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lukranit [14]
3 years ago
6

What is the electrical cable that may only be installed in a damp or wet location

Engineering
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

Standard Armored cable(AC) or Metal-clad cable(MC)

Explanation:

Standard Armored Cable (AC) or Metal-Clad Cable (MC) may be used based on the definition of words "wet or dry locations" in areas that may be temporarily exposed to the elements such as rain during construction.

Dry locations should not be confused with damp locations which are protected from these elements and water saturation or moisture. Some damp locations may include open porches, barns, under canopies, etc.

Wet locations occur in places where direct burial in the ground, in concrete, etc may occur. These are places where water or other liquid saturation is very possible, or in places that are always opened to these elements.

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A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40
tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/mm^{2}

c) 127.307 N/mm^{2}

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi*r^{2} = 78.55 mm^{2}

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give,  A2 = 0.1 A1 = 7.855  mm^{2}

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area  = F / A1 = 1000 / 78.55  

Engineering stress = 12.730 N / mm^{2}

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final  area  = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 N / mm^{2}

e) strain = change in length / original length

strain = 10 / 40  = 0.25

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3 years ago
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HOPE IT HELPS!!!!!!!!!

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