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sergiy2304 [10]
1 year ago
14

Question 9 of 10

Physics
1 answer:
romanna [79]1 year ago
5 0

A atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

<h3>What are nuclear forces?</h3>

The expression 'nuclear forces' makes reference to the attraction between different subatomic particles in a given nucleus of a particular atom, which represents the strongest force in the universe.

Subatomic particles include positively charged particles (i.e., protons), neutrally charged particles called neutrons, and electrically negative charged particles called electrons (e-).

In conclusion, a atomic neutron particle that is attracting another atomic neutron is an example of the strong nuclear force acting in an atom (Option B).

Learn more about nuclear forces here:

brainly.com/question/16959709

#SPJ1

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What is electrons ? explain it function<br>​
Neko [114]

Answer:

The electron is a subatomic particle, symbol e⁻ or β⁻ , whose electric charge is negative one elementary charge. Electrons belong to the first generation of the lepton particle family, and are generally thought to be elementary particles because they have no known components or substructure

Explanation:

functions of electrons

and electrons being the negatively charged particles of atom. Together, all of the electrons of an atom create a negative charge that balances the positive charge of the protons in the atomic nucleus

7 0
3 years ago
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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit
DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

So here we can say that the spear will end up above the fish relative to the actual position of the fish

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3 years ago
The amount of gravitational potential energy released as:_________.
nordsb [41]

Answer:

b.it depends on the distance it falls

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2 years ago
Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest
shutvik [7]

This is note the complete question, the complete question is:

One of the lousy things about getting old (prepare yourself!) is that you can be both near-sighted and farsighted at once. Some original defect in the lens of your eye may cause you to only be able to focus on some objects a limited distance away (near-sighted). At the same time, as you age, the lens of your eye becomes more rigid and less able to change its shape. This will stop you from being able to focus on objects that are too close to your eye (far-sighted). Correcting both of these problems at once can be done by using bi-focals, or by placing two lenses in the same set of frames. An old physicist instructor can only focus on objects that lie at distance between 0.47 meters and 5.4 meters.

Assume that the physics instructor would like to have normal visual acuity from 21 cm out to infinity and that his bifocals rest 2.0 cm from his eye. What is the refractive power of the portion of the lense that will correct the instructors nearsightedness?

Answer:  3.04 D

Explanation:

when an object is held 21 cm away from the instructor's eyes, the spectacle lens must produce 0.47m ( the near point) away.

An image of 0.47m from the eye will be ( 47 - 2 )

i.e 45 cm from the spectacle lens since the spectacle lens is 2cm away from the eye.

Also, the image distance will become negative

gap between lense and eye = 2cm

Therefore;

image distance d₁ = - 45cm = - 0.45m

object distance  d₀ = 21 - 2 = 19cm = 0.19m

P = 1/f = 1/ d = 1/d₀ + 1/d₁ = 1/0.19 + (-1/0.45)

P = 1/f =  5.26315789 - 2.22222222

P = 1/f = 3.04093567 ≈ 3.04 D

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3 years ago
b. Comparing and Contrasting Compare the change in atmospheric pressure with elevation to the change in water pressure
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