Answer:
Multiplying impulse response by t ( option D )
Explanation:
We can obtain The impulse response of strength 1 considering a unit step response by Multiplying impulse response by t .
When we consider the Laplace Domain, and the relationship between unit step and impulse, we can deduce that the Impulse response will take the inverse Laplace transform of the function ( transfer ) . Hence Multiplying impulse response by t will be used .
This determines whether two different String objects contain the same string.
a. the == operator
b. the = operator
c. the equals method
d. the stringCompare method
Answer: c. the equals method
Explanation:
There are more than one comparison that could be made about string objects. Hence, depending on the comparison we want to make, we could be in need of the == operator, equal() or the Stringcompare method. For the question posed above, we want to kno of two string objects containa the same string, hence we are concerned about comparing the content of each string. For this purpose, we make use of the equal(). In the case of making comparison about string reference, we use the == method while the StringCompare method comes in handy while making alphabetical or lexicographic comparison.
One kilogram of air as an ideal gas executes a Carnot power cycle having a thermal efficiency of 50%. The heat transfer to the air during the isothermal expansion is 50 kJ. At the end of the isothermal expansion, the pressure is 574 kPa and the volume is 0.3 m3. Determine
(a) the maximum and minimum temperatures for the cycle in K;
(b) the pressure and volume at the beginning of the isothermal expansion in bar and m3 respectively;
(c) the work and heat transfer for each of the four processes in kJ.
Answer:
(a) the maximum and minimum temperatures for the cycle in K;
Maximum temprature = T1 = 600.05K
Minimum temprature = 300.03K
(b) the pressure and volume at the beginning of the isothermal expansion in bar and m3 respectively;
Pressure (P4) = 0.67842 bar
Volume (V4) = 1.2693[tex]m^{3}[/tex]
(c) the work and heat transfer for each of the four processes in kJ.
[tex]W_{4-1}[/tex] = -215.25 kJ
The answers are clearly explained in the below mentioned document.
Explanation:
Kindly download the below mentioned document, I have explained in quite detail in it. I hope it will help. Thanks.
The Boeing 787 Dreamliner is billed to be 20% more fuel efficient than the comparable Boeing 767 and will fly at Mach 0.85. The midsize Boeing 767 has a range of 12,000 km, a fuel capacity of 90,000 L, and flies at Mach 0.80. For Boeing 787, assume the speed of sound is 700 mph and calculate the projected volumetric flow rate of fuel for each of the two Dreamliner engines in m3/s.
Answer:
the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
Explanation:
Given the data in the question;
First we determine the fuel economy of Boeing 767
Range = 12,000 km
fuel capacity = 90,000 L
so, fuel economy of Boeing 767 will be
η[tex]_f[/tex] = Range / fuel capacity
η[tex]_f[/tex] = ( 12,000 km / 90000 L ) ( 1000m / 1 km) ( 3.7854 L/gal × 264.2 gal/m² )
η[tex]_f[/tex] = 133,347.024 m/m³
Now, Boeing 787 is 20% more fuel efficient than Boeing 767
so fuel economy of Boeing 787 will be;
⇒ (1 - 20%) × fuel economy of Boeing 767
⇒ (1 - 0.2) × 133,347.024 m/m³
⇒ 0.8 × 133,347.024 m/m³
⇒ 106,677.6 m/m³
Hence, fuel economy of Boeing 787 dream line engine is
⇒ 106,677.6 m/m³ / 2 = 53,338.8 m/m³
Next, we find the velocity of Boeing 787
[tex]V_{787[/tex] = Mach number of 787 × speed of sound
given that; Mach number is 0.85 and speed of sound is 700 mph
we substitute
[tex]V_{787[/tex] = (0.85 × 700 mph) × ( 1 hr / 3600 s ) × ( 1609 m / 1 mile )
[tex]V_{787[/tex] = 265.9319 m/s
Now, to get the Volume flow rate for each dream liner engine { Boeing 787 };
Volumetric flow rate = velocity of flight / fuel economy
we substitute
= 265.9319 m/s / 53,338.8 m/m³
= 0.0049857 ≈ 0.005 m³/s
Therefore, the projected volumetric flow rate of fuel for each of the two Dreamliner engines is 0.005 m³/s
2 Select the correct answer from each drop-down menu. Identify the jobs with the type of risks they include. Michael is a/n engineer with the risk of inhaling very small crystalline particles that can cause silicosis, pulmonary diseases, or lung cancer. Tommy is a/n engineer with the risk of exposure to poisonous pollutants in production units when a certain reaction does not follow h. textile Reset Next chemical automobile
Answer:
2nd option is correct
Explanation:
as Tommy is an engineer only exposed to poisonous pollutants while Michael is an engineer exposed to small crystalline particles that can cause lung cancer and other diseases
I Hope You Got Your Answer
The 5-lb cylinder is falling from A with a speed vA = 10 ft/sonto the platform. Determine the maximum displacementof the platform, caused by the collision. The spring has anunstretched length of 1.75 ft and is originally kept incompression by the 1-ft long cables attached to the platform. Neglect the mass of the platform and spring and any energy lost during the collision.
Answer: hello some pictorial details related to your question is missing attached below is the missing detail
answer : 0.0735 ft
Explanation:
weight of cylinder = 5 IB
speed ( v ) = 10 ft/s
Calculate the maximum displacement of platform
Express the initial energy of the system as
potential energy of cylinder + potential energy of spring + Kinetic energy of cylinder
= mgh + 1/2 ky^2 + 1/2 mv^2
y = initial spring compression = 1.75 - 1 = 0.75
hence the initial energy can be expressed as
= 5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 ------------ ( 1 )
Next : determine the Final energy of the system
Final energy of the system = potential energy of the system
= 1/2 k ( x + 0.75 )^2 ----- ( 2 )
Equating equations ( 1 ) and ( 2 ) to determine the value of x
5 ( 3 + x ) + 1/2 (400)(0.75)^2 + 1/2 ( 5/32.2) (10)^2 = 1/2 k ( x + 0.75 )^2
solve for x
x ( max displace of platform ) = 0.0735 ft
An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression process, P1 = 90 kPa, T1 = 27°C, and V1 = 0.004 m3. The maximum cycle temperature is 1127°C. For each repetition of the cycle, calculate the heat rejection and the net work production. Also calculate the thermal efficiency and mean effective pressure for this cycle. Use constant specific heats at room temperature.
Answer:
i) Heat rejection = 1.0288 KJ
Network production = 1.212 kJ
ii) Thermal efficiency = 54.08%
mean effective pressure = 353.5 kPa
Explanation:
Given data :
Compression ratio ( r ) = 7
P1 ( initial pressure ) = 90 kPa,
T1 ( initial temperature ) = 27°C + 273 = 300 K
V1 = 0.004 m^3
Max cycle Temperature = 1127°C for each repetition of the cycle
i) Determine the heat rejection and net work production
considering that process 1-2 is an Isentropic compression
Find ; T2 = T1 ( r )^1.4-1
T2 = 300 ( 7 )^0.4 = 653.371 K
P2 = P1( r )^k
= 90 ( 7 )^1.4 = 1372.081 kPa
considering that process 3-4 is an Isentropic expansion
T4 = T3 / r^k-1
= 1400 / 7^(1.4 -1 ) = 642.82 K
Next ; Calculate the value of m
m = P1V1 / RT1 = 90(0.004) / 0.287 ( 300 )
m = 4.18 * 10^-3 kg
Finally :
amount of heat rejected = mCv ( T4 - T1 )
= 4.18 * 10^-3 ( 0.718 ) ( 642.82 - 300 )
Qout = 1.0288 KJ
amount of heat added ( Qin ) = mCv ( T3 - T2 )
= 4.18 * 10^-3 ( 0.718 ) ( 1400 - 653.371 )
hence Qin = 2.2408 kJ
Network production( Wnet) = Qin - Qout
= 2.2408 - 1.0288 ) KJ
= 1.212 kJ
ii) Determine the thermal efficiency and mean effective pressure of the cycle
Thermal efficiency = 1 - Qout / Qin
= 1 - ( 1.0288 / 2.2408 ) = 54.08%
mean effective pressure = Wnet / V1 ( 1 - 1/r )
= 1.212 / 0.004 ( 1 - 1/7 )
= 353.5 kPa
A series of end-milling cuts is currently used to produce an aluminum part that is an aircraft component. The purpose of the machining operation is to remove 95% of the part weight to create a structural frame. A total of 4.0 min is lost during the milling cycle due to tool repositioning. The part has a length = 1.6 m, width=0 m, and mm. The operation uses a four-tooth indexable-insert end mill with mm at a cutting speed 600 m/min, chip load 0.15 mm/tooth, and average cross-sectional area of cut-240 mm. High-speed machining has to replace the conventional milling process. The same chip load and average area of cut will be used, but the cutting speed will be increased to 3600 m/min , and the time lost for tool repositioning will be reduced to 2.0 min.
Determine :
(a) The cycle time of the current milling operation and
(b) The cycle time of the proposed HSM operation.
(c) Is this part a good candidate for high-speed machining?
Answer:
what-
Explanation:
I dont understand
List down 10 items made from plastic in your home. Describe why each item is made from plastic.
Answer:
each item is made from plastic are:
water bottlebucketjuggallonplastic bags as poly thin bagto cover wirehandle of knife,pressure cooker,etc plastic roofdustbinplastic plates ,cupsExplanation:
Items made from plastic in our home are
water bottlesjugbucketfood wrappersdisposable plastic cupshand sanitizer bottlesplastic grocery bagsstraws fruit basket Dustbinitems and their uses
Water bottles are used to drink waterfood wrappers are used to wrap the different types of food itemsplastic grocery bags are used to carry grocery itemshand sanitizer bottles are used to store sanitizerfruit basket is used to store and carry fruits .hope it is helpful to you
Before taking off a plane travels at a speed of 1/4 km per second. The runaway is 5 km. How many seconds does it take the plane to get to the end of the runaway?
Answer:
1 5segundos
Explanation:
Blood pressure is conventionally measured in the dimensions of millimeters in a column of mercury, and the readings are expressed as two numbers, for example, 120 and 80. The first number is called the systolic value, and it is the maximum pressure developed as the heart contracts. The second number (called the diastolic reading) is the pressure when the heart is at rest. In the units of kPa and psi, what is the difference in pressure between the given systolic and diastolic readings? The density of mercury is 13.54 Mg/m3.
Answer:
- the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa
- the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi
Explanation:
Given the data in the question;
blood pressure reading = 120 and 80 { systolic and diastolic }
To determine the difference in pressure between the two readings, we use the equation as follows;
change in pressure ΔP = p × g × h
where p is mercury density, g is acceleration due to gravity and h is difference of height in mercury column.
Frist,
difference of height in mercury column h = 120 - 80 = 40 mm = 0.04 m
given that; The density of mercury is 13.54 Mg/m³ = 13.54 × 10³ kg/m³
Not that Mg is Megagrams not Milligrams }
we know that g = 9.81 m/s²
so we substitute into our equation;
change in pressure ΔP = (13.54 × 10³) × 9.81 × 0.04
ΔP = 5313.096 kg/m-s² ≈ 5313.096 N/m²
ΔP = 5.313 KPa
Therefore, the difference in pressure between the given systolic and diastolic readings in KPa is 5.313 KPa.
In psi,
ΔP = 5.313 KPa
ΔP = 5313 Pa
ΔP = 5313 pa × ( 1.45 × 10⁻⁴ psi / 1 Pa )
ΔP = 0.770385 psi ≈ 0.77 psi
Therefore, the difference in pressure between the given systolic and diastolic readings in psi is 0.77 psi
4. Which of the following requires an endorsement on your CDL?
A. Air brakes
B. Double/triple trailers
C. Manual transmission
D. All of the above
The components which requires an endorsement on your Commercial Driver License (CDL) are: D. all of the above.
What is CDL?CDL is an acronym for Commercial Driver License and it can be defined as a category of driver's license that is issued to an individual, in order to indicate that he or she is qualified to operate and drive certain types of automobile vehicles or to use them in specific ways.
In the United States of America, the components which requires an endorsement on your Commercial Driver License (CDL) are:
Air brakesDouble/triple trailersManual transmissionRead more on CDL here: https://brainly.com/question/14326814
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What type of sensor is a crankshaft position sensor?
A qualified WPS is usable for
Answer:
It Is Usable for six months or until a process considered more efficient for a particular production weld is found, whichever comes first. Any format of the WPS is acceptable as long as it lists the parameters or variables listed by the code or specification.
Three single-phase, 10 kVA, 460/120 V, 60 Hz transformers are connected to form a three-phase 460/208 V transformer bank. The equivalent impedance of each transformer referred to HV side is 1.0 +j2.0 Ω. The transformer delivers 20 kW at 0.8 pf leading. Answer the following questions:
(a) Draw a schematic diagram showing the transformer connection.
(b) Determine the magnitude of transformer primary and secondary winding currents.
(c) Determine the primary voltage magnitude for this operating condition. Determine the voltage regulation
Answer:
A) attached below
B) I₁ = 18.1 A , I₂ = 69.39 A
C) V( magnitude) = 454.5 ∠ 5.04° V , Voltage regulation = ≈ -1.2%
Explanation:
A) Schematic diagram attached below
attached below
B) magnitude of primary and secondary winding currents
I₂ ( secondary current ) = P / √3 * VL * cos∅ ---------- ( 1 )
VL = Line voltage = 208
cos∅ ( power factor ) = 0.8
P = 20 * 10^3 watts
insert values into equation 1
I₂ = 69.39 A
I₁ ( primary current ) = I₂V2 / V1
I₁ = ( 69.39 * 120 ) / 460 = 18.1 A
C ) Calculate the Primary voltage magnitude and the Voltage regulation
V(magnitude ) = Vp + ( I₁ ∠∅ ) Req ( 1 + j2 = 2.24 ∠63.43° )
= 460 + ( 18.1 * cos^-1 (0.8) ) ( 1 + j2 )
= 460 + 40.544 ∠ 100.3°
∴ V( magnitude) = 454.5 ∠ 5.04° V
Voltage regulation
= ((Vmag - V1) / V1 )) * 100
= (( 454.5 - 460 / 460 )) * 100
= -1.195 % ≈ -1.2%
In addition to passing an ASE certification test, automotive technicians must have __________ year(s) of on the job training or __________ year(s) of on the job training and a two-year degree in automotive repair to qualify for certification.
An AISI 1040 steel rod of diameter 3 inches has a machined surface finish. It is heat-treated to a tensile strength of 150 kpsi and loaded in rotating bending. Determine the endurance strength of the rod, in kpsi.
Answer:
≈ 41.95 Kpsi
Explanation:
Note : We can Obtain parameters of Marin surface modification factor from Table 6-2
Given data:
Diameter of rod ( d ) = 3 inches
Tensile strength( Sut ) = 150 kpsi
Calculate the endurance strength
endurance strength ( Se ) = Ka*Kb*Kc*Kd *Ke*Kf * Se' -------- ( 1 )
Ka = aSut^b = ( 2.7 ) 150^-0.265 = 0.7157
Kb = 0.879 ( d )^0.107 = 0.879 ( 3 )^-0.107 = 0.7815
Kc, Kd , Ke , Kf = 1 , 1 , 1 , 1
Se' = 0.5 * 150 = 75
Back to equation 1
Se = 0.7157 * 0.7815 * 1 * -------- * 75
≈ 41.95 Kpsi
The following data represent the time of production (in hours) for two different factories for the same product. Which factory has the best average time of production? Which factory will you select and why?
Factory A
14, 10, 13, 10, 13, 10, 7
Factory B
9, 10, 14, 14, 11, 10, 2
Suppose you have two arrays: Arr1 and Arr2. Arr1 will be sorted values. For each element v in Arr2, you need to write a pseudo code that will print the number of elements in Arr1 that is less than or equal to v. For example: suppose you are given two arrays of size 5 and 3 respectively. 5 3 [size of the arrays] Arr1 = 1 3 5 7 9 Arr2 = 6 4 8 The output should be 3 2 4 Explanation: Firstly, you should search how many numbers are there in Arr1 which are less than 6. There are 1, 3, 5 which are less than 6 (total 3 numbers). Therefore, the answer for 6 will be 3. After that, you will do the same thing for 4 and 8 and output the corresponding answers which are 2 and 4. Your searching method should not take more than O (log n) time. Sample Input Sample Output 5 5 1 1 2 2 5 3 1 4 1 5 4 2 4 2 5
Answer:
The algorithm is as follows:
1. Declare Arr1 and Arr2
2. Get Input for Arr1 and Arr2
3. Initialize count to 0
4. For i in Arr2
4.1 For j in Arr1:
4.1.1 If i > j Then
4.1.1.1 count = count + 1
4.2 End j loop
4.3 Print count
4.4 count = 0
4.5 End i loop
5. End
Explanation:
This declares both arrays
1. Declare Arr1 and Arr2
This gets input for both arrays
2. Get Input for Arr1 and Arr2
This initializes count to 0
3. Initialize count to 0
This iterates through Arr2
4. For i in Arr2
This iterates through Arr1 (An inner loop)
4.1 For j in Arr1:
This checks if current element is greater than current element in Arr1
4.1.1 If i > j Then
If yes, count is incremented by 1
4.1.1.1 count = count + 1
This ends the inner loop
4.2 End j loop
Print count and set count to 0
4.3 Print count
4.4 count = 0
End the outer loop
4.5 End i loop
End the algorithm
5. End
Interpret the Blame responsibility and causation in your own words in the light of Columbia Accident.
Answer:
Proposed Improvements and Generic Lessons
Within 2 h of losing the signal from the returning spacecraft, NASA’s Administrator established the Columbia Accident Investigation Board (CAIB) to uncover the conditions that had produced the disaster and to draw inferences that would help the US space program to emerge stronger than before (CAIB, 2003). Seven months later, the CAIB released a detailed report that included its recommendations (Starbuck and Farjoun, 2005).
The CAIB (2003) report attempted to seek answers to the following four crucial questions:
1.
Why did NASA continue to launch spacecraft despite many years of known foam debris problems?
2.
Why did NASA managers conclude, despite the concerns of their engineers, that the foam debris strike was not a threat to the safety of the mission?
3.
How could NASA have forgotten the lessons of Challenger?
4.
What should NASA do to minimize the likelihood of such accidents in the future?
Although the CAIB’s comprehensive report raised important questions and offered answers to some of them, it also left many major questions unanswered (Starbuck and Farjoun, 2005).
1.
Why did NASA consistently ignore the recommendations of several review committees that called for changes in safety organization and practices?
2.
Did managerial actions and reorganization efforts that took place after the Challenger disaster contribute, both directly and indirectly, to the Columbia disaster?
3.
Why did NASA’s leadership fail to secure more stable funding and to shield NASA’s operations from external pressures?
By examining, with respect to the Columbia disaster, the case of NASA as an organization, one can try to extract generalizations that could be useful for other organizations, especially those engaged in high-risk activities—such as nuclear power plants, oil and gas, hospitals, airlines, armies, and pharmaceutical companies—and such generic principles may also be salutary for any kind of organization.
The CAIB (2003) report recommended developing a plan to inspect the condition of all RCC systems, the investigation having found the existing inspection techniques to be inadequate. RCC panels are installed on parts of the shuttle, including the wing leading edges and nose cap, to protect against the excessive temperatures of reentry. They also recommended that taking images of each shuttle while in orbit should be standard procedure as well as upgrading the imaging system to provide three angles of view of the shuttle, from liftoff to at least SRB separation. “The existing camera sites suffer from a variety of readiness, obsolescence, and urban encroachment problems.” The board offered this suggestion because NASA had had no images of the Columbia shuttle clear enough to determine the extent of the damage to the wing. They also recommended conducting inspections of the TPS, including tiles and RCC panels, and developing action plans for repairing the system. The report included 29 recommendations, 15 of which the board specified must be completed before the shuttle returned to flight status, and also made 27 “observations” (CAIB, 2005).
A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight of steel is 490 pcf. Compute:
a. the maximum tensile stress due to the weight of the wire
b. the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi.
Answer:
a) the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb
Explanation:
Given the data in the question;
Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in
Diameter d = 3/16 in
Unit weight w = 490 pcf
First we determine the area of the wire;
A = π/4 × d²
we substitute
A = π/4 × (3/16)²
A = 0.0276 in²
Next we get the Volume
V = Area × Length of wire
we substitute
V = 0.0276 × 4800
V = 132.48 in³
Weight of the steel wire will be;
W = Unit weight × Volume
we substitute
W = 490 × ( 132.48 / 12³ )
W = 490 × 0.076666
W = 37.57 lb
a) the maximum tensile stress due to the weight of the wire;
σ[tex]_w[/tex] = W / A
we substitute
σ[tex]_w[/tex] = 37.57 / 0.0276
= 1361.23 psi
Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi
b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi
Maximum load P that the wire can safely support its lower end will be;
P = ( σ[tex]_{all[/tex] - σ[tex]_w[/tex] )A
we substitute
P = ( 24000 - 1361.23 )0.0276
P = 22638.77 × 0.0276
P = 624.83 lb
Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb
When you see a school bus stop with its stop sign extended or its lights flashing, you must stop, except for cases when:
Answer:
Whenever you approach a school bus from any direction, which has stopped to pick up or let off passengers while operating its flashing red lights, you must stop your vehicle at least 25 feet from the school bus. The only time you do not have to stop is when you are on the other side of a divided highway. You must stay stopped until the bus has started again or the bus driver stops operating the flashing red lights.