A rock accelerates down a hill at 7 m/s2 with a force of 30.0 N. What is the mass of the rock?

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

melomori [17]

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

5 0

3 years ago

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To perform a drug lookup to ensure that the new compound has been added to the computer system properly , select in the toolbar

lutik1710 [3]

To perform a drug lookup to ensure that the new compound has been added to the computer system properly we have to select the new from the toolbar at the top of the screen

<h3>What is a computer system?</h3>

A computer system is a collection of computers, related hardware, and related software. The central processing unit (CPU), memory, input/output, and storage devices are the four main components of a computer system. To produce the desired result, all of these parts operate in concert as a single unit.

Selecting the new from the toolbar at the top of the screen will allow us to run a drug lookup to make sure the new compound has been properly added to the computer system.

Therefore the correct answer is the option C

Learn more about the computer system here

brainly.com/question/2612067

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3 0

2 years ago

The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m (2,716.5 feet) and has more than 160 stories.

Elena L [17]

Answer:

h = 599.5 m

Explanation:

Given,

height of structure = 828 m

weight of the tourist = 184 lb

= 184 x 0.45359 = 83.43 Kg

Potential energy = 187000 J

PE = m gd

$d = \frac{PE}{mg}$

$d = \frac{187000}{83.43\times 9.81}$

h = 228.5 m

Height of the room above the ground.

h = 828 - 228.5

h = 599.5 m

Height of the floor above ground is equal to 599.5 m.

4 0

3 years ago

Complex carbohydrates, such as _____ and _____, provide the body with long-lasting energy.

Sauron [17]

complex carbohydrates, such as starches and fiber, provide the body with long-lasting energy.

Hope this helps!

7 0

3 years ago

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The international Space Station (ISS) orbits the Earth once every 90 mins at an altitude of 409 km. How high would it have to be

Oksi-84 [34.3K]

It would have to be 36,719 Km high in order to be to be in geosynchronous orbit.

To find the answer, we need to know about the third law of Kepler.

<h3>What's the Kepler's third law?</h3>

It states that the square of the time period of orbiting planet or satellite is directly proportional to the cube of the radius of the orbit.
Mathematically, T²∝a³

<h3>What's the radius of geosynchronous orbit, if the time period and altitude of ISS are 90 minutes and 409 km respectively?</h3>

The time period of geosynchronous orbit is 24 hours or 1440 minutes.
As the Earth's radius is 6371 Km, so radius of the ISS orbit= 6371km + 409 km = 6780km.
If T1 and T2 are time period of geosynchronous orbit and ISS orbit respectively, a1 and a2 are radius of geosynchronous orbit and ISS orbit, as per third law of Kepler, (T1/T2)² = (a1/a2)³
a1= (T1/T2)⅔×a2

= (1440/90)⅔×6780

= 43,090 km

Altitude of geosynchronous orbit = 43,090 - 6371= 36,719 km

Thus, we can conclude that the altitude of geosynchronous orbit is 36,719km.

Learn more about the Kepler's third law here:

brainly.com/question/16705471

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6 0

2 years ago