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2 years ago

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

2 answers:

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]2 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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Predict the precipitate when the double replacement reaction occurs:<br> Ca(NO3)2(aq) + Na2CO3(aq) →

taurus [48]

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8 0

3 years ago

Read 2 more answers

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of $x= 5t^3+1$

Coordinate of position of $y=3t^2+2$

Time = 2.00 s

We need to calculate the acceleration

$a = \dfrac{d^2x}{dt^2}$

For x coordinates

$x=5t^3+1$

On differentiate w.r.to t

$\dfrac{dx}{dt}=15t^2+0$

On differentiate again w.r.to t

$\dfrac{d^2x}{dt^2}=30t$

The acceleration in x axis at 2 sec

$a = 60i$

For y coordinates

$y=3t^2+2$

On differentiate w.r.to t

$\dfrac{dy}{dt}=6t+0$

On differentiate again w.r.to t

$\dfrac{d^2y}{dt^2}=6$

The acceleration in y axis at 2 sec

$a = 6j$

The acceleration is

$a=60i+6j$

We need to calculate the net force

$F = ma$

$F = 3.00\times(60i+6j)$

$F=180i+18j$

The magnitude of the force

$|F|=\sqrt{(180)^2+(18)^2}$

$|F|=180.89\ N$

Hence, The net force acting on this object is 180.89 N.

3 0

3 years ago

When a roller coaster gets to the bottom of a descent, describe the energy transfers and changes to energy stores that happen if

Feliz [49]

Answer:

its B

Explanation:

It's B & A at the same time because A. a roller coaster uses brakes to slow down and stop. B is the most reasonable answer. Because all roller coasters go up and over a second time over the hill, but they also slow down. But go with B.

TELL ME IF I´M RITE

6 0

2 years ago

Two large, flat, horizontally oriented plates are parallel to each other, a distance d apart. Half-way between the two plates th

fgiga [73]

Answer:

Explanation:

Let the potential difference between the plate is V . Then in the first case

Electric field E between plate

E₁ = V / d

where d is separation between plate

When the plate separation becomes d / 2

Electric field E between plate

E₂ = V / d /2

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Or twice the earlier field

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3 years ago

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tatyana61 [14]

that is an example of negative acceleration because it is slowing down

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2 years ago