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3 years ago

2. It is a measuring instrument used to record the amount of

Engineering

1 answer:

lyudmila [28]3 years ago

3 0

Answer:

The correct option is;

D. Electric meter

Explanation:

An electric meter is a metering device that is used for the measurement of the electric power consumption of an electrical powered tools, a living space or a building

Electric meter readings are used by electric utility company to sell electric power to consumers at a given rate such that it allows the electric utility company to receive payment for the total power supplied, and for the consumer to regulate the amount of power consumed

The electric meters are usually calibrated in kilowatt hour (kWh) and prepaid meter displays the amount of units of power bought, while post paid meters are usually read once each billing period which is usually one month.

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WHAT IS A VACUOMETER?

Alex_Xolod [135]

It is a tool used to measure Low pressure

4 0

3 years ago

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A loss in value caused by an undesirable or hazardous influence offsite is which type of depreciation?

Lubov Fominskaja [6]

External depreciation may be defined as a loss in value caused by an undesirable or hazardous influence offsite.

<h3>What is depreciation?</h3>

Depreciation may be defined as a situation when the financial value of an acquisition declines over time due to exploitation, fray, and incision, or obsolescence.

External depreciation may also be referred to as "economic obsolescence". It causes a negative influence on the financial value gradually.

Therefore, it is well described above.

To learn more about Depreciation, refer to the link:

brainly.com/question/1203926

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5 0

2 years ago

What major advancement in machine tools occurred in the 1970s and what benefits did it provide? describe in your own words.

mixer [17]

Answer:

I'm just a seventh grader

4 0

3 years ago

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thermo A large container ship is propelled by a heat engine that uses fuel oil as a heat source and sea water as a heat sink. Th

TEA [102]

Answer:

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$ , $\dot Q_{out} = 162369.444\,kW$

Explanation:

The definition of thermal efficiency follows to this expression:

$\eta = \frac{\dot W}{\dot Q_{in}}$

$\eta = \frac{75\,kW}{\left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}$

$\eta = 4.617\times 10^{-4}\,(0.046\,\%)$

The rate of heat transfer to the ocean is:

$\dot Q_{out} = \dot Q_{in}-\dot W$

$\dot Q_{out} = \left(43000\,\frac{kJ}{L} \right)\cdot \left(13600\,\frac{L}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)-75\,kW$

$\dot Q_{out} = 162369.444\,kW$

3 0

4 years ago

Consider an ideal gas undergoing a constant pressure process from state 1 to state

Radda [10]

Answer:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

$ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}$

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

$\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,$

We obtain the expression to compute the specific entropy change:

$s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})$

Best regards.

6 0

4 years ago