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Nikitich [7]
2 years ago
10

La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a

Parita es de 12 km; exprese en pies ésta distancia
Physics
1 answer:
densk [106]2 years ago
7 0

Answer:

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Explanation:

La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.

Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:

a ⇒ b

c ⇒ x

Entonces x=\frac{c*b}{a}

La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?

1 km ⇒ 3280.84 pies

12 km ⇒ x

x=\frac{12 km*3280.84 pies}{1 km}

x= 39370.08 pies

<u><em>La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.</em></u>

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Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

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Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

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the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

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2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

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this expression holds for the points located at

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3) Point between the two spheres

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This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

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             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

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