As an engineer, you are requested to design a system to monitor the Covid-19 patients in the airport. The system must be able to detect the human temperature and if the temperature is more than 37.5°C, the system will isolate the human automatically and vaporize disinfection will be turned on as well. Identify the sensor and actuator for your design. (6 marks) With the aid of block diagram, describe the process as a feedback control system.

Answer:

\(\boxed{R_T_s=707 \Omega}\)

Explanation:

The following formula gives the total resistance of a series circuit:

\(R_T_s=R_1+R_2+R_3...+R_n\)

Therefore:

\(R_T_s= 200+86+91+180+150 \\R_T_s=707 \Omega\)

\(\text{-B$\mathfrak{randon}$VN}\)

The temperature of the system remains unchanged during expansion. There is heat exchange between both the system and the environment during the process. The system should be compressed or expanded at a slow pace so that the system allows appropriate heat exchange and keep the temperature constant.

The P-V curve in an isothermal process where:

PV = constant = C

⇒ dP / dV = – P / V

Work done in the isothermal process:

In this process, the gas follows Boyle’s law, and the work done by the gas is,

\(W = nRTln(\frac{V_2}{V_1} )\)

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Answer:

D) AND gate.

Explanation:

Given that:

A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print

These conditions are:

1. The printer's electronic circuits must be energized.

2. Paper must be loaded and ready to advance.

3. The printer must be "on line" with the microprocessor.

Now; if these conditions are met  the logic gate produces a HIGH output indicating readiness to print.

The objective here is to determine the basic logic gate used in this circuit.

Now;

For NOR gate;

NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.

For NOT gate.

NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.

Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".

Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.

Answer:

B the EOQ after setup reduction.

Through a program of setup reduction, thesetup cost is reduced to $20. Run costs are $2 per unit. CalculateThe EOQ after setup reduction. Hence, option B is correct.

The amount of an order known as a "economic order quantity" (EOQ) is one that lowers the overall cost of acquiring and maintaining inventories of goods or raw materials.

The supplier should be contacted to ascertain the optimal inventory size in order to lower the company's overall annual inventory cost. Economic order quantity, or EOQ, is the phrase. Supply, logistics, and operations management all use this measurement.

In its most basic form, EOQ is a technique for determining the quantity and frequency of orders required to satisfy a particular level of demand while reducing the cost per order.

Thus, option B is correct.

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Answer:

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Explanation:

From Thermodynamics we remember that thermal efficiency of the ideal Otto cycle (\(\eta_{th}\)), dimensionless, is defined by the following formula:

\(\eta_{th} = 1-\frac{1}{r^{\gamma-1}}\) (Eq. 1)

Where:

\(r\) - Compression ratio, dimensionless.

\(\gamma\) - Specific heat ratio, dimensionless.

Please notice that specific heat ratio under cold air standard conditions is \(\gamma = 1.4\).

If we know that \(r = 8.2\) and \(\gamma = 1.4\), then thermal efficiency of the ideal Otto cycle is:

\(\eta_{th} = 1-\frac{1}{8.2^{1.4-1}}\)

\(\eta_{th} = 0.569\)

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Improvements to brake systems are

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True. A metal-semiconductor junction can form a built-in potential due to the difference in work function between the metal and semiconductor materials.

What is metal-semiconductor junction?

A metal-semiconductor junction can also form a built-in potential and can act as a diode. This type of junction, called a Schottky barrier diode, is formed when a metal comes in contact with a semiconductor material. The built-in potential is created due to the difference in the work function of the metal and the semiconductor, which leads to a depletion region at the junction. The diode allows current to flow in one direction while blocking it in the other, similar to a conventional semiconductor diode.

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Crystalline and amorphous materials are two distinct classes of materials in terms of their thermal behavior and atomic arrangement.

a) Thermal behavior:

Crystalline materials have a well-defined and regular arrangement of atoms in a repeating pattern, which gives rise to their ordered atomic structure. As a result, crystalline materials have a characteristic melting point, where the atoms break free from their ordered structure and become disordered as the material transitions from solid to liquid state. In contrast, amorphous materials do not have a well-defined atomic arrangement and do not have a distinct melting point. Instead, they gradually soften and flow as the temperature increases, without undergoing a sharp phase transition.

b) Atomic arrangement:

Crystalline materials have a highly ordered arrangement of atoms, where the atoms are arranged in a regular pattern that repeats in all directions. This ordered structure gives rise to many of the characteristic properties of crystalline materials, such as their mechanical strength, optical properties, and electrical conductivity. In contrast, amorphous materials have a disordered atomic arrangement, where the atoms are arranged randomly without any repeating pattern. This lack of long-range order in amorphous materials gives rise to their unique properties, such as their flexibility, transparency, and lack of cleavage planes.

In summary, crystalline and amorphous materials differ in their thermal behavior and atomic arrangement. Crystalline materials have a well-defined atomic arrangement and a distinct melting point, while amorphous materials have a disordered atomic arrangement and do not have a well-defined melting point. Understanding the differences between these two classes of materials is important in engineering and materials science, as it can help to predict and control their properties and behavior.

Complete Question:

Stone has been used as a building material since ancient times because of its compressive strength, which is the?

Group of answer choices

A. ability to support pressure without breaking.

B. ability to press down and become solidly fixed in the ground.

C. relative weight of a block of stone to its size.

D. force a block exerts on the blocks around it.

Answer:

A. ability to support pressure without breaking.

Explanation:

Stone has been used as a building material since ancient times because of its compressive strength, which is the ability to support pressure without breaking.

Compressive strength can be defined as the ability of a structural element or particular material to withstand an applied, which is aimed at reducing the size of the structural element.

Simply stated, it is the ability of a structural element or material to withstand an applied load without deflections, fracture or having any crack.

In this context, a stone possesses the ability to resist or withstand compression loads.

Some examples of other materials or structural elements having good compressive strength are steel, bones, concrete etc. The standard unit of measurement of the compressive strength of a material is Mega Pascal (MPa) or pound-force per square inch (psi) in the United States of America.

The area occupied by one molecule of Red 40 dye on a 40 cm² wool piece can be found by dividing 0.004 m² by the number of dye molecules adsorbed on the surface.

To determine the area occupied by one molecule of Red 40 dye on the surface, we need to know the number of dye molecules that are present on the wool surface.

Let's assume that we have N molecules of Red 40 dye adsorbed on the surface of each 40 cm² wool piece.

The surface area of one wool piece is 40 cm² = 0.004 m².

The molecular weight of Red 40 dye is approximately 496 g/mol.

Assuming that the density of Red 40 dye is similar to that of water (1 g/cm³),

the volume occupied by one molecule of Red 40 dye is approximately 5.0 × 10⁻²³ m³ (496 g/mol divided by 6.02 × 10²³ molecules/mol).

To determine the area occupied by one molecule of Red 40 dye on the surface, we need to divide the surface area by the number of dye molecules, which gives:

Area occupied by one molecule = (0.004 m² / N)

Therefore, the area occupied by one molecule of Red 40 dye on the wool surface is (0.004 m² / N),

where N is the number of Red 40 dye molecules adsorbed on the surface of each 40 cm² wool piece.

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Answer:

False

Explanation:

first generation Computer uses vaccum tube.

Microprocessor was used by third generation Computer

(. ❛ ᴗ ❛.)

Only four of these—000, 001, 010, and 100—are used in the main sequence. Hence, the states 011, 101, 110, and 111 are the four unoccupied states.

What does JK flip flop decade counter entail?

Using JK flip flops and a NAND gate, a BCD or decade counter circuit is created. Because it is a 4-bit binary counter, the BCD counter architecture is fairly straightforward and calls for 4 JK flip flops. Below is a diagram showing how the decade counter is made. We can see from the diagram that J and K's outputs are connected to logic 1.

What does D flip flop-based 4-bit decade counter do?

An integer between 0 and 9 is displayed by a decade counter, which then resets to 0. Due to the The counter requires 4 bits of storage (since 23 10 24) to display 10 distinct numbers. This is achieved utilising four D-flip flops.

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Sorry sir i don't know the answer

Answer:

The power produced by the turbine is 23309.1856 kW

Explanation:

h₁ = 3755.39

s₁ = 7.0955

s₂ = sf + x₂sfg  =

Interpolating fot the pressure at 3.25 bar gives;

570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175

2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345

h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg

Power output of the turbine formula =

\(Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]\)

Which gives;

\(560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]\)

= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856

\(- \dot{W }\) = -22749.1856 - 560 = -23309.1856 kJ

\(\dot{W }\) = 23309.1856 kJ

Power produced by the turbine = Work done per second = 23309.1856 kW.

The response involves representation and interpretation of decimal numbers using a hypothetical floating-point format with a three-bit exponent and a four-bit significand, as well as the IEEE 754 single-precision floating-point format.

F- In a floating-point format with a three-bit exponent and a four-bit significand, (i) -12.5 would be 1 111 1000 and (ii) 13.0 would be 0 100 1100. G- Conversely, the decimal values represented by the patterns are (i) -1.5 and (ii) 1.5. H- In the IEEE 754 format, the hexadecimal representations are (i) BF800000 for -1.0, (ii) 80000000 for -0.0, and (iii) 43780000 for 256.015625. I- The binary scientific notations for these hexadecimal values are (i) 1.1011x2^3, (ii) 1.1111111111x2^127 (assuming this represents infinity), and (iii) -1.0x2^0 (assuming this is a negative zero). Floating-point format is a mathematical notation used in computer systems to represent real numbers.

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Answer:

Fourier coefficient, Ck:

\({ \rm{ c_{k}} = \int ^{ \infin} _{ - \infin } \{x(t) {e}^{jk \omega _{0} t} \} }\)

Fourier transform:

\({ \rm{x( \omega) = \int ^{ \infin} _{ - \infin} } \{x(t) {e}^{j \omega} t} \}\)

In responsible, accountable behavioral treatment, it is essential to follow the Evidence-Based Practice protocols.

Evidence-Based Practice (EBP) is a problem-solving process that is used in providing clinical care. EBP protocols include the following: 1. Clinician's judgment, clinical experience, and expertise should be included in the treatment process. 2. Evidence-Based Practice protocols provide the best evidence for the care of patients. 3. EBP considers the needs, preferences, and values of the patients. 4. The EBP includes the best research-based evidence to guide clinical decision-making. 5. The EBP helps healthcare professionals to provide optimal care that is consistent with their patients' values and expectations. 6. It helps clinicians in the critical evaluation of scientific research and in assessing the quality of research evidence. 7. EBP protocols include the best research-based evidence to guide clinical decision-making.Therefore, following the Evidence-Based Practice protocols in responsible, accountable behavioral treatment is crucial for optimal care.

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Therefore, the final pressure of the gas is 1.52 MPa (megaPascals).

We can use the ideal gas law to solve this problem:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to find the initial number of moles of gas:

n1 = (P1 V1)/(R T1)

= (300 x 10^3 Pa) x (0.3 m^3)/(8.31 J/(mol K) x (20 + 273) K)

= 4.97 mol

where R = 8.31 J/(mol K) is the gas constant.

Next, we can use the fact that the process is isothermal (i.e., at constant temperature) to find the final volume of the gas after it is compressed to a pressure of 800 kN/m^2:

P1 V1 = P2 V2

V2 = (P1 V1)/P2

= (300 x 10^3 Pa) x (0.3 m^3)/(800 x 10^3 Pa)

= 0.1125 m^3

Now we can use the fact that the process is adiabatic (i.e., no heat is exchanged with the surroundings) and that the initial and final volumes are the same to find the final pressure of the gas:

P1 V1^γ = P2 V2^γ

where γ is the adiabatic index (a property of the gas), which depends on the specific gas. For simplicity, we will assume that γ = 1.4, which is a reasonable value for diatomic gases such as nitrogen and oxygen.

P2 = P1 (V1/V2)^γ

= (300 x 10^3 Pa) x (0.3 m^3/0.1125 m^3)^1.4

= 1.52 x 10^6 Pa

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Unless otherwise authorized, two-way radio communications with Air Traffic Control are required for landings or takeoffs at all tower controlled airports only when weather conditions are less than VFR.

Two-way radio communications with Air Traffic Control (ATC) are required for landings or takeoffs at all tower controlled airports, but this requirement applies specifically when the weather conditions are less than Visual Flight Rules (VFR). VFR refers to weather conditions that allow pilots to operate the aircraft using visual references, such as clear visibility and suitable cloud clearance. When weather conditions are less than VFR, pilots are required to communicate with ATC to ensure safe operations and receive guidance or instructions.

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The correct combination of phase fields involved in a peritectic reaction is:

c. Two liquid and one solid.

In a peritectic reaction, two liquid phases combine to form a single solid phase. This occurs at a specific composition and temperature range, where the liquid phases and solid phase coexist. The reaction is characterized by the transformation of the two liquid phases into a solid phase, typically with a eutectic composition.

The other options listed do not accurately describe the phase fields involved in a peritectic reaction.

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Answer:

D

Explanation:

An ignition coil is an example of a Solenoid. Thus, the correct option for this question is D.

Solenoids may be characterized as a coil of wire that is usually in a cylindrical form and carries a current that acts like a magnet. Due to this, a migratory core is drawn into the coil when a current flows, and that is utilized especially as a switch or control for a mechanical device.

Similarly, an ignition coil also represents an example of a solenoid because it also consists of a coil of wire and carries a current that acts like a magnet. While the concept of the transformers is different as it stores a huge amount of electric current and spread it accordingly.

Therefore, an ignition coil is an example of a Solenoid. Thus, the correct option for this question is D.

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The best practice that Raven should follow is to use DC2 or DC3 as the Domain Naming Master of the following is a best practice that Raven should follow. The correct option is A.

The management of the addition or deletion of domains from the forest is the responsibility of the Domain Naming Master. For redundancy and fault tolerance, it is advised to split the FSMO roles among several domain controllers.

Since DC1 already has a copy of the global catalog, it is advantageous to choose a different domain controller (DC2 or DC3) as the Domain Naming Master to disperse the workload and guarantee high availability. This ensures that the forest's operations may continue even if one domain controller goes offline and prevents the creation of a single point of failure.

Thus, the ideal selection is option A.

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The safety technique that is recommended for this situation of moving a heavy industrial printer is to Bend at the knees when lifting.

This refers to the process of executing a task in a safe and acceptable way that does not put the user or others at risk

Hence, we can see that The safety technique that is recommended for this situation of moving a heavy industrial printer is to Bend at the knees when lifting.

This is to ensure that the person doing the lifting does not injure himself and has the proper form when lifting.

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Refrigerant 134a enters the evaporator of a refrigeration system operating at a steady-state at -8oC and a quality of 20% at a velocity of 5 m/s. The mass flow rate of the refrigerant is 0.0594 kg/s  and the velocity at the exit is 24.191  m/s.

From the information given, we can use the properties table of refrigerant-134a to determine the values of the specific volume of saturated liquid and vapor at -8° C.

Now the specific volume of the refrigerant at the inlet of the evaporator can be computed by using the formula;

\(\mathbf{v_1=v_f +x(v_g-v_f)}\)

where;

\(\mathbf{v_1=0.0007570 +0.2(0.092438-0.0007570)}\)

\(\mathbf{v_1=0.0007570 +0.2(0.091681)}\)

\(\mathbf{v_1=0.0007570 +0.0183362}\)

v₁ = 0.0191 m³/kg

The density of the refrigerant at the inlet of the evaporator is:

\(\mathbf{\rho_1 = \dfrac{1}{v_1}}\)

\(\mathbf{\rho_1 = \dfrac{1}{0.0191} }\)

\(\mathbf{\rho_1={52.356 \ kg/m^3}}\)

However, the density of the  refrigerant at the outlet of the evaporator is:

\(\mathbf{\rho_g = \dfrac{1}{v_g}}\)

\(\mathbf{\rho_g= \dfrac{1}{0.0092438} }\)

\(\mathbf{\rho_g={10.818 \ kg/m^3}}\)

Recall that:

Now, the mass flow rate of the refrigerant can be computed by using the formula:

\(\mathbf{m = \rho_1 \Big[ \dfrac{\pi}{4} \times d^2 \Big] v_1 }\)

\(\mathbf{m = 52.356 \Big[ \dfrac{\pi}{4} \times (0.017)^2 \Big] 5 }\)

mass flow rate (m) = 0.0594 kg/s

Also, the velocity of the refrigerant at the exit of the evaporator is determined by using the formula:

\(\mathbf{m=\rho_2 \Big[ \dfrac{\pi}{4} \times d^2 \Big] v_2}\)

\(\mathbf{ 0.0594 kg/s = 10.818 \Big[ \dfrac{\pi}{4} \times (0.017)^2 \Big] v_2 }\)

\(\mathbf{v_2 = 24.191 \ m/s}\)

Therefore, we can conclude that the mass flow rate of the refrigerant is 0.0594 kg/s  and the velocity at the exit is 24.191  m/s.

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