Answer:
The answer is 
.
Explanation:
From the law of gravity,
considering F as a conservative force, 
,
the general expression for gravitational potential energy is
,
where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.
However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as
,
(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is
,
which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).
When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is
,
then, the kinetik energy when the rock hits the surface is
,
so
where g is the gravitational acceleration generated by the Sun at R,
.
One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,
Where,
N = Number of loops
B = Magnetic Field
A = Cross-sectional Area
= Angular velocity
Re-arrange to find N,
Our values are given as,
Replacing at our equation we have:
Therefore the number of loops of wire should be wound on the square armature is 32 loops
Answer:
A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.
B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.
[D] = 0.90 M
Explanation:
With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.
This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.
- The first attached image contains the correct question.
- The solution to part A is presented in the second attached image.
- The solution to part B is presented in the third attached image.
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.
To put a finer point on it, let's give the car a direction. Say it's driving North.
a). From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .
b). From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.
c). A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.
The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.
Now follow me here . . .
The car and the pole are both seen to be moving south. BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.
That's what everybody on the train sees.
==============================================
Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:
You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ?
Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself ! Only of others.
The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else. And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.
And now I believe that I have adequately milked this one for 50 points worth.
