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jeyben [28]
3 years ago
15

What principle of light’s behavior can be used to make a far away object appear closer and also make a small object appear large

r?
Physics
2 answers:
Alex_Xolod [135]3 years ago
8 0

Explanation:

Refraction is the change in the direction of the wave passing from one medium to another.

The change in the direction of the propagation of any wave is due to the different speed at different points.

A far away object appears closer and a small object appears larger.

Take an example. Suppose, an object lies in water.

A far away object appears closer. The angle from which the rays of light reach the observer is larger than the angle that it makes in air.

This makes the angular size larger to the observer's eye which makes the object look larger relative what it looks in air. Similar but vice versa condition occurs in the case when a small object appears larger.

Therefore, refraction can be used to make a far away object appear closer and also make a small object appear larger.

4vir4ik [10]3 years ago
3 0
It would be refraction
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You place a chunk of naturally radioactive (it means not enriched for nuclear purposes) material on the not very exact scale and
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Answer:

he mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

Explanation:

A radioactive material is transformed into another material by the emission of some particular radioactive ones, the most common being alpha and beta rays, which is why in the transformation process a certain amount of mass is lost. The process is described by the expression

             

              N = No e^{- \alpha /t}

 

From this expression the quantity half life time (T_{1/2}) is defined with time so that half of the atoms have been transformed

           

            T_{1/2} = ln 2 /λ

in this case it does not indicate that T_{1/2}= 20 days is worth, for which periods have passed, in the first the number of radioactive atoms was reduced to half the number, leaving N´ and the second halved the number of nuclei that they were radioactive, leaving radioactive nuclei

first time of life

              N´ = ½ N

second time of life

              N´´ = ½ N´

              N´´ = ¼ N

consequently in the sample at the end of these two decay periods we have, assuming that after each emission the atom is stable (non-radioactive). After the first emission there are n₁ = N / 2 stable atoms, after the second emission n₂ = ¼ N stable atoms are added and there are still n₃ = ¼ N radioactive atoms, so the total number of atoms is

 

             n_total = n₁ + n₂ + n₃

Recall that the mass of the initial radioactive atoms is m₁, when transforming its mass of stable atoms is m₂ where

            m₂ < m₁

therefore mass of

 

             m_total = m₂ N / 2 + m₂ N / 4 + m₁ N / 4

             m_total = m₂ ¾ N + m₁ ¼ N

             m_total = N (  ¾ m₂ + ¼ m₁)

Since the mass of the emitted particles is small, it is slightly less than the initial 50 kg, so the correct answer is the first.

8 0
2 years ago
I cant belive im doing this- i will mark brainliest if this gets answered- mostly correct.
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I’m going to send a picture
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