In particular, a system may or may not be (1) Memoryless, (2) Time invariant, (3)Linear, (4) Casual, (5) Stable.

a. True
b. False

Answers

Answer 1

Answer:

a. True

Explanation:

A system may be sometimes casual, time invariant, memoryless, stable and linear in particular.

Thus the answer is true.

A system is casual when the output of the system at any time depends on the input only at the present time and in the past.

A system is said to be memoryless when the output for each of the independent variable at some given time is fully dependent on the input only at that particular time.

A system is linear when it satisfies the additivity and the homogeneity properties.

A system is called time invariant when the time shift in the output signal will result in the identical time shift of the output signal.

Thus a system can be time invariant, memoryless, linear, casual and stable.


Related Questions

can help me with this circuit question? ​
the last word is (cutoff)

Answers

Answer:

The answer is "[tex]25\times 10^{-9}[/tex]".

Explanation:

Using formula:

[tex]f_c=\frac{1}{2\pi RC}\\\\w_c= 4 \frac{krad}{sec}\\\\w_c=2\pi fc\\\\R=w\\\\c=\frac{1}{w_c\ R}\\\\[/tex]

  [tex]=\frac{1}{4 \times 10^3 \times 10\times 10^3}\\\\=\frac{1}{40 \times 10^6 }\\\\=0.025 \times 10^{-6 }\\\\=25\times 10^{-9}[/tex]

Can you use isentropic efficiency for a non-adiabatic compressor?

Can you use isothermal efficiency for an adiabatic compressor?

Answers

Mark brainliest please!

Isothermal work will be less than the adiabatic work for any given compression ratio and set of suction conditions. The ratio of isothermal work to the actual work is the isothermal efficiency. Isothermal paths are not typically used in most industrial compressor calculations.

Compressors

Compressors are used to move gases and vapors in situations where large pressure differences are necessary.

Types of Compressor

Compressors are classified by the way they work: dynamic (centrifugal and axial) or reciprocating. Dynamic compressors use a set of rotating blades to add velocity and pressure to fluid. They operate at high speeds and are driven by steam or gas turbines or electric motors. They tend to be smaller and lighter for a given service than reciprocating machines, and hence have lower costs.

Reciprocating compressors use pistons to push gas to a higher pressure. They are common in natural gas gathering and transmission systems, but are less common in process applications. Reciprocating compressors may be used when very large pressure differences must be achieved; however, since they produce a pulsating flow, they may need to have a receiver vessel to dampen the pulses.

The compression ratio, pout over pin, is a key parameter in understanding compressors and blowers. When the compression ratio is below 4 or so, a blower is usually adequate. Higher ratios require a compressor, or multiple compressor stages, be used.

When the pressure of a gas is increased in an adiabatic system, the temperature of the fluid must rise. Since the temperature change is accompanied by a change in the specific volume, the work necessary to compress a unit of fluid also changes. Consequently, many compressors must be accompanied by cooling to reduce the consequences of the adiabatic temperature rise. The coolant may flow through a jacket which surrounds the housing with liquid coolant. When multiple stage compressors are used, intercooler heat exchangers are often used between the stages.

Dynamic Compressors

Gas enters a centrifugal or axial compressor through a suction nozzle and is directed into the first-stage impeller by a set of guide vanes. The blades push the gas forward and into a diffuser section where the gas velocity is slowed and the kinetic energy transferred from the blades is converted to pressure. In a multistage compressor, the gas encounters another set of guide vanes and the compression step is repeated. If necessary, the gas may pass through a cooling loop between stages.

Compressor Work

To evaluate the work requirements of a compressor, start with the mechanical energy balance. In most compressors, kinetic and potential energy changes are small, so velocity and static head terms may be neglected. As with pumps, friction can be lumped into the work term by using an efficiency. Unlike pumps, the fluid cannot be treated as incompressible, so a differential equation is required:

Compressor Work
Evaluation of the integral requires that the compression path be known - - is it adiabatic, isothermal, or polytropic?
uncooled units -- adiabatic, isentropic compression
complete cooling during compression -- isothermal compression
large compressors or incomplete cooling -- polytropic compression
Before calculating a compressor cycle, gas properties (heat capacity ratio, compressibility, molecular weight, etc.) must be determined for the fluid to be compressed. For mixtures, use an appropriate weighted mean value for the specific heats and molecular weight.

Adiabatic, Isentropic Compression

If there is no heat transfer to or from the gas being compressed, the porocess is adiabatic and isentropic. From thermodynamics and the study of compressible flow, you are supposed to recall that an ideal gas compression path depends on:

Adiabatic Path
This can be rearranged to solve for density in terms of one known pressure and substituted into the work equation, which then can be integrated.
Adiabatic Work
The ratio of the isentropic work to the actual work is called the adiabatic efficiency (or isentropic efficiency). The outlet temperature may be calculated from
Adiabatic Temperature Change
Power is found by multiplying the work by the mass flow rate and adjusting for the units and efficiency.
Isothermal Compression

If heat is removed from the gas during compression, an isothermal compression cycle may be achieved. In this case, the work may be calculated from:

http://facstaff.cbu.edu/rprice/lectures/compress.html

An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier

Answers

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]

An aggregate blend consists of 65% of aggregate A and 35% of aggregate B. The bulk specific gravities of aggregate A and B are 2.45 and 3.25, respectively. What is the bulk specific gravity of the blend?
a) 2.45
b) 2.68
c) 2.73
d) 2.92

Answers

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

How to Cancel prescription

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settings
apple id
subscriptions

Engineers design products or processes to meet desired needs. Your desired need or goal (hopefully) is to graduate with your Bachelor of Science degree in engineering. But what is the process you need to apply to be successful in achieving this goal?

Answers

Components of a Success Civil Engineering Career: Skills, Tools, and Technology.

What is the key objective of data analysis

Answers

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.

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin

Answers

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

. En la facultad de Ingeniería Industrial se realizó una encuesta a 200 personas para saber que lenguaje de programación preferían para aprender al inicio, se obtuvo: 50 prefieren C, 65 prefieren C#, 77 prefieren Python, 100 prefieren C o C#, 105 prefieren C# o Python, 110 prefieren C o Python, 10 personas prefieren C y Python pero no C#.

Answers

Answer:

lalalalapumpe

Explanation:

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

Answers

Answer:

the output of the IF amplifier consist of 675 kHz

Explanation:

Given the data in the question;

AM signal carrier frequency [tex]_{RF[/tex]  = 460 kHz

Local oscillator frequency[tex]_{lo[/tex] = 1135 kHz

Now, The output of the IF amplifier consists of difference of local oscillator frequency & AM carrier signal frequency;

FREQUECY[tex]_{IF[/tex] = FREQUECY[tex]_{lo[/tex] - FREQUECY[tex]_{RF[/tex]

so we substitute in our given values

FREQUECY[tex]_{IF[/tex] = 1135 kHz - 460 kHz

FREQUECY[tex]_{IF[/tex] = 675 kHz

Therefore, the output of the IF amplifier consist of 675 kHz

Okay bro let’s go man yes yes

Answers

Answer:do me ti

Why not me

Why not me

do me ti

Why not me

Why not me

Do me ti

Why not me

Why not me

Explanation:

mitski

Convert the following indoor air quality standards, established by the U.S. Occupational Safety and Health Administration (OSHA), from ppmv to mg/m3 (at 25°C and 1atm) or vice versa.

a. Carbon dioxide (CO2), 5,000 ppmv
b. Formaldehyde (HCHO), 3.6 mg/m^3
c. Nitric oxide (NO), 25 ppmv

Answers

Wouldn’t this be science not engineering?

What is the per capita GDP of China? Be sure to indicate the calendar year that this information represents.

Answers

10,261.68 USD (2019)

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

What is GDP?  

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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to check for ripple voltage from the alternator, connect a digital multimeter and select

Answers

Answer:

isn't it summer? sjsushsiansudndd

Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)

Answers

Gf so 467 gah can-go FTC 346

The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension

Answers

Answer:

At the threads due to shear.

Explanation:

Given :

The allowable tensile stress = 207 MPa

The allowable shear stress = 103 MPa

If a tensile force is applied, the maximum shear stress occurs at the threads of  the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.

The portion of the bolt having threads has the minimum cross sectional area.

So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.

Resistance depends on which three properties of a wire?

Color and texture are not directly related to a wire’s resistance.

1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture

Answers

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.

A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.

Answers

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]

Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp

Answers

Complete Question:

Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by

C_p=10.03+0.0184T  C_p[=]Btu/lbmole- "F ; T[=] °F C,

Answer:

[tex]P'=0.377hp[/tex]

Explanation:

From the question we are told that:

Initial Temperature T_1=70 F

Final Temperature [tex]T_2=250pisa =114.94F[/tex]

Efficiency [tex]E=75\%=0.75[/tex]

Generally the equation for Work-done  is mathematically given by

 [tex]W=\int C_pT[/tex]

 [tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]

 [tex]W=527.21btu/ibmole[/tex]

 [tex]W=11.982btu/ibm[/tex]

Generally the equation for Efficiency  is mathematically given by

 [tex]E=\frac{isotropic Power}{Actual P'}[/tex]

 [tex]E=\frac{P}{P'}[/tex]

Since

Isotropic Power

  [tex]P=0.0167*11.982btu/ibm[/tex]

 [tex]P=0.2btu/s[/tex]

Therefore

 [tex]P'=\frac{0.2}{0.75}[/tex]

 [tex]P'=0266btu/s[/tex]

Since

 [tex]1btu/s=1.4148hp[/tex]

Therefore

 [tex]P'=0.377hp[/tex]

. (5 points) Based on the Reynolds number for each of the following objects, identify each flow as either inertial or viscous force dominant and in which cases the flows are laminar or turbulent. Flow Re A. A bee larva in honey 0.2 B. A ball tossed on Mars 500 C. An elderly driver on i95 1,2000,000

Answers

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.

If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding

Answers

Answer:

5,465.4165939453

Explanation:

formula

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

p=1000

r=0.09

n=4

t=5

A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft above that of the first. the total length of pipes required is 1000 ft. All pipes are 12 in. in diameter and are made of galvanized iron with relative roughness equal to 0.0005 (you may assume fully-rough flow). the pump performance curves suggest that the H-Q relationship is of the form: H_pump=665-0.051Q^2 (Q in ft) the expected flow rate the brake horsepower required to drive the pump (assume an efficiency of 78%). the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

Answers

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

For a pipe system with a pump (pumping uphill), the change in elevation is 400 feet and the total head loss is 408.5 feet. Assuming gage pressure at the entrance and exit and no difference in velocity between the entrance and exit, determine the total energy transferred to the water. Estimate the required power input if the pump efficie

Answers

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

A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?

Answers

Answer:

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter [tex]d=2m[/tex]

Length [tex]l=2.5m[/tex]

Weight [tex]W=22kN[/tex]

Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

[tex]W=(pwg)Vd[/tex]

Where

[tex]V_d=\pi/4(d)^2y[/tex]

Therefore

[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]

Therefore

Center of Bouyance B

[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]

Center of Gravity

[tex]G=\frac{I.B}{2}=2.6m[/tex]

Generally the equation for\BM is mathematically given by

[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]

Therefore

[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]

Therefore

[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]

Therefore

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D

Answers

Answer:

I think you might have forgotten to post the problems

The floor of a light storage warehouse is made of 6-in.-thick cinder concrete. The floor is a slab having a length of 16 ft and width of 14 ft.
A. Determine the resultant force caused by the dead load.
B. Determine the resultant force caused by the live load.

Answers

Explanation:

6/12 = 0.5ft

length = 16

width = 14

The volume of cinder concrete = 0.5 * 16 *14 = 112

the resultant force that is caused by the dead load

density of cinder concrete * volume

density is assumed to be 108

dead load = 108 * 112 = 12096 lb

resultant force caused by the live load

liveload = 125lb/ft2

= 125 * 14 * 16

= 28000 lb

A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)

Answers

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

Which of the following is NOT part of a car's drive train?
A axle
B rotor
C differential
D transmission

Answers

Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.

Which is not part of the drive train?

A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.

Therefore, the correct answer is option B rotors.

To learn more about the part of a car's drive train

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The purpose of the international residential code is to

Answers

Answer:

The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.

Explanation:

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.

Answers

Answer:

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Velocity in pipe section 1 = [tex]6.513m/s[/tex]

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = [tex]0.0251 m^3/s[/tex]

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Average velocity calculations:

Pipe section A:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]

Pipe section B:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]

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