Respuesta :
Answer:
1) 2,641,895.5 m
2) 1.85903476 × 10⁻⁹ kg
3) 9.32 C
4) 38.7 kg
Explanation:
1) The gravitational force of attraction between the object and the mass of the Earth is given by the following formula;
[tex]F} =G\dfrac{M_{1} \cdot m_{2}}{R^{2}} = m_2 \times g[/tex]
Where;
M₁ = The mass of the Earth = 5.97 × 10²⁴
m₂ = The mass of the object
G = The universal gravitational constant
R = The radius of the Earth = 6,378,100 m
g = The acceleration due to gravity = 9.8 m/s²
By comparison, we have;
[tex]\dfrac{G \cdot M_{1}}{R^{2}} = g[/tex]
When the object is r meters away, g is halved, therefore, we have;
[tex]\dfrac{G \cdot M_{1}}{r^{2}} = \dfrac{g}{2} =\dfrac{G \cdot M_{1}}{\left (2\times R^{2}\right )}[/tex]
Therefore, r² = 2·R² = 2 × (6,378,100 m)² = 8.13603192 × 10¹³ m²
r = √(8.13603192 × 10¹³ m²) = 9019995.52217 m
The distance of the object from the surface of the Earth = 9,019,995.52217 - 6,378,100 = 2,641,895.52217 m ≈ 2,641,895.5 m
The object needs to be approximately 2,641,895.5 m from the Earth's surface to experience half of the acceleration due to gravity experienced on the Earth's surface
2) The electrostatic force between the charges = k×q²/r²
Where;
q = 1.602 × 10⁻¹⁹ C
r = Distance between the charges
k = Coulomb constant = 8.9875517923 × 10⁹ kg·m³·s⁻²·C⁻²
We have;
The electrostatic force between the charges = 8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²
The gravitational force between the charges = 6.67408 × 10⁻¹¹ × m²/r²
Given that the ratio between the two forces = 1, we have;
(8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²)/(6.67408 × 10⁻¹¹ × m²/r²) = 1
∴ m² = (8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²)/(6.67408 × 10⁻¹¹) = 3.4560102×10⁻¹⁸ kg²
The mass m = √(3.4560102×10⁻¹⁸ kg²) = 1.85903476 × 10⁻⁹ kg
The mass of each particle will be 1.85903476 × 10⁻⁹ kg
3) F = k×q²/r²
q² = F × r²/k
Which gives;
q² = 50 × 125000²/(8.9875517923 × 10⁹) ≈ 86.93 C²
q = √86.93 ≈ 9.32 C
The charges of each object = 9.32 C
4) [tex]F} =G\dfrac{m^2 }{r^{2}}[/tex]
r = 0.25 cm = 0.0025 m
F = 0.016 N
G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²
Therefore;
m² = F·r²/G = 0.016 × (0.0025)²/(6.67408 × 10⁻¹¹) ≈ 1498.334 kg²
m = √(1498.334 kg²) ≈ 38.7 kg
The mass of each object = 38.7 kg.
The Coulomb and gravitational laws together with Newton's second law allow to find the results for the questions are:
1) The distance where the acceleration of gravity measures half its value on earth is; r’= 2.64 106 m
2) The mass of the particle is: m = 1.86 10-9 kg
3) The charge of the particles is: q = 9.32 C
4) The mass is: m = 38.7 kg
Part 1
The Universal Gravitational Law says that the force between two objects is proportional to their masses and inversely proportional to the square of their distance
F =[tex]-G \frac{Mm}{r^2}[/tex]
Where F is the gravitational force, G the gravitational constant, M and m the mass of the two bodies and r the distance between them.
They ask to calculate the distance to the point where the acceleration is half that of the gravitation acceleration at the earth's surface.
a = g / 2
If we use Newton's second law that gives the relationship between the net force, the mass and the acceleration of the body.
F = m a
Let's substitute
[tex]-G \frac{Mm}{r^2} = m \frac{g}{2}[/tex]
r² = [tex]\frac{2GM}{g}[/tex]
Let's calculate
r² = [tex]\frac{2 \ 6.67 \ 10^{-11} \ 5.97 \ 10^{24} }{9.8}[/tex]
r = [tex]\sqrt{81.360 \ 10^{12}}[/tex]
r = 9.02 10⁶ m
This is the distance from the center of the earth, the distance measured from the surface of the planet is
r = R + r '
r ’= r- R
r ’= 9.02 10⁶ - 6.378 10⁶
r’= 2.64 10⁶ m
Part 2
Coulomb's law states that the electric force between two charged bodies is:
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
Where q₁ and q₂ are the charges of the leather and k is the Coulomb constant
They ask what mass the particles must have so that the electrostatic force (Fe) and the gravitational force (Fg) have been equal
Indicate the charges of the particles q₁ = q₂ = 1.602 10⁻¹⁹ C
Let's equalize the two forces
[tex]F_e = F_g \\k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{ r^2 } \\k q^2 = G m^2[/tex]
[tex]m = \sqrt{\frac{k}{G} } \ q[/tex]
m = [tex]\sqrt{\frac{8.99 \ 10^9 }{ 36.67 \ 10^{-11}} } \ 1.602 \ 10^{-19}[/tex]
m = [tex]\sqrt{1.3478 \ 10^{20} } \ 1.602 \ 10^{-19}[/tex]
m = 1.86 10⁻⁹ kg
Part 3
They indicate that the electrostatic force between two objects is F = 50 N and the distance between them is r = 125 km = 125 103 m, they ask what is the charge of the objects.
Let's use Coulomb's law
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
q² = [tex]\frac{F}{k} \ r^2[/tex]
q² = [tex]\frac{50 \ (125 \ 10^3)^2 }{8.99 \ 10^9 }[/tex]
q = [tex]\sqrt{86.93}[/tex]
q = 9.32 C
Part 4
Indicate the gravitational force between two objects is F_g = -0.016 N when they are separateda distance r = 0.25 cm = 0.25 10⁺² m, ask the mass of the bodies.
Let's use the law of universal gravitation.
[tex]-G \frac{Mm}{r^2} \\m^2 = - \frac{F}{G} \ r^2[/tex]
Let's calculate
m² = [tex]- \frac{(-0.016) }{6.67 \ 10^{-11}} \ (0.25 \ 10^{-2})^2[/tex]
m = [tex]\sqrt{1498.33}[/tex]
m = 38.7 kg
In conclusion using the Coulomb and gravitational laws together with Newton's second law we can find the results for the questions are:
1) The distance where the acceleration of gravity measures half its value on earth is; r’= 2.64 106 m
2) The mass of the particle is: m = 1.86 10-9 kg
3) The charge of the particles is: q = 9.32 C
4) The mass is: m = 38.7 kg
Learn more here: brainly.com/question/12356144
