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Andrews [41]
2 years ago
10

PLS HELP! which statement is the best example of how society affects the useof technology? A. Farmers in hilly areas grow crops

on terraces built on the hillsides. B. Farmers make more money using modern farming machines. C. Farmers produce the most desirable vegetables by using pesticides. D. Farmers work harder than bankers to make a living. PLS HELP!!
Engineering
1 answer:
Zielflug [23.3K]2 years ago
7 0

Answer:

A

Explanation:

because  Farmers in hilly areas grow crops on terraces built on the hillsides.

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3 years ago
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
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To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

V = 690V

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P_{real} = 3V_{ph}I_{ph} Cos\theta

Now the Phase Voltage is,

V_{ph} = \frac{V}{\sqrt{3}}

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P_{real} = 3V_{ph}I_{ph} Cos\theta

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I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}

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3 years ago
Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting t
Natasha2012 [34]

Answer:

\dot W_{in} = 273.69\,kW

Explanation:

The pump is modelled after the First Law of Thermodynamics. A reversible process means that fluid does not report any positive change in entropy:

\dot W_{in} + \dot m \cdot (h_{in}-h_{out}) = 0

The properties of the fluid at entrance and exit are, respectively:

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P = 20\,kPa

T = 60.06\,^{\textdegree}C

h = 251.42\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

Outlet (Subcooled Liquid)

P = 6000\,kPa

T = 60.06\,^{\textdegree}C

h = 257.502\,\frac{kJ}{kg}

s = 0.8320\,\frac{kJ}{kg\cdot K}

The power input to the pump is computed hereafter:

\dot W_{in} = \left(45\,\frac{kg}{s} \right)\cdot \left(257.502\,\frac{kJ}{kg} -251.42\,\frac{kJ}{kg} \right)

\dot W_{in} = 273.69\,kW

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