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tigry1 [53]
3 years ago
7

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:
vova2212 [387]3 years ago
8 0

Answer:

A continuity test

Explanation:

A continuity test is used to verified that current will flow in an electrical circuit, it performed by placing a small voltage across the chosen path. continuity test ensure that the equipment grounding conductor is electrically continuous and this test is perform on all the 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. example of equipment used in testing current flow in continuity test are Analog multi-meter, voltage/continuity tester etc.

Continuity test and terminal connection test are the two test required by OSHA on all electrical equipment

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Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

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If an older multimeter is set to display its reading in megohms, what must you do to determine the correct value? Please help.
Gekata [30.6K]

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Explanation:

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