Implement Breadth First Search and Depth First Search in Trees Using c/c++. Your programs should take input from the user and ask user to enter the values to insert in Tree and then perform BFS and DFS
Sample Output:
1.Insert values in tree
2.Perform BFS
3.Perfrom DFS
4.Exit
Answers
The program creates a tree based on the user's inputs and performs BFS or DFS according to their choice. The BFS traversal outputs the nodes in breadth-first order, while the DFS traversal uses the in-order approach.
Here's an implementation of Breadth First Search (BFS) and Depth First Search (DFS) in C++. The program allows the user to insert values into a tree and then perform BFS or DFS on the tree based on their choice.
cpp
Copy code
#include <iostream>
#include <queue>
#include <stack>
using namespace std;
// Tree node structure
struct TreeNode {
int data;
TreeNode* left;
TreeNode* right;
TreeNode(int value) {
data = value;
left = nullptr;
right = nullptr;
}
};
// Function to insert a value into a tree
TreeNode* insert(TreeNode* root, int value) {
if (root == nullptr) {
return new TreeNode(value);
} else {
if (value <= root->data) {
root->left = insert(root->left, value);
} else {
root->right = insert(root->right, value);
}
return root;
}
}
// Breadth First Search (BFS) traversal of a tree
void BFS(TreeNode* root) {
if (root == nullptr) {
return;
}
queue<TreeNode*> q;
q.push(root);
cout << "BFS traversal: ";
while (!q.empty()) {
TreeNode* current = q.front();
q.pop();
cout << current->data << " ";
if (current->left) {
q.push(current->left);
}
if (current->right) {
q.push(current->right);
}
}
cout << endl;
}
// Depth First Search (DFS) traversal (inorder) of a tree
void DFS(TreeNode* root) {
if (root == nullptr) {
return;
}
stack<TreeNode*> s;
TreeNode* current = root;
cout << "DFS traversal: ";
while (current != nullptr || !s.empty()) {
while (current != nullptr) {
s.push(current);
current = current->left;
}
current = s.top();
s.pop();
cout << current->data << " ";
current = current->right;
}
cout << endl;
}
int main() {
TreeNode* root = nullptr;
int choice, value;
do {
cout << "1. Insert values in tree" << endl;
cout << "2. Perform BFS" << endl;
cout << "3. Perform DFS" << endl;
cout << "4. Exit" << endl;
cout << "Enter your choice: ";
cin >> choice;
switch (choice) {
case 1:
cout << "Enter the value to insert: ";
cin >> value;
root = insert(root, value);
break;
case 2:
BFS(root);
break;
case 3:
DFS(root);
break;
case 4:
cout << "Exiting program." << endl;
break;
default:
cout << "Invalid choice. Please try again." << endl;
}
cout << endl;
} while (choice != 4);
return 0;
}
This program provides a menu-driven interface where the user can choose to insert values into the tree, perform BFS, perform DFS, or exit the program. The BFS and DFS algorithms are implemented using a queue and a stack, respectively. The program creates a tree based on the user's inputs and performs BFS or DFS according to their choice. The BFS traversal outputs the nodes in breadth-first order, while the DFS traversal uses the in-order approach.
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Related Questions
Should transistors used in switching circuits be biased in the active region? Why or why not?
Answers
Answer:
no
Explanation:
No. More power is dissipated when the transistor is in its active region. In general, transistors in switching circuits are biased either "on" or "off". Time spent in the active region is minimized.
_____
On the other hand, speed can be enhanced if the transistors are active. So, it's a speed/power trade-off. Usually power is of more interest, particularly when there are millions of switching circuits. However, in certain applications, speed may be the priority, so the transistor will be biased in its active region.
1. Two aluminium strips and a steel strip are to be bonded together to form a composite bar. The modulus of elasticity of steel is 200 GPa and 75 GPa for aluminium. The allowable normal stress in steel is 220 MPa and 100 MPa in aluminium. Determine the largest permissible bending moment when the composite bar is bent about horizontal axis. a
Answers
Answer:
1.933 KN-M
Explanation:
Determine the largest permissible bending moment when the composite bar is bent horizontally
Given data :
modulus of elasticity of steel = 200 GPa
modulus of elasticity of aluminum = 75 GPa
Allowable stress for steel = 220 MPa
Allowable stress for Aluminum = 100 MPa
a = 10 mm
First step
determine moment of resistance when steel reaches its max permissible stress
next : determine moment of resistance when Aluminum reaches its max permissible stress
Finally Largest permissible bending moment of the composite Bar = 1.933 KN-M
attached below is a detailed solution
A total of 42 gallons of a glycol solution meeds to be added to a heating water loop to ensure the water does not freeze if there is a heating equipment failure. The glycol solution comes in 1.75 galloncans. How many cans will be needes? A total of how many cann of glycol will be required
Answers
Answer:
24 cans
Explanation:
The number of glycol solution needed to prevent water from freezing is 42 galloons. The glycol solution comes in cans of 1.75 gallons.
Therefore 1 can of glycol = 1.75 gallon.
Number of cans of glycol needed = Total gallons of glycol needed / number of gallon present in each can
Number of cans of glycol needed = 42 gallons / (1.75 gallon per can)
Number of cans of glycol needed = 24 cans
Therefore a total of 24 cans are needed.
[3] (25%) Superheated steam at 9 MPa and 480oC leaves the steam generator of a vapor power plant. Heat transfer and frictional effects in the line connecting the steam generator and the turbine reduce the pressure and temperature at the turbine inlet to 8.6 MPa and 440oC, respectively. The pressure at the exit of the turbine is 10 kPa, and the turbine operates adiabatically. Liquid leaves the condenser at 8 kPa, 35 oC. The pressure is increased to 9.6 MPa across the pump. The turbine and pump isentropic efficiencies are 85%. The mass flow rate of steam is 80 kg/s. Determine (a) the net power output, in kW. (b) the thermal efficiency (c) the rate of heat transfer from the line connecting the steam generator and the turbine, in kW. (d) the mass flow rate of the condenser cooling water, in kg/s, if the cooling water enters at 15oC and exits at 35oC with negligible pressure change.
Answers
The condenser in the Rankine cycle rejects heat to the surrounding or the flowing fluid at the constant pressure
.. You should
an eyewash station periodically.
A)
Remove
B) O Paint
C) Inspect
DO Move
Answers
Answer:
C.
Explanation:
You want to make sure it still works. You don't want to move it periodically though in case of an emergency.
What use is a TPM when implementing full disk encryption?
Answers
A TPM, or Trusted Platform Module, is useful when implementing full disk encryption because it enhances security and protects sensitive data. The TPM is a hardware component that stores encryption keys securely, preventing unauthorized access to the encrypted disk.
Here's a step-by-step explanation of how a TPM works with full disk encryption:
1. Full disk encryption software encrypts the entire disk, including the operating system, applications, and data, to protect against unauthorized access.
2. The TPM generates and securely stores the encryption key needed to decrypt the disk. This key is never exposed to the operating system or any applications, reducing the risk of it being compromised.
3. When the computer starts up, the TPM checks the system's integrity and releases the encryption key only if the system passes the integrity check.
4. The operating system then uses the encryption key to decrypt the disk, allowing it to boot and granting access to the encrypted data.
By using a TPM in conjunction with full disk encryption, you ensure that the encryption key is stored securely and that your encrypted disk remains protected from unauthorized access.
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In which type of scan does an attacker scan only ports that are commonly used by specific programs?Select one:a. vanilla scanb. strobe scanc. random scand. ping sweep
Answers
The type of scan in which an attacker scans only ports that are commonly used by specific programs is called a strobe scan.
Strobe scan is a type of port scanning technique that involves scanning a specific range of ports on a target system to find open ports that are commonly used by specific applications or services. This technique helps attackers to identify vulnerable services or applications that can be exploited for unauthorized access or other malicious activities.
Unlike vanilla scans that scan all ports, strobe scans are targeted and faster, as they only scan for specific ports. However, they are also more easily detected by intrusion detection systems (IDS) or firewalls because they follow a predictable pattern.
As such, security experts recommend implementing security measures such as firewalls and intrusion detection systems to detect and prevent strobe scans and other types of port scanning techniques.
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please hurry i’ll give you 15 points
Answers
Answer:
measures, dissolves, liquid, carbon dioxide, evaporates, water vapor, mold, decompose
Social Engineering as Art and Science The logic behind social engineering is simple - it can be easy to get all the information and access that one needs from any person as long as you know how to trick a person into giving you the data you need with the least resistance possible. By being able to pull off a social engineering trick, you will be able to get your hands on to a device, account, or application that you need to access in order to perform bigger hacks or hijack an identity altogether. That means that if you are capable of pulling of a social engineering tactic before attempting to go through all other hijacking tactics up your sleeve, you do not need to make additional effort to penetrate a system. To put this entire concept into simpler terms, social engineering is a form of hacking that deals with manipulation of victims through social interaction, instead of having to break right away into a computer system. What makes social engineering difficult is that it is largely based on being able to secure trust, which is only possible by getting someone's trust. For this reason, the most successful hackers are capable of reading possible responses from a person whenever they are triggered to perform any action in relation to their security system. Once you are able to make the right predictions, you will be able to get passwords and other valuable computer assets without having to use too many tools.
Answers
Social engineering is considered as both an art and a science. It is a form of hacking that involves the manipulation of victims through social interaction instead of directly breaking into a computer system.
The logic behind social engineering is simple, if one knows how to trick a person into giving out the data they need, they can easily access all the information and access they need with the least resistance possible. This makes social engineering a crucial part of hacking since it allows hackers to gain access to devices, accounts, or applications without making any additional effort.
By using social engineering tactics, a hacker can access a system without having to go through all the other hijacking tactics up their sleeve.The most challenging part of social engineering is securing trust, which is only possible by getting someone's trust. Hackers use various tactics to predict possible responses from a person whenever they are triggered to perform any action in relation to their security system.
The ability to read possible responses from a person is a significant skill for hackers since it enables them to predict passwords and other valuable computer assets without having to use too many tools. Successful hackers use social engineering as a powerful tool to penetrate a system.
In conclusion, social engineering is an essential component of hacking, and a significant part of its success lies in the art of manipulation.
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Makine yüzeysel gemi ünvan değişikliği teknolojisi
Answers
I inferred you meant emerging technologies we see today.
Explanation:
1. 3D Printing
A three dimensional printing allows a digital model to be printed (constructed) into a physical object.
A box, an industrial design and many more could be printed within minutes.
2. AI voice recognition devices
Another trend in the tech world is the rise in artificial intelligence been used in voice recognition devices that not only recognize one's voice but also take commands from the user.
This technology allows one to listen to the internet, such as listening to music online.
The tech world is ever changing with new technology beyond one's consumption on the increase daily.
Influence of specimen size and fiber content on mechanical properties of ultra-high-performance fiber-reinforced concrete
Answers
Specimen size and fiber content have a significant influence on the mechanical properties of ultra-high-performance fiber-reinforced concrete (UHPFRC).
The mechanical properties of UHPFRC, such as compressive strength, flexural strength, and toughness, are affected by both the specimen size and the fiber content.
Firstly, the specimen size plays a crucial role in determining the mechanical properties of UHPFRC. Generally, smaller specimens tend to exhibit higher strength values compared to larger specimens. This phenomenon, known as the size effect, is attributed to the increased probability of having fewer defects and a higher fiber-to-matrix bond in smaller specimens. As the size of the specimen increases, the potential for the presence of defects, such as voids or cracks, also increases, leading to a reduction in the mechanical properties.
Secondly, the fiber content in UHPFRC directly influences its mechanical properties. The addition of fibers enhances the tensile and flexural strength, as well as the toughness of the concrete. Higher fiber contents result in a greater amount of reinforcement within the material, improving its ability to withstand cracking and deformation under load. However, there is a critical fiber content beyond which further increases may not provide significant improvements or may even lead to reduced mechanical performance. This is because excessive fiber content can cause clustering, inhibiting the effective transfer of stress within the matrix.
In summary, the mechanical properties of UHPFRC are influenced by the specimen size and fiber content. Smaller specimens tend to exhibit higher strength values due to the reduced presence of defects, while the addition of fibers enhances the overall performance. However, there is an optimum fiber content that provides the best balance between reinforcement and matrix properties.
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Process synchronization can be done on
Answers
Answer:
it is c) both (a) and (b)
Explanation:
A salvage part is considered aftermarket? true or false
Answers
Please give Brainliest.
A full wave bridge rectifier is used to charge a battery of 24 v and 120 w-h. if the supply voltage with 220 v is used to feed the full-wave rectifier with 4:1 rectifier transformer solutions
Answers
A full wave bridge rectifier is an electronic circuit that converts an alternating current (AC) signal into a direct current (DC) signal. This is done by using a rectifier transformer to convert the high-voltage AC signal.
In this scenario, the full wave bridge rectifier is being used to charge a battery of 24 volts and 120 watt-hours. The supply voltage being used to feed the rectifier is 220 volts, and a 4:1 rectifier transformer solution is being used. This means that the voltage across the secondary winding of the transformer is one-fourth of the voltage across the primary winding.
To calculate the output voltage of the rectifier, we first need to determine the voltage across the secondary winding of the transformer. Since the voltage across the primary winding is 220 volts, the voltage across the secondary winding will be 220/4 = 55 volts.
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The moisture content of a saturated clay is 160 %. The specific gravity of the soil solids Gs is 2.40. What are the wet and dry densities of the saturated clay? hints: what is the degree of saturation of a saturated soil?)
Answers
Answer: 162.4
Sorry if you get it wrong :(
Explanation:
What is the correct order of steps in the scientific method
Answers
Answer:
1) Make a hypothesis, about what u think will happen if you do something
2) conduct the experiment
3) analyze and see what happens
4) do it several more times and compare ur answers
5) make a conclusion from ur answers
6) celebrate bc u done
Design a Deterministic Finite State Machine to accept the language.
L = { w Î {0, 1}*: w ends with 101}
Answers
We will prove that the machine accepts w by demonstrating that it ends in state D when w is input. Because w ends with 101, it can be decomposed into three parts: w = xyz, where x and y are arbitrary (possibly empty) strings over {0,1}, and z = 101.
A deterministic finite state machine (DFSM) can be used to recognize the language L, which consists of all strings over the alphabet {0,1} that end with 101. The machine should have at least one accepting state, indicating that the input string is in the language.
For the language L, the following is the deterministic finite state machine (DFSM):The machine has four states, which are labeled A, B, C, and D. The machine starts in state A, which is the initial state. The machine accepts the input string if it ends in state D.
A transition from state A to state A is possible on input 0 or 1. A transition from state A to state B is possible only on input 1. A transition from state B to state B is possible on input 0 or 1. A transition from state B to state C is possible only on input 0. A transition from state C to state D is possible only on input 1. All other transitions are impossible.
We can now show that this machine accepts the language L.Let w be any string in L, so w ends with 101. We will prove that the machine accepts w by demonstrating that it ends in state D when w is input. Because w ends with 101, it can be decomposed into three parts: w = xyz, where x and y are arbitrary (possibly empty) strings over {0,1}, and z = 101.
The machine starts in state A. After reading x, it is still in state A. After reading y, it enters state B. After reading 10 (the first two digits of z), it is still in state B. Finally, after reading the digit 1, it enters state C.
After reading z, it is in state D, which is the accepting state. This completes the proof that the machine accepts w, so L is recognized by the machine. Provides the steps to design a deterministic finite state machine to accept the language L.
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Vivian's Violins has sales of $326,000, contribution margin of $184,000 and fixed costs total $85,000. Vivian's Violins net operating income is Blank______. Multiple choice question. $241,000 $57,000 $99,000
Answers
Based on the calculations, Vivian's Violins net operating income is equal to: D. $99,000.
How to calculate net operating income?Mathematically, the net operating income of an individual or business firm can be calculated by using this formula:
Net operating income = Contribution margin - Total fixed costs
Given the following data:
Sales = $326,000Contribution margin = $184,000Fixed costs total $85,000Substituting the given parameters into the formula, we have;
Net operating income = $184,000 - $85,000
Net operating income = $99,000.
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Q-) please give me a reference about Tack coat? Pleae i need it please??!!
Answers
Answer:
Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.
Explanation:
How 3D printer works?
Answers
They use an inklike plastic that goes through a fine needle or dispenser that make the shape you want with a cierten code by the computer.this is a simple way of saying it.
A 75-hp, 850 r/min, 500-V series motor has an efficiency of 90.2 percent
when operating at rated conditions. The series field has 30 turns/pole, and
the motor parameters are:
Armature
0.1414 ohms
IP + CW
0.0412 ohms
Series
0.0189 ohms
Resistance,
The motor has a maximum safe speed of 1700 r/min. Is it safe to operate the
motor with a 10-hp shaft load? Use the magnetization curve in Figure
11.16, and show all work. Neglect changes in windage and friction
Answers
It is not safe to operate the motor with a 10-hp shaft load.
Explanation :
Solution-
First, determine the rated speed and current of the motor:
Rated speed = 850 r/min
Rated current = 500 V / (0.1414 ohms + 0.0412 ohms + 0.0189 ohms) = 30.22 A
Next, determine the speed and current for a 10-hp shaft load:
Speed = 1700 r/min
Current = 10 hp / (75 hp * 0.902) = 0.1343 A
Finally, determine whether it is safe to operate the motor with the 10-hp shaft load using the magnetization curve. we can see that the current at 1700 r/min is 0.1264 A, which is less than the current for the 10-hp shaft load (0.1343 A).
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A total station has an ISO 17123-3 specified accuracy of ± 1.6 ′′. What is the estimated precision of an angle observed with two repetitions?
Answers
Where the above ISO is given, the estimated precision of an angle observed with two repetitions is: 1.13
Why is this so?To estimate the precision of an angle observed with two repetitions we must use the following formula:
Precision = accuracy/ √n
In this case, n is number of repetitions
So
Precision = 1.6 / √2
Precision = 1.1313708499
So, It is to be noted that the estimated precision of an angle observed with two repetitions is 1.13 (when rounded to 3 significance figures)
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Determine the final temperature when air is expanded isentropically from 1425 kPa and 477°C to 100 kPa in a piston-cylinder device. Use the average specific heats from the ideal gas property tables. The final temperature is __ K.
Answers
The final temperature of air when expanded isentropically from 1425 kPa and 477°C to 100 kPa in a piston-cylinder device using the average specific heats from the ideal gas property tables is 315.5 K.
To determine the final temperature of air when expanded isentropically from 1425 kPa and 477°C to 100 kPa in a piston-cylinder device, we need to use the ideal gas law and the equation for isentropic expansion.
Firstly, we can use the ideal gas law to calculate the initial specific volume of air:
V1 = R*T1/P1
where V1 is the specific volume, R is the specific gas constant, T1 is the initial temperature (477°C = 750 K), and P1 is the initial pressure (1425 kPa).
V1 = 0.1989 m^3/kg
Next, we can use the equation for isentropic expansion to calculate the final specific volume of air:
V2 = V1*(P1/P2)^(1/gamma)
where V2 is the final specific volume, P2 is the final pressure (100 kPa), and gamma is the ratio of specific heats (1.4 for air).
V2 = 1.0265 m^3/kg
Finally, we can use the ideal gas law again to calculate the final temperature:
T2 = P2*V2/R
T2 = 315.5 K
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using the following data for july, calculate the cost of goods manufactured: beginning finished goods inventory 150,475. Ending finished goods inventory 145,750. sales 400,000. Gross Margin 120,000. The cost of goods manufactured was
Answers
Answer:
The correct response is "$275,275".
Explanation:
The given values are:
Sales,
= 400,000
Gross margin,
= 120,000
Beginning Inventory goods,
= 150,475
Finished inventory goods,
= 145,750
Now,
The cost of goods sold will be:
= \(Sales-Gross \ margin\)
On substituting the values, we get
= \(400,000-120,000\)
= \(280,000\)
As we know,
⇒ \(Cost \ of \ goods \ sold=Beginning \ inventory \ goods+ cost \ of \ goods \ manufactured-Ending \ inventory \ goods\)
⇒ \(280,000=150,475+ cost \ of \ goods \ manufactured-145750\)
⇒ \(280,000=cost \ of \ goods \ manufactured+4,725\)
⇒ \(Cost \ of \ goods \ manufactured=280,000-4,725\)
⇒ \(=275,275\) ($)
You are the engineer for farmer Anderson who is planning on a residential housing development in his former apple orchard in Massachusetts . Soil samples analyzed for arsenic show following results :
sample concentration in soil
sample 1 ..................... 52.7 mg/kg
sample 2 ..................... 11.3 mg /kg
sample 3 ....................... 4.9 mg /kg
what do you advice your client based on this information ...
Answers
The advice that will be given to the client based on the information is that the samples have arsenic levels within the limits.
What is a soil?Soil, also known as earth or dirt, is a complex mixture of organic matter, minerals, gases, liquids, and organisms that support life. Soil is the loose surface material that covers the majority of the land. It is made up of both inorganic and organic particles. Soil provides structural support to agricultural plants as well as a source of water and nutrients. The chemical and physical properties of soils vary greatly.
All of the samples collected have arsenic levels that are within the limits. The concentration of arsenic in samples 2 and 3 is well within the limit. The average value of 21 mg/kg has shown no adverse effects on residents and animals living in already-established residential areas.
Garden vegetable consumption has also not been linked to higher urinary arsenic levels. As a result, there will be no effects of such concentration for residential purposes.
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How would you increase the size of the base unit of length in the metric system
Answers
2. A silo (base not included) is to be constructed in the form of a cylinder surmounted by a hemisphere. The cost of construction per square unit
of surface area is twice as great for the for the hemisphere as it is for the cylindrical side wall. Determine the dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction.
Answers
An extremum is a point where the function has its highest or lowest value, and at which the slope is zero.
The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum is as follows;
Height of cylinder = 2 × Radius of cylinderReason:
The given parameters are;
Form of silo = Cylinder surmounted by a hemisphere
Cost of construction of hemisphere per square unit = 2 × The cost of of construction of the cylindrical side wall
Required:
The dimensions to be used if the volume is fixed and the cost of construction is to be kept to a minimum
Solution:
The fixed volume of the silo, V, can be expressed as follows;
\(V = \pi \cdot r^2 \cdot h + \dfrac{2}{3} \cdot \pi \cdot r^3\)
\(h = \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 }\)Where;
h = The height of the cylinder
r = The radius of the cylinder
The surface area is, Surface Area = 2·π·r·h + 2·π·r²
The cost, C = 2·π·r·h + 2×2·π·r² = 2·π·r·h + 4·π·r²
Therefore;
\(C = 2 \times \pi \times r\times \dfrac{V - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } + 4 \cdot \pi \cdot r^2 = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}\)
\(C = \dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r}\)At the minimum value, we have;
\(\dfrac{dC}{dr} =0 = \dfrac{d}{dr} \left(\dfrac{8 \cdot r^3 \cdot \pi + 6 \cdot V}{3 \cdot r} \right) = \dfrac{16 \cdot r^3 \cdot \pi - 6 \cdot V}{3 \cdot r^2}\)Which gives;
16·π·r³ = 6·V
\(V = \dfrac{16 \cdot \pi \cdot r^3}{6} = \dfrac{8 \cdot \pi \cdot r^3}{3}\)Which gives;
\(h = \dfrac{\dfrac{8 \cdot \pi \cdot r^3}{3} - \dfrac{2}{3} \cdot \pi \cdot r^3 }{\pi \cdot r^2 } = 2 \cdot r\)
h = 2·r
The height of the silo, h = 2 × The radius, 2
Therefore, the dimensions to be used if the volume is fixed and the cost is to be kept to a minimum is, height, h = 2 times the radius
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In which type of operations does lean production work best? check all that apply.
Answers
Lean production works best with standardized and repeatable processes. Lean production is a manufacturing technique that frequently aims to cut down on wait times for both suppliers' and customers' .
Lean manufacturing seeks to improve processes so that they use less energy, time, and resources overall. For some businesses than others, lean tools may be a better fit. But among the most helpful lean tools are Kaizen, 5S, Kanban, Value Stream Mapping, and Focus PDCA. Value, the value stream, flow, pull, and perfection are the five cornerstones of lean manufacturing, respectively. These are now the cornerstones on which lean is implemented. 1. Value: Value is established from the viewpoint of the customer and relates to the price point at which they are willing to purchase goods or services. Lean production is a manufacturing strategy that frequently aims to cut down on wait times for both suppliers' and customers' responses as well as those inside the manufacturing machine.
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A specimen of steel 100mm diameter with a guage length of 100mm tested a distruction it has an extension of 0.50mm under a load of 70KN and the load at elastic limit is 10KN the maximum load is 140KN the total extension of fracture is 58mm and the diameter at the neck is 16mm . find the Stress of elastic limit, young modulus, percentage enlogation, percentage reduction in area , and ultimate tensile stress?
Answers
The stress at elastic limit is 1.27 N/mm^2, the Young's modulus is 254 N/mm^2, the percentage elongation is 57.5%, the percentage reduction in area is 97.44%, and the ultimate tensile stress is 696.67 N/mm^2.
How to calculate the valuesStress = Force / Area
Young's modulus = Stress / Strain
Percentage elongation = (extension / gauge length) x 100%
Percentage reduction in area = [(original area - area at neck) / original area] x 100%
Ultimate tensile stress = Maximum load / Area
Diameter of specimen = 100 mm
Gauge length = 100 mm
Extension at 70 KN load = 0.50 mm
Load at elastic limit = 10 KN
Maximum load = 140 KN
Total extension at fracture = 58 mm
Diameter at neck = 16 mm
We can calculate the area of the specimen as follows:
Area = π/4 x d^2
Area = π/4 x (100 mm)^2
Area = 7853.98 mm^2
The stress at elastic limit can be calculated as:
Stress = Load / Area
Stress = 10 KN / 7853.98 mm^2
Stress = 1.27 N/mm^2
The Young's modulus can be calculated as:
Strain = Extension / Gauge length
Strain = 0.50 mm / 100 mm
Strain = 0.005
Stress = Load / Area
Load = Stress x Area
Load = 1.27 N/mm^2 x 7853.98 mm^2
Load = 9982.16 N
Young's modulus = Stress / Strain
Young's modulus = 1.27 N/mm^2 / 0.005
Young's modulus = 254 N/mm^2
The percentage elongation can be calculated as:
Percentage elongation = (extension / gauge length) x 100%
Percentage elongation = (58 mm - 0.50 mm) / 100 mm x 100%
Percentage elongation = 57.5%
The percentage reduction in area can be calculated as:
Original area = π/4 x (100 mm)^2 = 7853.98 mm^2
Area at neck = π/4 x (16 mm)^2 = 201.06 mm^2
Percentage reduction in area = [(original area - area at neck) / original area] x 100%
Percentage reduction in area = [(7853.98 mm^2 - 201.06 mm^2) / 7853.98 mm^2] x 100%
Percentage reduction in area = 97.44%
The ultimate tensile stress can be calculated as:
Area at neck = π/4 x (16 mm)^2 = 201.06 mm^2
Ultimate tensile stress = Maximum load / Area
Ultimate tensile stress = 140 KN / 201.06 mm^2
Ultimate tensile stress = 696.67 N/mm^2
Therefore, the stress at elastic limit is 1.27 N/mm^2, the Young's modulus is 254 N/mm^2, the percentage elongation is 57.5%, the percentage reduction in area is 97.44%, and the ultimate tensile stress is 696.67 N/mm^2.
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When identifying online resources, which web address endings should a student look for? Check all that apply.
Answers
Answer:
1 .edu
and
3 .gov
Explanation:
The other two are commercial sites.
When identifying online resources, .edu and .gov are some of the web address endings that should a student look for consistently.
What do you mean by Web address?Web address may be defined as a type of network address that significantly contains information about the location of the webpage. It is also known as the URL (uniform resource locator).
Like the address for your home, a web address organizes information about a web page's location in a predictable way. It is an interconnected system of public web pages accessible through the Internet. Every webpage, image, and video has its own unique Uniform Resource Locator (URL), which is also known as a web address.
Therefore, when identifying online resources, .edu and .gov are some of the web address endings that should a student look for consistently.
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Your question seems incomplete. The most probable complete question is as follows:
When identifying online resources, which web address endings should a student look for? Check all that apply.
.gov.in.edu.inc