two equations I'd most likely need for this situation, the kinetic energy equation and the potential energy equation.
Because the energy right at the top of the swing motion is equal to the energy right in the "bottom" of the swing's motion (due to conservation of energy), i made the kinetic energy equal to the potential energy as indicated by Ek = Ep.
I also noted the "initial" and "final" height of the swing with hi and hf respectively.
So initially looking at this i thought, what the heck, there's no mass. Then i figured that using the conservation of energy law i could take the mass value from the Ek equation and use it in the Ep equation. So what i did was take the Ek equation and rearranged it for m as you can hopefully see. Then i substituted the rearranged Ek equation into the Ep equation.
So then the equation reads something like Ep = (rearranged Ek equation for m) × g (which is -9.81) × change in height (hf - hi).
Then i simplify the equation a little. When i multiply both sides by v^2 i can clearly see that there is one E on each side (at that stage i don't need to clarify which type of energy it is because Ek = Ep so they're just the same anyway). So i just canceled them out and square rooted both sides.
The answer i got was that the max velocity would be 4.85m/s 3sf, assuming no losses (eg energy lost to friction).
I do hope I'm right and i suppose it's better than a blank piece of paper good luck my dude xx
