To determine how far apart the 2 kg book and the 5 kg vase are based on the gravitational force between them, we can use Newton's Law of Universal Gravitation, which is given by:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the force of gravity between the objects,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \; \text{N} \cdot (\text{m}^2) / \text{kg}^2)\)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given the problem's parameters:
- [tex]\( F = 7.41 \times 10^{-11} \; \text{N} \)[/tex]
- [tex]\( m_1 = 2 \; \text{kg} \)[/tex]
- [tex]\( m_2 = 5 \; \text{kg} \)[/tex]
- [tex]\( G = 6.67430 \times 10^{-11} \; \text{N} \cdot \text{(m}^2) / \text{kg}^2 \)[/tex]
We need to solve for [tex]\( r \)[/tex]. Rearranging the formula to solve for [tex]\( r \)[/tex], we get:
[tex]\[ r = \sqrt{G \frac{m_1 m_2}{F}} \][/tex]
Substitute the provided values into the formula:
[tex]\[ r^2 = \frac{G \cdot m_1 \cdot m_2}{F} \][/tex]
[tex]\[ r^2 = \frac{6.67430 \times 10^{-11} \cdot 2 \cdot 5}{7.41 \times 10^{-11}} \][/tex]
[tex]\[ r^2 \approx 9.00715249662618 \][/tex]
Now take the square root of both sides to find [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{9.00715249662618} \][/tex]
[tex]\[ r \approx 3.00 \; \text{m} \][/tex]
Therefore, the distance between the 2 kg book and the 5 kg vase is approximately 3.00 meters.
So the correct answer is:
D. 3.00 m
