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AleksAgata [21]
3 years ago
8

The gravitational force between two objects increases as​

Physics
1 answer:
Misha Larkins [42]3 years ago
8 0

Answer: Their masses increases or the distance <u>between</u> them decreases

Explanation:

According to Newton's law of Gravitation, the gravitational force F exerted between two bodies of masses m1 and m2  and separated by a distance r  is equal to the <u>product of their masses</u> and inversely proportional to the <u>square of the distance</u>:

F=G\frac{(m1)(m2)}{r^2}   (1)

Where G is the Gravitational Constant

As we can see, if <u>one body increases its mass</u> or if <u>both bodies increase their mass</u>, the gravitational force will<u> increase</u> as well.

In addition, is the <u>distance between the two bodies is decreased</u>, the  <u>gravitational force will increase</u>.

Therefore:

The gravitational force between two objects increases as​ <u>their mass increases or the distance between them decreases.</u>

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A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin
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Answer:-24,5m/s

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Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

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Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

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Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

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