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Lady_Fox [76]
3 years ago
5

1. Calcula la fuerza de atracción electrostática entre dos cuerpos de cargas q1 = -18 C y q2 = +5 mC, separados entre sí por una

distancia de 25 cm. ¿Qué ocurre si se duplica la distancia?
Physics
1 answer:
romanna [79]3 years ago
6 0

Answer:

A) F=-20.16×10⁹N

B) if the distance doubles, force is 4 times smaller.

Explanation:

q1=-28C

q2=5mC=0.005C

d=25cm=0.25m

Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.

Thus:

F=9×10⁹×(-28)×0.005/0.25²

F=-20.16×10⁹N

The minus sign indicates attraction.

If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.

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Calculate the tension on the rope.<br> A. 57.89 N<br> B. 32.73 N<br> C. 69.84 N<br> D. 12.55 N
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Hi there!

\large\boxed{\text{B. 32.73N}}

To calculate the tension, we must calculate the acceleration of the system.

Begin with a summation of forces:

∑F = -M₁gsinФ + T - T + M₂g

Simplify and solve for acceleration: (Tensions cancel out)

a = \frac{-M_1gsin\theta + T - T + M_2g}{M_1+M_2}

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