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dexar [7]
3 years ago
15

Oxides of nitrogen contribute to the formation of?

Engineering
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

B. Acid rain.

C. Photochemical smog.

Explanation:

Oxides of nitrogen contribute to the formation of photochemical smog and acid rain. Photochemical smog is a type of smog produced when ultraviolet light from the sun reacts with nitrogen oxides in the atmosphere while on the other hand, when nitrogen oxide react with the water vapor in the atmosphere forming nitric acid which falls on the earth surface with the help of precipitation.

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You first get a new job, and make a new company and then by amazon to traumatize Jeff Bezos after his divorce

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Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost m
Ilia_Sergeevich [38]

Answer:

ordinary bulb total cost is $39.54

fluorescent bulb total cost is $13.05

amount save = 39.54 - 13.05 = $26.49

resistance = 626.1 ohm

Explanation:

in the 1st part

bulb on time = 3 year = 4380 hours

life of bulb = 750 h

so number of bulb required = \frac{4380}{750}

number of bulb required = 6

cost of 6 bulb is = 6 × 0.75 = $4.5

so

cost of operation is = 100 × 4380 × \frac{0.08}{1000}

cost of operation = $35.04

so total cost will be = $4.5 + $35.04  = $39.54

and

when compare with florescent bulb

time = 3 year = 4380 h

life of bulb = 10000 h

so number of bulb required = \frac{4380}{10000}

number of bulb required = 0.43 = 1

cost of 6 bulb is = 1 × 5 = $5

so

cost of operation is = 23 × 4380 × \frac{0.08}{1000}

cost of operation = $8.05

so total cost will be = $5 + $8.05  = $13.05

in part 2nd

total amount save while compare bulb is

amount save = 39.54 - 13.05 = $26.49

and in part 3rd

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resistance = \frac{v^2}{P}

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6 0
3 years ago
A brake caliper is considered a suspension item.<br> True<br> False
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True


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5 0
3 years ago
A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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