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alexgriva [62]
3 years ago
13

Willa, the cartoon witch, dusts her crystal ball with her silk scarf, causing it to become charged with 5.0 x 10^ -9 C. Willa th

en stares into the crystal ball and the wart on the end of her nose experiences an electric field strengths of 2200 N/C. How far is the trip of her nose from the center of the crystal ball?
Physics
1 answer:
DIA [1.3K]3 years ago
3 0

Answer:

The distance between the tip of her nose from the center of the crystal ball is 14.3 cm.

Explanation:

Given that,

Charge, q=5\times 10^{-9}\ C

The end of her nose experiences an electric field strengths of 2200 N/C

It is required to find the distance between the tip of her nose from the center of the crystal ball. The formula of the electric field is given by :

E=\dfrac{kq}{r^2}

r is distance

r=\sqrt{\dfrac{kq}{E}} \\\\r=\sqrt{\dfrac{9\times 10^9\times 5\times 10^{-9}}{2200}} \\\\r=0.143\ m\\\\r=14.3\ cm

So, the distance between the tip of her nose from the center of the crystal ball is 14.3 cm.

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4 years ago
Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,
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Answer:

a) La magnitud del desplazamiento es de 5 m

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b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}

Lo que da;

\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}

Lo que da;

\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

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c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

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