The kinematic viscosity of oxygen at 20◦c and a pressure of 150 kpa (abs) is 0. 104 stokes. Determine the dynamic viscosity of oxygen at this temperature and pressure

Starting a synchronous motor with DC excitation applied to the rotor can cause the rotor to lock into a fixed position, preventing it from synchronizing with the rotating magnetic field and causing damage to the motor.

A synchronous motor operates by synchronizing its rotating magnetic field with the rotating magnetic field of the stator. The rotor of a synchronous motor is typically excited by a DC current to create a magnetic field. However, if the motor is started with DC excitation still applied to the rotor, the rotor will not be able to synchronize with the stator's magnetic field. This can cause the rotor to lock into a fixed position, resulting in excessive current flow and potentially causing damage to the motor. Therefore, it is crucial to ensure that DC excitation is removed from the rotor before starting a synchronous motor to prevent this issue from occurring.

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Resistors limit the amount of current flow in the circuit. For devices connected to the circuit, some currents can damage their integrity. Therefore, resistors are often used to protect devices connected to the circuit that follow it.

Answer:

The view factor for radiation emitted by surface 2 to surface 1 is 0.1

Explanation:

Given

\(F_{12} = 0.4\)

\(A_1 = 0.01m^2\)

\(A_2 = 0.04m^2\)

Required

Determine \(F_{21}\)

To do this, we make use of the following equivalent ratio

\(A_1 * F_{12} = A_2 * F_{21}\)

Make \(F_{21\) the subject

\(F_{21} = \frac{A_1 * F_{12}}{ A_2}\)

Substitute values into the equation

\(F_{21} = \frac{0.01m^2 * 0.4}{0.04m^2}\)

\(F_{21} = \frac{0.01 * 0.4}{0.04}\)

\(F_{21} = \frac{0.004}{0.04}\)

\(F_{21} = 0.1\)

The Ada 83 designers' decision to allow the imple-mentor to choose between implementing in out mode parameters by copy or by reference is a valuable choice.

which provides significant advantages to the implementor as well as to the code executed.In Out Mode Parameters:In Ada 83, the implementation of in-out parameters could be selected by the implementor, which means that the imple-mentor can choose whether to use the call-by-value technique or the call-by-reference technique.

When calling a subroutine, both in and out parameters are copied to the callee’s frame. The called subprogram might utilize this to make modifications in the copy.

The result must then be copied back to the caller. These copying and copying back steps can be avoided by using reference parameters.

If the in-out parameter is implemented by reference, changes made to the parameter in the callee are automatically reflected in the caller.

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From the AD711 data sheets, the typical maximum output current limit is 20 mA. This implies that the maximum voltage at the output allowed for a load resistance can be calculated by multiplying the load resistance by the maximum current limit.

This can be expressed mathematically as \(Vout = Iload × Rload\), where Vout is the maximum voltage at the output allowed for a load resistance, Iload is the maximum output current limit, and Rload is the load resistance.Therefore, the maximum voltage at the output allowed for a load resistance is \(Vout = 20 mA × Rload.\)

This means that for a load resistance of 500 Ω, the maximum voltage at the output allowed is\(Vout = 20 mA × 500 Ω = 10 V\). Hence, the typical maximum output current limit is 20 mA, and the maximum voltage at the output allowed for a load resistance of 500 Ω is 10 V. This information can be found in the AD711 data sheets.

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Answer:

  c.  25 μA

Explanation:

The dependent current source means that 31 times i1 will flow through the 6kΩ resistor, effectively increasing its voltage drop to 31 times that which would result from i1 alone. In effect, the 6kΩ resistor behaves in the left-side circuit loop as though it were 31×6kΩ = 186kΩ (with no dependent current source).

Then the current i1 is equivalent to that created by a 5+1 = 6V source through a 54kΩ +186kΩ = 240kΩ circuit impedance.

  (6V)/(240kΩ) = 25 μA

_____

Additional comment

The voltage across the 6kΩ resistor is (186/240)·6V = 4.65V, and the 25 μA current generates a voltage of 30·(25 μA)(1.8kΩ) = 1.35V across the 1.8kΩ resistor. This means the voltage source at the right side of the diagram needs to be at least 4.65 +1.35 = 6.0V in order to support the calculated voltage drops.

Answer:

you might be facing some difficulty in observing this point of division between your question and me

Answer:

Static Equilibrium.....

Explanation:

When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium.....

Answer:

Hope this will help you

Have a good day

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

The peak frequency deviation, minimum bandwidth, and baud for a binary FSK signal for the given frequencies are respectively;

a) 0.5 kHz

b) 9 kHz

c) 4000

1) The peak frequency deviation is gotten from the formula;

∆f = |f_m - f_s|/f_b

where;

Thus;

∆f = |38 - 40|/4

∆f = 0.5 kHz

2) The minimum bandwidth is given by the formula;

B = 2(∆f + f_b)

B = 2(0.5 + 4)

B = 9 kHz

3) For FSK signal, N = 1, and the baud is gotten from the Equation;

baud = f_b/1

f_b = 4 kbps = 4000 bps

Thus; baud = 4000/1 = 4000

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The employee consistently demonstrated professionalism and dedication to their role. They arrived punctually and often went above and beyond their assigned tasks. Their reliability and self-motivation were impressive, showing a strong sense of responsibility in achieving goals and meeting deadlines.

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The split() method splits a string into an array of substrings. The split() method returns the new array. The split() method does not change the original string. If (" ") is used as separator, the string is split between words.

Following are the program in Java Programming language

import java. util.*; //import package.

public class Main //define class main

{

public static Array List<Character>to Character Array(String s) //define function  

{

Array List<Character> o b = new Array List<Character>(); //create Array List object.

for(char ch1: s.to Char Array())// iterating over the loop

{

ob. add(ch1); //add Character in list.

}

return o b; //return the object I. e return the value

}

public static void main(String arcs[]) // main method.

{

String str; // variable declaration.

int I; //define variable.

Scanner ob1 = new Scanner(System.in); //creating scanner class object.

System .out. print("Enter any string: "); //message

str = ob1.nextLine();//input value in str variable.

Array List<Character> ob2 = to Character Array(str); //create Array List reference variable and call to Character Array() function

System. out. print("Elements of the character array list: ");//message

for(I = 0;i<ob2.size();I++) //loop

{

Output:

Enter any string: Stony Brook

Elements of the character array list: S t o n y   B r o o k  

Explanation:

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A bend is defined as a change in the shape of the part between the damaged and undamaged area that is smooth and continuous.

What is a kink?

A kink can be defined as a sharp bend with a small radius over a short distance.

So when any part is kinked it must be replaced without any doubt. A part is kinked if it just doesn't work on the repair.

What is a bend?

Unlike a kink, a bend can be restored. That is after a bend also  a part can be bought back to its original position.

When the part is straightened, it is returned to proper shape and state without any areas of permanent deformation.

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When a level 1 technician is unable to solve an issue, they should escalate the problem to a higher level of support or a more experienced technician for further assistance.

Level 1 technicians typically handle basic technical issues and provide initial troubleshooting and support. However, there are cases where they may encounter complex or challenging problems that go beyond their expertise. In such situations, it is important for the level 1 technician to recognize their limitations and escalate the issue to a higher level of support. This can involve seeking assistance from a more experienced technician, a specialized team, or a supervisor. By escalating the problem, the level 1 technician ensures that the issue is addressed by someone with the necessary skills and knowledge to effectively resolve it. This helps prevent prolonged downtime, minimizes potential mistakes, and ensures the problem is handled by the appropriate resources, ultimately leading to a more efficient and effective resolution.

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The ceramic compound that is most responsible for the initial, most significant, and most rapid hardening of cement within concrete during the induction stage of curing is calcium silicate hydrate (C-S-H).

During the hydration process, cement particles react with water molecules to form C-S-H, which then begins to precipitate and fill the spaces between the cement particles, forming a strong and durable material. This process occurs quickly and is responsible for the initial hardening of the concrete.

In addition to C-S-H, other ceramic compounds such as calcium hydroxide and ettringite also contribute to the hardening of concrete. However, these compounds form at a slower rate and are less significant in terms of initial hardening.

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Implementation of a Product Backlog item is considered complete when the item has no work remaining in order to be potentially released.

When the item has no work remaining in order to be potentially released. In the context of Agile software development, the Product Backlog represents a prioritized list of requirements or user stories that need to be implemented. Each item in the Product Backlog should be refined, estimated, and broken down into tasks during Sprint Planning. The implementation of a Product Backlog item is considered complete when all the necessary work has been done to potentially release it to the users or stakeholders.

This includes development, testing, quality assurance, and any other tasks required to ensure the item meets the defined acceptance criteria and is ready for release. The completion of a Product Backlog item is not necessarily tied to the end of a Sprint or the completion of other unrelated work in the Sprint Backlog.

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MIPS assembly program is an assembly language that is popularly used in computers and other devices that use Reduced Instruction Set Computing (RISC) architecture.

MIPS Assembly language is a low-level language that provides significant control over the computer hardware. The following MIPS assembly program calculates the sum of the first hundred integers:Explanation:The following MIPS assembly program calculates the sum of the first hundred integers: .text main: move $a0, $0 li $t0, 100 loop: add $a0, $a0, $t0 addi $t0, $t0, -1 bez $t0, loop li $v0, 1 syscall li $v0, 10 syscall There are two bugs in this program. Please find and fix them.Bug 1: The initial value of $t0 should be 99 instead of 100.Bug 2: The main answer of the program is not being printed to the console. The main answer should be printed to the console using system call 1 (li $v0, 1) before the second syscall (li $v0, 10) to exit the program.

Therefore, the corrected code is shown below: .text main: move $a0, $0 li $t0, 99 # The initial value of $t0 should be 99 loop: add $a0, $a0, $t0 addi $t0, $t0, -1 bez $t0, loop li $v0, 1 # Print the result to the console using system call 1 syscall li $v0, 10 # Exit the program using system call 10 syscall The main answer is the sum of the first hundred integers. The sum is stored in $a0 after the loop ends. The corrected code correctly prints the main answer to the console using system call 1 (li $v0, 1) before the second syscall (li $v0, 10) to exit the program. Therefore, the main answer is 5050.

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The resources and instructions that should be used for the procedures of performing PMCS are:

PMCS is an acronym for preventive maintenance checks and services and it can be defined as the maintenance, checks, and services that are typically performed before, during, and after the use of any type of military equipment such as:

Basically, the resources and instructions that should be used for the procedures of performing PMCS are:

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Answer:

true

Explanation:

Answer:Referencing is a standardised method of acknowledging the sources of information and ideas you have used.

Explanation:Depending on the way in which they record sources, scholarly reference styles can be divided into three main categories: documentary notes styles, parenthetical (or author-date) styles, and numbered styles. Within each category there are several, slightly different reference styles.

Referencing allows you to acknowledge the contribution of other writers and researchers in your work. Any university assignments that draw on the ideas, words or research of other writers must contain citations. Referencing is also a way to give credit to the writers from whom you have borrowed words and ideas.

Answer:

i) Heat rejection =  1.0288 KJ

  Network production = 1.212 kJ

ii) Thermal efficiency = 54.08%

   mean effective pressure = 353.5 kPa

Explanation:

Given data :

Compression ratio ( r )  = 7

P1 ( initial pressure ) = 90 kPa,

T1 ( initial temperature ) = 27°C + 273 = 300 K

V1 = 0.004 m^3

Max cycle Temperature = 1127°C for each repetition of the cycle

i) Determine the heat rejection and net work production

considering that process 1-2 is an Isentropic compression

Find ; T2 = T1 ( r )^1.4-1

      T2 = 300 ( 7 )^0.4  = 653.371 K

    P2 = P1( r )^k

          = 90 ( 7 )^1.4  = 1372.081 kPa

considering that process 3-4 is an Isentropic expansion

T4 = T3 / r^k-1

    = 1400 / 7^(1.4 -1 ) = 642.82 K

Next ; Calculate the value of m

m = P1V1 / RT1   = 90(0.004) / 0.287 ( 300 )

                      m = 4.18 * 10^-3 kg

Finally :

amount of heat rejected = mCv ( T4 - T1 )

                                          = 4.18 * 10^-3 ( 0.718 ) ( 642.82 - 300 )

                                    Qout  = 1.0288 KJ

amount of heat added ( Qin ) =  mCv ( T3 - T2 )

                                                = 4.18 * 10^-3 ( 0.718 ) ( 1400 - 653.371 )

                   hence           Qin  = 2.2408 kJ

Network production( Wnet) = Qin - Qout

                                  = 2.2408 - 1.0288 ) KJ

                                  = 1.212 kJ

ii) Determine the thermal efficiency and mean effective pressure of the cycle

Thermal efficiency = 1 - Qout / Qin

                               = 1 - ( 1.0288 / 2.2408 )  = 54.08%

mean effective pressure = Wnet / V1 ( 1 - 1/r )

                                         = 1.212 / 0.004 ( 1 - 1/7 )

                                         = 353.5 kPa

Answer:

Sure! Here's how to draw the shear and moment diagrams for the beam:

To draw the shear diagram, start by finding the shear force at the left end of the beam (V1). This will be equal to M0, since there is no external force acting on the left end of the beam. Then, draw a horizontal line segment to represent the shear force at that point.

Next, move to the right end of the beam (at x=8m). The shear force at this point (V2) will be equal to the shear force at the left end, minus the total load on the beam. So, if the total load on the beam is W, then V2 = M0 - W. Draw a horizontal line segment to represent the shear force at this point.

Finally, connect the two line segments with a smooth curve. This curve represents the variation of the shear force along the length of the beam.

To draw the moment diagram, start by finding the moment at the left end of the beam (M1). This will be equal to 0, since there is no external force acting on the left end of the beam and the shear force is constant (M0) along the length of the beam. Then, draw a horizontal line segment to represent the moment at that point.

Next, move to the right end of the beam (at x=8m). The moment at this point (M2) will be equal to the moment at the left end, plus the total load on the beam multiplied by the length of the beam (WL). So, M2 = M1 + WL = W*L. Draw a horizontal line segment to represent the moment at this point.

Finally, connect the two line segments with a smooth curve. This curve represents the variation of the moment along the length of the beam.

I hope this helps! Let me know if you have any questions.

Explanation:

A change in ambient temperature or air density can have a significant effect on gas turbine engine performance. These factors influence the engine's efficiency, power output, and fuel consumption.

1)Firstly, an increase in ambient temperature causes a decrease in air density. As air density decreases, the mass of air entering the engine per unit time (mass flow rate) also decreases. This leads to a reduction in the engine's power output since less air is available for combustion with fuel. Conversely, a decrease in ambient temperature increases air density, resulting in a higher mass flow rate and increased power output.

2)Secondly, a change in ambient temperature affects the engine's thermal efficiency. Higher temperatures cause an increase in the temperature difference between the inlet air and the hot gases exiting the combustion chamber, which can lead to decreased thermal efficiency. Lower ambient temperatures, on the other hand, increase thermal efficiency as the temperature difference is greater.

3)Lastly, variations in air density can also affect fuel consumption. When air density is low, the engine requires more fuel to maintain a specific power output, leading to increased fuel consumption. Conversely, higher air density allows the engine to achieve the desired power output with less fuel, resulting in improved fuel efficiency.

4)In conclusion, changes in ambient temperature and air density significantly impact gas turbine engine performance by influencing power output, thermal efficiency, and fuel consumption. Understanding these effects is crucial for optimal engine operation and management.

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Order the parts needed is NOT a duty of the General Service Technician.

General service technicians are knowledgeable workers who are in charge of carrying out repair tasks on cars and other light vehicles. Customers must interact with these specialists in order to discuss pricing estimates for undertaking car maintenance services. They are required to undertake any required repairs or part replacements as well as visual vehicle inspections. They must possess the necessary skills to operate equipment that mounts and balances vehicle tyres. General service personnel should be able to install car batteries and inspect the electrical systems of vehicles.

Therefore, order the parts needed is NOT a duty of the General Service Technician.

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The corresponding stream function is

psi = 1/6 x^6 - 5/4 x^4y^2 + 5/6 x^2

We can make it incompressible by adding a harmonic function to the potential function. A harmonic function satisfies Laplace's equation, which states that the sum of the second partial derivatives with respect to x and y is zero. Adding a harmonic function to the potential function will not change the velocity field, but it will make the divergence zero.

One way to find a harmonic function to add is to look for a function u(x,y) that satisfies Laplace's equation and that makes the mixed partial derivatives of u and phi equal. That is:

d^2u/dx^2 + d^2u/dy^2 = 0

d^2u/dxdy = d^2phi/dxdy

The second equation implies that:

d^2u/dxdy = -d^2u/dydx = 20x^3 - 20xy^2 + 10y^3

Integrating once with respect to x gives:

du/dy = 5x^4y - 5x^2y^2 + 5/2 y^4 + g(y)

where g(y) is a constant of integration that depends only on y. Taking the derivative with respect to x, we get:

d^2u/dxdy = 20x^3y - 10xy^2 + g'(y)l

Adding this to the original potential function, we get:

phi = x^5 − 10x^3y^2 + 5xy^4 − x^2 + y^2 - 5/2 y^5 + x(5/5 x^4y - 5/3 x^2y^2 + 5/4 y^4)

This potential function gives an incompressible flow, with velocity field:

Vx = - dphi/dy = 20x^3y - 5y^3 - 2x + x(5x^3 - 10xy^2 + 5y^4)

Vy = dphi/dx = 5x^4 - 20x^2y + 10xy^3 + 2y + y(5x^3 - 10xy^2 + 5y^4)

The corresponding stream function can be found by solving the equations:

dpsi/dx = Vy

dpsi/dy = -Vx

This gives: psi = 1/6 x^6 - 5/4 x^4y^2 + 5/6 x^2

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