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Klio2033 [76]
3 years ago
13

Someone please help. Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign a

re placed near each other.
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

The answer is based on the conservation of energy law; something you should really understand by now.  

For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.  

So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.  

It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.  

It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.  

But there's more.  

Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.  

And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.

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What is a travelling wave and a standing wave? What are the differences between both of them?

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Which of the following is an example of an allocation decision?|
Oksana_A [137]

Answer:

The answer is D- Coal is set aside to burn as heating fuel

Explanation:

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¿Qué cantidad de calor absorbió una masa de 4 g de cinc al pasar de 20 °C a<br> 180 °C?
Alborosie

(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,

Ce = 0.093 cal/g. °c

Q = m C ΔT

Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )

Q= 4×0.093 × 160

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I hope I helped you^_^

7 0
2 years ago
Uma carga puntiforme de + 3,0uC é colocada em um ponto P de um campo elétrico gerado por uma partícula eletrizada com carga desc
expeople1 [14]

Responda:

1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5

Explicação:

Dado o seguinte:

Carga (q) = 3uC = 3 × 10 ^ -6C

Força elétrica (Fe) = 18N

Intensidade do campo elétrico (E) =?

1)

Lembre-se:

Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)

Fe = qE; E = Fe / q

E = 18N / (3 × 10 ^ -6C)

E = 6N / 10 ^ -6C

E = 6 × 10 ^ 6NC ^ -1

2)

Lembre-se:

E = kQ / r ^ 2

E = intensidade do campo elétrico

Q = carga de origem

r = distância de espera = 30cm = 30/100 = 0,3m

K = 9,0 × 10 ^ 9

6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2

9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09

Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9

Q = 0,06 × 10 ^ (6-9)

Q = 0,06 × 10 ^ -3

Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC

7 0
3 years ago
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