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Marysya12 [62]
3 years ago
7

Make a proposal to add a small pizza shop to a historical part of town. How could it be designed to “fit” into the area?

Engineering
2 answers:
spin [16.1K]3 years ago
7 0
It could be designed because if we had a small little pizza shop it would save our problems of going through all the trouble just for pizza
Nookie1986 [14]3 years ago
4 0
The small pizza shop could maybe add a special item recognizing one of the history of the town. Or, the small pizza shop could design their building to fit the historical theme.
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Bridge A is the longest suspension bridge in a Country. Bridge B is 5555 feet shortershorter than Bridge A. If the length of Bri
MatroZZZ [7]

Answer:

the length of the bridge B is L = m - 5555 ft

Explanation:

Denoting the length of the bridge B as L we get

length of bridge B = length of bridge A - 5555 ft

since length of bridge A=m

L = m - 5555 ft

4 0
3 years ago
Showing or hiding records in a database is called “filtering.”<br> True<br> False
agasfer [191]

Answer:

TRUE

Explanation:

4 0
2 years ago
Read 2 more answers
A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is mainta
strojnjashka [21]

Answer:

\dot W_{out} = 399.47\,kW

Explanation:

The turbine is modelled after the First Law of Thermodynamics:

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

The work done by the turbine is:

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

The properties of the water are obtained from property tables:

Inlet (Superheated Steam)

P = 10\,MPa

T = 520\,^{\textdegree}C

h = 3425.9\,\frac{kJ}{kg}

Outlet (Superheated Steam)

P = 1\,MPa

T = 280\,^{\textdegree}C

h = 3008.2\,\frac{kJ}{kg}

The work output is:

\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW

\dot W_{out} = 399.47\,kW

5 0
3 years ago
A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of he
hichkok12 [17]

Answer:

ΔT=  11.94 °C

Explanation:

Given that

mass of water = 10 kh

Time t= 15 min

Heat lot from water = 400  KJ

Heat input to the water = 1  KW

Heat input the water= 1 x 15 x 60

                                =900 KJ

By heat balancing

Heat supply - heat rejected = Heat gain by water

As we know that heat capacity of water

C_p=4.187 \frac{KJ}{kg-K}

Q=mC_p\Delta T

Now by putting the values

900 - 400 = 10 x 4.187 x ΔT

So  rise in temperature of water ΔT=  11.94 °C

6 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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