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Sagot :
To find the gravitational force between two objects, you can use Newton's law of universal gravitation. The formula is:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two 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 two objects (in kilograms),
- [tex]\( r \)[/tex] is the distance between the centers of the two objects (in meters).
Given:
- The mass of each bowling ball, [tex]\( m_1 = m_2 = 8 \, \text{kg} \)[/tex],
- The distance between the two bowling balls, [tex]\( r = 2 \, \text{m} \)[/tex].
Now plug these values into the formula:
[tex]\[ F = \left( 6.67 \times 10^{-11} \, \text{N} \cdot \left(\text{m}^2 / \text{kg}^2\right) \right) \frac{8 \, \text{kg} \times 8 \, \text{kg}}{(2 \, \text{m})^2} \][/tex]
First, calculate [tex]\( (2 \, \text{m})^2 \)[/tex]:
[tex]\[ (2 \, \text{m})^2 = 4 \, \text{m}^2 \][/tex]
Next, calculate [tex]\( 8 \, \text{kg} \times 8 \, \text{kg} \)[/tex]:
[tex]\[ 8 \, \text{kg} \times 8 \, \text{kg} = 64 \, \text{kg}^2 \][/tex]
Now the formula looks like this:
[tex]\[ F = \left( 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2 / \text{kg}^2) \right) \frac{64 \, \text{kg}^2}{4 \, \text{m}^2} \][/tex]
Divide [tex]\( 64 \, \text{kg}^2 \)[/tex] by [tex]\( 4 \, \text{m}^2 \)[/tex]:
[tex]\[ \frac{64 \, \text{kg}^2}{4 \, \text{m}^2} = 16 \, \text{kg}^2 / \text{m}^2 \][/tex]
Now multiply:
[tex]\[ F = 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2 / \text{kg}^2) \times 16 \, \text{kg}^2 / \text{m}^2 \][/tex]
[tex]\[ F = 6.67 \times 10^{-11} \times 16 \][/tex]
Calculate [tex]\( 6.67 \times 16 \)[/tex]:
[tex]\[ 6.67 \times 16 = 106.72 \][/tex]
Now add the exponent:
[tex]\[ 106.72 \times 10^{-11} = 1.0672 \times 10^{-9} \, \text{N} \][/tex]
So, the gravitational force between the two bowling balls is:
[tex]\[ F = 1.0672 \times 10^{-9} \, \text{N} \][/tex]
Referring to the given options:
- A. [tex]\( 2.14 \times 10^{-9} \, \text{N} \)[/tex]
- B. [tex]\( 3.21 \times 10^{-8} \, \text{N} \)[/tex]
- C. [tex]\( 2.68 \times 10^{-10} \, \text{N} \)[/tex]
- D. [tex]\( 1.07 \times 10^{-9} \, \text{N} \)[/tex]
The closest and correct answer is:
D. [tex]\( 1.07 \times 10^{-9} \, \text{N} \)[/tex]
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the two 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 two objects (in kilograms),
- [tex]\( r \)[/tex] is the distance between the centers of the two objects (in meters).
Given:
- The mass of each bowling ball, [tex]\( m_1 = m_2 = 8 \, \text{kg} \)[/tex],
- The distance between the two bowling balls, [tex]\( r = 2 \, \text{m} \)[/tex].
Now plug these values into the formula:
[tex]\[ F = \left( 6.67 \times 10^{-11} \, \text{N} \cdot \left(\text{m}^2 / \text{kg}^2\right) \right) \frac{8 \, \text{kg} \times 8 \, \text{kg}}{(2 \, \text{m})^2} \][/tex]
First, calculate [tex]\( (2 \, \text{m})^2 \)[/tex]:
[tex]\[ (2 \, \text{m})^2 = 4 \, \text{m}^2 \][/tex]
Next, calculate [tex]\( 8 \, \text{kg} \times 8 \, \text{kg} \)[/tex]:
[tex]\[ 8 \, \text{kg} \times 8 \, \text{kg} = 64 \, \text{kg}^2 \][/tex]
Now the formula looks like this:
[tex]\[ F = \left( 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2 / \text{kg}^2) \right) \frac{64 \, \text{kg}^2}{4 \, \text{m}^2} \][/tex]
Divide [tex]\( 64 \, \text{kg}^2 \)[/tex] by [tex]\( 4 \, \text{m}^2 \)[/tex]:
[tex]\[ \frac{64 \, \text{kg}^2}{4 \, \text{m}^2} = 16 \, \text{kg}^2 / \text{m}^2 \][/tex]
Now multiply:
[tex]\[ F = 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2 / \text{kg}^2) \times 16 \, \text{kg}^2 / \text{m}^2 \][/tex]
[tex]\[ F = 6.67 \times 10^{-11} \times 16 \][/tex]
Calculate [tex]\( 6.67 \times 16 \)[/tex]:
[tex]\[ 6.67 \times 16 = 106.72 \][/tex]
Now add the exponent:
[tex]\[ 106.72 \times 10^{-11} = 1.0672 \times 10^{-9} \, \text{N} \][/tex]
So, the gravitational force between the two bowling balls is:
[tex]\[ F = 1.0672 \times 10^{-9} \, \text{N} \][/tex]
Referring to the given options:
- A. [tex]\( 2.14 \times 10^{-9} \, \text{N} \)[/tex]
- B. [tex]\( 3.21 \times 10^{-8} \, \text{N} \)[/tex]
- C. [tex]\( 2.68 \times 10^{-10} \, \text{N} \)[/tex]
- D. [tex]\( 1.07 \times 10^{-9} \, \text{N} \)[/tex]
The closest and correct answer is:
D. [tex]\( 1.07 \times 10^{-9} \, \text{N} \)[/tex]
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