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4 years ago

5

If, when a relay is energized, you measure 0.05 VAC across a set of normally open contacts, the A. relay is operating properly B

. contacts must be burnt C. contacts must be pitted D. coil is open

1 answer:

LiRa [457]4 years ago

3 0

Answer: A

Explanation:

It means the relay is working properly.

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when $T_{c}$ is decreased by 't',

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$\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)$

$\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}$

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3 years ago

Which of the following would require the filing of an EIS? a. expansion of an airport in a national park b. allowing snowmobiles

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d. All of the above would require an EIS.

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6. Find the heat flow in 24 hours through a refrigerator door 30.0" x 58.0" insulated with cellulose fiber 2.0" thick. The tempe

Ilia_Sergeevich [38]

Answer:

The heat flow in 24 hours through the refrigerator door is approximately 1,608.57 BTU

Explanation:

The given parameters are;

The duration of the heat transfer, t = 24 hours = 86,400 seconds

The area of the refrigerator door, A = 30.0" × 58.0" = 1,740 in.² = 1.122578 m²

The material of the insulator in the door = Cellulose fiber

The thickness of the insulator in the door, d = 2.0" = 0.0508 m

The temperature inside the fridge = 38° F = 276.4833 K

The temperature of the room = 78°F = 298.7056 K

The thermal conductivity of cellulose fiber = 0.040 W/(m·K)

By Fourier's law, the heat flow through a by conduction material is given by the following formula;

$\dfrac{Q}{t} = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d}$

$Q = \dfrac{k \cdot A \cdot (T_2 - T_1) }{d} \times t$

Therefore, we have;

$Q = \dfrac{0.04 \times 1.122578 \times (298.7056 - 276.4833 ) }{0.0508} \times 86,400 =1,697,131.73522$

The heat flow in 24 hours through the refrigerator door, Q = 1,697,131.73522 J = 1,608.5705140685 BTU

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3 years ago