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Zinaida [17]
2 years ago
9

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

e moon to orbit the earth. The force of gravitational attraction is represented by the equation
F=Gm1m2r2
where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units of force are kg⋅m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation
,
where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following?
A. kg3m⋅s2
B. kg⋅s2m3
C. m3kg⋅s2
D. mkg⋅s2

Physics
2 answers:
faltersainse [42]2 years ago
8 0

The unit of the gravitational constant is \boxed{{{{{\text{m}}^{\text{3}}}} \mathord{\left/{\vphantom {{{{\text{m}}^{\text{3}}}} {{\text{kg}} \cdot {{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{\text{kg}} \cdot {{\text{s}}^{\text{2}}}}}}.

Further Explanation:

The Newton’s law of gravitation states that the force of attraction experienced by two bodies is directly proportional to the product of their mass and inversely proportional to the square of the distance between the two bodies.

The mathematical expression for the Newton’s law of gravitation is as shown below.

F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}  

Here, G is the gravitational constant, {m_1}\& {m_2} are the mass of two bodies and r is the distance of the bodies.

Simplify the expression for the gravitational constant G.

G = \dfrac{{F{r^2}}}{{{m_1}{m_2}}}

The SI unit of force is {{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}, the unit of distance is  and the unit of mass is {\text{kg}}.

Substitute the SI units of force, distance and mass in the above expression.

\begin{aligned}G&=\dfrac{(\text{kg}\cdot\text{m/s}^2)(\text{m})^2}{(\text{kg})(\text{kg})}\\&=\frac{\text{m}^2}{\text{kg}\cdot\text{s}^2}\\&=\text{m}^3/\text{kg}\cdot\text{s}^2\end{aligned}  

Thus, the unit of the gravitational constant is  \boxed{{{{{\text{m}}^{\text{3}}}} \mathord{\left/{\vphantom {{{{\text{m}}^{\text{3}}}} {{\text{kg}} \cdot {{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{\text{kg}} \cdot {{\text{s}}^{\text{2}}}}}}.

Learn More:

  1. The amount of kinetic energy an object has depends on its brainly.com/question/137098
  2. Which of the following are units for expressing rotational velocity, commonly denoted by ω brainly.com/question/2887706
  3. A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line joining earth's center of mass and the meteor's center of mass brainly.com/question/6536722

Answer Details:

Grade: High School

Chapter: Gravitation

Subject: Physics

Keywords: Gravity, attracts, SI units, two bodies, one another, force, masses of the body, magnitude of force, gravitational constant, distance, units.

melomori [17]2 years ago
5 0

The units of G must be C. m³ / ( kg s² )

<h3>Further explanation</h3>

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }

<em>F = Gravitational Force ( Newton )</em>

<em>G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )</em>

<em>m = Object's Mass ( kg )</em>

<em>R = Distance Between Objects ( m )</em>

Let us now tackle the problem !

To find unit of Gravitational Constant can be carried out in the following way:

F = G \frac{m_1 ~ m_2}{R^2}

{[N]}= G\frac{{[kg]}{[kg]}}{{[m^2]}}

{[kg ~ m / s^2]}= G \frac{{[kg^2]}}{{[m^2]}}

G = \frac{{[kg ~ m / s^2]}{[m^2]}} {{[kg^2]} }

G = \frac{{[kg ~ m^3 / s^2]}} {{[kg^2]} }

G = \frac{{[m^3 / s^2]}} {{[kg]} }

\boxed {G = \frac{{[m^3]}} {{[kg ~ s^2]} }}

The unit of G must be \large {\boxed {\frac{m^3} {kg ~ s^2 }}}

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

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Current density of the fuse when it melts is 620 A/cm^2

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8 0
2 years ago
Determine the torque applied to the shaft of a car that transmits 225 hp
Arisa [49]

Incomplete question.The complete question is here

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Answer:

Torque=0.51 Btu

Explanation:

Given Data

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Answer:

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4 0
2 years ago
Determine the force P required to maintain the 200-kg engine in the position for which θ = 30°. The diameter of the pulley at B
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Answer:

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Explanation:

The missing diagram is seen in the first image below.

From the second image, we can see the schematic diagram of the engine hanging over the pulley.

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Then;

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\alpha = tan^{-1} \bigg(\dfrac{CD}{AB-AD} \bigg )

replacing their respective values, where;

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\alpha = tan^{-1} \bigg(\dfrac{2 \ sin \ 30^0}{2-2 \ cos \ 30^0} \bigg )

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\alpha \simeq 75^0

From the third diagram attached below:

The tension occurring in the thread BC is equal to force P

T_{BC} = P

Using the force equilibrium expression along the horizontal direction.

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replacing the value of \alpha \simeq 75^0

-T_{AC} \  cos 30^0 + P cos 75^0  = 0

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P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0} \ \ \ - - -  (1)

Along the vertical direction, the force equilibrium equation can be expressed as:

\sum F_y =0

-W + P \ sin \alpha + T_{AC} \ sin \ 30^0  = 0

W = P \ sin \ \alpha + T_{AC} \ sin \ 30^0

replacing \alpha \simeq 75^0 and P  =\dfrac{ T_{AC} \ cos \ 30^0}{\ cos \ 75^0}

W =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

Also, replacing W for (200 × 9.81) N

200 \times 9.81 =\dfrac{T_{AC} \ cos \ 30^0}{cos \ 75^0}\times sin \ 75^0 + T_{AC} \ sin \ 30^0

200 \times 9.81 = T_{AC} \ cos \ 30^0 \ tan \ 75^0 + T_{AC} \ sin \ 30^0

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1 year ago
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The relationship is

    R₂ / R₁ = 2W²/L

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