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jok3333 [9.3K]
3 years ago
10

Landslides are most common in

Physics
1 answer:
nikitadnepr [17]3 years ago
3 0

shorelines of the southeast U.S.

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What might happen if water molecules did not have a slight negative charge on one end and a slight positive charge on the other?
Katena32 [7]

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.

5 0
3 years ago
The law of inertia to both moving and no moving objects. True or false
Zepler [3.9K]
True, the law of inertia effects both moving and non-moving objects.
3 0
2 years ago
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Please answer me!!! asappp​
Anastasy [175]

Answer:

omg this is hard

Explanation:

Yes it is lol

5 0
2 years ago
An ammeter with a resistance of 5.0 ohm is connected in series with a 3.0V cell and a lamp rated at 300 mA, 3V. Calculate the cu
Alex

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

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Which of the following is a strength training option?
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D. All of the above

PLZ MARK ME AS BRAINLEIST ;)

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