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enyata [817]
3 years ago
10

list out main types of material used in design and Manufacture of product give one example for each in engineering application ?

​
Engineering
1 answer:
cluponka [151]3 years ago
7 0
One of the types of product design is interface design. Another type of product design is called visual product design.
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b

Explanation:

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Whats the best used for cable -stayed bridge
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a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

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A reversible process and an irreversible process both have the same________ between the same two states. a. Internal energy b. W
Vlad1618 [11]

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a) Internal energy

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As we know that internal energy is a point function so it did not depends on the path ,it depends  at the initial and final states of process.All point function property did not depends on the path.Internal energy is a exact function.

Work and heat is a path function so these depend on the path.They have different values for different path between two states.Work and heat are in exact function.

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3 years ago
What are the benefits of NSTP CWTS?​
geniusboy [140]

Explanation:

The purpose of this program is to recognize the Youth's vital role in nation- building, promote consciousness among youth and develop their physical, moral, spiritual, intellectual and social well-being. It shall inculcate in the youth patriotism, nationalism, and advance their involvement in public and civic affairs.

8 0
3 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
3 years ago
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