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timofeeve [1]
2 years ago
6

1. Use Coulomb’s Law (equation below) to calculate the approximate force felt by an electron at point A in the schematic below.

For the sake of simplicity, assume that the charge reservoirs, Q- and Q+ , can be treated as point charges each 2.5 cm away from point A and 5.0 cm away from each other. Make sure to denote which direction the force will act. The red and blue regions indicated in the schematic are conductors. Here, Q- is equal to the charge of six electrons, and Q+ is equal to the charge of six protons.
Physics
1 answer:
Amiraneli [1.4K]2 years ago
8 0

Answer:

Explanation:

From the data it appears that A is the middle point between two charges.

First of all we shall calculate the field at point A .

Field due to charge -Q ( 6e⁻ ) at A

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

Its direction will be towards Q⁻

Same field will be produced by Q⁺ charge . The direction will be away

from Q⁺  towards Q⁻ .

We shall add the field  to get the resultant field  .

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

Force on electron put at A

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

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Un niño de 25 kg corre por un jardín con una velocidad de 2.5 m/s de forma que su trayectoria es tangente al borde de un carruse
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Answer:

La velocidad angular del niño y del carrusel cuando se mueven juntos es 0.208 radianes por segundo.

Explanation:

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Donde:

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v - Velocidad lineal inicial del niño, medida en metros por segundo.

R - Radio máximo del carrusel, medida en metros.

I - Momento de inercia del carrusel, medida en kilogramo-metros cuadrados.

\omega - Velocidad angular final del sistema niño-carrusel, medida en radianes por segundo.

Si sabemos que m = 25\,kg, v = 2.5\,\frac{m}{s}, R = 2\,m y I = 500\,kg\cdot m^{2}, tenemos que la velocidad angular final es:

\omega = \frac{m\cdot v\cdot R}{m\cdot R^{2}+I}

\omega = \frac{(25\,kg)\cdot \left(2.5\,\frac{m}{s} \right)\cdot (2\,m)}{(25\,kg)\cdot (2\,m)^{2}+500\,kg\cdot m^{2}}

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Answer:

The acceleration of the rocket is 10 m/s².

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Mass of the rocket is, m=50\ kg

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W=mg=50\times 9.8=490\ N

Now, net force acting on the rocket in upward direction is given as:

F_{net}=F_{th}-W\\F_{net}=990-490=500\ N

Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.

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Answer:

35 m/s down

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v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.

3 0
2 years ago
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