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mixas84 [53]
3 years ago
12

Q¹=0,07Mc Q²=2C r=1,08 cm F=.......?

Physics
1 answer:
Nostrana [21]3 years ago
5 0
The force (F) of attraction or repulsion between two point charges (Q1 and Q2) is given by the following rule:
F = <span>(k * q1 * q2) / (r^2)  where:
</span>q1 and q2 are the charges
k is coulomb's constant = 9 x 10^9<span> N. m</span>2/ C<span>2
</span>r is the distance between the two charges.

Applying the givens in the mentioned equation, we find that:
F = (9 x 10^9<span> x 0.07 x 10^6 x 2) / (0.0108)^2 = 1.08 x 10^19 n </span> 
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ruslelena [56]

Answer:

  r = 0.765 m

Explanation:

For this exercise let's use the law of universal gravitation

        F = G \ \frac{m_1m_2}{r^2}

        r = \sqrt{G \frac{m_1m_2}{F} }

let's reduce the magnitudes to the SI system

         m₁ = 250 g (1 kg / 1000g) = 0.250 kg

         m₂ = 650 g (1 kg / 1000g) = 0.650

         F = 1.85 10-6 dynes (1 N / 10⁵ dynes) = 1.85 10-11 N

let's calculate

        r = \sqrt{ \frac{ 6.67 \ 10^{-11} \ 0.250 \ 0.650}{1.85 \ 10^{-11}} }

        r = √0.58587

        r = 0.765 m

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