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Andru [333]
3 years ago
10

MCQ : What are the main the constituents of acid rains?

Engineering
2 answers:
IrinaVladis [17]3 years ago
4 0

Answer:

c is correct

Explanation:

nexus9112 [7]3 years ago
3 0

Answer:

a. Nitrogen Oxide

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Determine the period of each of the following discrete-time signals (if a signal is not periodic, denote its period by infinity)
sergiy2304 [10]

Answer:

a) it is periodic

N = (20/3)k = 20 { for K =3}

b) it is Non-Periodic.

N = ∞

c) x(n) is periodic

N = LCM ( 5, 20 )

Explanation:

We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.

then the period of the signal is given as

N = ( 2π/w₀)K

k is least integer for which N is also integer

Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2

now

a) cos(2π(0.15)n)

w₀ = 2π(0.15)

Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3

so, it is periodic

N = (20/3)k = 20 { for K =3}

b) cos(2n);

w₀ = 2

Now, 2π/w₀ = 2π/2) = π

so, it is Non-Periodic.

N = ∞

c)  cos(π0.3n) + cos(π0.4n)

x(n) = x1(n) + x2(n)

x1(n) = cos(π0.3n)

x2(n) =  cos(π0.4n)

so

w₀ = π0.3

2π/w₀  = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3

∴ N1 = 20

AND

w₀ = π0.4

2π/w₀  = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5

∴ N² = 5

so, x(n) is periodic

N = LCM ( 5, 20 )

6 0
3 years ago
2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m
Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

4 0
3 years ago
Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 20 C.?
SIZIF [17.4K]

Answer:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

Explanation:

Yes,If we use river water which is entering at 20⁰ C in the condenser then it is possible to maintain the pressure of 10 KPa in condenser.

The saturation temperature of steam is 45.81⁰ C at the pressure of 10 KPa which is higher than 20⁰C of river water. So river water at 20⁰C can be used to maintain the condenser pressure to 10 KPa.

7 0
3 years ago
The power supply converts the wall outlet AC power into DC power. T or F
Alika [10]

Answer:

True

Explanation:

All computer parts require DC power to operate, and wall outlets provide AC Power.

7 0
3 years ago
A lake with an area of 525 [acre] was monitored during a one-month period. The average inflow was 30 [cfs] for the month and the
Alex73 [517]

Answer:

\Delta S = 1581663.5ft^3

Explanation:

We need to calculate the change in storage through the changes given,

That is,

\Delta S = P+I-O-O_{seepage}-E

Where the loss are representing by,

P= Precipitation\\I= Inflow\\O= Outflow\\O_{Seepage}= Outflow by seepage\\E=Evaporation

So calculating the values we have

\Delta S = P+(I-0)-O_{seepage}-E

\Delta S = 4.25+ (30ft^3/s-27ft^3/s)-1.5in-6in

The values inside the are parenthesis need to be konverted as I note here.

(30days(24hr/1day)(3600s/1hr)(1acre.ft/43560ft^3)(1/525acres)(12in/1ft)

That is,

\Delta S = 0.83in\Delta S= (0.83in*1ft/12in)(525acres)\\\Delta S=36.31 acres.ft\\\Delta S=36.31acres.ft*(43560ft^3/acrees.ft)\\\Delta S = 1581663.5ft^3

3 0
3 years ago
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