To determine how far apart the [tex]$4\, \text{kg}$[/tex] book and the [tex]$7\, \text{kg}$[/tex] lamp are, given that the gravitational force between them is [tex]$2.99 \times 10^{-10}\, \text{N}$[/tex], we will 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,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\( 6.67430 \times 10^{-11}\, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the book [tex]\( 4\, \text{kg} \)[/tex],
- [tex]\( m_2 \)[/tex] is the mass of the lamp [tex]\( 7\, \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the distance between the book and the lamp.
First, we will rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{G m_1 m_2}{F}} \][/tex]
Now, substitute the given values into the equation:
[tex]\[ r = \sqrt{\frac{(6.67430 \times 10^{-11}\, \text{m}^3 \text{kg}^{-1} \text{s}^{-2})(4 \, \text{kg})(7 \, \text{kg})}{2.99 \times 10^{-10} \, \text{N}}} \][/tex]
Performing the calculation:
[tex]\[ r = \sqrt{\frac{(6.67430 \times 10^{-11})(4)(7)}{2.99 \times 10^{-10}}} \][/tex]
[tex]\[ r \approx 2.50 \, \text{m} \][/tex]
Thus, the distance between the book and the lamp is approximately [tex]\( 2.50 \, \text{m} \)[/tex].
Therefore, the correct answer is:
C. [tex]\( 2.50 \, \text{m} \)[/tex]
