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3 years ago

What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?

Physics

1 answer:

Katena32 [7]3 years ago

5 0

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 ( $H_{2}O$ ) 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.

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Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M $_S$ = 1.99 × 10³⁰ kg

Mass of the neutron star

M $_N$ = 2( M $_S$ )

M $_N$ = 2( 1.99 × 10³⁰ kg )

M $_N$ = ( 3.98 × 10³⁰ kg )

Radius of neutron star R $_N$ = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω $_N$ .

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM $_N$ = / R $_N$ ² = mR $_N$ ω $_N$ ²

ω $_N$ ² = GM $_N$ = / R $_N$ ³

ω $_N$ = √(GM $_N$ = / R $_N$ ³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω $_N$ = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω $_N$ = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω $_N$ = √ 120831133.3636777

ω $_N$ = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0

2 years ago

The tent meteorite, found in 1897 near cape York, on the west coast of Greenland, is the largest meteorite exhibited by any muse

sweet-ann [11.9K]

Answer:
3.0883 x 10^10mg

Explanation:
1 kilogram = 1000 000 milligrams
So, 30 883 x 1000 000 = 30 883 000 000mg

6 0

3 years ago

A pronghorn, the fastest North American animal, is capable of running for at 18 m/s (40 mph) for 10 minutes. The time limit isn’

Otrada [13]

Explanation:

The increase in the body temperature of pronghorn, the fastest North American animal, results from the chemical energy of the pronghorn converting into kinetic energy with efficiency less than 100%. The remaining energy is converted into heat energy. Thus, raising the temperature of pronghorn.

Due to the chemical energy it gains both kinetic and heat energy.

3 0

2 years ago

What is the magnitude of the momentum of a 3400 kg airplane traveling at a speed of 450 miles per hour?

Vitek1552 [10]

Answer:

The magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ .

Explanation:

Given that,

Mass of the airplane, m = 3400 kg

Speed of the airplane, v = 450 miles per hour

Since, 1 mile per hour = 0.44704 m/s

v = 201.16 m/s

We need to find the magnitude of momentum of the airplane. It is given by the product of mas and velocity such that,

$p=mv$

$p=3400\ kg\times 201.16 \ m/s$

$p=683944\ kg-m/s$

$p=6.83\times 10^5\ kg-m/s$

So, the magnitude of momentum of the airplane is $6.83\times 10^5\ kg-m/s$ . Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

A periodic wave with wavelength 2m has a speed of 4m/s. What is the waves frequency?.

Gekata [30.6K]

The wave frequency is 2 Hz.

What is wave frequency ?

The number of waves that pass through a fixed point in a given amount of time is referred to as the wave frequency. The hertz is the SI unit for wave frequency (Hz).

$f = v / w$