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In-s [12.5K]
2 years ago
9

¿Por que una una tapa pequeña de agua purificada se mantiene quieta por más tiempo sobre una rampa, en comparación al tiempo red

ucido que permanece un carrito del supermercado sobre la misma superficie?
Physics
1 answer:
shusha [124]2 years ago
5 0

Answer:

Esto sucede principalmente por el movimiento rotacional de las ruedas del carrito.

Como sabemos, existen dos tipos de coeficientes de fricción, el estacionario y el cinético.

En principio, la tapa se mantendrá quieta por que no tendrá la suficiente fuerza como para vencer al coeficiente de fricción estático, por lo que no podrá moverse hasta que esta reciba un pequeño impulso, como puede ser una pequeña corriente de aire. (asumo que la superficie curva de la tapita es la que esta en contacto con la rampa)

(Voy a simplificar el movimiento rotacional, pero creo que es suficiente para explicar la situación)

Ahora, en el carrito, las ruedas están conectadas a un eje.

La fuerza gravitatoria podemos pensarla que esta arriba de este eje, mas o menos en el centro del carrito, y esta fuerza intentara que el carrito baje de la rampa (lo mismo pasa para la tapita), pero a diferencia de la tapita, esta fuerza es aplicada a una distancia dada del eje (o del centro de las rueditas) lo que podemos pensar que funciona como una palanca, amplificando así la fuerza, por lo que esta vez el rozamiento estático podrá detener la superficie de la rueda que esta en contacto con el suelo, pero cuanto el centro de masa se mueva un poco, el punto de contacto entre la rueda y el suelo se moverá, de esta manera aparece lo que llamamos movimiento rotacional (si no hubiera fricción, las ruedas simplemente deslizarían sobre la rampa) que le permite al carrito ignorar en parte al coeficiente de fricción estático.

Otro factor importante, es que las tapitas suelen tener una superficie rugosa y un centro de masa desfazado, lo que no las hace buenas rotadoras.

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Answer:

             E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

       E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element  and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

         λ = Q / L

If we derive from the length we have

        λ = dq/dx       ⇒    dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

        dE = k dq / x²2

        dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

        E = k \int\limits^{d+L}_d {\lambda/x^{2}} \, dx

We take out the constant magnitudes and perform the integral

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Evaluating

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Using   λ = Q/L

        E = k Q/L [ 1/d  - 1/ (d+L)]

 

let's use a bit of arithmetic to simplify the expression

     [ 1/d  - 1/ (d+L)]   = L /[d(d+L)]

The final result is

     E = k Q / [d(d+L)]

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Answer:

A

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Begin as protostars, which fire up when they collapse and become denser and hotter.

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3 years ago
Use the five rules to determine the validity of this categorical syllogism:
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Question:

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An argument whose conclusion is  implied based on the logical stipulation of the premise is a deductive argument

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Here, the distributed term in the conclusion are "extremely dense objects" is not distributed in the premise, therefore the categorical syllogism is invalid.

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