Answer :
Final Answer:
Magnitude of the net gravitational force exerted by these objects on the 61.0-kg object placed midway between them: 1.89 x 10⁻⁹N
Explanation:
To calculate the net gravitational force [tex](\(F\))[/tex] exerted by the 195-kg and 495-kg objects on the 61.0-kg object placed midway between them, we can use Newton's law of universal gravitation:
[tex]\[F = \frac{{G \times m_1 \times m_2}}{{r^2}}\][/tex]
Where:
[tex]\(F\)[/tex] = gravitational force (in newtons, N)
[tex]\(G\)[/tex] = gravitational constant [tex](\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\))[/tex]
[tex]\(m_1\)[/tex] = mass of the first object (in kilograms, kg)
[tex]\(m_2\)[/tex] = mass of the second object (in kilograms, kg)
[tex]\(r\)[/tex] = distance between the centers of the two objects (in meters, m)
In this case,[tex]\(m_1 = 195 \, \text{kg}\), \(m_2 = 495 \, \text{kg}\), and \(r = 3.20 \, \text{m}\).[/tex] Plugging in these values, we get:
[tex]\[F = \frac{{(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2) \times (195 \, \text{kg}) \times (495 \, \text{kg})}}{{(3.20 \, \text{m})^2}}\][/tex]
Solving for [tex]\(F\), we find \(F = 1.89 \times 10^{-9} \, \text{N}\).[/tex]
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