Two concrete spans of a 250-m-long bridge are placed end to end so that no room is allowed for expansion. If a temperature incre
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Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;
where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:
The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:
Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
The blocks must be pushed with a force higher than 359 Newtons horizontally in order to accomplish this friction levitation feat.
Explanation:
The first step in resolving any physics problem is to draw the given scenario (if possible), see the attached image to have an idea of the objects and forces involved.
The large cube in red is being pushed from the left by a force whose value is to be found. That cube has its own weight
, and it is associated with the force of gravity which points downward. Newton's third law stipulates that the response from the floor is an upward pointing force on the cube, and it's called the normal force
.
A second cube is being pushed by the first, and since the force is strong enough it is able to keep such block suspended as if it were glued to the first one, due to friction. As in the larger cube, the smaller one has a weight
pointing downwards, but the normal force in this block doesn't point upwards since its 'floor' isn't below it, but in its side, therefore the normal force directs it to the right as it is shown in the picture. Normal forces are perpendicular to the surface they contact. The final force is the friction between both cubes, that sets a resistance of one moving parallel the other. In this case, the weight of the block its the force pointing parallel to the contact surface, so the friction opposes that force, and thus points upwards. Friction forces can be set as
, where
is the coefficient of static friction between the cubes.
Now that all forces involved are identified, the following step is to apply Newton's second law and add all the forces for each block that point in the same line, and set it as equal its mass multiplied by its acceleration. The condition over the smaller box is the relevant one so its the first one to be analyzed.
In the vertical component: Since the idea is that it doesn't slips downwards, the vertical acceleration should be set to zero
, and making explicit the other forces:
[N]. In the last equation gravity's acceleration was rounded to 10 [m/s
].
In its horizontal component: , this time the horizontal acceleration is not zero, because it is constantly being pushed. However, the value of the normal force and the mass of the block are known, so its horizontal acceleration can be determined:
[m/s
]. Notice that this acceleration is higher than the one of gravity, and it is understandable since you should be able to push it harder than gravity in order for it to not slip.
Now the attention is switched to the larger cube. The vertical forces are not relevant here, since the normal force balances its weight so that there isn't vertical acceleration. The unknown force comes up in the horizontal forces analysis: , since the force
is not only pushing the first block but both, the mass involved in this equation is the combined masses of the blocks, the acceleration is the same for both blocks since they move together;
[N]. The resulting force is quite high but not impossible to make by a human being, this indicates that this feat of friction suspension is difficult but feasable.
