Sure, let's solve this step-by-step.
The gravitational force between two objects can be calculated using Newton's law of universal gravitation, which is given by the formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force between the objects
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot ( \text{m}^2 / \text{kg}^2) \)[/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
Here we have:
- [tex]\( m_1 = 0.1 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 0.1 \, \text{kg} \)[/tex]
- [tex]\( r = 0.15 \, \text{m} \)[/tex]
Step-by-step solution:
1. We need to substitute the given values into the formula.
2. The formula becomes:
[tex]\[ F = 6.67 \times 10^{-11} \, \text{N} \cdot \frac{(0.1 \, \text{kg}) \cdot (0.1 \, \text{kg})}{(0.15 \, \text{m})^2} \][/tex]
3. First, calculate the product of the masses:
[tex]\[ 0.1 \, \text{kg} \times 0.1 \, \text{kg} = 0.01 \, \text{kg}^2 \][/tex]
4. Next, calculate the square of the distance:
[tex]\[ (0.15 \, \text{m})^2 = 0.0225 \, \text{m}^2 \][/tex]
5. Now, substitute these values back into the formula:
[tex]\[ F = 6.67 \times 10^{-11} \, \frac{0.01 \, \text{kg}^2}{0.0225 \, \text{m}^2} \][/tex]
6. Simplify the fraction:
[tex]\[ F = 6.67 \times 10^{-11} \times \frac{0.01}{0.0225} \][/tex]
[tex]\[ F = 6.67 \times 10^{-11} \times 0.4444 \][/tex]
7. Finally, multiply the values:
[tex]\[ F \approx 2.9644 \times 10^{-11} \, \text{N} \][/tex]
Thus, the gravitational force between the shoes is approximately [tex]\( 2.9644 \times 10^{-11} \, \text{N} \)[/tex].
Among the given choices:
A. [tex]\( 2.96 \times 10^{-10} \, \text{N} \)[/tex]
B. [tex]\( 296 \times 10^{-11} \, \text{N} \)[/tex]
C. [tex]\( 4.44 \times 10^{-11} \, \text{N} \)[/tex]
D. [tex]\( 4.44 \times 10^{-12} \, \text{N} \)[/tex]
The correct answer is not exactly listed among the choices, but the closest one, considering significant figures and rounding, would be:
B. [tex]\( 296 \times 10^{-11} \, \text{N} \)[/tex]
This is equivalent to [tex]\( 2.96 \times 10^{-10} \, \text{N} \)[/tex] when rewritten in standard form.
