According to Coulomb’s Law, the force will quadruple.
However…
Let’s assume, for the sake of argument, that both proton and electron behave entirely like particles.
The trajectory of the electron will stretch, and its distance from the nucleus will oscilate until is estabilizes again at the original value.
They get it from the experiment.
The maximum amount of work performed is
Explanation:
The efficiency of a real heat engine is given by the equation:
(1)
where
is the temperature of the cold reservoir
is the temperature of the hot reservoir
However, the efficiency of a real heat engine can be also written as:
where
is the maximum work done
is the heat absorbed from the hot reservoir
can be written as
where
is the heat released to the cold reservoir
So the previous equation can be also written as
(2)
By combining eq.(1) and (2) we get
And re-arranging the equation and solving for , we find
Learn more about work and heat:
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<span>There are two possible arrangements of q1,q2,and q3 in this problem. They are:
q3, 2 cm gap, q1, 2 cm gap, q2
or
q1, 2/3 cm gap, q3, 4/3 cm gap, q2
We really don't care about the absolute magnitude of q, so the fact that it's 1.00 nano Coulombs is totally irrelevant to this problem. The only thing important is the relative charge and distances between the particles.
The force exerted between two particles is expressed as
F = q1*q2/r^2.
q1,q2 = charges on the particles.
r = distance between the particles.
Depending upon the relative charge (positive or negative) the force may be either attraction, or repulsion. But since the signs of all the charges mentioned are the same, I'll assume that the force will be repulsive.
For the distance between q1 and q3 I'll use the value "r". And since q1 and q2 are 2 cm apart, for the distance between q3 and q2, I'll use the value (2-r). So we have the following equations.
Force between q1 and q3
F = q1*q3/r^2
Force between q2 and q3
F = q2*q3/(2-r)^2
Set the 2 equations equal to each other
q1*q3/r^2 = q2*q3/(2-r)^2
Substitute the known values and solve for r.
q1*q3/r^2 = q2*q3/(2-r)^2
q*q/r^2 = 4q*q/(2-r)^2
q^2/r^2 = 4q^2/(4 - 4r + r^2)
1/r^2 = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4r^2/(r^2(4 - 4r + r^2))
0 = (4r^2-(4 - 4r + r^2))/(r^2(4 - 4r + r^2))
0 = (4r^2 - 4 + 4r - r^2)/(r^2(4 - 4r + r^2))
0 = (3r^2 - 4 + 4r)/(r^2(r-2)(r-2))
Now let's look at the numerator and denominator of the expression and see where we can get a value of 0. The denominator is allowed to have any value EXCEPT 0 and that will occur at r = 2, or r=0. And nothing else in the denominator will help the expression become 0. But if the numerator is 0, then the expression is 0. So let's see at what values the numerator is 0. Using the quadratic formula with A=3, B = 4 and C=-4, we get zeros at r = -2 and r = 2/3. Both of those values make sense. If r = -2, that means that the charges are arranges q3, q1, q2 with q1 being 2 cm from q1 and 4 cm from q2. And for r = 2/3, that also makes sense with the charges being arranged q1, q3, q2 and q3 is 2/3cm from q1 and 4/3cm from q2. In both cases, q3 is twice as from from q2 as it is from q1.</span>
-Velocity is the speed of any moving object in a given direction, whilst Speed is the rate of an object's ability to move.
-Velocity can change if the direction or time is changed, the basic equation of velocity is: v = d/t
v - velocity
d - distance
t - time
If one of these factors change, it affects the other.
Hope this answers the question!