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Sagot :
To find the distance between a [tex]$2 \, \text{kg}$[/tex] laptop and a [tex]$4 \, \text{kg}$[/tex] jar of pennies, given that the gravitational force between them is [tex]$3.42 \times 10^{-10} \, \text{N}$[/tex], we can use Newton's law of universal gravitation. The formula for 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:
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the universal gravitational constant [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex].
Let's isolate [tex]\( r \)[/tex] in the formula:
[tex]\[ r^2 = \frac{G \cdot m_1 \cdot m_2}{F} \][/tex]
Taking the square root of both sides will give us [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}} \][/tex]
Plugging in the given values:
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
- [tex]\( m_1 = 2 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 4 \, \text{kg} \)[/tex]
- [tex]\( F = 3.42 \times 10^{-10} \, \text{N} \)[/tex]
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times 2 \, \text{kg} \times 4 \, \text{kg}}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
Simplifying the expression under the square root:
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times 8 \, \text{kg}^2}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{\frac{5.33944 \times 10^{-10} \, \text{m}^3 \, \text{s}^{-2}}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{1.56122 \, \text{m}^2} \][/tex]
[tex]\[ r \approx 1.25 \, \text{m} \][/tex]
Therefore, the distance between the laptop and the jar of pennies is closest to option A.
So, the correct answer is:
A. [tex]$1.25 \, \text{m}$[/tex].
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the universal gravitational constant [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex].
Let's isolate [tex]\( r \)[/tex] in the formula:
[tex]\[ r^2 = \frac{G \cdot m_1 \cdot m_2}{F} \][/tex]
Taking the square root of both sides will give us [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{G \cdot m_1 \cdot m_2}{F}} \][/tex]
Plugging in the given values:
- [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
- [tex]\( m_1 = 2 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 4 \, \text{kg} \)[/tex]
- [tex]\( F = 3.42 \times 10^{-10} \, \text{N} \)[/tex]
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times 2 \, \text{kg} \times 4 \, \text{kg}}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
Simplifying the expression under the square root:
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times 8 \, \text{kg}^2}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{\frac{5.33944 \times 10^{-10} \, \text{m}^3 \, \text{s}^{-2}}{3.42 \times 10^{-10} \, \text{N}}} \][/tex]
[tex]\[ r = \sqrt{1.56122 \, \text{m}^2} \][/tex]
[tex]\[ r \approx 1.25 \, \text{m} \][/tex]
Therefore, the distance between the laptop and the jar of pennies is closest to option A.
So, the correct answer is:
A. [tex]$1.25 \, \text{m}$[/tex].
Answer:
To find the distance between a
2
kg
2kg laptop and a
4
kg
4kg jar of pennies, given that the gravitational force between them is
3.42
×
1
0
−
10
N
3.42×10
−10
N , we can use Newton's law of universal gravitation. The formula for the gravitational force
F between two masses
1
m
1
and
2
m
2
separated by a distance
r is:
=
⋅
1
⋅
2
2
F=
r
2
G⋅m
1
⋅m
2
where
G is the universal gravitational constant
6.67430
×
1
0
−
11
m
3
kg
−
1
s
−
2
6.67430×10
−11
m
3
kg
−1
s
−2
.
Let's isolate
r in the formula:
2
=
⋅
1
⋅
2
r
2
=
F
G⋅m
1
⋅m
2
Taking the square root of both sides will give us
r :
=
⋅
1
⋅
2
r=
F
G⋅m
1
⋅m
2
Plugging in the given values:
-
=
6.67430
×
1
0
−
11
m
3
kg
−
1
s
−
2
G=6.67430×10
−11
m
3
kg
−1
s
−2
-
1
=
2
kg
m
1
=2kg
-
2
=
4
kg
m
2
=4kg
-
=
3.42
×
1
0
−
10
N
F=3.42×10
−10
N
=
6.67430
×
1
0
−
11
m
3
kg
−
1
s
−
2
×
2
kg
×
4
kg
3.42
×
1
0
−
10
N
r=
3.42×10
−10
N
6.67430×10
−11
m
3
kg
−1
s
−2
×2kg×4kg
Simplifying the expression under the square root:
=
6.67430
×
1
0
−
11
m
3
kg
−
1
s
−
2
×
8
kg
2
3.42
×
1
0
−
10
N
r=
3.42×10
−10
N
6.67430×10
−11
m
3
kg
−1
s
−2
×8kg
2
=
5.33944
×
1
0
−
10
m
3
s
−
2
3.42
×
1
0
−
10
N
r=
3.42×10
−10
N
5.33944×10
−10
m
3
s
−2
=
1.56122
m
2
r=
1.56122m
2
≈
1.25
m
r≈1.25m
Therefore, the distance between the laptop and the jar of pennies is closest to option A.
So, the correct answer is:
A.
1.25
m
1.25m
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