Answer :
To find the distance between two buckets of paint given their masses and the gravitational force between them, we can use the formula for gravitational force:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses.
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects.
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( m_1 = 9 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 9 \, \text{kg} \)[/tex]
- [tex]\( F = 6.0 \times 10^{-10} \, \text{N} \)[/tex]
We need to find [tex]\( r \)[/tex]. First, rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
[tex]\[ r = \sqrt{G \frac{m_1 m_2}{F}} \][/tex]
Now plug in the values:
[tex]\[ G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \][/tex]
[tex]\[ m_1 = 9 \, \text{kg} \][/tex]
[tex]\[ m_2 = 9 \, \text{kg} \][/tex]
[tex]\[ F = 6.0 \times 10^{-10} \, \text{N} \][/tex]
Now calculate:
[tex]\[ r = \sqrt{(6.67430 \times 10^{-11}) \frac{(9 \, \text{kg} \times 9 \, \text{kg})}{6.0 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{(6.67430 \times 10^{-11}) \frac{81 \, \text{kg}^2}{6.0 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \times 81}{6.0 \times 10^{-10}}} \][/tex]
[tex]\[ r = \sqrt{\frac{5.406183 \times 10^{-9}}{6.0 \times 10^{-10}}} \][/tex]
[tex]\[ r = \sqrt{9.010305} \][/tex]
[tex]\[ r \approx 3.0017170086468843 \][/tex]
Therefore, the distance between the two buckets of paint is approximately [tex]\( 3.0 \, \text{m} \)[/tex].
So, the correct answer is:
D. [tex]\( 3.0 \, \text{m} \)[/tex]
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses.
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex].
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects.
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( m_1 = 9 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 9 \, \text{kg} \)[/tex]
- [tex]\( F = 6.0 \times 10^{-10} \, \text{N} \)[/tex]
We need to find [tex]\( r \)[/tex]. First, rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
[tex]\[ r = \sqrt{G \frac{m_1 m_2}{F}} \][/tex]
Now plug in the values:
[tex]\[ G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \][/tex]
[tex]\[ m_1 = 9 \, \text{kg} \][/tex]
[tex]\[ m_2 = 9 \, \text{kg} \][/tex]
[tex]\[ F = 6.0 \times 10^{-10} \, \text{N} \][/tex]
Now calculate:
[tex]\[ r = \sqrt{(6.67430 \times 10^{-11}) \frac{(9 \, \text{kg} \times 9 \, \text{kg})}{6.0 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{(6.67430 \times 10^{-11}) \frac{81 \, \text{kg}^2}{6.0 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \times 81}{6.0 \times 10^{-10}}} \][/tex]
[tex]\[ r = \sqrt{\frac{5.406183 \times 10^{-9}}{6.0 \times 10^{-10}}} \][/tex]
[tex]\[ r = \sqrt{9.010305} \][/tex]
[tex]\[ r \approx 3.0017170086468843 \][/tex]
Therefore, the distance between the two buckets of paint is approximately [tex]\( 3.0 \, \text{m} \)[/tex].
So, the correct answer is:
D. [tex]\( 3.0 \, \text{m} \)[/tex]