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KIM [24]
3 years ago
10

What happens to the charge on the conductive sphere when it is connected to a source of charge such as the electrostatic voltage

source?
Physics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer and Explanation:

The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.

Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.

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Compare the gravitational force on a 33 kg mass at the surface of the Earth (with ra-
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The formula we will be using is;F = G m1 m2 / r^2 
F earth = 308N 
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4 years ago
Problema 1. Un Clavadista se lanza desde un trampolín a diferentes alturas 10 m, 3 m, y 1 m, Calcular:
Salsk061 [2.6K]

La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

<h3 /><h3>Velocidad </h3>

La velocidad es la relación entre la distancia total recorrida y el tiempo total empleado. Está dado por:

Velocidad = distancia/tiempo

Para 10m de altura:

  • Velocidad = 10 m/1.42 s = 7.04 m/s

Para 3m de altura:

  • Velocidad = 3 m/0.78 s = 3.84 m/s

Para 1m de altura:

  • Velocidad = 1 m/0.44 s = 2.27 m/s

La velocidad del buzo es de 7.04 m/s, 8.84 m/s y 2.27 m/s respectivamente.

Obtenga más información sobre la velocidad en: brainly.com/question/4931057

4 0
3 years ago
S.I unit for distance =______
garri49 [273]

Answer:

opinion a

Explanation:

the si units of distance is metre (m)

3 0
3 years ago
Read 2 more answers
Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12
Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

F_C = k\frac{C_1C_2}{R^2}

where k = 8.99\times10^9 nm^2/C^2 is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}

C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}

Since their total charge is 200C:

C_1 + C_2 = 200 or C_1 = 200 - C_2

We can substitute the above equation

C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}

-C_2^2 +200C_2 - 2.13\times10^{-9} = 0

C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}

C= \frac{-200\pm200}{-2}

C = 1.06 \times 10^{-11} or C \approx 200

So the larger charge is C = 200 C

8 0
3 years ago
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