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
1. I think you should compare diagrams of moon phases from the textbook to diagrams of moon phases online. Because if you pick D it will take to long and C will help you out whith 3 different things to look at.
2. The moon changes in appearances from the perspective of people on earth because it's revolving around the planet and the earth is revolving around the sun, so A. Hoped this helped.
Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
No because you don’t learn about synthetic inventions yet in your first year
Answer:
16.96 W
Explanation:
Power: This can be defined as the rate at which work is done by an object. The S.I unit of power is Watt(W).
From the question,
P = (F×d)/t....................... Equation 1
Where P = power, F = force, d = distance, t = time.
Given: F = 75 N, d = 42 m, t = 3.1 min = 3.1×60 = 186 s
Substitute these values into equation 1
P = (75×42)/186
P = 16.94 W
Hence the average power delivered by the child = 16.96 W
