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alexira [117]
3 years ago
11

Two small plastic spheres each have a mass of 1.2 g and a charge of -56.0 nC . They are placed 3.0 cm apart (center to center).

Part A What is the magnitude of the electric force on each sphere? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
butalik [34]3 years ago
3 0

Answer:

F_e = +/- 0.013133 N

Explanation:

Given:

- charge on each sphere q = -56 nC

- separation of spheres r = 3.0 cm

- charge constant k = 8.99*10^9

Find:

- Magnitude of Electric Force F_e on each sphere.

Solution:

The magnitude of electric Force F_e of one sphere on other is:

                                F_e = k*q^2 / r^2

- Plugging the given values :

                                F_e = (8.99*10^9) * (56*10^-19)^2 / (0.03)^2

                                F_e = 0.013133 N

- An equal and opposite force is experienced on the other sphere. Hence, F_e = + / - 0.013133 N

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A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
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This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

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6 0
2 years ago
Una prenda de 320gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene 40 cm y gira con una frecuencia de 4
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Answer:

Período del tambor: T = 0.25\,s, fuerza sobre la prenda: F \approx 80.852\,N, velocidad lineal del tambor: v \approx 10.053\,\frac{m}{s}, velocidad angular del tambor: \omega \approx 25.133\,\frac{rad}{s}.

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

<em>"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle </em><em>a)</em><em> el período, </em><em>b) </em><em>la velocidad angular, </em><em>c) </em><em>la fuerza con la que gira la prenda y </em><em>d) </em><em>la velocidad lineal de la lavadora."</em>

El tambor gira a velocidad angular constante (\omega), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga (a), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor (T), en segundos:

T = \frac{1}{f} (1)

Donde f es la frecuencia, en hertz.

(f = 4\,hz)

T = \frac{1}{4\,hz}

T = 0.25\,s

Ahora determinamos la fuerza aplicada sobre la prenda (F), en newtons:

F = m\cdot a (2)

F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}} (2b)

Donde:

m - Masa de la prenda, en kilogramos.

r - Radio interior del tambor, en metros.

(m = 0.32\,kg, r = 0.4\,m, T = 0.25\,s)

F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}

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v = \frac{2\pi\cdot r}{T} (3)

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v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}

v \approx 10.053\,\frac{m}{s}

Y la velocidad angular del tambor de la lavadora:

\omega = \frac{2\pi}{T}

(T = 0.25\,s)

\omega = \frac{2\pi}{0.25\,s}

\omega \approx 25.133\,\frac{rad}{s}

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