Given that L = {w € {a,b,c,d)* \ #.(w) = #r(w) = #c(w) = #a(w)}
This means that L contains all strings in the alphabet {a,b,c,d} with the property that the number of occurrences of each symbol a, b, c and d is equal to each other.
For example, aabbcdd belongs to L because there are two occurrences of a, two of b, two of c and two of d, hence #a(w) = #b(w) = #c(w) = #d(w) = 2.
Theorem: L is not a context-free language.
Proof: Let n be a natural number greater than 1. Consider the string s = an bn cn dn. Observe that s is a member of L. We prove by contradiction that s cannot be generated by a context-free grammar.
Suppose that s can be generated by a context-free grammar G. Let p be the constant in the pumping lemma. Consider the substring x = an−p bn−p cn−p dn−p. We show that x satisfies the conditions of the pumping lemma.
If x can be written as uvwxy with |vwx| ≤ p and |vx| ≥ 1, then vwx must contain at most three different symbols, say a, b and c. Observe that vwx cannot contain both a and c because then uvvwxxy would contain a different number of a's and c's. Similarly, vwx cannot contain both b and d. Therefore, vwx can be one of the following strings: a, b, c, ab, ac, bc, abc. We consider each case separately and show that the string uv2wx2y violates at least one of the conditions of L.
Case 1: vwx = a.
If we choose u = ε, v = a, w = ε, x = ε, and y = bn−p cn−p dn−p, then uv2wx2y has more a's than b's, c's and d's, hence uv2wx2y ∉ L.
Case 2: vwx = b.
Similar to Case 1, we choose u = ε, v = b, w = ε, x = ε, and y = an−p cn−p dn−p, then uv2wx2y has more b's than a's, c's and d's, hence uv2wx2y ∉ L.
Case 3: vwx = c.
Similar to Case 1, we choose u = ε, v = c, w = ε, x = ε, and y = an−p bn−p dn−p, then uv2wx2y has more c's than a's, b's and d's, hence uv2wx2y ∉ L.
Case 4: vwx = ab.
Similar to Case 1, we choose u = ε, v = a, w = b, x = ε, and y = cn−p dn−p, then uv2wx2y has more a's than c's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = ab, w = ε, x = ε, and y = cn−p dn−p, then uv2wx2y has more a's than c's and d's, hence uv2wx2y ∉ L.
Case 5: vwx = ac.
Similar to Case 1, we choose u = ε, v = a, w = c, x = ε, and y = bn−p dn−p, then uv2wx2y has more a's than b's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = ac, w = ε, x = ε, and y = bn−p dn−p, then uv2wx2y has more a's than b's and d's, hence uv2wx2y ∉ L.
Case 6: vwx = bc.
Similar to Case 1, we choose u = ε, v = b, w = c, x = ε, and y = an−p dn−p, then uv2wx2y has more b's than a's and d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = bc, w = ε, x = ε, and y = an−p dn−p, then uv2wx2y has more b's than a's and d's, hence uv2wx2y ∉ L.
Case 7: vwx = abc.
Similar to Case 1, we choose u = ε, v = a, w = b, x = c, and y = dn−p, then uv2wx2y has more a's than d's, hence uv2wx2y ∉ L. Similarly, if we choose u = ε, v = abc, w = ε, x = ε, and y = dn−p, then uv2wx2y has more a's than d's, hence uv2wx2y ∉ L.
In all cases, we obtain a contradiction. Therefore, the assumption that s can be generated by a context-free grammar is false. Hence, L is not a context-free language.
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A specimen made from a brittle material with a cross-section area of 0.004 m2 was gradually loaded in tension until it yielded at a load of 380 kN and fractured slightly after the yield point. If the specimen’s material observed elastic deformation until fracture, determine the material’s toughness in terms of the energy absorbed, in kJ. Take E = 200
Note that the toughness of the material is 0.0226 kJ.
How is this so?
Toughness = (Area of triangle * Cross-sectional area) / 1,000
= (0.5 * 380 kN * 200 GPa * 0.004 m2) / 1,000
= 0.0226 kJ
Toughness is important in physics as it measures a material's ability to absorb energy and withstand deformation or fracture.
It helps determine the material's resistance to cracking and breaking under stress, making it crucial in applications where durability and reliability are required.
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An Ethernet star LAN is implemented using
O continuous polling of all hosts to detect transmission before they can use coalises
O a self-programming switch at the center of the star
O required coaxial cables
O all of these
O none of these
An Ethernet star LAN is implemented using: B. a self-programming switch at the center of the star.
What is a network topology?In Computer technology, a network topology can be defined as the graphical representation of the various networking devices that are used to create and manage a network connection.
What are the types of topology?In Computer networking, there are six (6) main types of network topology and these include the following:
Ring topologyMesh topologyTree topologyHybrid topologyStar topologyBus topologyGenerally speaking, an Ethernet star local area network (LAN) simply refers to a type of network topology in which all of the network devices are directly connected to a common hub (centralized computer) or hub-and-spoke and it can be implemented by using a self-programming switch that is installed at the center of the star.
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an experimental study on the air side heat transfer performance of the perforated fin-tube heat exchangers under the frosting conditions
By conducting this study, researchers aim to gain insights into the behavior of heat exchangers under frosting conditions. This knowledge can help in designing more efficient heat exchangers for applications such as air conditioning, refrigeration, and heat pump systems.
The phrase "an experimental study on the air side heat transfer performance of the perforated fin-tube heat exchangers under the frosting conditions" refers to a scientific investigation that examines how well perforated fin-tube heat exchangers transfer heat on the air side when they are subjected to frosting conditions.
In this study, researchers conduct experiments to understand how the performance of the heat exchangers is affected when frost accumulates on their surfaces.
The frosting conditions simulate real-world scenarios where heat exchangers may operate in cold environments and experience frost formation.
To evaluate the heat transfer performance, the researchers may measure parameters such as the heat transfer rate, temperature differences, and pressure drops across the heat exchangers.
They may also compare the performance of different designs or configurations of perforated fin-tube heat exchangers.
By conducting this study, researchers aim to gain insights into the behavior of heat exchangers under frosting conditions.
This knowledge can help in designing more efficient heat exchangers for applications such as air conditioning, refrigeration, and heat pump systems.
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What is the function and role of product tear down charts, and how do engineers utilize them in the reverse engineering process?
Answer:
Product Teardown 28 pieces (1) Plastic packaging: protect and display product for purchase. (4) Exterior screws: hold case halves together. (1) Right case half: acts as part of a handle and contains the rest of the parts. (1) Left case half: acts as part of a handle and contains the rest of the parts.
Explanation:
A product teardown process is an orderly way to know about a particular product and identify its parts, system functionality to recognize modeling improvement and identify cost reduction opportunities. Unlike the traditional costing method, tear down analysis collects information to determine product quality and price desired by the consumers.
Answer:
?
Explanation:
the stock room technician gave you 6.0 g of naf. how many milliliters of a 300 mm solution can you make?
To determine the amount of milliliters of a 300 mM solution that can be made from 6.0 g of Naf, we first need to calculate the number of moles of Naf.
How to calculate the amount of solution can you make?
The molar mass of Naf can be found using its chemical formula, which is NaF. The molar mass of Na is approximately 23 g/mol and the molar mass of F is approximately 19 g/mol, so the molar mass of NaF is approximately 42 g/mol.
Using the molar mass of NaF, we can calculate the number of moles of NaF in 6.0 g:
6.0 g ÷ 42 g/mol = 0.143 mol
Next, we can use the molarity of the solution (300 mM) to find the number of milliliters of the solution that can be made from 0.143 moles of NaF:
0.143 mol ÷ 300 mM = 0.000476 L = 476 mL
So, you can make 476 mL of a 300 mM solution from 6.0 g of Naf.
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A series RLC circuit is driven by an ac source with a phasor voltage Vs=10∠30° V. If the circuit resonates at 10 3 rad/s and the average power absorbed by the resistor at resonance is 2.5W, determine that values of R, L, and C, given that Q =5.
Answer:
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F
Explanation:
Given the data in the question;
Vs = 10∠30°V { peak value }
V"s\(_{rms\) = 10/√2 ∠30° V
resonance freq w₀ = 10³ rad/s
Average Power at resonance Power\(_{avg\) = 2.5 W
Q = 5
values of R, L, and C = ?
We know that;
Power\(_{avg\) = |V"s\(_{rms\)|² / R
{ resonance circuit is purely resistive }
we substitute
2.5 = (10/√2)² × 1/R
2.5 = 50 × 1/R
R = 50 / 2.5
R = 20Ω
We also know that;
Q = w₀L / R
we substitute
5 = ( 10³ × L ) / 20
5 × 20 = 10³ × L
100 = 10³ × L
L = 100 / 10³
L = 0.1 H
Also;
w₀ = 1 / √LC
square both side
w₀² = 1 / LC
w₀²LC = 1
C = 1 / w₀²L
we substitute
C = 1 / [ (10³)² × 0.1 ]
C = 1 / [ 1000000 × 0.1 ]
C = 1 / [ 100000 ]
C = 0.00001 ≈ 1 × 10⁻⁵ F
Therefore;
R = 20Ω
L = 0.1 H
C = 1 × 10⁻⁵ F
When forming online contracts, ____________ is a cyber-technique that preserves electronic data of offer and acceptance without using local hard drives and flash drives.
cloud computing
Cloud computing is a cyber-technique used in the formation of online contracts that preserves electronic data of offer and acceptance without the use of local hard drives and flash drives.
What is Cloud Computing?
Cloud computing refers to the delivery of various services over the Internet. These resources include data storage, databases, servers, networking, and software.
Instead of storing files on the proprietary hard drive and local storage device, cloud-based storage allows them to be to saved to a remote database. Any electronic device with internet access has access to data and the software programs required to run it.
Individuals and businesses are increasingly turning to cloud computing for a variety of reasons, including cost savings, increased productivity, speed and efficiency, performance, and security.
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Configuring an Active Directory Domain Controller
1.Why would an administrator want to use the MAP Toolkit?
2.Based on the results of the MAP inventory you performed in the lab, which operating system was installed on the TargetWindows02b server?
3.Based on the results of the MAP inventory you performed in the lab, which desktop and server software was installed on the TargetWindows02b server?
4.Which tasks, other than the ones performed in this exercise, can administrators use the MAP Toolkit to perform?
5.Which utility is used to transform a standalone Windows Server 2012 R2 system into an Active Directory domain controller?
6.What is the importance of SafeModeAdministratorPassword when using PowerShell to install and configure Active Directory?
7.What considerations should you take into account when choosing a domain name?
Answer:
wat
Explanation:
1. Asbestos can be dangerous but is not a known carcinogen.
A) O True
B) O False
Asbestos may cause cancer by inhalation. Moreover, asbestos is also known to cause different respiratory diseases (especially asbestosis) by affecting the mucosa of the bronchi.
The statement "Asbestos can be dangerous but is not a known carcinogen" is FALSE.Asbestos is a group of materials that were widely used in building constructions until 1990.Asbestos is a toxic substance that may cause mesothelioma and lung cancer.Mesothelioma is a type of cancer that develops on the outer surface that covers different internal organs.Moreover, asbestosis is a chronic disease in the lungs caused by long-term exposure to asbestos.Learn more in:
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Answer:
False
Explanation:
padded dashboards in cars are safer in an accident than non padded ones because passengers hitting the dashboard encounter a _____
Padded dashboards in cars are safer in an accident than non padded ones because passengers hitting the dashboard encounter a: (A) increased time of impact.
What is an automobile?In Engineering, an automobile is also referred to as a car or motorcar and it can be defined as a four-wheeled vehicle that is designed and developed to be propelled by an internal-combustion (gasoline) engine, especially for the purpose of transportation from one location to another.
Generally speaking, a padded dashboards in automobile vehicles (cars) are safer in the course of an accident than non-padded dashboards because a passenger who hits the dashboard is most likely to encounter a lengthened time of contact or an increased time of impact, which may eventually lead to casualties.
In conclusion, it is very important to purchase automobile vehicles (cars) with a padded dashboard by the manufacturer.
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Complete Question:
Padded dashboards in cars are safer in an accident than non padded ones because passengers hitting the dashboard encounter a _____?
(a) increased time of impact
(b) decreased impulse
(c) increased force
(d) increased impulse.
A process has the following transfer functions: Process g= 1 / (20s+1)(10s+1) Control valve: gv = -0.25 / 0.5s+1 PI controller: gc = Kc (1 + 1/Tis) i. Design a feedback PI controller for the above process, use Cohen-Coon tuning rule (Table 15.3). Please remember to submit all programs that you use to design the controller. For example, the program to find the approximate model and/or spreadsheet to calculate the controller parameters ii. In Simulink, develop a block diagram with the designed PI controller. Then generate the output response, for a step change of magnitude 3 at time 4s in the setpoint. Before time 4, the setpoint and the output stays at the value of 1.
i. The feedback PI controller for the process G is \($$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$\)
ii. It can be observed that the output eventually stabilizes at the new setpoint.
i. Designing a feedback PI controller using Cohen-Coon tuning rule:
Cohen-Coon tuning rule is used to tune PI controllers. This method has an approximate model of the system and is not suitable for tuning PID controllers.
Cohen-Coon tuning rule:
\($$\begin{aligned} &K_c = \frac{1}{K_p}\left[ {\frac{28}{13} + \frac{{3{{\tau }}_p }}{{{{{\left( {3{{\tau }}_p +{{\tau }}_d } \right)}}}}}} \right] \\ &\frac{1}{{{T_i}}} = \frac{1}{\theta }\left[ {\frac{4}{13} + \frac{{{{\tau }}_d }}{{3{{\tau }}_p +{{\tau }}_d }}} \right] \\ \end{aligned}$$\)
Given: Process G = 1/(20s+1)(10s+1)
Control Valve: gv = -0.25 / 0.5s+1
We need to find out the feedback PI controller for the process G.
Approximate the model and determine the process parameters. Using the given transfer functions, we can determine the time constant and the time delay.
\($$G = \frac{1}{(20s + 1)(10s + 1)}$$$$G = \frac{1}{200s^2 + 30s + 1}$$$$\tau_p \\= \frac{1}{\omega_p} \\= \frac{1}{\sqrt{200}} \\= 0.0707$$\\$$\omega_d = 0.1$$\)
Therefore, from the Cohen-Coon tuning rule, we can determine the values of Kc and Ti.
\($$K_c = 1.5416$$\\$$Ti = 0.4183$$\)
Hence the feedback PI controller for the process G is \($$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$\)
ii. Developing a Simulink block diagram with the designed PI controller
Here is the Simulink block diagram with the designed PI controller. As mentioned above, the setpoint and the output stays at the value of 1 before time 4, and after 4s there is a step change of magnitude 3 in the setpoint.
The block diagram is designed such that it simulates the response for the next 20 seconds. The controller output is shown in red and the process variable in blue.
The final output response is shown below. The output response, after 4 seconds of time and a setpoint change of 3, is similar to the response of a standard PI controller.
It can be observed that the output eventually stabilizes at the new setpoint.
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Which two great lakes are connected by the straits of mackinac
The two great lakes that are connected by the Straits of Mackinac are Lake Michigan and Lake Huron.
The Straits of Mackinac is a narrow waterway that separates the Upper and Lower Peninsulas of Michigan, and connects Lake Michigan to Lake Huron. It is approximately 5 miles wide and 30 miles long. The Mackinac Bridge, also known as the "Mighty Mac," spans the Straits of Mackinac and is one of the longest suspension bridges in the world.Lake Michigan and Lake Huron are two of the five Great Lakes of North America.
Lake Michigan is the third largest Great Lake by volume and the sixth largest freshwater lake in the world. It is located entirely within the United States, bordered by Wisconsin, Illinois, Indiana, and Michigan. Lake Huron is the second largest Great Lake by surface area and the third largest by volume. It is shared by the United States and Canada, with the Canadian province of Ontario to the north and the state of Michigan to the south.
The Straits of Mackinac is an important waterway for shipping and transportation, as well as a popular tourist destination. It is also home to many species of fish and wildlife, and has played a significant role in the history and culture of the Great Lakes region.
The Straits of Mackinac connect two of the Great Lakes, specifically Lake Michigan and Lake Huron. This important waterway serves as a natural boundary between Michigan's Upper and Lower Peninsulas and plays a crucial role in the region's transportation and commerce.
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Assume that two stars are in circular orbits about a mutual center of
mass and are separated by a distance a. Assume also that the angle of
inclination i and their stellar radii are r1 and r2.
(a) Find an expression for the smallest angle of inclination that will pro-
duce an eclipse. Hint: Refer to Figure 7. 8 of your text.
(b) If a = 2 AU, r1 = 10 R⊙, and r2 = 1 R⊙, what minimum value of i
will result in an eclipse?
Eclipsing stars have orbital planes that are parallel to the sky's plane. Since the length of ingress and the length of the eclipse rely on the disparity in the diameters of the stars
The orbital inclination over which eclipses can occur depends on the relative diameters of the stars, but in general, only a tiny percentage of the known binary systems will be observed to experience eclipses. The varying light has two functions. Since the length of ingress and the length of the eclipse rely on the disparity in the diameters of the stars, it enables the estimation of the respective radii of the stars. That is, if Δt1 is the total time between first and last contact, and Δt2 is the period of totality, assuming that the eclipse is annular or total, then.
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The displacement at C (dc) of the beam DCE (see the figure) is equal to: a) The deflection at C of the beam DCE. b) S, plus the elongation of the steel rod. c) Displacement of the beam AB at A (OA). d) The elongation of the steel rod.
The displacement at point C (dc) of the beam DCE can be determined by considering various factors contributing to the overall displacement of the beam. Let's examine the given options to determine the correct answer.
a) The deflection at C of the beam DCE: The deflection at C refers to the bending or flexing of the beam at point C. While deflection does contribute to the overall displacement of the beam, it is not the sole factor determining the displacement at C. Therefore, option a) alone is not sufficient to describe the displacement at C.
b) S, plus the elongation of the steel rod: This option suggests that the displacement at C is equal to S (presumably a distance) added to the elongation of the steel rod. It implies that both the original displacement S and the elongation of the steel rod contribute to the displacement at C. However, without further information on the relationship between S and the steel rod, we cannot definitively conclude that option b) is correct.
c) Displacement of the beam AB at A (OA): This option implies that the displacement at C is equal to the displacement of the beam AB at point A (OA). It suggests that the movement of the beam at A is directly transferred to the displacement at C. However, this assumption might not hold true depending on the structural configuration and loading conditions of the beam. Therefore, option c) cannot be considered as the definitive answer.
d) The elongation of the steel rod: This option proposes that the elongation of the steel rod alone determines the displacement at C. While the elongation of the steel rod may have an effect on the overall displacement, it is unlikely to be the sole factor contributing to the displacement at C.
In conclusion, based on the given options, it is not possible to determine the exact relationship between the displacement at C (dc) and the other variables mentioned. Additional information regarding the specific structural configuration and loading conditions of the beam is required to provide a more accurate answer.
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Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Answer:
3 m/s²
Explanation:
Acceleration is calculated as :
a= Δv/ t
where ;
Δv = change in velocity
Δv = 45 - 0 = 45 m/s
t= 15 s
a= 45 /15
a= 3 m/s²
The primary characteristic of oil that affects a service technician is its_________________.
Volatility
Lubricity
Brand
Viscosity
Answer:
ABCD
Explanation:
(b) Briefly explain how the following three (3) technological advancements, have revolutionized the field of Mechanical Engineering by mentioning the deviations from the traditional practices. Computer Aided Design (CAD), i) ii) 3D printing and iii) Simulation
Here’s a brief explanation of how these three technological advancements have revolutionized the field of Mechanical Engineering.
What is the explanation for the above response?Computer Assisted Design (CAD): One of the most widely utilized software design tools is CAD. It is used by engineers and designers to model, validate, and convey ideas prior to production. CAD software models are frequently utilized as inputs to various mechanical engineering and design tools1.
ii) 3D printing: Extra tools for producing goods on a CNC machine or 3D printer are available and are occasionally incorporated into the CAD program. This has enabled quick prototyping and the creation of complicated geometries that were previously impossible with typical manufacturing methods1.
iii) Simulation: Computer-Aided Engineering (CAE) encompasses a wide variety of studies. Before building physical prototypes, it conducts complicated tasks like as finite element analysis (FEA) and computational fluid dynamics (CFD).
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(20 points) {brainliest} pls helpp
Manufacturing processes involve several types of waste. Which methodology seeks to reduce all types of waste to improve efficiency?
A. Six Sigma
B. Just-in-time production
C. Agile project management
D. Lean manufacturing
technician a says transistors are semiconductors. technician b says transistors function as one-way valves for current flow. who is correct?
Technician A is correct. Transistors are semiconductors that are used to amplify or switch electronic signals and electrical power. Technician B's statement is more applicable to diodes, which function as one-way valves for current flow.
Technician A is correct. Transistors are indeed semiconductors, which are materials that can conduct electricity under certain conditions but not others. They are commonly used in electronic devices as amplifiers or switches. Technician B's statement is not entirely accurate. While transistors can control the flow of current, they do not function as one-way valves. Instead, they rely on the properties of semiconductors to regulate the flow of electrons.
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Air at 40°C flow steadily through the pipe shown in Fig. 1 below. If P1 = 40 kPa (gage), P2 = 10 kPa (gage), D = 3d, Patm ≅ 100 kPa, the average velocity at section 2 is V2 = 25 m/s, and air temperature remains nearly constant, determine the average speed at section 1.
Based on the average velocity at section 2, and the absolute pressures at both sectors, the average speed at section 1 is 2.226 m/s.
What is the average speed at section 1?Density at P₁:
= (40 + 100) / (0.287 x (40 + 273))
= 1.5585 kg/m³
Density at P₂:
= (10 + 100) / (0.287 x (40 + 273))
= 1.2245 kg/m³
The average speed at section 1 is:
= (Density at P₂ x d² x 25.5) / (Density at P₁ x 9d²)
= 2.226 m/s
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Consider the flow of mercury (a liquid metal) in a tube. How will the hydrodynamic and thermal entry lengths compare if the flow is laminar
Answer:
Explanation:
Considering the flow of mercury in a tube:
When it comes to laminar flow of mercury, the thermal entry length is quite smaller than the hydrodynamic entry length.
Also, the hydrodynamic and thermal entry lengths which is given as DLhRe05.0= for the case of laminar flow. It should be noted however, that Pr << 1 for liquid metals, and thus making the thermal entry length is smaller than the hydrodynamic entry length in laminar flow, like I'd stated in the previous paragraph
Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.
This question is incomplete, the complete question is;
A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N.
Determine the minimum required wire radius assuming a factor of safety of 3 and a yield strength of 1500 MPa.
answer in mm please
Answer:
the minimum required wire radius is 5.3166 mm
Explanation:
Given that;
Load F = 11100N
N = 3
∝y = 1500 MPa
∝workmg = ∝y / N = 1500 / 3 = 500 MPa
now stress of Wire:
∝w = F/A
500 × 10⁶ = 11100 / A
A = 22.2 × 10⁻⁶ m²
so
(π/4)d² = A
(π/4)d² = 22.2 × 10⁻⁶
d² = 2.8265 × 10⁻⁵
d = 5.3165 7 × 10⁻³ m³
now we convert to mm(millimeters)
d = 5.3166 mm
Therefore the minimum required wire radius is 5.3166 mm
a weak entity set can always be made into a strong entity set by adding to its attributes the primary-key attributes of its identifying entity set. outline what sort of redundancy will result if we do so.
A weak entity set can always be converted into a strong entity set by simply adding to its attributes the primary-key attributes of its identifying entity set. But doing so will result in the redundant storage of the primary key.
Although a weak entity set can be transformed into a strong entity set by adding appropriate primary-key attributes, but this strategy results in the redundant storage of the primary key. Since the primary key of the weak entity set can be deduced from its relationship with the strong entity set, if we add primary-key attributes to the weak entity set, then they will be present in both the relationship set and the entity set that need to be the same, resulting in the redundant storage of the primary key.
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describe how to check vehicle ride height
Answer:
Vehicle manufacturers specify various reference points on the frame, body, fender, bumper or suspension. On some applications, ride height may even be specified as a distance between two points on the vehicle's suspension, such as the distance from the frame rail to the axle or a suspension control arm.
If a transformer is operated at rated frequency but voltage higher then the rated value, how do you expect the following quantities to change:-
A) No-load current.
B) Hysterics loss.
C) Eddy current loss.
Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point)
Answer:
Most methods of transportation rely on some sort of infrastructure to drive, steer, navigate, or direct at some point or another in a journey. Which category of transportation system is least reliant on infrastructure?(1 point). road
Explanation:
If current flowing through a conductor is 10 mA. how many electrons travel through it in 10 s?
6.25×10⁻¹⁸ electrons will flow through a copper wire.
What are electrons?Electrons can be defined as the sum of atomic particles that is the career of negative charge, they are responsible for the chemical property of an element, the electrons are present in the shell.
Use formula ( I = ne/t )
Here,
number of electron
electron =1.6 x 10¯¹9 C
We have a relation between charge and current as
Charge = Current×time i.e
Q = i×t.
Now we have current = 10 amp & time = 10sec from given data.
Therefore charge Q = 10×1 = 100 coulombs.
Again we have a relation between charge and no of electrons
I.e no of electrons= total charge / 1.6×10⁻¹⁹
So, no of electrons = ( 100C/1.6×10⁻¹⁸ )
= 6.25×10⁻¹⁸ electrons.
Therefore, 6.25×10⁻¹⁸ electrons will flow through a copper wire carrying 10 amps current for 10 sec
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what was the main drawback of ford's assembly line
A.)the cars broke down quickly
B.)production of the cars required many hours of labor
C.) The cars required a lot of fuel to run
D.)production was not flexible
Answer:
D. Im pretty sure at least. you're welcome
A rigid pavement on a new interstate with 3 lanes in each direction has been conservatively designed with a 12-inch slab, an Ec of 5.5 million psi, a concrete modulus of rupture of 700 psi, a load transfer coefficient of 3.0, an initial PSI of 4.5, and a TSI of 2.5. The overall standard deviation is 0.35, the modulus of subgrade reaction is 290 lb/in3, and the drainage coefficient is 0.9. The pavement was designed for 600 30-kip tandem axles per day and 1400 20-kip single axle loads per day total for all 3 lanes. If the desired reliability is 90%, how long will the pavement last in years
The pavement will last between 41 and 50 years with a reliability of 90%.
A road surface, also known as pavement, is a durable surface material that is laid down on an area intended to support vehicular or pedestrian traffic, such as a road or walkway.
First, calculate the number of equivalent single axle loads (ESALS):
ESALS = 600*30 + 1400*20 = 55600.
Next, calculate the design life (L) using the formula: L =
(ESALS/18000)^(1/TSI) * (Ec/PSI)^(-1/2*SD).
Plugging in the values,
L = (55600/18000)^(1/2.5) * (5.5 x 10^6/4.5)^(-1/0.7) = 41.8.
Lastly, calculate the reliability (R) using the formula:
R = (1-LT^(-1/2))*100.
Calculating for a pavement life of 41 years and 50 years gives a reliability of 90% for both values.
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Why is it important that the orifices and the passages of the cutting tip be free of dirt, scratches,
and burrs?
Answer: It is very important that the orifices and passages be kept clean and free of burrs to permit free gas flow and to form a well-shaped flame.
Explanation: