(b) (10 pts.) Consider a linear time-invariant system with H(e) = tude response |H(ejw)|. 1+e-jw (1-ae-jw)2 Determine the magni- 1000/101100² 5b. a = 6

As the sheet metal stock hardness increases in a blanking operation, the clearance between the punch and die should be increased.

This is because harder materials require more force to cut through, and a larger clearance allows for more room for the material to deform and flow during the cutting process. However, it is important to note that increasing the clearance too much can lead to burrs and a lower quality cut, so the clearance should be carefully adjusted based on the specific material and cutting conditions.In a blanking operation, as the sheet metal stock hardness increases, the clearance between the punch and die should be increased. As the sheet metal stock hardness increases in a blanking operation, the clearance between the punch and die should be increased.This is because harder materials require more space to prevent excessive wear on the tooling and to facilitate proper shearing of the metal.

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The material property does the Charpy test determine is:(d) Toughness.

The Charpy impact test is utilized to determine the toughness of a material, specifically its ability to absorb energy when subjected to an impact load. Toughness, in this case, is defined as the ability of a material to resist the propagation of cracks or other flaws when subjected to dynamic or impact loading. In materials science, toughness is often described as the area under the stress-strain curve of a material up to its point of failure.  A pendulum hammer is used to hit the sample at the notched part with a specific amount of force.

The amount of energy absorbed by the sample is calculated by measuring the height difference between the point at which the pendulum starts to fall and the point where it swings up to after striking the sample. The energy absorbed during the fracture is proportional to the reduction in the height of the pendulum and is used to determine the toughness of the material. Therefore, the Charpy test is mainly used to determine the toughness of a material, making the correct option (d) Toughness.

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The Charpy test determines the toughness of a material.What is the Charpy test.The Charpy test is a test that measures a material's resistance to brittle fracture.

The test's results are critical in determining a material's serviceability in cold temperatures or environments where impact loading is possible. It is mostly used to assess the resistance of a material to impact loading. It involves striking a notched specimen with a swinging pendulum to produce a fracture and then measuring the amount of energy absorbed by the specimen.What is toughness,Toughness is the ability of a material to resist cracking when subjected to high-stress conditions. Toughness is defined as the amount of energy a material can absorb before it breaks. In general, materials with higher toughness are more durable. Toughness is critical in the construction of buildings, bridges, airplanes, ships, and other structures subjected to sudden and unpredictable impact forces.What material property does the Charpy test determine.The Charpy test determines the toughness of a material. The Charpy test is a common method for determining a material's toughness. The test is frequently used in construction and engineering to assess the impact resistance of materials. The toughness of a material, measured in joules, can be determined using the Charpy test.

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The  first thing a technician need to check if a new drive is not recognized by the operating system are:

The issues on how to fix it are:

Hence, The  first thing a technician need to check if a new drive is not recognized by the operating system are:

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When designing a set of simple test programs to determine the type compatibility rules of a C compiler to which you have access, it is important to consider the different data types that are used in C programming. An example of a set of test programs that can be used to determine the type compatibility rules of a C compiler:

Integer Test the compatibility of the C compiler with integer data types. It declares two variables of type int, initializes them with values, and then adds them together. The result is printed to the screen. If the program compiles and runs without any errors, then the C compiler is compatible with integer data types.

Floating-Point Test  the compatibility of the C compiler with floating-point data types. It declares two variables of type float, initializes them with values, and then adds them together. The result is printed to the screen. If the program compiles and runs without any errors, then the C compiler is compatible with floating-point data types.

By running the set of simple test programs described above, you can determine the type compatibility rules of a C compiler to which you have access. If any of the programs do not compile or run without errors, then you can determine which data types are not compatible with the C compiler and adjust your code accordingly.

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Answer:

b

Explanation:

i did it yeater dayajsbs

Answer: To show his father what he had done

Explanation:

Percy Jackson is the son of the great Greek god of the seas, Poseidon. He did not however, know this for a long time. After he finds out, his life becomes more exciting as he embarks on many adventures.

According to Grover who is his best friend and guardian, Percy sent Medusa's head to Mount Olympus to show his father what he had done so that Poseidon would acknowledge and be proud of him.

Answer:

Part A

The thickness of the compacted soil is approximately 4.3467 × 10⁻¹ m

Part B

The weight of water to be added is approximately 19886.\(\overline{36}\) kN, the volume of the water added is approximately 2,027.77 m³

Explanation:

The parameters of the soil are;

The volume of sol the excavator excavates, \(V_T\) = 10,000 m³

The moist unit weight, W = 17.5 kN/m³

The moisture content = 10%

The area of the project, A = 20,000 m²

The required dry unit weight = 18.3 kN/m³

The required moisture content = 12.5%

Part A

Therefore, we have;

The moist unit weight = Unit weight = (\(W_s\) + \(W_w\))/\(V_T\)

The moisture content, MC = 10% = (\(W_w\)/\(W_s\)) × 100

∴ \(W_w\) = 0.1·\(W_s\)

∴ The moist unit weight = 17.5 kN/m³ = (\(W_s\) + 0.1·\(W_s\))/(10,000 m³)

1.1·\(W_s\) = 10,000 m³ × 17.5 kN/m³ = 175,000 kN

\(W_s\) = 175,000 kN/1.1 = 159,090.\(\overline{09}\) kN

For the required soil, we have;

The required dry unit weight = 18.3 kN/m³ = \(W_s\)/\(V_T\) = 159,090.\(\overline{09}\) kN/\(V_T\)

\(V_T\) = 159,090.\(\overline{09}\) kN/(18.3 kN/m³) ≈ 8,693.4923 m³

The total volume of the required soil ≈ 8,693.4923 m³

Volume \(V_T\) = Area, A × Thickness, d

∴ d =  \(V_T\)/A

d = 8,693.4923 m³/(20,000 m²) ≈ 4.3467 × 10⁻¹ m

The thickness of the compacted soil ≈ 4.3467 × 10⁻¹ m

Part A

The moisture content, MC = 12.5% = (\(W_w\)/\(W_s\)) × 100

\(W_w\) = \(W_s\) × MC/100 = 159,090.\(\overline{09}\) kN × 12.5/100 = 19886.\(\overline{36}\) kN

The weight of water to be added, \(W_w\) = 19886.\(\overline{36}\) kN

Where the density of water, ρ = 9.807 kN/m³

Therefore, we have;

The volume of water, V = \(W_w\)/ρ

∴ V = 19886.\(\overline{36}\) kN/(9.807 kN/m³) ≈ 2027.77 m³

The volume of water, V ≈ 2027.77 m³

Both cultures imposed rigid moral standards. Sometimes works of art and architecture challenge our perceptions of what was formerly thought to be feasible and what our forefathers were capable of.

An excellent example is the defunct megalithic capital of Nan Madol, which is situated in a lagoon next to the island of Pohnpei in the Federated States of Micronesia in the Pacific Ocean. By stating that over a period of four centuries, the people of Pohnpei moved an average of 1,850 tons of basalt per year—and no one is quite sure how they did it—archaeologist Rufino Mauricio helps us put these enormous amounts into perspective.

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An example program that fulfills the given requirements:

python

Copy code

def display_names(players):

   sorted_names = sorted(players.keys())

   print("Player names:")

   for name in sorted_names:

       print(name)

def display_stats(players):

   name = input("Enter the player name: ")

   if name in players:

       stats = players[name]

       print("Stats for", name)

       print("Wins:", stats["wins"])

       print("Losses:", stats["losses"])

       print("Ties:", stats["ties"])

   else:

       print("No such player:", name)

def main():

   players = {

       "Player A": {"wins": 10, "losses": 5, "ties": 3},

       "Player B": {"wins": 7, "losses": 8, "ties": 1},

       "Player C": {"wins": 12, "losses": 2, "ties": 4}

   }

   display_names(players)

   while True:

       choice = input("Do you want to view player stats? (y/n): ")

       if choice.lower() != "y":

           break

       display_stats(players)

   print("Program terminated.")

if __name__ == "__main__":

   main()

In this program, a dictionary of dictionaries named players is used to store the stats for each player. The display_names function prints an alphabetical list of player names. The display_stats function allows the user to enter a player name and displays the corresponding stats if the player exists. The program loops until the user chooses to stop, and finally, it prints a termination message.

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The air velocity is approximately 6.2 m/s. A Pitot-static probe is a device that is used to measure the velocity of fluid, in this case, air.

The probe consists of two tubes, a Pitot tube and a static tube. The Pitot tube is positioned in the airflow, and it measures the total pressure of the fluid, which includes the static pressure and the dynamic pressure. The static tube, on the other hand, is positioned perpendicular to the airflow and measures only the static pressure.

In this particular scenario, the Pitot tube is connected to a water manometer, which is a device that measures the difference in fluid levels between the two arms of a U-shaped tube. The difference in fluid levels is proportional to the difference in pressure between the two points being measured.

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Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

Input

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

Processing

Assign variable to the input number (x)

Calculate the square (x = 5 * 5)

Display the result (25) ----> This will be passed to the output module

Output

Display 25

To monitor production SQL databases against injection attacks, several security countermeasures can be implemented such as Input Validation and Sanitization, WAF, Security Auditing and Penetration Testing etc.

1)Input Validation and Sanitization: Implement strict input validation and sanitization techniques to ensure that user-supplied data is properly validated and sanitized before being used in SQL queries.

2)Web Application Firewalls (WAF): Deploy a WAF to analyze incoming web traffic and detect and block potential SQL injection attempts.

3)Principle of Least Privilege: Ensure that database accounts and application service accounts have the minimum required privileges.

4)Regular Patching and Updates: Keep the database management system and associated software up to date with the latest security patches.

5)Log Monitoring and Analysis: Implement a centralized log management system to monitor and analyze logs from the database server.

6)Security Auditing and Penetration Testing: Conduct regular security audits and penetration tests on the production SQL databases.

To ensure that an organization incorporates penetration testing and web application testing as part of its implementation procedures, the following steps can be taken:

1)Develop a Security Testing Policy: Establish a comprehensive security testing policy that outlines the requirements, methodologies, and frequency of security testing, including penetration testing and web application testing.

2)Incorporate Security Testing in Development Lifecycle: Integrate security testing into the software development lifecycle (SDLC).

3)Engage Professional Security Testing Services: Partner with reputable security firms or hire skilled penetration testers to conduct thorough assessments of the organization's infrastructure, applications, and systems.

4)Regularly Review and Update Testing Procedures: Continuously review and update the security testing procedures to adapt to new threats and emerging vulnerabilities.

5)Implement Remediation Processes: Establish a systematic approach for addressing identified vulnerabilities and weaknesses.

By implementing these measures, organizations can ensure that penetration testing and web application testing are integrated into their implementation procedures, leading to improved security and reduced risks of successful attacks.

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The entropy changes in the system when there are constant specific heats and variable specific heats are found out to be

a) ASsys= 2.20 kJ/K

b) ASsys= 1.56 kJ/K respectively.

(a) Constant Specific Heat

To solve the problem, use the following formula:

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

ASsys = [Cv ln(T₂/T₁)] + [R ln(V₂/V₁)]

where: ASsys= system entropy change

R = 0.287 kJ/kg.K

Cv = 0.717 kJ/kg.K

T₁ = 17 + 273 = 290 K (initial temperature)

P₁ = 120 kPa (constant pressure)

P₂ = 120 kPa (constant pressure)

V₁ = 300 L

= 0.3 m³ (initial volume)

V₂ = V₁ [since V is constant]

= 0.3 m³

T₂ = T₁ + q/Cp

where q= amount of heat added

Cp= specific heat of air at constant pressure

From the given data,

Q= P x V

= 120 kPa x 0.3 m³

= 36 kJ

Q/Δt = 200 W

= 200 J/s

Cp= 1.005 kJ/kg.K

[Table A-2, at 17°C, for air]

T₂ = (q/Cp) Δt + T₁

= (200/1.005) x 900 + 290

= 692 K

Now apply the formula for system entropy change.

ASsys = Cv ln(T₂/T₁)+R ln(V₂/V₁)

= 0.717 x ln(692/290) + 0.287 x ln(0.3/0.3)

= 2.20 kJ/K

(b) Variable Specific Heat

To solve the problem, use the following formula:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

whereASsys= system entropy change

ΔT= T₂ - T₁

= (692-290)

= 402 K

R= 0.287 kJ/kg.K

V₁ = V₂

= 0.3 m³

Using the data from Table A-2, Cps and T can be tabulated as:

Cp (kJ/kg.K)T (K)1.0052731.005284.16 (Note: T₂)1.005692

Apply the trapezoidal rule for the integral to get:

ASsys = ∫(Cp/T)dT- R ln(V₂/V₁)

= 0.287[(1.005273/290) + (1.005284.16/273) + (1.005692/692)]- 0.287 ln(0.3/0.3)

= 1.56 kJ/K

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Answer:

D. Fire clay bricks

Explanation:

Just like in the steel mills with the big furnaces that's what causes all the smoke to come out of the stacks

Answer:

d is correct

Explanation:

Answer:

Anything you want to do in Hootsuite can be found in the ___ Sidebar_____, with the main workspace in the ___center______

Sidebar; center

Explanation:

The main workspace of the Hootsuite is located in the center.  The sidebar is where the core access to the Hootsuite functionality, like Streams, Inbox, Planner, Analytics, Publisher, and the App Directory, is obtained.  Hootsuite is a media management platform for curating content, scheduling posts, managing team members, and measuring performances.

Answer: Aerosol Cans

Explanation: I just did the quiz

To determine the switching frequencies required to obtain the two output voltages (3.3V and 5V) in the buck converter, we need to consider the voltage conversion ratio and the on-time of the switch.

In a buck converter, the voltage conversion ratio is given by:

Voltage Conversion Ratio = Output Voltage / Input Voltage

For the 3.3V output, the conversion ratio is:

Conversion Ratio (3.3V) = 3.3V / 10V = 0.33

For the 5V output, the conversion ratio is:

Conversion Ratio (5V) = 5V / 10V = 0.5

The on-time of the switch is fixed at 0.1 ms.

The switching frequency can be calculated using the formula:

Switching Frequency = (Conversion Ratio * Input Voltage) / On-time

For the 3.3V output:

Switching Frequency (3.3V) = (0.33 * 10V) / 0.1 ms = 330 kHz

For the 5V output:

Switching Frequency (5V) = (0.5 * 10V) / 0.1 ms = 500 kHz

Therefore, to obtain the desired output voltages of 3.3V and 5V, the switching frequencies should be 330 kHz and 500 kHz, respectively.

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❎❎❎❎❎❎❎ sorry but that didn't help me that much

Answer:

Each wattmeter measures a line-to-line voltage between two of the three power supply lines. In this configuration, the total power, watts is accurately measured by the algebraic sum of the two wattmeter values. Pt = P1 + P2. This holds true if the system is balanced or unbalanced.

Many outsourced projects operate in a virtual environment in which people are linked by computers, faxes, computer-aided design systems, and video teleconferencing.

Tele conferencing is defined as the use of a telecommunications equipment by two or more persons to communicate. Teleconferencing facilitates team communication by making it simple to schedule meetings and keep in touch.

A business practice known as outsourcing entails hiring a third party to carry out duties, manage operations, or offer services on the company's behalf. By outsourcing them, you may maintain your current team accessible for internal work while paying a project manager a fixed price to execute the job.

Thus, many outsourced projects operate in a virtual environment in which people are linked by computers, faxes, computer-aided design systems, and video teleconferencing.

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Fault location techniques are used in power systems for accurate pinpointing of the fault position.

This paper presents a comparative study between two fault location methods in distribution network with Distributed Generation (DG). Both methods are based on computing the impedance using fundamental voltage and current signals. The first method uses one-end information and the second uses both ends

You can try several things to make your balaclava tighter if it's too loose or doesn't fit properly. Some suggestions are as follows :

1. By Adjust the drawstring: You can adjust the fit of most balaclavas by adjusting the drawstring.  Try tightening the drawstring on your balaclava if it has one.

2. Wear a hat or beanie underneath Adding some extra bulk to your balaclava can help fill out the space if it is too loose around the head.

3. Use a hair tie: The excess fabric on your balaclava can be gathered up with a hair tie if it's too loose around your neck.

4. Sew in some elastic: To create a tighter fit, you can sew some elastic into the balaclava if you are handy with a needle and thread

5. Buy a smaller size:   The balaclava may need to be purchased in a smaller size if none of these suggestions work.

Thus, you can try several things to make your balaclava tighter.

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With this in mind, the answer to the provided student question is as follows:When parking a car on a level street, it is important to turn your front wheels towards the curb.

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As a result, Day(1,n+1) will be a list of days to buy a new automobile and sell the old one for the least amount of money. The code is given below.

Dynamic programming is mostly an improvement over straightforward recursion. Dynamic Programming may be used to optimize any recursive solution that has repeated calls for the same inputs. To avoid having to recompute the outcomes of sub-problems later, the idea is to just store them. With this straightforward improvement, time complexity is reduced from exponential to polynomial.

To write a program on dynamic program on car ownership:

By using the given data,

we have to write the program,

OPT(i, j) = minimum cost possible of purchasing car at year i, maintaining from i to j and then selling at j

OPT(i, i+1) = p(i)+m(i, i+1)-v(i, i+1)

OPT(i, j) =(min k) OPT(i, k)+OPT(k, j) for  i ≤ k ≤ j

For each pair of (i, j) where i<j, Day(i, j) will contain the list of days to buy car between i to j.

To see the code, see in the attached image.

Time complexity = O(n4).

Therefore, Day(1,n+1) will be a list of days to purchase a new automobile and sell the old one for the lowest price.

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They don't; instead, they seek out any and all bits of evidence, regardless of whether they confirm or refute their theories, impartially and with an open mind.

A scientist is someone who methodically acquires data and uses it to form hypotheses, test those assumptions, and further knowledge and understanding. A scientist's methodology, such as the use of statistics (statisticians) or data, can also be used to further characterize them.

A chemist is an expert in the subject. They most likely focus on a particular area of biology, such as biochemistry, which examines the chemical composition of living things. A cytologist is an expert in cell research. Ecologists research ecosystems, or the surroundings in which we live.

An expert who conducts and gathers research to advance knowledge in a particular field is referred to as a scientist. Making hypotheses, testing them using diverse techniques like statistics and data, and drawing conclusions are all things that scientists can do.

Thus, They don't; instead, they seek out any and all bits of evidence.

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(a) The active part of the inflow expands its cross-section as it passes the tidal turbine due to conservation of mass and energy.

(b) Several reasons contribute to the theoretical maximum efficiency not being achieved in practical tidal turbine deployments, including turbulence, non-uniform flow, and mechanical losses.

(c) Estimating the total fluid loading on the Alstom 1.4 MW OceadeTM turbine requires considering the flow velocity, rotor and tower dimensions, and making assumptions about the flow characteristics and structural properties.

(d) The estimates of forces obtained from the fluid loading calculations are essential for designing the turbine and tower structure by ensuring that they can withstand the anticipated loads and stresses.

(a) The active part of the inflow expands its cross-section as it passes the tidal turbine due to the principle of conservation of mass and energy. As the tidal flow encounters the turbine rotor, some of the kinetic energy of the flow is converted into mechanical energy to drive the turbine. To satisfy the conservation of mass, the cross-sectional area of the flow must increase to compensate for the reduction in flow velocity caused by energy extraction.

(b) Achieving the theoretical maximum efficiency in practical tidal turbine deployments is challenging due to several reasons. First, tidal flows are often characterized by turbulence, which disrupts the uniformity of the flow and reduces overall efficiency. Second, tidal flow itself is not uniformly distributed, and the flow characteristics vary with tidal cycles, further impacting efficiency. Lastly, mechanical losses in the turbine's components, such as friction and resistance, reduce the efficiency of energy conversion.

(c) Estimating the total fluid loading on the Alstom 1.4 MW OceadeTM turbine involves considering the flow velocity, rotor diameter, and tower dimensions. Assuming a tidal flow velocity of 3 m/s, the fluid loading can be estimated by considering the momentum change and forces acting on the rotor and tower surfaces. Assumptions may include a simplified flow model, neglecting factors such as turbulence and non-uniform flow, and assuming a stationary tower. These assumptions simplify the calculation while providing a reasonable estimate of the fluid loading.

(d) The estimates of forces obtained from the fluid loading calculations are crucial for designing the turbine and tower structure. These estimates help engineers determine the required structural strength, material selection, and design considerations to ensure that the turbine and tower can withstand the anticipated fluid forces and mechanical stresses. By considering the estimated forces, designers can optimize the structural integrity, stability, and reliability of the turbine and tower, ensuring safe and efficient operation in tidal environments.

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Total unit weight, γ = 1.0825 kg/m³Void ratio, e = 3Porosity, n = 0.75

Water content (w%) = 12%Degree of saturation (Sr) = 75%Specific gravity of solids (Gs) = 2.65We have to calculate the total unit weight, void ratio (e), and porosity (n).Calculation:

First, we have to calculate the dry unit weight, γd

Dry unit weight, γd = γw / (1 + w%)

Let's calculate γw

Water content (w%) = 12%

So, moisture content (w) = w% * γd / 100= 12/100 * 1.65= 0.198 kg/kg of dry soil

Total weight of soil (Wt) = Weight of solids (Ws) + Weight of water (Ww)

Weight of solids (Ws) = Volume of solids (Vs) * Gs * γw/Vs = 1

Total volume of soil = Volume of solids (Vs) + Volume of voids (Vv)

Let's calculate Vv Degree of saturation (Sr) = 75%Vv/Vs = Sr / (1 - Sr)= 0.75 / 0.25= 3

Total volume of soil = Volume of solids (Vs) + Volume of voids (Vv)= 1 + 3= 4m^3

Let's calculate the weight of water,

WwWw = Wt - WsWw = 4 * 0.12 = 0.48 kg

Weight of solids, Ws = Gs * γs * Vsγs = (Wt - Ww) / Vst = (1 - 0.12) * 1.65= 1.452 kg/ m^3

Ws = Gs * γs * Vs= 2.65 * 1.452= 3.85 kg

Total unit weight, γ= Wt / Vt= (Ws + Ww) / Vt= (3.85 + 0.48) / 4= 1.0825 kg/m^3

Now, calculate the void ratio, eVv/Vs = 3Vv / Vt = Vv / (Vs + Vv) = 3 / 4e = Vv / Vs = 3 / 1 = 3

Porosity, n= Vv / Vt= 3 / 4= 0.75

Answer: Total unit weight, γ = 1.0825 kg/m³Void ratio, e = 3Porosity, n = 0.75

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In the radio communication system, multipath is the propagation phenomenon that causes diffusion or reflection in multiple different directions of a signal.

Multipath is a propagation mechanism that impacts the propagation of signals in radio communication. Multipath results in the transmission of data to the receiving antenna by two or more paths. Diffusion and reflection are the causes that create multiple paths for the signal to be delivered.

Diffraction occurs when a signal bends around sharp corners; while reflection occurs when a signal impinges on a smooth object. When a signal is received through more than one path because of the diffraction or reflection, it creates phase shifting and interference of the signal.

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