The natural variation of a process relative to the variation allowed by the design specifications is known as

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2.  The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.  

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

Answer:

[tex]w_{out}=319.1\frac{BTU}{lbm}[/tex]

Explanation:

Hello,

In this case, for the inlet stream, from the steam table, the specific enthalpy and entropy are:

[tex]h_1=1456.0\frac{BTU}{lbm} \ \ \ s_1=1.6413\frac{BTU}{lbm*R}[/tex]

Next, for the liquid-vapor mixture at the outlet stream we need to compute its quality by taking into account that since the turbine is adiabatic, the entropy remains the same:

[tex]s_2=s_1[/tex]

Thus, the liquid and liquid-vapor entropies are included to compute the quality:

[tex]x_2=\frac{s_2-s_f}{s_{fg}}=\frac{1.6313-0.39213}{1.28448}=0.965[/tex]

Next, we compute the outlet enthalpy by considering the liquid and liquid-vapor enthalpies:

[tex]h_2=h_f+x_2h_f_g=236.14+0.965*933.69=1136.9\frac{BTU}{lbm}[/tex]

Then, by using the first law of thermodynamics, the maximum specific work is computed via:

[tex]h_1=w_{out}+h_2\\\\w_{out}=h_1-h_2=1456.0\frac{BTU}{lbm}-1136.9\frac{BTU}{lbm}\\\\w_{out}=319.1\frac{BTU}{lbm}[/tex]

Best regards.

Answer:

30 feet

Explanation:

Given data :

design speed = 30 miles/h

super elevation = 0.08

determine the width of the turning roadway

calculate the value of R = V^2 / 15( e + p)

e = 0.08 , p = 0.2 , v = 30

R = (30)^2 / 15 ( 0.08 + 0.2 )

  = 900 / 15 ( 0.28 )

  ≈ 215 ft

pavement width from the calculation above = 28 ft

width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )

Answer:

Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.

Explanation:

In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about  1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.

There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.

Answer:

A)  ( N ) = 1.54

B)  N ( Goodman ) = 1.133,  N ( Morrow) = 1.35

Explanation:

width of steel bar = 25-mm

thickness of steel bar = 10-mm

diameter = 6-mm

load on plate = between 12 kN AND 28 kN

notch sensitivity = 0.83

A ) Fatigue factor of safety based on yielding criteria

= δa + δm = [tex]\frac{Syt}{n}[/tex]   =  91.03 + 227.58 = 490 / N

therefore Fatigue number of safety ( N ) = 1.54

δa (amplitude stress ) = kf ( Fa/A) = 2.162 * ( 8*10^3 / 190 ) = 91.03 MPa

A = area of steel bar = 190 mm^2 , Fa = amplitude load = 8 KN , kf = 2.162

δm (mean stress ) = kf ( Fm/A ) = (2.162 * 20*10^3 )/ 190 = 227.58 MPa

Fm = mean load  = 20 *10^3

B) Fatigue factor of safety based on Goodman and Morrow criteria

δa / Se + δm / Sut = 1 / N

= 91.03 / 183.15 + 227.58 / 590 = 1 /N

Hence N = 1.133 ( based on Goodman criteria )

note : Se = endurance limit (calculated) = 183.15 , Sut = 590

applying Morrow criteria

N =   1 / ( δa/Se) + (δm/ δf )

   = 1 / ( 91.03 / 183.15 ) + (227.58 / 935 )  

   = 1.35

Answer:

D.  Resistance = 30 ohms

Explanation:

Using Ohm's law

V = I times R

Given:

V = 60 V

I = 2 A

Resistance = V / I = 60 V / 20 A

Resistance = 30 ohms

Answer:

IV > [tex]I^{2} R[/tex]

Explanation:

The current in the power line = I

The voltage in the power line = V

The resistance of the power line = R

Power supplied from the power house = P

power delivered to the households = [tex]p[/tex]

We know that the power supplied to a power line system is proportional to

P = IV    ....1

we also know that according to Ohm's law, the relationship between the voltage, resistance, and current through an electrical system is given as

V = IR    ....2

substituting equation 2 into equation 1, the power delivered to the households is proportional to the square of the current.

[tex]p[/tex] = [tex]I^{2} R[/tex]    ....3

The problem is that when power is delivered across a transmission line, some of the power is loss due to Joules heating effect of the power lines. This energy and power loss is proportional to [tex]I^{2}[/tex] therefore, the electrical power delivered to the households will be less than the electrical power supplied from the power station. This means that

P > [tex]p[/tex]

equating these two powers from equations 1 and equation 3, we have

IV > [tex]I^{2} R[/tex]

Answer:

the torque required to RAISE the load is Tr = 18.09 Nm

the torque required to LOWER the load is Tl = 10.069 ≈ 10.07 Nm

the Overall Efficiency e = 0.2199 ≈ 0.22

Explanation:

Given that; F = 5 kN, p = 5mm, d = 35mm

Dm = d - p/2

Dm = 35 - ( 5/2) = 35 - 2.5

DM = 32.5mm

So the torque required to RAISE the load is

Tr = ( 5 × 32.5)/2 [(5 + (π × 0.09 × 32.5)) / ( (π × 32.5) - ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tr = 81.25 × (14.1892 / 101.6518) + 6.75

Tr = 11.3414 + 6.75

Tr = 18.09 Nm

the torque required to LOWER the load is

Tl =  ( 5 × 32.5)/2 [(π × 0.09 × 32.5) - 5) / ( (π × 32.5) + ( 0.09 × 5))] + [( 5 × 0.06 × 45)/2]

Tl = 81.25 × 4.1892 / 102.5518 + 6.75

Tl = 3.3190 + 6.75

Tl = 10.069 ≈ 10.07 Nm

So since torque required to LOWER the load is positive

that is, the thread is self locking

Therefore the efficiency is

e = ( 5 × 5 ) / ( 2π × 18.09 )

e = 25 / 113.6628

e = 0.2199 ≈ 0.22

Answer:

A) Flow depth = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

Velocity before jump ( v1 ) = 7 m/s

flow depth before jump ( y1 ) = 0.8 m

g = 9.81 m/s

Esi = 3.3 m ( calculated )

attached below is a detailed solution of the problem

Answer:

V2= 1666.6 volts

Explanation:

Given data

primary turns N1= 100 turns

secondary turns N2= 1500 turns

primary voltage V1= 110 volts

secondary voltage V2= ?

We can solve for the output voltage using the turns ration sated below

  Turns Ratio = N1 / N2 = V1 / V2

Substituting our given data into the expression we have

100/1500= 110/V2

Making V2 subject of formula we have

V2= 110/(100/1500)=  1666.6 volts

V2= 1666.6 volts

Hence the secondary voltage is 1666.6 volts

Answer:

As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.

Answer: hello attached below is the diagram which is part of your question

Total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k  it violates Clausius increase of entropy which is Sgen > 0

Explanation:

Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .  

applying the increase in entropy principle to prove this

temp of cold reservoir (t hot)= 600 k

temp of hot reservoir(t cold) = 1220 k

energy (q) = 100 kj

total entropy change  = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k

entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k

entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k

hence it violates  Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0

Answer:

a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min

b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Explanation:

Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²

first we find the volume and Area;

Volume V = πD²t / 4

Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³

Area A = 2πD²/ 4 + πDt

Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}

Area A = 392,699.08 + 31,415.93

Area A = 424,115 mm²

a)

Chvorinov’s rule

T(aluminium) = Cm (V/A) ²

T(aluminium) =  2.0 × (3,926,991 / 424,115) ²

T(aluminium) = 171.5 s = 2.86 min

∴ the minimum time (minutes) for the casting to solidify is 2.86 min

b)

For cast iron

Cm (mold constant = 1.488 sec/mm²)

Chvorinov’s rule

T(iron) = Cm (V/A) ²

T(iron) = 1.488 × (3,926,991 / 424,115) ²

T(iron) = 127.5720s = 2.13 min

Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)

Answer:

thread depth = 2.5 mm

thread width = 2.5 mm

mean diameter = 27.5 mm

root diameter = 25 mm

lead of screw = 5 mm

Explanation:

given data

power screw diameter D = 30 mm

thread pitch  P = 5 mm

solution

First, we get here thread depth fr square thread

thread depth = [tex]\frac{P}{2}[/tex]   ......................1

thread depth = [tex]\frac{5}{2}[/tex]

thread depth = 2.5 mm

and

thread width for square thread

thread width = [tex]\frac{P}{2}[/tex]   ......................2

thread width = [tex]\frac{5}{2}[/tex]

thread width = 2.5 mm

and

mean diameter is

mean diameter = D - [tex]\frac{P}{2}[/tex]    ................3

mean diameter = 30 - [tex]\frac{5}{2}[/tex]

mean diameter = 27.5 mm

and

root diameter is

root diameter = D - P   ....................4

root diameter = 30 - 5

root diameter = 25 mm

and

lead of screw for single thread so n = 1

so lead of screw = 1 × 5

lead of screw = 5 mm

Answer:

Average heat transfer coefficient =  31 kw/m^2 k

Heat transfer rate per meter length of pipe =  116.808 KW

Explanation:

water temperature = 20⁰c,  

free-stream velocity = 1.5 m/s

circular pipe diameter = 2.0 cm = 0.02 m

surface temperature = 80⁰c

A) calculate average heat transfer coefficient

we apply the formula below :

m = αAv

A (area) = [tex]\pi /4 (d)^2[/tex]

m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5

   = 10^3 * 0.7857( 0.0004) * 1.5

   = 0.4714 kg/s

Average heat transfer coefficient  

h = [tex]\frac{m(cp)}{A}[/tex]  ,  A = [tex]\pi DL[/tex]

L = 1 m , m = 0.4714 kgs , cp = 4.18

back to equation

h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex]   = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k

B) Heat transfer rate per meter length of pipe

Q = ha( ΔT ),  a = [tex]\pi DL[/tex]

   = 31 * 0.0628 * ( 80 - 20 )

  = 31 * 0.0628 * 60 = 116.808 KW

Answer:

Explanation:

The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.

At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.

Then link CE has 15 N tension.

The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.

In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.

At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.

At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.

The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.

In summary, the link forces are ...

_____

Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.

Answer: A prototype

Explanation:

The scale model that proves the initial concept is called a domain model.

A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.

A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.

A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.

Therefore, a domain model is the scale model that proves the initial concept.

To learn more about the scale model, refer to the below link:

https://brainly.com/question/14341149

#SPJ2

Answer:

pay off the parking tickets

Explanation:

In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.

Answer:

The answer is below

Explanation:

An AC voltage is represented by the relation v= 12 sin(500πt). Determine the:

The equation of an AC voltage is given as:

[tex]V=V_msin(2\pi ft)[/tex]

Where Vm is the maximum value of voltage and f is the frequency

From V= 12 sin(500πt), Vm = 12, 2πft = 500πt

(a) The peak to peak voltage is total amplitude (both the negative and positive amplitude) of the voltage, it is the difference between the positive amplitude and the negative amplitude. The peak to peak voltage ([tex]V{p-p}[/tex]) is given as:

[tex]V_{p-p}=2V_m=2*12=24\ V[/tex]

b) The frequency is the number of oscillation per second. The frequency (f) is gotten from:

2πft = 500πt

2f = 500

f = 500/2

f = 250 Hz

c) The root mean square voltage is the dc value of the voltage. It is given by:

[tex]V_{rms}=\frac{V_m}{\sqrt{2} }=\frac{12}{\sqrt{2} }=8.5\ V[/tex]

d) The period (T) is the time taken to complete one oscillation, it is given by:

[tex]T=\frac{1}{f}\\ \\T=\frac{1}{250} =0.004\ s[/tex]

Answer:

A) 251.8 kj/kg

B) 0.9150 kj/kg-k

C) 155.4 kj/kg

F) 1.50

G) 3.95 kw

H) 2.6 kw

Explanation:

Given conditions :

air conditioner : R -134a

compressor efficiency (nc) = 90%.

T1 = 9⁰c,  T3 = 39⁰c, mass flow rate = 0.027 kg/s

A) Specific enthalpy at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

from the R-134a property table

h1 = 251.8 kj/kg

B ) specific entropy ( kj/kg-k) at the compressor inlet

at T = 9⁰c the saturated vapor (x) = 1

s = 0.9150 kj/kg-k ( from the R-134a property table )

C) specific enthalpy at the compressor exit

at T3 = 39⁰c , s2 = s1

has = 165.12 kj/kg

h2 = 155.4 kj/kg

attached below is the remaining solution to some of the problems

Answer:

  resistance

Explanation:

A strain gauge changes resistance with applied strain.

Answer:

maximum temperature = 1322 k

rate of heat addition = 212 kw

Explanation:

compression ratio = 17

cut off ratio = 1.3

power produced = 140 Kw

state of air at the beginning of the compression = 90 kPa and 578 c

Determine the maximum temperature of air

attached below is the detailed solution

Answer:

efficiency

Explanation:

Answer:

The correct answer is Efficiency.

Explanation:

I got it right on the plato test.

Answer:

C) Decreases with an increase in temperature

Explanation:

As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.

With increased heating due to heavy current flow, the transistor is damaged.

Answer:

The rate of heat transfer has increased.

Explanation:

Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.

Answer:

Explanation:

MOSFET: The acronym for  "metal oxide semiconductor field-effect transistor"  are mostly used in switching circuits.

There are two classes of MOSFET

1. Depletion mode

2. Enhancement mode

          Generally a  MOSFET is a kind of transistor, it is actually a field effect transistor with tree terminals gate, source and drain terminals, also the MOSFET can be used as an amplifier for the amplification of electronic signals in the electronic circuit/devices

Answer:

Explanation:

f = 50.0 Hz, L = 0.650 H, π = 3.14

C = 4.80 μF, R = 301 Ω resistor. V = 120volts

XL = wL = 2πfL

= 2×3.14×50* 0.650

= 204.1 Ohm

Xc= 1/wC

Xc = 1/2πfC

Xc = 1/2×3.14×50×4.80μF

= 1/0.0015072

= 663.48Ohms

1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2

√ 90601 + (459.38)^2

√ 90601+211029.98

√ 301630.9844

= 549.209

Z = 549.21Ohms

2. I=V/Z = 120/ 549.21Ohms =0.218Ampere

3. P=V×I = 120* 0.218 = 26.16Watt

Note that

I rms = Vrms/Xc

= 120/663.48Ohms

= 0.18086A

4. I(max) = I(rms) × √2

= 0.18086A × 1.4142

= 0.2557

= 0.256A

5. V=I(max) * XL

= 0.256A ×204.1

=52.2496

= 52.250volts

6. V=I(max) × Xc

= 0.256A × 663.48Ohms

= 169.85volts

7. Xc=XL

1/2πfC = 2πfL

1/2πfC = 2πf× 0.650

1/2×3.14×f×4.80μF = 2×3.14×f×0.650

1/6.28×f×4.8×10^-6 = 4.082f

1/0.000030144× f = 4.082×f

1 = 0.000030144×f×4.082×f

1 = 0.000123f^2

f^2 = 1/0.000123048

f^2 = 8126.922

f =√8126.922

f = 90.14 Hz

Answer:

There are five (5) unknown support reactions in this problem.

Explanation:

A roller joint rotates and translates along the surface on which the roller rests. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. This allows the roller to move in a single plane along the surface where it rests.

A cable support provides support in one direction, parallel, and in opposite direction to the load on it. There exists a single reaction from the cable pointed upwards.

A ball-and-socket joint have  reaction forces in all 3 cardinal  directions. This allows it to move in the x-y-z plane.

The total unknown reactions on the member are five in number.

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )