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Ne4ueva [31]
3 years ago
11

Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.

Engineering
1 answer:
coldgirl [10]3 years ago
7 0

Answer:

so hard it is

Explanation:

I don't know about this

please mark as brainleast

byýyy

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MANUFACTURING QUESTION PLEASE HELP​
Nat2105 [25]

Answer:

"He then pours the metal into a mold and allows it to solidify."

3 0
3 years ago
A cylindrical metal specimen having an original diameter of 10.55 mm and gauge length of 54.5 mm is pulled in tension until frac
kumpel [21]

Answer:

(a) 53.94%

(b) 26.61%

Explanation:

Change in area will be given by

\triangle A=\frac {\pi(R_o^{2}-r_n^{2})}{\pi R_o^{2}} where \triangle A represent change in area R is radius and subscripts O and n represent original and new respectively.

Substituting 10.55/2 for original radius and 7.16/2 for new radius then

\triangle A=\frac {\pi(5.275^{2}-3.58^{2})}{\pi 5.275^{2}}\times 100\approx 53.94

(b)

Similarly, percentage elongation will be found by dividing the change in length by the the original length. In this case, rhe original length was 54.5mm and goes to 69 mm so the change in length is given by subtracting the final length from the original length

Percentage elongation is \frac {69-54.5}{54.5}\times 100\approx 26.61

6 0
4 years ago
Source 1 can supply energy at the rate of 11000 kJ/min at 310°C. A second Source 2 can supply energy at the rate of 110000 kJ/mi
VladimirAG [237]

Answer:

Source 2.

Explanation:

The efficiency of the ideal reversible heat engine is given by the Carnot's power cycle:

\eta_{th} = 1 - \frac{T_{L}}{T_{H}}

Where:

T_{L} - Temperature of the cold reservoir, in K.

T_{H} - Temperature of the hot reservoir, in K.

The thermal efficiencies are, respectively:

Source 1

\eta_{th} = 1 - \frac{311.15\,K}{583.15\,K}

\eta_{th} = 0.466 \,(46.6\,\%)

Source 2

\eta_{th} = 1 - \frac{311.15\,K}{338.15\,K}

\eta_{th} = 0.0798 \,(7.98\,\%)

The power produced by each device is presented below:

Source 1

\dot W = (0.466)\cdot (11000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 85.433\,kW

Source 2

\dot W = (0.0798)\cdot (110000\,\frac{kJ}{min})\cdot (\frac{1\,min}{60\,s} )

\dot W = 146.3\,kW

The source 2 produces the largest amount of power.

8 0
3 years ago
For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft
ivanzaharov [21]

Answer:

total time  = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - e^{(-kt^n)}   ...........1    

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}    

solve it we get

k = 0.003786  = 37.86 × 10^{4}

so here

y = 1 - e^{(-kt^n)}

1 - y  =  e^{(-kt^n)}

take ln both side

ln(1-y) = -k × t^n  

so

t = \sqrt[n]{-\frac{ln(1-y)}{k}}    .............2

now we will put the value of y = 87% in equation  with k and find out t

t = \sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}

total time  = 304.21 s

7 0
3 years ago
In the video "Understanding Flight', What is the main benefit of flying to the people in the airport?
ivolga24 [154]

Answer:

a

Explanation:

common sense

6 0
3 years ago
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