How would you achieve the linear convolution of a 100 sample time series and a 20 tap filter in the frequency domain?

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]

[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = [tex]T_o [ \Delta S][/tex]

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW

Answer:

hmmm.........

Explanation:

Answer:

0.2

Explanation:

Since the span and chord of the rectangular wing is missing, due to it being from the other question, permit me to improvise, or assume them. While you go ahead and substitute the ones from your question to it, as it's both basically the same method.

Let the span of the rectangular wing be 0.225 m

Let the chord of the rectangular wing be 0.045 m.

Then, the area of any rectangular chord is

A = chord * span

A = 0.045 * 0.225

A = 0.010 m²

And using the weight of the glider given to us from the question, we can find the LER for the wing.

LER = Area / weight.

LER = 0.010 / 0.05

LER = 0.2.

Therefore, using the values of the rectangular wing I adopted, and the weight of the glider given, we can see that the LER of the glider is 0.2

Please mark brainliest...

Answer: 0.2025

Explanation: I got it correct

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

Answer: the air flow rate a is 90 kg/sec; Option b) 90 kg/sec is the correct answer

Explanation:

Given that;

product of combustor flow rate m = 100 kg/s

air-fuel = 9

Airflow rate = ?

⇒We know that in the combustor, air fuel are mixed and then ignited,

⇒air fuel products are exited at the combustor

let air and fuel be a and b respectively

⇒ a + b = 100 kg/sec ----- let this be equation 1

now

⇒ air / fuel = 9

a / b = 9

a = 9b -----------let this be equation 2

now input a = 9b in equation 1

9b + b = 100 kg/sec

10b = 100 kg/sec

b = 10 kg/sec

we know that

a = 9b

so a = 9 × 10 = 90 kg/sec

Therefore the air flow rate a is 90 kg/sec

Answer: The 8-core machine saves  87.5% of the dynamic power.

Explanation:

Let Fold = f , Vold = V , Cold = Capacitance

so

Old Dynamic power = Cold × (Vold × Vold) × f

therefore for the 8-core machine

 Fnew / Fold = 1/4

Fnew = Fold/4

we were told that Voltage decreases proportional to frequency,

so

Vnew / Vold = 1/4

Vnew = V / 4

So New Capacitance will be;

Cnew = Cold

Thus, New Dynamic power = 8 × Cnew × ( Vnew × Vnew ) ×  Fnew

= 8 × Cold × (Vold × Vold/16) × ( f/4 )

=  8 × ( Cold ) × ( Vold × Vold ) × ( f ) / 64

= (Old Dynamic Power) / 8

therefore

Old Dynamic Power / New Dynamic Power = 8

Thus, Percentage of power saved will be;

Percentage power saved = 100 × ( Old Dynamic Power - New Dynamic Power ) / Old Dynamic Power

=   100 × (8-1) / 8

= 87.5 %

Therefore The 8-core machine saves  87.5% of the dynamic power.

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

Answer: attached below is the power triangles

7.13589 MVAR

Explanation:

Power ( P1 ) = 10 MW

power factor ( cos ∅ ) = 0.6 lagging

New power factor = 0.85

Calculate the reactive power of a capacitor to be connected in parallel

Cos ∅ = 0.6

therefore ∅ = 53.13°

S = P1 / cos ∅ = 16.67 MVA

Q1 = S ( sin ∅ ) = 13.33 MVAR  ( reactive power before capacitor was connected in parallel )

note : the connection of a capacitor in parallel will cause a change in power factor and reactive power while the active power will be unchanged i.e. p1 = p2

cos ∅2 = 0.85 ( new power factor )

hence ∅2 =  31.78°

Qsh ( reactive power when power factor is raised to 0.85 )

= P1 ( tan∅1 - tan∅2 )

= 10 ( 1.333 - 0.6197 )

= 7.13589 MVAR

Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

Answer:

Super critical

1.2 m

Explanation:

Q = Flow rate = [tex]12\ \text{m}^3/\text{s}[/tex]

w = Width = 3 m

d = Depth = 90 cm = 0.9 m

A = Area = wd

v = Velocity

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]Q=Av\\\Rightarrow v=\dfrac{Q}{wd}\\\Rightarrow v=\dfrac{12}{3\times 0.9}\\\Rightarrow v=4.44\ \text{m/s}[/tex]

Froude number is given by

[tex]Fr=\dfrac{v}{\sqrt{gd}}\\\Rightarrow Fr=\dfrac{4.44}{\sqrt{9.81\times 0.9}}\\\Rightarrow F_r=1.5[/tex]

Since [tex]F_r>1[/tex] the flow is super critical.

Flow is critical when [tex]Fr=1[/tex]

Depth is given by

[tex]d=(\dfrac{Q^2}{gw^2})^{\dfrac{1}{3}}\\\Rightarrow d=(\dfrac{12^2}{9.81\times 3^2})^{\dfrac{1}{3}}\\\Rightarrow d=1.2\ \text{m}[/tex]

The depth of the channel will be 1.2 m for critical flow.

Explanation:

D. B. C. A. E. Is this a good idea

Answer:Total Voltage = 14V

Explanation: it is possible that a circuit  can contain more than one source of electromotive force which can cause flow of current in the same or opposite direction . When the  connection to  voltage sources  allows for current  from the voltage sources to flow in  same direction,it is termed  Series aiding  Thus, the  Total/effective voltage in a series aiding circuit is  computed as the sum of series aiding voltages .

 Here we have the series aiding voltages to be 5V and 9V ,

therefore,

Total Voltage = 5V + 9V

= 14V

Answer:

the motor input power is 19.42 KW

the Reactive power is 17.65 KVAR

Current is 31.56 A

Explanation:

Given that;

V = 480V

h.p = 25 hp

p.f = 0.74 lagging

n_motor = 96%

so output = 25hp

and we know that;

1hp = 746 watt

watt = hp × 1hp

so output in watt = 25 × 746 = 18650 Watt = 18.65 KW

n_motor = (output / input) × 100

96 = 1865 / Input

96Input = 1865

Input = 1865 / 96

Input = 19.42 KW

Therefore the motor input power is 19.42 KW

P = √( 3 × V × I × cos∅)

19.42 = √( 3 ×480 × I × 0.74)

I = 31.56 A

Therefore Current is 31.56 A

Q = √( 3 × V × I × sin∅)

we know that

cos∅ = 0.74

so ∅ = cos⁻¹(0.74) = 42.26

so we substitute

Q = √( 3 × 480 × 31.56 × sin(42.26))

 = 17.65 KVAR

Therefore the Reactive power is 17.65 KVAR

Answer:

C/C++

Explanation:

C/C++

Answer:

406.140 KHz

Explanation:

Given data:

Rsig = 100 kΩ

Rin = 100kΩ

Cgs = 1 pF,

Cgd = 0.2 pF,  and   etc.

Determine the expected 3-dB cutoff frequency

first find the CM miller capacitance

CM = ( 1 + gm*ro || RL )( Cgd )

     = ( 1 + 5*10^-3 * 25 || 20 ) ( 0.2 )

     = ( 11.311 ) pF

now we apply open time constant method to determine the cutoff frequency

Th = 1 / Fh

hence : Fh = 1 / Th = [tex]\frac{1}{(Rsig +Rin) (Cm + Cgs )}[/tex]

                               = [tex]\frac{1}{( 200*10^3 ) ( 12.311 * 10^{-12} )}[/tex] =  406.140 KHz

Answer:

The appropriate solution will be "6.4 cm".

Explanation:

The given values are:

Length,

l = 0.2 m

Thermal conductivity,

K₁ = 1.82 W/m-K

K₂ = 0.095 W/m-K

Temperature,

T = 950 K

T = 300 K

Heat transfer rate,

Q = 830 W/m²

Now,

⇒  [tex]Q = \frac{\Delta T}{\frac{L_1}{K_1 A} +\frac{L_2}{K_2 A} }=\frac{A \Delta T}{\frac{L_1}{K_1 } +\frac{L_2}{K_2 } }[/tex]

⇒ [tex]\frac{Q}{A} =\frac{\Delta T}{\frac{L_1}{K_1} +\frac{L_2}{K_2} }[/tex]

On substituting the above given values in the equation, we get

⇒ [tex]830=\frac{(980-300)}{\frac{0.2}{1.82} +\frac{x}{0.095} }[/tex]

On applying cross-multiplication, we get

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{950-300}{830}[/tex]

⇒ [tex]\frac{0.2}{1.82} +\frac{x}{0.095} =\frac{650}{830}[/tex]

⇒                [tex]x =0.639 \ m[/tex]

⇒                [tex]x=6.345 \ i.e., 6.4 \ m[/tex]  

Answer: its A

Explanation:

Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr