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Rus_ich [418]
3 years ago
5

What is the relationship between distance and mass in termsnof gravitational force

Physics
2 answers:
Softa [21]3 years ago
7 0

There is no relationship between distance and mass.  But the gravitational force
between two objects is related to both of those quantities.

-- The gravitational force between two objects is directly proportional to
the product of their masses.

-- The force is also inversely proportional to the square of the distance
between their centers.


tensa zangetsu [6.8K]3 years ago
5 0
Well,

I'm pretty sure you wanted how the distance between objects or the mass of two objects affects the gravitational force.

The Second Universal Law of Gravitation states that the gravitational force between two objects is directly proportional to the mass of the two objects. (mass)

The Third Universal Law of Gravitation is a bit more complex: The gravitational force between two objects is inversely proportional to the square of the distances between those two objects.  This means that if the initial distance between two objects is 1 unit, and the initial gravitational force between those two objects is 1 unit, and then the distance between the two objects is halved, then the gravitational force will increase by the square of 2, or 4 units.  So the final distance will be 0.5 units and the final gravitational force will be 5 units.
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The diameter of Earth (to two significant figures) is 7900 miles. Calculate its circumference.
Bess [88]

Answer:

The circumference of the Earth is 24818.58 miles

Explanation:

Analysis conceptual : The formula of the circumference is the following:

L= π*D Formula (1)

Where:

L : is the length of the circumference in miles (mi)

π : is the constant

D : is the diameter of the circumference in miles (mi)

Known data

π = 3.1416

D= 7900 miles:  Diameter of the Earth

Problem development

We apply the formula 1 to calculate the circumference of the Earth  (L):

L= π*7900 miles

L= 24818.58 miles

8 0
4 years ago
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Mariana [72]

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5 0
3 years ago
A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.

F ∝ e

F = ke

where;

k is the spring constant

k = F / e

k = 40 / 0.04

k = 1000 N/m

Therefore, the value of the spring constant of this spring is 1000 N/m

7 0
3 years ago
What is normal for a spring that obeys hook's law ?
Bezzdna [24]
A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm
7 0
3 years ago
Read 2 more answers
A car of mass m, traveling at constant speed, rides over the top of a round hill. How do the normal force of the road on the car
dybincka [34]

Answer:

The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.

Explanation:

Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:

At the top of a round hill

Normal force = Gravitational force - centripetal force

At the foot of a round hill

Normal Force = centripetal force + Gravitational force

4 0
3 years ago
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