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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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Suppose the average mass of each of 20,000 asteroids in the solar system is 1017 kg. Compare the total mass of these asteroids t
blondinia [14]

Answer:

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

39929.4542466 m

Explanation:

Total mass of the asteroids

m_a20000\times 10^{17}=2\times 10^{21}\ kg

m_e = Mass of Earth = 5.972\times 10^{24}\ kg

The ratio is

\dfrac{m_a}{m_e}=\dfrac{2\times 10^{21}}{5.972\times 10^{24}}\\\Rightarrow \dfrac{m_a}{m_e}=0.000334896182184

The mass of the asteroids is 0.000334896182184 times the mass of the Earth.

Volume is given by

V=\dfrac{m}{\rho}\\\Rightarrow \dfrac{4\pi}{3\times 8} d^3=\dfrac{m}{\rho}\\\Rightarrow d^3=\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{m}{\rho})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{3\times 8}{4\pi}\dfrac{10^{17}}{3000})^{\dfrac{1}{3}}\\\Rightarrow d=39929.4542466\ m

The diameter is 39929.4542466 m

6 0
2 years ago
Describe the relationship of attractive forces between molecules
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The game was played by swinging the big mallet down hard enough to cause the bell to ring. If it took a 44 newton force to ring
nata0808 [166]
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3 0
2 years ago
Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and
Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

v_f^2=v_i^2+2ax

Where,

v_f = Final velocity

v_i = Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

F_f = \mu_k (mg)  \rightarrowFrictional Force

F = ma \rightarrow Force by Newton's second Law

Where,

m = mass

a= acceleration

\mu_k = Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

F_f = F

\mu_k mg=ma

a = \mu_k g

Therefore,

v_f^2=v_i^2+2ax

0=v_i^2+2(\mu_k g)x

Re-arrange to find x,

x = \frac{v_i^2}{2(-\mu_k g)}

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0
2 years ago
Match the following
Fofino [41]

Answer:

1. b

2. d

3. e

4. c

5. a

Explanation:

These are just basic definitions. Let me know if you need further clarification.

7 0
3 years ago
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