3 years ago

Who want to play "1v1 lol unblocked games 76"

Engineering

2 answers:

alina1380 [7]3 years ago

7 0

Answer:

Corey go save jade her ex is being annoying

Explanation:

kotegsom [21]3 years ago

3 0

Answer:I want to play what game?

Explanation:

You might be interested in

In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t

Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0

3 years ago

Is an example of an electrical device.

Yuki888 [10]

I think that it is all of the above

5 0

3 years ago

Read 2 more answers

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

Which of the following is not an electronic device ?

PolarNik [594]

Answer:

Explanation:

it's does not transmit any energy

6 0

2 years ago

How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?

topjm [15]

F=m*a

F=80*20

F =1600 ans"

7 0

3 years ago

Read 2 more answers