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
No answer is possible until we know the number that belongs after the words "... angular speed of ".
A material must readily accept electron flow to be a good conductor of electricity. Electrical conductors are electrical charge carriers with electrons that move with ease from atom to atom when charged with voltage. Examples of good conductors are copper, brass, steel, gold, and aluminum.
Answer:
4A
Explanation:
Current is charge per time. So if the second wire delivers twice as much charge in the same amount of time, it must have twice the current.
2 × 2A = 4A
The actual diameter of the heart is 12.25 cm. Given : Heart measure = 14.7 cm Magnification factor = 1.2
14.7 / 1.2 = 12.25
