Will give brainliest to correct answer. A particle has a charge of 1.9x10-6C. What is the charge on a second particle if the electrical force acting on it is 25N at a distance of 0.32m away?

1.5X10^-4C

Explanation:

Expression for the electric force between the two charges is given by -

F = (k*q1*q2) / r^2

Here, k = constant = 9 x 10^9 N*m^2 / C^2

q1 = unknown

F=25N

q2 = 1.9x10^-6 C

r = 0.32 m

Substitute the given values in the above expression -

25=9x10^9 *(q1) * 1.9 x10^ -6/ (0.32m)^2

25= 17,100* (q1)/ 0.1024

multiply both sides by 0.1024 to get rid of the denominator

2.56=17,100*(q1)

Divide both sides by 17,100 to isolate q1

q1=1.5x10^-4C

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

when it gains or loses electrons