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Sagot :
To determine the distance between the [tex]$2 \, \text{kg}$[/tex] laptop and the [tex]$4 \, \text{kg}$[/tex] jar of pennies, given the gravitational force of [tex]$3.42 \times 10^{-10} \, \text{N}$[/tex], we can 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]\( (6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}) \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( F = 3.42 \times 10^{-10} \, \text{N} \)[/tex],
- [tex]\( m_1 = 2 \, \text{kg} \)[/tex],
- [tex]\( m_2 = 4 \, \text{kg} \)[/tex].
First, we rearrange the gravitational force formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
Next, substitute the known values into the equation:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}\right) \frac{(2 \, \text{kg}) (4 \, \text{kg})}{3.42 \times 10^{-10} \, \text{N}} \][/tex]
Simplify the expression inside the fraction:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}\right) \frac{8 \, \text{kg}^2}{3.42 \times 10^{-10} \, \text{N}} \][/tex]
Calculate the numerical result of the fraction:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11}\right) \frac{8}{3.42 \times 10^{-10}} \][/tex]
[tex]\[ r^2 \approx 1.561239766081871 \, \text{m}^2 \][/tex]
Now, take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{1.561239766081871} \, \text{m} \][/tex]
[tex]\[ r \approx 1.2494958047476075 \, \text{m} \][/tex]
Rounding to two decimal places, the distance [tex]\( r \)[/tex] is approximately [tex]\( 1.25 \, \text{m} \)[/tex].
Therefore, the correct answer is:
A. [tex]\(\boxed{1.25 \, \text{m}}\)[/tex]
[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]\( (6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}) \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( F = 3.42 \times 10^{-10} \, \text{N} \)[/tex],
- [tex]\( m_1 = 2 \, \text{kg} \)[/tex],
- [tex]\( m_2 = 4 \, \text{kg} \)[/tex].
First, we rearrange the gravitational force formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
Next, substitute the known values into the equation:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}\right) \frac{(2 \, \text{kg}) (4 \, \text{kg})}{3.42 \times 10^{-10} \, \text{N}} \][/tex]
Simplify the expression inside the fraction:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1}\right) \frac{8 \, \text{kg}^2}{3.42 \times 10^{-10} \, \text{N}} \][/tex]
Calculate the numerical result of the fraction:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11}\right) \frac{8}{3.42 \times 10^{-10}} \][/tex]
[tex]\[ r^2 \approx 1.561239766081871 \, \text{m}^2 \][/tex]
Now, take the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{1.561239766081871} \, \text{m} \][/tex]
[tex]\[ r \approx 1.2494958047476075 \, \text{m} \][/tex]
Rounding to two decimal places, the distance [tex]\( r \)[/tex] is approximately [tex]\( 1.25 \, \text{m} \)[/tex].
Therefore, the correct answer is:
A. [tex]\(\boxed{1.25 \, \text{m}}\)[/tex]
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