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8090 [49]
4 years ago
8

What is the latin name for the hydra constellation

Physics
1 answer:
slava [35]4 years ago
6 0
 The latin name for hydra constellation is "Water snake"
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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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As you enter the decision-making proccess you would have to see the different options that you have, after this you just have to consider and evaluate the consequences of the different options that you have, in this phase you have to weight the consequences and what it would take to make those decisions.

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