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3 years ago

Which quality refers to the ability of an individual to guide other people, teams, or organizations?

Engineering

2 answers:

sukhopar [10]3 years ago

8 0

the answer is

C. leadership

kykrilka [37]3 years ago

6 0

Answer:

The answer is C. Leadership

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A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat

jeka94

Answer:

the maximum thermal efficiency is 29%

Explanation:

the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by

Maximum efficiency= 1 - T2/T1

where

T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293

T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413

therefore

Maximum efficiency= 1 - T2/T1 = 1 - 293/413 = 0,29 =29%

3 0

3 years ago

Good night. I need to go to bed. Byeeeeeeeeeeeee.

dimaraw [331]

Answer:

BYEEEEEEEEEEEE3EEEEEEEEEE

Explanation:

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3 years ago

Use the map to answer the question.

djverab [1.8K]

Answer:

Captain falcon

Explanation:

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2 years ago

The aluminum rod AB (G 5 27 GPa) is bonded to the brass rod BD (G 5 39 GPa). Knowing that portion CD of the brass rod is hollow

Temka [501]

Answer:

Qcd=0.01507rad

QT= 0.10509rad

Explanation:

The full details of the procedure and answer is attached.

7 0

4 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago