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Kobotan [32]
3 years ago
7

Im passed due someone help meeeeeee

Engineering
2 answers:
vovangra [49]3 years ago
8 0

Answer:

how are supposed to help when you can't do anything?

tamaranim1 [39]3 years ago
4 0
Earthquakes mostly happen in California
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I believe number 4 I could be wrong but I think it’s 4
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What kind of value should an employee possess when employees are expected to be responsible and fair?
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E self-confidence

Explanation:

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g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial
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0.05 mg / gallon

Explanation:

mass of chemecila coming in per minute = 50*10 = 500 mg/min

at a time t min , M = mass of chemical = 500*t mg

conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon

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3 years ago
An old refrigerator consumes 247 W of power. Assuming that the refrigerator operates for 19 hours everyday, what is the annual o
german

Answer:

The annual operating cost of the refrigerator is $102.78.

Explanation:

Power consumed by the refrigerator = 247 W = 247/1000 = 0.247 kW

Daily operation of the refrigerator = 19 hours

Annual operation of the refrigerator = 365 × 19 = 6,935 hours

Annual energy consumed = 0.247 kW × 6,935 hours = 1712.945 kWh

1 kWh of electricity cost $0.06

1712.945 kWh will cost 1712.945 × $0.06 = $102.78

Annual operating cost = $102.78

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3 years ago
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The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A
adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

COP=\frac{Q_{in}}{W}

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}

So we equate the COP of our heater with COP of Carnot heater

\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}

Rearrange the equation

\frac{1.25}{4}(24-T_{out})^2-24=0

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

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