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An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying a negative charge of 15.0 μC. What is the net force the asteroids exert upon each other? (G = 6.673 x 10-11 N∙m2/kg2 and k = 9.0 x 109 N∙m2/C2.) An asteroid of mass 5.7×104 kg carrying a negative charge of 14.0 μC is 190 m from a second asteroid of mass 5.5×104 kg carrying a negative charge of 15.0 μC. What is the net force the asteroids exert upon each other? (G = 6.673 x 10-11 N∙m2/kg2 and k = 9.0 x 109 N∙m2/C2.) 5.6×105 N 5.4×105 N 5.7×10−3 N 4.7×10−5 N

Respuesta :

Answer:

repulsion force es  R = 5,224 10⁻³ N

Explanation:

We use Newton's second law to add forces, the elective force is repulsive because the asteroids have the same type of charge and gravitational force is always attractive.

Gravitational force

      Fg = G m1 m2 / r²

      Fg = 6,673 10⁻¹¹ 5.7 10⁴ 5.5 10⁴/190²

      Fg = 5,795 10⁻⁶ N

Electric force

     Fe = k q1 q2 / r²

     Fe = 8.99 10⁹ 14.0 10⁻⁶ 15.0 10⁻⁶ / 190²

     Fe = 5.2296 10⁻³ N

The resulting force is

     R = Fe - Fg

     R = 5.2296 10⁻³ - 5.795 10⁻⁶=  5229.6 10⁻⁶ - 5,795 10⁻⁶

     R = 5,224 10⁻³ N

this is a repulsive force and the asteroids move away from each other

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