Answer:
It would not be possible the cohesion among water molecules by the polar covalent bonding.
Well, to understand this in a better way, let's begin by explaining that water is special due to its properties, which makes this fluid useful for many purposes and for the existence of life.
In this sense, one of the main properties of water is cohesion (molecular cohesion), which is the attraction of molecules to others of the same type. So, water molecule (
) has 2 hydrogen atoms attached to 1 oxygen atom and can stick to itself through hydrogen bonds.
How is this possible?
By the polar covalent bonding, a process in which electrons are shared unequally between atoms, due to the unequal distribution of electrons between atoms of different elements. In other words: slightly positive and slightly negative charges appear in different parts of the molecule.
Now, it can be said that a water molecule has a negative side (oxygen) and a positive side (hydrogen). This is how the oxygen atom tends to monopolize more electrons and keeps them away from hydrogen. Thanks to this polarity, water molecules can stick together.
Hello, love! The answer is True, or T, on Edge2020.
Hope this helped!
~ V.
The answer is: [C]: "elasticity" .
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Answer:
42244138.951 m
Explanation:
G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²
r = Radius of orbit from center of earth
M = Mass of Earth = 5.98 × 10²⁴ kg
m = Mass of Satellite
The satellite revolves around the Earth at a constant speed
Speed = Distance / Time
The distance is the perimeter of the orbit
![v=\frac{2\pi \times r}{24\times 3600}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D)
The Centripetal force of the satellite is balanced by the universal gravitational force
![m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m](https://tex.z-dn.net/?f=m%5Cfrac%7Bv%5E2%7D%7Br%7D%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cfrac%7B%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%7D%7Br%7D%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20%5Cleft%28%5Cfrac%7B2%5Cpi%20%5Ctimes%20r%7D%7B24%5Ctimes%203600%7D%5Cright%29%5E2%3D6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5C%5C%5CRightarrow%20r%5E3%3D%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5C%5C%5CRightarrow%20r%3D%5Cleft%28%5Cfrac%7B6.667%5Ctimes%2010%5E%7B-11%7D%5Ctimes%205.98%5Ctimes%2010%5E%7B24%7D%5Ctimes%20%2824%5Ctimes%203600%29%5E2%7D%7B%282%5Cpi%29%5E2%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5CRightarrow%20r%3D42244138.951%5C%20m)
The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m