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2 years ago

The annual average of solar photovoltaic energy in Phoenix is 6,720

Engineering

1 answer:

Dovator [93]2 years ago

3 0

Answer:

The amount of money saved is $22,075.2

Explanation:

The solar voltaic energy in Phoenix, S $_E$ = 6,720 Wh/m²/day

The area on the roof available for solar panels, A = 500 m²

The percentage of the energy received by the solar panels converted into electricity, η = 15% = 0.15

The cost of electricity, C = $0.12/kWh

The number of days in a year, t = 365 days

Therefore, the maximum area covered by the solar panel = A = 500 m²

The total solar photovoltaic energy received by the solar panels, 'E', in a year is given as follows;

E = S $_E$ × A × t

By plugging in the values of the variables, we get;

E = 6,720 Wh/m²/day × 500 m² × 365 days = 1,226,400,000 Wh

The electrical energy generated, $E_e$ = η × E

∴ $E_e$ = 0.15 × 1,226,400,000 Wh = 183960000 Wh = 183,960 kWh

The amount of money saved = The cost of electricity generated = C × $E_e$

C × $E_e$ = $0.12/kWh × 183,960 kWh = $22,075.2

∴ The amount of money saved = C × $E_e$ = $22,075.2

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the voltage across a 5mH inductor is 5[1-exp(-0.5t)]V. Calculate the current through the inductor and the energy stored in the i

Serggg [28]

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

$V(t) = 5(1-e^{-0.5t})$

The current flowing through the inductor is given by

$i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)$

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

$i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\$

$i(t) = 1000t +2000e^{-0.5t} -2000\\$

So the current at t = 2 seconds is

$i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A$

The energy stored in the inductor at t = 2 seconds is

$W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J$

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A computer maintenance company wants to 'capture' the knowledge that employees carry around in their heads by creating a databas

galina1969 [7]

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3 years ago

Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture

alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

b) 2.0 MΩ, +/- 20%

Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

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2) Black = 2nd digit = 0

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MaRussiya [10]

Answer:

The correct/closest option is b

Explanation:

Restriction enzymes are enzymes (endonucleases) that cut short DNA strands at specific sites. Hence, each restriction enzyme has it's own specific site (between two bases) it cuts at. There are two types of end that can be produced by this cut; the blunt end and the sticky end.

A restriction enzyme recognizes (palindromic sequence) and cut in it's own specific end.

For example, if a restriction enzyme cuts between a guanine (G) and an adenine (A), and it cuts a palindromic double stranded DNA in the manner below, it produces a sticky end.

G║AATTC

CTTAA║G

And if a restriction enzyme cuts between guanine (G) and cytosine (C) in the manner below, it produces a blunt end.

GGG║CCC

CCC║GGG

Hence, from the question, restriction enzymes (although chosen by the scientist based on desired sequence to be cut) recognize the sticky or blunt ends itself.

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True
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