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k0ka [10]
3 years ago
9

A business owned by one people is a

Engineering
2 answers:
Whitepunk [10]3 years ago
6 0

Answer:

A sole proprietorship, also known as the sole trader, individual entrepreneurship or proprietorship, is a type of enterprise that is owned and run by one person and in which there is no legal distinction between the owner and the business entity.

Explanation:

Damm [24]3 years ago
3 0

Answer:

A business owned by one people is a-Sole proprietorship

Explanation:

This is correct i did it for my electives class which was career reaserch and decision making.

brainliest?

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An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum
Licemer1 [7]

Answer:

Base temperature is 46.23 °C

Explanation:

I've attached explanations

6 0
3 years ago
Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with
Scilla [17]

Answer:

DIAMETER  = 9.797 m

POWER = \dot W = 28.6 kW

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

v =\frac{RT}{P}

  where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

v =  \frac{0.287\times 293}{100}

v = 0.8409 m^3/ kg

from continuity equation

A_1 V_1 = A_2 V_2

\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2

D_2 = D_1 \sqrt{\frac{V_1}{V_2}}

D_2 = 8 \times \sqrt{\frac{12}{8}}

D_2 = 9.797 m

mass flow rate is given as

\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}

\dot m = 717.309 kg/s

the power produced \dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]

\dot W = 28.6 kW

8 0
3 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

8 0
3 years ago
Going green means: increasing one's initiatives toward a concern for the environment. increasing one's bottom line, before any o
Tcecarenko [31]

Answer:

Going green means increasing one's initiatives toward a concern for the environment.

Explanation:

Going green involves all the knowledge and practices that can lead to more environmentally friendly and ecologically responsible decisions and lifestyles, which would protect and sustain the natural resources present in the environment for both present and future generations.

8 0
3 years ago
Five kilograms of air at 427°C and 600 kPa are contained in a piston–cylinder device. The air expands adiabatically until the pr
son4ous [18]

Answer:

The entropy change of the air is 0.240kJ/kgK

Explanation:

T_{1} =427+273K,T_{1} =700K\\P_{1} =600kPa\\P_{2} =100kPa

T_{2}  is unknown

we can apply the following expression to find T_{2}

-w_{out} =mc_{v} (T_{2} -T_{1} )

T_{2} =T_{1} -\frac{w_{out } }{mc_{v} }

now substitute

T_{2} =700K-\frac{600kJ}{5kg*0.718kJ/kgK} \\T_{2}=533K

To find entropy change of the air we can apply the ideal gas relationship

Δs_{air}=c_{p} ln\frac{T_{2} }{T_{1} } -Rln\frac{P_{2} }{P_{1} }

Δs_{air} =1.005*ln(\frac{533}{700})-0.287* in(\frac{100}{600} )

Δs_{air} =0.240kJ/kgK

4 0
3 years ago
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