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Maru [420]
2 years ago
15

All of the following are examples of capital intensive industries EXCEPT: *

Engineering
2 answers:
OverLord2011 [107]2 years ago
8 0

i believe it is soft drink production

Vadim26 [7]2 years ago
7 0

Answer:

hospital

Explanation:

  • Soft drink production is a capital intensive industry as it's depends upon the economy and trade.
  • Oil extraction is a national industry and it also involves the intention of trade.
  • Manufacturing is similar to soft drink production and has intense of trade.
  • Hospital doesn't has the intention of trade so it's the answer
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An empty metal can is heated to 908C and sealed. It is then placed in a room to cool to 208C. What is the pressure inside the ca
Natali5045456 [20]

The pressure inside the can upon cooling is 0.4 atm.

<u>Explanation:</u>

Given -

Initial Temperature, T1 = 908°C = 908 + 273 K = 1181 K

Final Temperature, T2 = 208°C = 208 + 273 K = 481 K

Pressure upon cooling, P2 = ?

Using Gay Lussac's law:

P1/T1 = P2/T2

P2 = P1 X T2 / T1

P2 = 1 atm X 481 / 1181

P2 = 0.4 atm

Therefore, the pressure inside the can upon cooling is 0.4 atm.

3 0
3 years ago
A ____ is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does no
Alika [10]

Answer:

A relief valve is either in the pressure reducer or in the downstream side of the system to ensure that the control air pressure does not exceed about 30 psig.

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2 years ago
A PMOS device with VT P = −1.2 V has a drain current iD = 0.5 mA when vSG = 3 V and vSD = 5 V. Calculate the drain current when:
Ksju [112]
The answer is b ! Hope I helped
7 0
3 years ago
At a festival, spherical balloons with a radius of 140.cm are to be inflated with hot air and released. The air at the festival
Tpy6a [65]

Answer:

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Explanation:

5 0
3 years ago
A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
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