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spayn [35]
2 years ago
11

Explain crystallographic defects.

Engineering
1 answer:
Sliva [168]2 years ago
5 0

Answer:

interruptions of regular patterns in crystalline solids. They are common because positions of atoms or molecules at repeating fixed distances determined by the unit cell parameters in crystals, which exhibit a periodic crystal structure, are usually imperfect.

Explanation:

Cus am pro

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Many farms and ranches use electric fences to keep animals from getting into or out of specific pastures. When switched on, an e
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Answer:

Aluminum

Explanation:

The best material to use when creating an electric fence would be Aluminum. Aluminum wiring is incredibly durable and can be easily obtained. Since aluminum is a non-magnetic metal its conducting capabilities far exceed other metallic options in the market and is also why companies choose aluminum for their high tension cable wiring. Aside from being more expensive than other feasible options its durability and conducting capabilities make it easily the best option.

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You can help build a safe work environment by using your knowledge of violence prevention strategies to spot what?
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warning signs

Explanation:

give directions on your surroundings

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2 years ago
A steel pipe of 400-mm outer diameter is fabricated from 10-mm-thick plate by welding along a helix that forms an angle of 20° w
Verdich [7]

Explanation:

Outer di ameter d_{0}=400 \mathrm{mm}[tex] Thickness of the cylinder [tex]t=10 \mathrm{mm}

\therefore[tex] Inner diam eter [tex]d_{i}=d_{0}-2 t=400-2 \times 10

d_{1}=380 \mathrm{mm}

Given loading on the cylinder P=300 \mathrm{kN} Helix an gle of the weld form \theta=20^{\circ}

(i) Normal stress on the plane at angle \theta=20^{\circ} is

\sigma=\frac{P \cos ^{2} \theta}{A_{0}}

\text { Where } A_{0}=\frac{\pi}{4}\left(d_{0}^{2}-d_{1}^{2}\right)

\quad=\frac{\pi}{4}\left(400^{2}-380^{2}\right)

=12252.21 \mathrm{mm}^{2}

=12.25221 \times 10^{-9} \mathrm{m}^{2}

\sigma=\frac{-300 \times 10^{2} \times \cos ^{2} 20}{12.25221 \times 10^{-1}}

=-21.6 \mathrm{MPa}

(ii) Shear stress along an angle of \theta=20^{\circ} is \tau=\frac{P}{A_{0}} \cos \theta \sin \theta

=\frac{-300 \times 10^{-1} \times \cos 20 \times \sin 20}{12.25221 \times 10^{-3}}

=-7.86 \mathrm{MPa}

3 0
2 years ago
What can be the main disadvantage of pulse amplitude modulation?​
Feliz [49]

Answer:

transmission bandwidth required is very large.

Explanation:

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2 years ago
Bessy's calf weighed 99.19 pounds when calved on March 7th. Her calf gained an average of 1.85 pounds per day for 266 days. What
Zepler [3.9K]

Answer:

591.3

Explanation:

99.19 + (1.85 × 266) = 591.29

rounded = 591.3

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3 years ago
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