Answer:
q2 = -4.35*10^-9C
Explanation:
In order to find the values of the second charge, you use the following formula:
(1)
V: electric potential = 1.14 kV = 1.14*10^3 kV
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1: charge 1 = 8.60*10^-9 C
q2: charge 2 = ?
r1: distance to the first charge = 20.7mm = 20.7*10^-3 m
r2: distance to the second charge = 15.1mm
You solve the equation (1) for q2, and replace the values of the other parameters:
![q_2=\frac{r_2}{k}[V-k\frac{q_1}{r_1}]=\frac{Vr_2}{k}-\frac{q_1r_2}{r_1}\\\\q_2=\frac{(1.14*10^3V)(15.1*10^{-3}m)}{8.98*10^9Nm^2/C^2}-\frac{(8.60*10^{-9}C)(15.1*10^{-3}m)}{20.7*10^{-3}m}\\\\q_2=-4.35*10^{-9}C](https://tex.z-dn.net/?f=q_2%3D%5Cfrac%7Br_2%7D%7Bk%7D%5BV-k%5Cfrac%7Bq_1%7D%7Br_1%7D%5D%3D%5Cfrac%7BVr_2%7D%7Bk%7D-%5Cfrac%7Bq_1r_2%7D%7Br_1%7D%5C%5C%5C%5Cq_2%3D%5Cfrac%7B%281.14%2A10%5E3V%29%2815.1%2A10%5E%7B-3%7Dm%29%7D%7B8.98%2A10%5E9Nm%5E2%2FC%5E2%7D-%5Cfrac%7B%288.60%2A10%5E%7B-9%7DC%29%2815.1%2A10%5E%7B-3%7Dm%29%7D%7B20.7%2A10%5E%7B-3%7Dm%7D%5C%5C%5C%5Cq_2%3D-4.35%2A10%5E%7B-9%7DC)
The values of the second charge is -4.35*10^-9C
It'll speed up in the direction it's being pushed unbalanced. Or, it could slow down if applied the right amount of unbalanced power. And finally if the unbalanced push is pushed it could change the direction of it's current motion it was traveling.<span>
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<span>to absorb heat flow from air molecules circulating around the refrigerant tubing i hope this helps you</span>
(4) Elevations:
Point A: 900 ft
Point B: 900 ft
Point C: 1500 ft
Point D: 1000 ft
(5) Contour interval: 100 ft
(6) Plot #1 best fits the profile seen in the contour map.
