In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 × )(1.2 × )
PE = 5.0 x joules
None of the options shown is correct.
I can't see that cube from here.
But if the length of the side of the cube is ' K ' units,
then the surface area of the cube is 6K² units², and
the volume of the cube is K³ units³.
The ratio of the surface area to the volume is
(6K² units²) / (K³ units³) = (6) / (K units) .
So for example, if the side of the cube is 2 inches, then
the ratio of surface area to volume is "3 per inch".
That's the answer. I did the whole thing in order to earn
the points, but I don't expect you to understand much of it,
because I see from your username that you suck at math.
I'm sorry you decided that. Now that you've put up the
brick wall, it'll be even harder for any math to find its way
in there, and you'll miss out on a lot of the fun.
This theory was first proposed by Nicolaus Copernicus. Copernicus was a Polish astronomer. He first published the heliocentric system in his book: De revolutionibus <span>orbium </span>coelestium<span> , "On the revolutions of the heavenly bodies," which appeared in 1543.</span>
Answer:
Density =mass/volume 20/10=2
Answer:
25000 V
Explanation:
The formula for potential is
V = Kq/r
Potential at B due to the charge placed at origin O
V1 = K q / OB
V1 = 10000 V
Potential at B due to the charge placed at A
V2 = K q / AB
V2 = 15000 V
Total potential at B
V = V1 + V2 = 10000 + 15000 = 25000 V