Answer:
a) The electric field at that point is
newtons per coulomb.
b) The electric force is
newtons.
Explanation:
a) Let suppose that electric field is uniform, then the following electric field can be applied:
(1)
Where:
- Electric field, measured in newtons per coulomb.
- Electric force, measured in newtons.
- Electric charge, measured in coulombs.
If we know that
and
, then the electric field at that point is:


The electric field at that point is
newtons per coulomb.
b) If we know that
and
, then the electric force is:



The electric force is
newtons.
Answer:
The answer to your question is below
Explanation:
If Oxygen has an atomic number of 8, we can conclude that:
- If neutral, it has 8 protons and 8 electrons
- It has 8 neutrons and its atomic mass is 16 (A = 8 + 8)
- It must be located in the group VI A
- Its valence number must be 6
- Its Oxidation number is -2
The force on the tool is entirely in the negative-y direction.
So no work is done during any moves in the x-direction.
The work will be completely defined by
(Force) x (distance in the y-direction),
and it won't matter what route the tool follows to get anywhere.
Only the initial and final y-coordinates matter.
We know that F = - 2.85 y². (I have no idea what that ' j ' is doing there.)
Remember that 'F' is pointing down.
From y=0 to y=2.40 is a distance of 2.40 upward.
Sadly, since the force is not linear over the distance, I don't think
we can use the usual formula for Work = (force) x (distance).
I think instead we'll need to integrate the force over the distance,
and I can't wait to see whether I still know how to do that.
Work = integral of (F·dy) evaluated from 0 to 2.40
= integral of (-2.85 y² dy) evaluated from 0 to 2.40
= (-2.85) · integral of (y² dy) evaluated from 0 to 2.40 .
Now, integral of (y² dy) = 1/3 y³ .
Evaluated from 0 to 2.40 , it's (1/3 · 2.40³) - (1/3 · 0³)
= 1/3 · 13.824 = 4.608 .
And the work = (-2.85) · the integral
= (-2.85) · (4.608)
= - 13.133 .
-- There are no units in the question (except for that mysterious ' j ' after the 'F',
which totally doesn't make any sense at all).
If the ' F ' is newtons and the 2.40 is meters, then the -13.133 is joules.
-- The work done by the force is negative, because the force points
DOWN but we lifted the tool UP to 2.40. Somebody had to provide
13.133 of positive work to lift the tool up against the force, and the force
itself did 13.133 of negative work to 'allow' the tool to move up.
-- It doesn't matter whether the tool goes there along the line x=y , or
by some other route. WHATEVER the route is, the work done by ' F '
is going to total up to be -13.133 joules at the end of the day.
As I hinted earlier, the last time I actually studied integration was in 1972,
and I haven't really used it too much since then. But that's my answer
and I'm stickin to it. If I'm wrong, then I'm wrong, and I hope somebody
will show me where I'm wrong.
D. Sea level has risen more than 300 centimeters. This is true because as the earth gets warmer, ice will melt and turn into water which goes into the ocean making it rise.
The positive charge is strongest in the middle, because that is were the charges are going off from.