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guapka [62]
3 years ago
10

A - 28.4-μC charge is placed 16.4 cm from a charge q, the force between the two charges is 1240 N. What is the value of q?

Physics
1 answer:
anastassius [24]3 years ago
3 0

Answer:

1.31×10⁻⁴ C or 131 μC

Explanation:

From the question,

Applying Coulomb's Law,

F = kqq'/r²..................... Equation 1

Where F = Force between the charges, q = first charge, q' = second charge, r = distance between the charges, k = coulomb's constant.

make q' the subject of the equation

q' = F×r²/(kq)

q' = Fr²/kq................ Equation 2

Given: F = 1240 N, r = 16.4 cm = 0.164 m, q = 28.4 μC = 2.84×10⁻⁵ C,

Constant: k = 8.98×10⁹ Nm/C²

Substitute these values into equation 2

q' = (1240×0.164²)/[(2.84×10⁻⁵)×(8.98×10⁹)

q' = (33.35104)/(25.5032×10⁴)

q' = 1.31×10⁻⁴ C

q' = 131 μC

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Answer:

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        y = y₀ + v_{oy} t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

       0 = y₀ – ½ g t²

       t= √ (2 y₀/g)

calculemos

       t= √ ( 2 12 / 9,8)

       t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida  

        x = v₀ₓ t

        x = 80 1,56

        x= 124,8 m

La velocidad de impacto cuando toca el suelo

        vx = v₀ₓ = 80 ms

        v_{y} = v_{oy} – gt

        v_{y} = - 9,8 1,56

        v_{y} = - 15,288 m/s

la velocidad es

       v = (80 i^ - 15,288 j ) m/s

Traducttion  

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

        y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero (v_{oy} = 0)

       0 =y₀I - ½ g t²

       t = √ (2y₀ / g)

let's calculate

       t = √ (2 12 / 9.8)

       t = 1.56 s

Projectile range is the horizontal distance traveled

        x = v₀ₓ t

        x = 80 1.56

        x = 124.8 m

Impact speed when it hits the ground

        vₓ = v₀ₓ = 80 ms

        v_{y} = v_{oy} - gt

        v_{y} = - 9.8 1.56

        v_{y} = - 15,288 m / s

the speed is

       v = (80 i ^ - 15,288 j) m / s

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