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Irina18 [472]
2 years ago
10

Is there gravity in water?

Physics
2 answers:
Snowcat [4.5K]2 years ago
6 0

Gravity works the same way in water that it works in air or a vacuum -- but you have to consider the force of gravity on the water as well as on the object you put into it

Alexus [3.1K]2 years ago
3 0

Answer: Yes.

Explanation:

There is gravity in water because How does the water go up in the cycle? the garavity with the wind pull it up.

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The chart lists four different kinds of atoms.
scZoUnD [109]

They cannot be divided and still keep the properties of the element true about all of these atoms. Option D is corect.

<h3>What is an atom?</h3>

The atom consists of matter that may be split without releasing electrical charges.

It's also the smallest unit of matter with chemical element features. As a result, the atom is the fundamental unit of science.

In the nucleus proton and the neutron is exists. The condition of the atom to be electrically neutral is that the number of the proton and electron should be the same.

They cannot be separated while still maintaining the element's characteristics for each and every one of these atoms.

Hence, option D is corect.

To learn more about the atom refer to the link;

brainly.com/question/1566330

#SPJ1

3 0
1 year ago
Please help me! i cant find this one anywhere!
Vitek1552 [10]

Answer:

I think the answer is C.

Explanation:

hope this works

7 0
3 years ago
A woman has a mass of 55 kg on earth. what would be the women's weight on the moon? (The moon’s gravity is 1.6 N).
Alex17521 [72]

Answer:

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Explanation:

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8 0
3 years ago
Can somebody help please?
Art [367]

Answer:

x positive and y negative I think

Explanation:

6 0
2 years ago
What is the magnitude of the electric field 17.1 cm directly above an isolated 1.83Ã10â5 C charge?
11111nata11111 [884]

Answer:

Electric field, E=5.63\times 10^{16}\ N/C

Explanation:

Given that,

Charge, q=1.83\times 10^5\ C

We need to find the magnitude of electric field 17.1 cm (0.171 m) above an isolated charge. Electric field at a point is given by :

E=\dfrac{kq}{r^2}

E=\dfrac{9\times 10^9\times 1.83\times 10^5\ C}{(0.171\ m)^2}

E=5.63\times 10^{16}\ N/C

So, the electric field is 5.63\times 10^{16}\ N/C. Hence, this is the required solution.

5 0
3 years ago
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