Answer:
Total electric potential, 
Explanation:
It is given that,
First charge, 
Second charge, 
Distance of first charge from origin, 
Distance of second charge from origin, 
We need to find the total electric potential at the origin. The electric potential at the origin is given by :



V = 1321826.08 V
or

So, the total electric potential at the origin is
. Hence, this is the required solution.
Answer:

Explanation:
<u>Horizontal Launch
</u>
It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:
- The horizontal speed is constant and equal to the initial speed

- The vertical speed is zero at launch time, but increases as the object starts to fall
- The height of the object gradually decreases until it hits the ground
- The horizontal distance where the object lands is called the range
We have the following formulas




Where
is the initial horizontal speed,
is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.
If we know the initial height of the object, we can compute the time it takes to hit the ground by using

Rearranging and solving for t



We then replace this value in

To get



The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing 

The launch velocity is

Answer:
B/4
Explanation:
The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:
The field at twice the distance is B/4.
Explanation:
Below is an attachment containing the solution.
Answer:
analog-to-digital
Explanation:
An analog-to-digital converter, or ADC as it is more commonly called, is a device that converts analog signals into digital signals.