KE=1/2*m*v^2
So 480=1/2 * m * 8^2
480=1/2 * m * 64
480=32 * m
480/32=m
15=m
The collision of the two waves would bring force and that would make the new wave smaller
(AB) & (CD)
Most machines are fueled by gasoline and electricity
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
Answer:
a) We could not see it at all.
Explanation:
The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.
