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Sagot :
To determine the mass of the lighter canoe, we will use Newton's law of universal gravitation, which 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 the formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/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]\( F \)[/tex] is the gravitational force ([tex]\( 2.378 \times 10^{-13} \, \text{N} \)[/tex])
- [tex]\( r \)[/tex] is the separation distance ([tex]\( 1,500 \, \text{m} \)[/tex])
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two canoes
We are given that one canoe is twice as massive as the other. Let's denote the mass of the lighter canoe as [tex]\( m \)[/tex]. Therefore, the mass of the heavier canoe will be [tex]\( 2m \)[/tex].
Using the given values in the gravitational force formula:
[tex]\[ 2.378 \times 10^{-13} = G \frac{m \cdot 2m}{(1500)^2} \][/tex]
Substitute [tex]\( G \)[/tex] and solve for [tex]\( m \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \frac{2m^2}{(1500)^2} \][/tex]
Let's isolate [tex]\( m^2 \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \frac{2m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \cdot \frac{2m^2}{2250000} \][/tex]
Simplify the equation:
[tex]\[ 2.378 \times 10^{-13} = \frac{6.67430 \times 10^{-11} \cdot 2m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = \frac{1.33486 \times 10^{-10} m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = \frac{1.33486 \times 10^{-10} m^2}{2250000} \][/tex]
Multiply both sides by [tex]\( 2250000 \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} \times 2250000 = 1.33486 \times 10^{-10} m^2 \][/tex]
[tex]\[ 5.3505 \times 10^{-7} = 1.33486 \times 10^{-10} m^2 \][/tex]
Now divide by [tex]\( 1.33486 \times 10^{-10} \)[/tex] to isolate [tex]\( m^2 \)[/tex]:
[tex]\[ m^2 = \frac{5.3505 \times 10^{-7}}{1.33486 \times 10^{-10}} \][/tex]
[tex]\[ m^2 \approx 4008.285513087515 \][/tex]
Take the square root of both sides to solve for [tex]\( m \)[/tex]:
[tex]\[ m \approx \sqrt{4008.285513087515} \][/tex]
[tex]\[ m \approx 63.311022050568056 \][/tex]
Rounded to two significant figures:
[tex]\[ m \approx 63.31 \, \text{kg} \][/tex]
Thus, the mass of the lighter canoe is approximately [tex]\( 63 \, \text{kg} \)[/tex].
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/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]\( F \)[/tex] is the gravitational force ([tex]\( 2.378 \times 10^{-13} \, \text{N} \)[/tex])
- [tex]\( r \)[/tex] is the separation distance ([tex]\( 1,500 \, \text{m} \)[/tex])
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two canoes
We are given that one canoe is twice as massive as the other. Let's denote the mass of the lighter canoe as [tex]\( m \)[/tex]. Therefore, the mass of the heavier canoe will be [tex]\( 2m \)[/tex].
Using the given values in the gravitational force formula:
[tex]\[ 2.378 \times 10^{-13} = G \frac{m \cdot 2m}{(1500)^2} \][/tex]
Substitute [tex]\( G \)[/tex] and solve for [tex]\( m \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \frac{2m^2}{(1500)^2} \][/tex]
Let's isolate [tex]\( m^2 \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \frac{2m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = 6.67430 \times 10^{-11} \cdot \frac{2m^2}{2250000} \][/tex]
Simplify the equation:
[tex]\[ 2.378 \times 10^{-13} = \frac{6.67430 \times 10^{-11} \cdot 2m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = \frac{1.33486 \times 10^{-10} m^2}{2250000} \][/tex]
[tex]\[ 2.378 \times 10^{-13} = \frac{1.33486 \times 10^{-10} m^2}{2250000} \][/tex]
Multiply both sides by [tex]\( 2250000 \)[/tex]:
[tex]\[ 2.378 \times 10^{-13} \times 2250000 = 1.33486 \times 10^{-10} m^2 \][/tex]
[tex]\[ 5.3505 \times 10^{-7} = 1.33486 \times 10^{-10} m^2 \][/tex]
Now divide by [tex]\( 1.33486 \times 10^{-10} \)[/tex] to isolate [tex]\( m^2 \)[/tex]:
[tex]\[ m^2 = \frac{5.3505 \times 10^{-7}}{1.33486 \times 10^{-10}} \][/tex]
[tex]\[ m^2 \approx 4008.285513087515 \][/tex]
Take the square root of both sides to solve for [tex]\( m \)[/tex]:
[tex]\[ m \approx \sqrt{4008.285513087515} \][/tex]
[tex]\[ m \approx 63.311022050568056 \][/tex]
Rounded to two significant figures:
[tex]\[ m \approx 63.31 \, \text{kg} \][/tex]
Thus, the mass of the lighter canoe is approximately [tex]\( 63 \, \text{kg} \)[/tex].
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