Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:
lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:
To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):
=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached
Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
The more arms it has the less of a chance the prey has to swim away.
Vi = 2m/s
a= 4.5 m/s
d= 340 m
vf= ?
use this equation ... vf^2=vi<span>^2+2ad
you should get vf = 55.3
hope this helps </span>
Answer: It is both B and D
Select all that apply.
At night, thermal energy moves _____.
from space to the atmosphere
from the land to the atmosphere
from the atmosphere to the land
from the atmosphere to space
