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Mamont248 [21]
2 years ago
5

What electric force would a stationary 3.8 C charge experience if it were far away from any other charges

Physics
1 answer:
MAVERICK [17]2 years ago
3 0

Answer:

The electric force will be  0 N

Explanation:

From the question we are told that

   The magnitude of the charge is  q_1 = 3.8 \ C

   

Generally from Coulombs law the electric force  between two charges is mathematically represented as

         F = \frac{ k  *  q_1 q_2 }{r^2}

Here r is the distance of separation between that two charges.

  Now from the question we are told that the charge is far away from any other charge hence we can say that the distance between the charge and any other charge is  r = \infty

So

       F = \frac{ k  *  3.8  * q_2 }{\infty^2}

=>    F =0 \ N

Hence the electric force will be  0 N

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Determine the ratio of the flow rate through capillary tubes A and B (that is, Qa/Qb).
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To solve this problem we can use the concepts related to the change of flow of a fluid within a tube, which is without a rubuleous movement and therefore has a laminar fluid.

It is sometimes called Poiseuille’s law for laminar flow, or simply Poiseuille’s law.

The mathematical equation that expresses this concept is

\dot{Q} = \frac{\pi r^4 (P_2-P_1)}{8\eta L}

Where

P = Pressure at each point

r = Radius

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Of all these variables we have so much that the change in pressure and viscosity remains constant so the ratio between the two flows would be

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

From the problem two terms are given

R_A = \frac{R_B}{2}

L_A = 2L_B

Replacing we have to

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_A^4}{r_B^4}\frac{L_B}{L_A}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{r_B^4}{16*r_B^4}\frac{L_B}{2*L_B}

\frac{\dot{Q_A}}{\dot{Q_B}} = \frac{1}{32}

Therefore the ratio of the flow rate through capillary tubes A and B is 1/32

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Answer:

In a closed system, the total energy is conserved or remains the same as energy transformations take place.

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