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saveliy_v [14]
3 years ago
6

Use the data table and below to answer

Physics
1 answer:
almond37 [142]3 years ago
4 0
City 4 because it is closer to the equator
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\sf{ \huge}{ \colorbox{pink}{  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:}}

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7 0
3 years ago
Can someone help meeeeee... show how to solve it plzzzzzzzz
liubo4ka [24]
<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have a gravitation force F_{1}=16units, given by the formula written at the beginning. Let’s rename the distance r as d:

F_{1}=G\frac{m_{1}m_{2}}{d^2}     (1)

And we are asked to find the gravitation force F_{2} with a given distance of \frac{d}{2}:

F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}      

F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}     (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate G m_{1}m_{2} in both equations:

From (1):

Gm_{1}m_{2}=F_{1}{d}^{2}     (3)

From (2):

Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}     (4)

If (3)=(4):

F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}     (5)

Now we have to find F_{2}:

F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}      

F_{2}=4F_{1}     (6)

If F_{1}=16 units:

F_{2}=(4)(16 units)        

F_{2}=64 units>>>>This is the new force of attraction     

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They do because if one thing is moving and the is there both moveing in once
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3 years ago
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