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bonufazy [111]
1 year ago
5

This test should be performed on all cord sets, receptacles that aren't part of a building or structure's permanent wiring, and

cord-and-plug connected equipment required to be grounded. The point of this test is to make sure that the equipment grounding conductor is electrically continuous.
Engineering
1 answer:
Ray Of Light [21]1 year ago
5 0

Answer:

That would be Continuity test

Explanation:

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Sinks must be used for the correct intended purpose to prevent
irakobra [83]

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... spilling water or getting anything cascading onto the floor

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3 years ago
Zachary finished an internship as a Software Quality Assurance Engineer. For which job is he best qualified?
astra-53 [7]

Answer:

B. A software development firm needs someone to find and fix bugs on multiple computer platforms.

Explanation:

A software quality assurance engineer is someone who monitors every phase of the software development process so as to ensure design quality, making sure that the software adheres to the standards set by the development company. Finding bugs would make this intern a amazing bug finder

6 0
3 years ago
Read 2 more answers
If particleboard is used as wall sheathing, the grade mark with type _____ or _____ should be stamped on it.
kramer

If particleboard is used as wall sheathing, the grade mark with type M1 or M2 should be stamped on it.

<h3>What is particle board?</h3>

Particle board is notably used as floors underlayment or as a base for parquet floors, timber floors, or for carpets. For this purpose, the particle forums are dealt with with unique chemical compounds and resins to cause them to water-resistant or termite proof.

Waferboard, OSB, and composite plywood, while carried out as wall sheathing, offer a nail base for software of shingle siding.

Read more about the sheathing:

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8 0
2 years ago
A cylindrical resistor element on a circuit board dissipates 0.6 W of power. The resistor is 1.5 cm long, and has a diameter of
Burka [1]

Answer:

a. 51.84Kj

b. 2808.99 W/m^2

c. 11.75%

Explanation:

Amount of heat this resistor dissipates during a 24-hour period

= amount of power dissipated * time

= 0.6 * 24 = 14.4 Watt hour

(Note 3.6Watt hour = 1Kj )

=14.4*3.6 = 51.84Kj

Heat flux = amount of power dissipated/ surface area

surface area = area of the two circular end  + area of the curve surface

=2*\frac{\pi D^{2} }{4} + \pi DL\\=2*\frac{\pi *(\frac{0.4}{100} )^{2} }{4} + \pi *\frac{0.4}{100} *\frac{1.5}{100}

= 2.136 *10^-4 m^{2}

Heat flux =\frac{0.6}{2.136 * 10^{-4} } = 2808.99 W/m^{2}

fraction of heat dissipated from the top and bottom surface

=\frac{\frac{2*\pi D^{2} }{4} }{\frac{2*\pi D^{2}}{4} + \pi DL } \\\\=\\\frac{\frac{2*\pi *(\frac{0.4}{100} )^{2} }{4} }{\frac{2*\pi *(\frac{0.4}{100}  )^{2} }{4} +\pi *\frac{0.4}{100} *\frac{1.5}{100} } \\\\=\frac{2.51*10^{-5} }{2.136*10^{-4} } \\\\\= 0.1175

=11.75%

8 0
3 years ago
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Dahasolnce [82]

Answer:

crosscut

Explanation:

8 0
1 year ago
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