<span>
In layman's term: </span>like charges don't attract while opposite charges do<span>electrostatic forces between point A( which is charged) and point B (which is also charged) are proportional to the charge of point A and point B. </span><span>there is also something else about this law that I don't quite remember.</span>
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<span />Here is the formula:
<span>F = k x Q1 x Q2/d^<span>2</span></span>
<span>What the formula means:</span>
F=force between charges
Q1 and Q2= amount of charge
d=distance between these two charges
k= Coulombs constant (proportionally constant)
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I think that about covers it and hopefully this helped.
C) Acceleration is the rate of change of velocity of an object. Velocity is the speed and direction of an object so acceleration is used to describe the rate of change. I hope this helps!!
Answer:
326149.2 KJ
Explanation:
The heat transfer toward and object that suffered an increase in temperature can be calculated using the expression:
Q = m*cv*ΔT
Where m is the mass of the object, cv is the specific heat capacity at constant volume, which basically means the amount of heat necessary for a 1kg of water to increase 1C degree in temperatur, and ΔT is the change in temperature.
A 65000 L swimming pool will have a mass of:
65000L *
= 65000 kg
The specific heat capacity at constant volume of water is equal to 4.1814 KJ/KgC.
We replace the data and get:
Q = m*cv*ΔT = 65000 kg * 4.1814 KJ/KgC * 1.2°C = 326149.2 KJ
Answer:
An equipotential surface is circular in the two-dimensional. Since the electric field lines are directed radially away from the charge, hence they are opposite to the equipotential lines. Therefore, the electric field is perpendicular to the equipotential surface.
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.
