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Over [174]
3 years ago
14

Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie

ld at the midpoint between the two charges?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
3 0
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
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How does an atom of potassium-41 become a potassium ion with a +1 charge? 19 K 39.10
stiv31 [10]

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Movement of a molecule against its concentration gradient can occur through Select one or more: a. facilitated diffusion b. pass
uranmaximum [27]

Answer:

The answer to your question is:

Explanation:

There are two kinds of cell transport passive transportation and active transportation.

Passive transportation does not need energy because molecules move from higher concentration to lower concentration.

Active transportation needs energy because molecules moves against concentration.  

a. facilitated diffusion It's an example of passive transportation so this answer is wrong.

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3 years ago
A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the
Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁  = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

    m₁v₁ + m₂v₂  = v(m₁ + m₂)

Insert parameters and solve for v;

   5 x 1  + 15 x 0 = v (5 + 15 )

          5  = 20v

          v = \frac{5}{20}   = 0.25m/s

5 0
3 years ago
Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

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