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Sagot :
To determine the gravitational force between two cars, each with a mass of 1050 kg, at various distances, we use Newton's law of universal gravitation. This law states that the gravitational force [tex]\(F\)[/tex] between two masses [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] separated by a distance [tex]\(r\)[/tex] is given by:
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [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] 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.
Let's plug in the given values and calculate the gravitational force for each of the distances.
### Calculation for Distance A: 5.6 meters
[tex]\[ F_A = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{5.6^2} \][/tex]
### Calculation for Distance B: 18 meters
[tex]\[ F_B = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{18^2} \][/tex]
### Calculation for Distance C: 33 meters
[tex]\[ F_C = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{33^2} \][/tex]
### Calculation for Distance D: 21 meters
[tex]\[ F_D = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{21^2} \][/tex]
Based on the calculations, we obtain the following gravitational forces:
- For Distance A (5.6 meters): [tex]\(2.34643359375 \times 10^{-6} \, \text{N}\)[/tex]
- For Distance B (18 meters): [tex]\(2.2711159722222221 \times 10^{-7} \, \text{N}\)[/tex]
- For Distance C (33 meters): [tex]\(6.757039256198347 \times 10^{-8} \, \text{N}\)[/tex]
- For Distance D (21 meters): [tex]\(1.668575 \times 10^{-7} \, \text{N}\)[/tex]
Thus, the gravitational forces between the two cars at the given distances are:
- For distance A (5.6 m): [tex]\(2.34643359375 \times 10^{-6} \, \text{N}\)[/tex]
- For distance B (18 m): [tex]\(2.2711159722222221 \times 10^{-7} \, \text{N}\)[/tex]
- For distance C (33 m): [tex]\(6.757039256198347 \times 10^{-8} \, \text{N}\)[/tex]
- For distance D (21 m): [tex]\(1.668575 \times 10^{-7} \, \text{N}\)[/tex]
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [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] 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.
Let's plug in the given values and calculate the gravitational force for each of the distances.
### Calculation for Distance A: 5.6 meters
[tex]\[ F_A = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{5.6^2} \][/tex]
### Calculation for Distance B: 18 meters
[tex]\[ F_B = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{18^2} \][/tex]
### Calculation for Distance C: 33 meters
[tex]\[ F_C = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{33^2} \][/tex]
### Calculation for Distance D: 21 meters
[tex]\[ F_D = \frac{6.67430 \times 10^{-11} \times 1050 \times 1050}{21^2} \][/tex]
Based on the calculations, we obtain the following gravitational forces:
- For Distance A (5.6 meters): [tex]\(2.34643359375 \times 10^{-6} \, \text{N}\)[/tex]
- For Distance B (18 meters): [tex]\(2.2711159722222221 \times 10^{-7} \, \text{N}\)[/tex]
- For Distance C (33 meters): [tex]\(6.757039256198347 \times 10^{-8} \, \text{N}\)[/tex]
- For Distance D (21 meters): [tex]\(1.668575 \times 10^{-7} \, \text{N}\)[/tex]
Thus, the gravitational forces between the two cars at the given distances are:
- For distance A (5.6 m): [tex]\(2.34643359375 \times 10^{-6} \, \text{N}\)[/tex]
- For distance B (18 m): [tex]\(2.2711159722222221 \times 10^{-7} \, \text{N}\)[/tex]
- For distance C (33 m): [tex]\(6.757039256198347 \times 10^{-8} \, \text{N}\)[/tex]
- For distance D (21 m): [tex]\(1.668575 \times 10^{-7} \, \text{N}\)[/tex]
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