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gizmo_the_mogwai [7]
3 years ago
5

Find the electric potential VP at point P. [Hint: To input a natural logarithm into the answer box, simply type the letters "ln"

(for example ln(x) is the natural logarithm of x).] Express your answer in terms of d, L, Q, and ϵ0.

Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

After finding the electric potential VP at point P = Q/Чπϵ₀L ㏑(1+\frac{L}{d})

Explanation:

I believe it is a part C question.  

The derivative of V and P  will be directly proportional to the differential dq and the inverse of Чπϵ₀δ........

Please find detailed solution in the attached picture as i believe that is the answer to the part C question you are seeking for.

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Answer: 9.91 * 10^-5

Explanation: acellus

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3 years ago
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What is the relationship between earths spheres and matter?
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4 years ago
En un instante pasa por A un cuerpo con MRU de 20 m/seg, 5 segundos después pasa en su persecución, por el mismo punto A otro cu
sukhopar [10]

Answer:

Los cuerpos se encuentran luego de 15 segundos a los 300 metros.

Explanation:

El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante. Es decir, que en este caso el movimiento es lineal en una única dirección  y la velocidad de desplazamiento es constante.

La posición del cuerpo después de un tiempo se calcula a partir de la posición inicial y de la velocidad del cuerpo mediante la expresión:

x=x0+v⋅t

donde:

  • x0 es la posición inicial.
  • v es la velocidad que tiene el cuerpo a lo largo del movimiento.
  • t es el intervalo de tiempo durante el cual se mueve el cuerpo.

En este caso, si el tiempo empleado por el primer cuerpo es t, el del segundo que sale 5 segundos más tarde será t-5. Siendo la velocidad del primer cuerpo 20 m/s y la del segundo cuerpo 30 m/s, entones la posición de cada uno será:

x1 = 20 m/s* t

x2 = 30 m/s* (t - 5 s)

Ambos se encuentran cuando sus posiciones son iguales:

x2=x1

30*(t - 5) = 20*t

30*t - 30*5= 20*t

30*t - 150 = 20 t

30*t - 20*t= 150

10*t= 150

t= 150÷10

t=15 segundos

Reemplazando en la expresiones de posición obtienes:

x1 = 20 m/s* 15 s= 300 m

x2 = 30 m/s* (15 s - 5 s)= 300 m

<u><em>Los cuerpos se encuentran luego de 15 segundos a los 300 metros.</em></u>

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-10.267

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