Answer:
q = 0.0392 / V, for V= 0.1V q = 0.392 C
Explanation:
For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy
starting point
Em₀ = U = q V
final point
Em_f = K = ½ m v²
Em₀ = Em_f
q V = ½ m v²
q =
let's calculate
q = ½ 0.40 10⁻³ 14² / V
q = 0.0392 / V
The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is
V = 0.1 V
q = 0.0392 / 0.1
q = 0.392 C
Uh I think it is inertia but I could be wrong
Hi there!
We can use the following equation to derive acceleration:
vf = vi + at, where:
vi = 0 m/s (starts from rest)
a = acceleration
time = time (sec)
Rewrite:
vf = at
Solve for acceleration:
vf/t = a
30/6 = 5 m/s²
I’m not sure I’m just gonna assume that the flashlight closed too the wall is gonna be brighter because most of it’s light is directed on the wall rather then the one further back is kinda pointed everyone else & I think it relates to stars as if the brighter the star the closed it is to earth ?
<span>The correct
answer between all the choices given is the third choice or letter C. I am
hoping that this answer has satisfied your query and it will be able to help
you in your endeavor, and if you would like, feel free to ask another question.</span>