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Stolb23 [73]
2 years ago
5

characterizing vibration-assisted atomic force microscopy (afm)-based nanomachining via perception of acoustic emission phenomen

a using a sensor-based real-time monitoring approach
Physics
1 answer:
DaniilM [7]2 years ago
8 0

The study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

<h3>What is nanomachining?</h3>

The expression nanomachining makes reference to the study of nanometric machines (nanomachines) and related materials, which can be achieved by different approaches including sensor-based strategies related to acoustic auditive phenomena.

In conclusion, the study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

Learn more about nanomachines here:

brainly.com/question/20875598

#SPJ1

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lesya692 [45]

Answer:

All but 4 I believe

Explanation:

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3 years ago
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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
what causes ocean water near the equator to be warmer than ocean water farther north? a. upwellings spew warm water toward the e
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The structure and curvature of the Earth results in beams of sunlight glancing off the equator and reaching other areas of the Earth. This means that the areas at the equator receive more energy as sun's rays hit them directly.
Therefore, the answer is C.
8 0
3 years ago
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FOR FLVS STUDENTS ONLY///NO BIT LINKS, SCAM LINKS, OR GIBBERSH
Marysya12 [62]

Answer:

whats thw question tho

Explanation:

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Please help I’ll give brainliest
BigorU [14]

Answer:

The answer are as followed: B is the correct answer, hope that helps :)

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