An AM signal having a carrier frequency of 460 kHz is to be mixed with a local oscillator signal at a frequency of 1135 kHz. What does the output of the IF amplifier consist of

Answer:

I think you might have forgotten to post the problems

Answer:

The answer is " [tex]\Delta p = f(V1, p, V2, d, D, L)[/tex]"

Explanation:

Please find the complete question in the attached file.

Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.

The temperature change thus relies on V1 v2 p d D L.

Answer:

The correct solution is "20 volt".

Explanation:

Given that:

Current,

I = 10 mA

or,

 = [tex]10\times 10^{-3} \ A[/tex]

Resistance,

R = 2 K ohm

or,

  = [tex]1\times 10^3 \ ohm[/tex]

Now,

The voltage will be:

⇒ [tex]V=IR[/tex]

By putting the values, we get

        [tex]=10\times 10^{-3}\times 2\times 10^{3}[/tex]

        [tex]= 20 \ volt[/tex]

Answer:

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Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H[tex]_{pump[/tex] = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D  = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H[tex]_p[/tex] = (p/ρq)₂ + (v²/2g)₂ + z₂ + h[tex]_L[/tex]

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[  ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH[tex]_p[/tex]Q / η

where; H[tex]_p[/tex] = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ([tex]P_{atom[/tex] / ρg) - h[tex]_s[/tex] - ( P[tex]_v[/tex] / ρg )

we substitute

25  = ( 2116 / 62.42 ) - h[tex]_s[/tex] - ( 30 / 62.42 )

h[tex]_s[/tex] = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

Answer:

Explanation:

From the given information;

There is no change or any difference in velocity in between the inlet and the outlet.

Therefore by using Bernoulli's equation, we have:

[tex]\dfrac{V_1^2}{2g}+ \dfrac{P_1}{\gamma}+ z_1 + Epump= \dfrac{V_2^2}{2g}+ \dfrac{P_2}{\gamma}+ z_2+ H_L[/tex]

By dividing like terms on both sides, the equation is reduced to:

[tex]z_1 + E_{pump} = z_2+H_L \\ \\ E_{pump} =(z_2-z_1)+H_L[/tex]

where;

[tex]\Delta z = 400[/tex]

[tex]\Delta z = z_2-z_1[/tex]

[tex]\text{total head loss}= 408.5[/tex]

[tex]E_{pump} =(400)+408.5[/tex]

[tex]E_{pump} = 808.5 \ ft[/tex]

The required power input can be determined by using the formula:

[tex]P= \dfrac{\gamma_wQH_{pump}}{\eta}[/tex]

Assuming the missing pump efficiency = 70% and the flow rate Q= 1.34

Then:

[tex]P= \dfrac{62.40\times 1.34 \times 808.5}{0.7}[/tex]

[tex]P = \dfrac{96576.48 \ ft.lb/s}{550\dfrac{ ft*lb/s}{hp}}[/tex]

P = 175.594 hp

The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.

GDP termed Gross Domestic Product, has been evaluated with the value producing the economy of the region with the values added with the used products formed to be the less of the economy produced. It has been termed as the measure of the income of a region and not the wealth.

The per capita GDP has been the total income earned by a person in a region during a specified period of time. The calculation has been made by dividing the total gross income of the region by the total population.

China has been the world's most populous country in the East Asian region. It has been found that the per capita GDP of China is low because of its large population. In the calendar year 2021, the per capita GDP of China was 12,359 U.S. dollars.

Learn more about the GDP, here:

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Answer:

i) σ1 = 133.5 MPa

  σ2 = -2427 MPa

ii) 78.89 MPa

Explanation:

Given data:

ε1 = 0.0020 and ε2 = –0.0010

E = 71 GPa

v = 0.35

i) Determine the biaxial stresses  σ1 and σ2 using the relations below

ε1 = σ1 / E - v (σ2 / E)   -----( 1 )

ε2 = σ2 / E - v (σ1 / E)  -------( 2 )

resolving equations 1 and 2

σ1 = E / 1 - v^2 {  ε1 + vε2 } ---- ( 3 )

σ2 = E / 1 - v^2 {  ε2 + vε1 } ----- ( 4 )

input the given data into equation 3 and equation 4

σ1 = 133.5 MPa

σ2 = -2427 MPa

ii) Calculate the value of the maximum shear stress ( Zmax )

Zmax = ( σ1 - σ2 ) / 2

         = 133.5 - ( - 2427 ) / 2

         = 78.89 MPa

Answer:

5,465.4165939453

Explanation:

formula

A=P(1+r/n)^n(t)

p=1000

r=0.09

n=4

t=5

Answer:

The generator delivers current of 500.11 A

Explanation:

Given the data in the question;

mechanical energy delivered to the generator = 1500 hp

efficiency η = 80.0 %

terminal potential difference of the generator = 1790 V

we know that;

1 hp = 746 W

so

the mechanical energy delivered to the generator will be

Generator Input = ( 1500 × 746 )W = 1119000 W

So the generator output will be;

Generator Output = Generator Input × η

we substitute

Generator Output = 1119000 W × 80.0 %

Generator Output = 1119000 W × 0.8

Generator Output = 895200 W

So the Current will be;

[tex]I[/tex] = Generator Output / terminal potential difference of the generator

we substitute

[tex]I[/tex] =  895200 W / 1790 V

[tex]I[/tex] =  500.11 A

Therefore, The generator delivers current of 500.11 A

Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.

Answer:

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

Explanation:

The given values are,

σ=1.65 MPa

γs=0.60 J/m2

E= 2.0 GPa

The maximum possible length is calculated as:

[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

a) Determine temperature at end of fin

m = √ hp/Ka

   = √( 140*2 ) / ( 12 * 10^-3 * 55 )

   = √ 280 / 0.66 = 20.60

Attached below is the remaining answers

Answer:

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Explanation:

Answer:

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Answer:

a) the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air is 1065 kJ/min

Explanation:

Given the data in the question;

[ Outdoor ] ←  Q[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]

Rate of heat removed from the house; Q[tex]_L[/tex]  = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW

Net-work input; W[tex]_{net, in[/tex] = 5.25 kW

a) The coefficient of performance of the air conditioner; COP.

COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]

we substitute

COP = 12.5 kW / 5.25 kW

COP = 2.38

Therefore, the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air.

Q[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]

we substitute

Q[tex]_H[/tex] = 12.5 kW + 5.25 kW

Q[tex]_H[/tex] = 17.75 kW

Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min

Q[tex]_H[/tex] = 1065 kJ/min

Therefore, the rate of heat transfer to the outside air is 1065 kJ/min

Answer:

"192.3 watt" is the right answer.

Explanation:

Given:

Efficient amplifier,

= 65%

or,

= 0.65

Power,

[tex]P_c=250 \ watt[/tex]

As we know,

⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]

By putting the values, we get

        [tex]=P_c(1+\frac{1}{2} )[/tex]

        [tex]=1.5 \ P_c[/tex]

Now,

⇒ [tex]P_i=(P_t-P_c)[/tex]

        [tex]=1.5 \ P_c-P_c[/tex]

        [tex]=\frac{P_c}{2}[/tex]

DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".

hence,

The DC input power will be:

= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]

= [tex]\frac{125}{0.65}[/tex]

= [tex]192.3 \ watt[/tex]

Answer:

2.68

Explanation:

Percentage by Mass of each Aggregate :

Pa = 65% ; Pb = 35%;

Bulk Specific gravity of each aggregate :

Ga = 2.45 ; Gb = 3.25

Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)

Gsb = (65 + 35) / (65/2.45 + 35/3.25)

Gsb = (65 + 35) / 37.299843

Gsb = 100 / 37.299843

Gsb = 2.68

Answer:

33.56 ft^3/sec.in

Explanation:

Duration = 6 hours

drainage area = 185 mi^2

constant baseflow = 550 cfs

Derive the unit hydrograph using the inverse procedure

first step : calculate for the volume of direct runoff hydrograph using the details in table 2 attached below

Vdrh = sum of drh *  duration

        = 29700 * 6 hours ( 216000 secs )

        = 641,520,000 ft^3.

next step : Calculate the volume of runoff in equivalent depth

Vdrh / Area = 641,520,000  / 185 mi^2

                    = 1.49 in

Finally derive the unit hydrograph

Unit of hydrograph = drh /  volume of runoff in equivalent depth

                                = 50 ft^3 / 1.49 in  =  33.56 ft^3/sec.in

Answer:

The answer is "22.501,-22.899"

Explanation:

Just as in the previous problems find the angle the velocity makes with the  x-axis and radius of curvature.

[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]

For the radius of curvature, we can use the expression from the last two  problems, but first express the position and derivatives as y(x).

[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]

The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)

[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]

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Explanation:

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Answer:

Part A

The flow of the bee lava is inertia dominant and laminar

Part B

The flow of a baseball in Mars is inertia dominated and the flow is laminar

Part C

The flow of an elderly on i95 is viscous force dominated and turbulent

Explanation:

The Reynold's number when inertia is dominant is low, and the flow is laminar

When viscous force is dominant, the Reynold's number is high, and we have turbulent flow

The Reynold's number of laminar flow is Re < 2,000

The Reynold's of unstable or intermediate flow is 2,000 < Re < 4,000

The Reynold's number of turbulent flow is Re > 4,000

Therefore, we have;

Part A

The Reynold's number of a bee lava, Re = 0.2, therefore, the flow is inertia and laminar

Part B

The Reynold's number of the ball in Mars is Re = 500, therefore, given that the Reynold's number is less than 2,000, the inertia is dominant, and the flow is laminar flow

Part C

The Reynold's number of the driver on i95 = 1,200,000 which is larger than 4,000, therefore, the flow is viscous force dominated and the flow is turbulent.

Answer:

2

Explanation:

From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.