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mamaluj [8]
3 years ago
6

When the distance between two objects doubles, what happens to the gravitational attraction between these two objects?

Physics
1 answer:
Feliz [49]3 years ago
8 0

Answer:

As indicated by Newton's law of attraction each article or body in the universe draws in every single item towards one another and that power of fascination is straightforwardly relative to the result of their masses and contrarily corresponding to the square of the distance between them.  

The power of gravity between two articles will diminish as the distance between them increments. The two most significant elements influencing the gravitational power between two items are their mass and the distance between their focuses. As mass increments, so does the power of gravity, however an increment in distance mirrors a reverse proportionality, which makes that power decline dramatically.  

At that point by Newton's All inclusive Law of Attractive energy;  

F=GMm/R^2  

Mm= result of the majority  

R=Distance Between the two masses by focus.  

On the off chance that R is multiplied, new force=GMm/(2R)^2  

=GMm/4R^2  

Unique Power/New Force=4/1  

F/4=New Power

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If sunspots are active, more solar flares will result creating an increase in geomagnetic storm activity for Earth.

Here, the sun's magnetic field changes polarity approximately every 11 years. So, for reversal polarity of the sun, 22 year solar cycle is characterized by a variation in the number of sunspots and a reversal of the polarity of the Sun as a whole.

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5 0
2 years ago
Which question could be tested in a scientific manner?
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3 years ago
A train is moving on a flat, continuous track. label all forces acting upon the object
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6 0
3 years ago
With an initial velocity of 20km per hour a car accelerated at 8m/s2 for 10s,what is the position of the car at the end of 10s
Dima020 [189]
  • initial velocity=u=20km/h=5.5m/s
  • Acceleration=a=8m/s^2
  • Time=t=10s

\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2

\\ \sf\longmapsto s=5.5(10)+\dfrac{1}{2}(8)(5.5)^2

\\ \sf\longmapsto s=55+4(5.5)^2

\\ \sf\longmapsto s=55+4(30.25)

\\ \sf\longmapsto s=55+121

\\ \sf\longmapsto s=176m

4 0
2 years ago
If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun
inessss [21]

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

s=-8t^2+32t...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

\dfrac{ds}{dt}=0

\dfrac{d(-8t^2+32t)}{dt}=0

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

s=-8(2)^2+32(2)

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

6 0
4 years ago
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