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nekit [7.7K]
3 years ago
10

When must an Assured Equipment Grounding Conductor Program (AEGCP) be in place?

Engineering
1 answer:
sasho [114]3 years ago
8 0

Answer:

Only when working around power lines

Explanation:

Assured Equipment Grounding Conductor Program (AEGCP) is applicable mainly in sites dealing with power. Therefore, the most appropriate worksite to have Assured Equipment Grounding Conductor Program is when  working around power lines.

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A 36 ft simply supported beam is loaded with concentrated loads 16 ft inwards from each support. On the left side, the dead load
lana66690 [7]

Answer:

1st part: Section W18X76  is adequate

2nd part: Section W21X62 is adequate

Explanation:

See the attached file for the calculation

8 0
3 years ago
Air expands through a turbine operating at steady state. At the inlet p1 = 150 lbf/in^2, T1 = 1400R and at the exit p2 = 14.8 lb
Paraphin [41]

Answer:

The power developed in HP is 2702.7hp

Explanation:

Given details.

P1 = 150 lbf/in^2,

T1 = 1400°R

P2 = 14.8 lbf/in^2,

T2 = 700°R

Mass flow rate m1 = m2 = m = 11 lb/s Q = -65000 Btu/h

Using air table to obtain the values for h1 and h2 at T1 and T2

h1 at T1 = 1400°R = 342.9 Btu/h

h2 at T2 = 700°R = 167.6 Btu/h

Using;

Q - W + m(h1) - m(h2) = 0

W = Q - m (h2 -h1)

W = (-65000 Btu/h ) - 11 lb/s (167.6 - 342.9) Btu/h

W = (-65000 Btu/h ) - (-1928.3) Btu/s

W = (-65000 Btu/h ) * {1hr/(60*60)s} - (-1928.3) Btu/s

W = -18.06Btu/s + 1928.3 Btu/s

W = 1910.24Btu/s

Note; Btu/s = 1.4148532hp

W = 2702.7hp

5 0
3 years ago
Gtjffs
grandymaker [24]

the required documents is 3000

4 0
2 years ago
Explain the term electric current as used in engineering principles​
vodomira [7]

Answer:

<em>Electric current is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.</em><em>Current is directly proportional to voltage, inversely proportional to resistance. One of the most common electrical measurements you'll use is the watt, a unit of electrical power: W (Watts) = E (Volts) x I (Amperes). The quantity of electric charge is measured in coulombs.</em><em>They can even pass through bones and teeth. This makes gamma rays very dangerous. They can destroy living cells, produce gene mutations, and cause cancer.</em>

Explanation:

hey mate this is the best answer if you're studying engineering!

8 0
3 years ago
Read 2 more answers
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W
cupoosta [38]

Answer:

It will not  experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

<em>σc = KIc / (Y*√(π*a))</em>

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0
3 years ago
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