1 year ago

On what kind of sectional drawing would you find the maximum length of a building?

Engineering

1 answer:

Genrish500 [490]1 year ago

5 0

Answer:

A longitudinal section

Explanation:

A longitudinal section is a section drawn along the length of an object, as opposed to a cross section, which is drawn across the width or diameter of an object.

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At what stage of development is an engineering team the most productive?

Alexxx [7]

Answer:

Explanation:

A because you are continuing to keep moving and thinking.

3 0

2 years ago

Read 2 more answers

Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s

ioda

Answer: r = 0.8081; s = -0.07071

Explanation:

A = (150i + 270j) mm

B = (300i - 450j) mm

C = (-100i - 250j) mm

R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

270r - 450s - 250 = 0

150r + 300s = 100

270r - 450s = 250

solving simultaneously,

r = 0.8081 and s = -0.07071

QED!

5 0

3 years ago

The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye

Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0

3 years ago

Argue the importance to society of incorporating green building into an engineer’s designs, with at least two examples.

klio [65]

It is important because now a days we all need help from engineers

3 0

2 years ago

A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength

Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0

2 years ago