A positive charge of 0.00047 C is 15 m from a negative charge of 0.00089 C. What is the force of one of the charges due to the other charge in units of Newtons?

16.732 N

Explanation:

Given:

q1 = 0.00047 C = 4.7 x 10^-4 C

q2 = 0.00089 C = 8.9 x 10^-4 C

d = 15 m

k = 9 x 10^9 N m^2 / C^2

To Find:

F = ?

Solution:

F = k x q1 x q2/d^2

F = 9 x 10^9 x 4.7 x 10^-4 x 8.9 x 10^-4 / 15 x 15

F = 9 x 4.7 x 8.9 x 10^9 x 10^-4 x 10^-4 / 225

F = 9 x 4.7 x 8.9 x 10^9 x 10^-8 / 225

F = 9 x 4.7 x 8.9 x 10 / 225

F = 418.3/25

F = 1673.2/100

Therefore, F = 16.732 N

PLZ MARK ME AS BRAINLIEST!!!

probably not if net force is zero

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.