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Fudgin [204]
3 years ago
14

Are front-end engineers starting to decline in China?

Engineering
1 answer:
laiz [17]3 years ago
7 0

Yes. They are declining in China. Very fast

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Sort the following alphabets using MergeSort and give required steps. [2 Marks]
OlgaM077 [116]

Answer:

I'm afraid i can't visualise it to you but visit the site below to help you out <3

Explanation:

https://opendsa-server.cs.vt.edu/embed/mergesortAV

4 0
3 years ago
A tension test is carried out on an Al alloy specimen which has an original diameter of 0.505 in and an original gauge length of
Contact [7]

Answer:

Detailed solution is given in attached image

5 0
3 years ago
What is the pressure at the bottom of a 25 ft volume of hydraulic fluid with a weight density of 55 lb/ft3 a. 114.6 psi b. 1375p
Assoli18 [71]

Answer:

d) 9.55 psi

Explanation:

pressure at the bottom is =ρgh

weight density is ρg=55 lb/ft³

h=25ft

pressure at the bottom is =55\times 25

                                  =1375psf

1 ft = 12 inch

pressure at bottom =\frac{1375}{12^2}

                                = 9.55 psi

so, answer will be option (d) which is 9.55 psi

3 0
2 years ago
The most important element of green construction is that it is a(n) __________ approach to building. *
madam [21]
Environmentally friendly


Since it focuses on are sustainable and efficient with and are made with the future in mind.
5 0
3 years ago
An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass
slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia = \frac{\pi}{64 (D_O^2 -D_i^2)}

I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4

Area = \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2

radius = \sqrt{\frac{I}{A}}

r = \sqrt{\frac{3.894}{3.015}

r = 1.136 in

slenderness ratio = \frac{L}{r}

                              = \frac{14 *12}{1.136} = 147.8

buckling load = P_cr = \frac{\pi^2 EI}}{l^2}

P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}

P_{cr} = 13.62 kips

3 0
2 years ago
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