Physics, 18.09.2019 03:40 deannajd03

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
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

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Answer from: salihanegawo

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

Further explanation

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

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

F = Gravitational Force ( Newton )

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

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

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 $\large {\boxed {\frac{m^3} {kg ~ s^2 }}}$

Learn moreImpacts of Gravity : linkEffect of Earth’s Gravity on Objects : linkThe Acceleration Due To Gravity : linkAnswer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

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

Gravity causes objects to be attracted to one another. this attraction keeps our feet firmly planted

Answer from: bovalle

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

Simplify the expression for the gravitational constant .

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

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:

The amount of kinetic energy an object has depends on its link Which of the following are units for expressing rotational velocity, commonly denoted by ω link 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 link

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.

Answer from: Quest

Idon't know it i just need my points lol

Answer from: Quest

the unit of the measure = $\frac{j}{kgk}$

explanation:

the property that is quantified by the energy needed per of material to raise the temperature by one.

si unit of energy = joule (j)

si unit of temperature = kelvin (k)

si unit of mass of material = kilogram (kg)

the property that is quantified by the energy needed per of material to raise the temperature by one is called specific heat.

the unit of specific heat is $\frac{j}{kgk}$

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