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Juliette [100K]
3 years ago
8

Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70

cm. Determine the magnitude of the electrical force of repulsion between them.
Physics
1 answer:
blagie [28]3 years ago
8 0

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

F=k\frac{q_{1} q_{2} }{r^{2}}

where k is a proportionality constant known as the Coulomb's law constant. Its value is 9 \times 10^{9} Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = 4 \times 10^{-9} C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN

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Answer:

Los gases siempre se adaptaran al volumen del contenedor en el que estén.

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P*V = n*k*T

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(Notar que si el frasco no estuviera cerrado, entonces el numero n variaría de forma incontrolable, y este problema no se podría resolver de forma sencilla)

Entonces, si la presión baja, también baja la temperatura, pero el volumen se mantendrá constante, eso es lo importante.

Como el volumen se mantiene constante, el volumen que tomara el gas va a ser igual al volumen del frasco, sabemos que el volumen de un cilindro es:

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r = radio

pi = 3.14

h = altura.

en este caso, r = 5'' = 5 in

                       h = 7 in

Entonces el volumen sera:

V = 3.14*(5in)^2*7in = 549.5 in^2

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<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
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</span>
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