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IRISSAK [1]
2 years ago
13

A TV USE 75 WATTS WHILE IN USED ASSMING THAT ITIS USED 4 HOURS EVERY DAY HOW MUCH ENERGY IN 4 IN KWH WOULD THE TV CONSUME ANNUAL

LY?.
Engineering
2 answers:
prohojiy [21]2 years ago
7 0

Answer:

i don't think i understand the question

Explanation:

alex41 [277]2 years ago
4 0

Answer:

Explanation:

The wattage listed is the maximum power drawn by the ... If John uses a window fan (200 watts) 4 hours a day for 120 ... TV, 120V, 75W.

                                                  or

75 w 4 hrs 30 days. So your total consumption is 75*4*30= 9000watt hr.

1 unit = 1000 watt hr.

9000 watt hr = 9 unit.

Total cost 9*2.5= RS 22.5

pls mark me as brainliest and this is a usefull answer pls dont delete

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Comparison of copper and aluminium conductors looking at their properties
Alexeev081 [22]

Answer:

The density of the copper is higher than aluminium. Hence it is heavier compared to aluminium conductors it requires strong structures and hardware to bear the weight. More ductile and has high tensile strength.

...

Aluminium & Copper properties.

Property Copper (Cu) Aluminium (Al)

Density (g/cm3) 8.96 2.70

7 0
3 years ago
The products of combustion from burner are routed to an industrial application through a thin-walled metallic duct of diameter D
slamgirl [31]

Answer:

Start by calculating the heat lost falling from Tm,i to Tm,o

As a first approximation use an average Cp of (Tm,i + Tm,o)/2 and an average density at (Tm,i + Tm,o)/2

Cp air at (Tm,i + Tm,o)/2 ----(1)

Density of air at (Tm,i + Tm,o)/2 : PV=nRT, V=T1/T2, air density at 293K = 1.204kg/m^3

Air density at (Tm,i + Tm,o)/2 1.204×293/(Tm,i + Tm,o)/2

Energy lost per kg for (Tm,i - Tm,o)K drop: (Tm,i - Tm,o)K × (1) =

Time of travel: t

Energy lost per kg drop/t seconds = 24.3 kW

Volume occupied by 1kg air at at mean tempera ture = 1/Air density at mean temperature

Length of pipe ( Di m diameter) needed to hold Volume occupied by 1 kg of air at mean temperature :

Cross section = π/4 Di^2, Volume occupied by 1 Kg of air at mean temp. ÷ π/4Di^2

Surface area of pipe Di m diameter by L m long = Length of pipe to hold Volume of air in m × π*Di

Q/A=k (delta T)/ thickness,

Thickness of insulation = Area × k ×dT / Q

Explanation:

6 0
3 years ago
Please. my brain isn’t working right now
Alex787 [66]

Answer:

10kQ is to 36......... D2 _ D1 D4

7 0
3 years ago
What is the built-in pollution control system in an incinerator called
Kobotan [32]

Explanation:

hbyndbnn☝️

7 0
3 years ago
In a wind-turbine, the generator in the nacelle is rated at 690 V and 2.3 MW. It operates at a power factor of 0.85 (lagging) at
Juli2301 [7.4K]

To solve this problem we will apply the concepts related to real power in 3 phases, which is defined as the product between the phase voltage, the phase current and the power factor (Specifically given by the cosine of the phase angle). First we will find the phase voltage from the given voltage and proceed to find the current by clearing it from the previously mentioned formula. Our values are

V = 690V

P_{real} = 2.3MW

Real power in 3 phase

P_{real} = 3V_{ph}I_{ph} Cos\theta

Now the Phase Voltage is,

V_{ph} = \frac{V}{\sqrt{3}}

V_{ph} = \frac{690}{\sqrt{3}}

V_{ph} = 398.37V

The current phase would be,

P_{real} = 3V_{ph}I_{ph} Cos\theta

Rearranging,

I_{ph}=\frac{P_{real}}{3V_{ph}Cos\theta}

Replacing,

I_{ph}=\frac{2.3MW}{3( 398.37V)(0.85)}

I_{ph}= 2.26kA/phase

Therefore the current per phase is 2.26kA

6 0
3 years ago
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