the plots in problem 1 are to be sketched by hand. you can check your work with matlab but just submit the hand sketched plots. plot them approximate but neat and label the axes and scales. yi 1 s 10 transfer fcn sum 1/s integrator k gain yo yoa. write the function required to determine the root locus (letting k vary) for the system shown and plot the root locus as k varies. i. is there any value of k for which this system is unstable? if so, what is the value? ii. indicate on your plot the gain value, k, where the damping ratio is 0.707. b. solve for the gain, k, required to achieve an open loop crossover of 10 rad/si. what gain is required? ii. what are the phase and gain margins for this design?

Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

Answer:

a. The rate of loss of entropy from the high-temperature reservoir is 1.2 MJ/s x (1 - 0.35) = 0.78 MJ/s, and the rate of gain of entropy by the low-temperature reservoir is 1.2 MJ/s x 0.35 = 0.42 MJ/s.

b. The work done by the engine is 1.2 MJ/s x 0.35 = 0.42 MW.

c. The total entropy gain of the system is 0.42 MJ/s x (1000 K - 308 K) = 124.8 MJ/s.

d.  = 83 MJ/s

Explanation:

Answer: 0

Explanation: i need to add it up to get what u want

Explanation:

I think is SAE standard J2810

Shear and moment diagrams for the loaded beam are shown below: {image}Let the two different forces of the beam be F1 = 16kN,

w1 = 3kN/m, and

w2 = 5kN/m. Therefore, the reactions at point A are

RA = 3.25kN and

MA = 3.25kNm. At the same time, the reactions at point B are

RB = 18.75kN and

MB = 78.25kNm. For x > 0, the value of V is equal to (RA - w1x) and for x > 6, the value of V is equal to (RB - w1x).

Therefore, the shear force diagram for the given loaded beam is shown below: {image}The moment diagram for the given loaded beam is shown below:

Answers to the given questions are as follows:

When x = 10,

V = -4.25 kN, 1kN ≤ Absolute Vmax ≤ 10kN,

Absolute Vmax = 13.5 kN. When

x = 10,

M = -10.5 kNm, 1kNm ≤ Absolute Mmax ≤ 10kNm,

Absolute Mmax = 78.25 kNm.

Thus, the shear and moment diagrams of the loaded beam have been drawn and the given questions have been answered as well.

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The minimum clock period for this CPU should be at least 345 ps.

To determine the latencies and clock period requirements for different instructions in the given exercise, we will consider the provided values for the logic block latencies.

4.7.1:

The latency of an R-type instruction refers to the time required for the instruction to complete its execution. In this case, the R-type instruction consists of register read, ALU operation, and register write. From the given values, we can determine the total latency by summing the latencies of the logic blocks involved:

Latency = Register Read + ALU Adder + Register Write

Latency = 150 ps + 25 ps + 150 ps

Latency = 325 ps

Therefore, the clock period should be at least 325 ps to ensure the correct execution of an R-type instruction.

4.7.2:

The latency of ld (load) instruction represents the time required to complete the load operation, which involves register read, sign extension, ALU operation, and register write. Adding up the latencies of the involved logic blocks:

Latency = Register Read + Sign Extend + ALU Adder + Register Write

Latency = 150 ps + 20 ps + 25 ps + 150 ps

Latency = 345 ps

Thus, the clock period should be at least 345 ps for the correct execution of the ld instruction.

4.7.3:

Similar to the ld instruction, the sd (store) instruction involves register read, sign extension, ALU operation, and register write. Adding up the latencies:

Latency = Register Read + Sign Extend + ALU Adder + Register Write

Latency = 150 ps + 20 ps + 25 ps + 150 ps

Latency = 345 ps

The clock period should be at least 345 ps for the correct execution of the sd instruction.

4.7.4:

The latency of beq (branch equal) instruction involves register read, ALU operation, and control logic. Summing up the latencies:

Latency = Register Read + ALU Adder + Control

Latency = 150 ps + 25 ps + 50 ps

Latency = 225 ps

A clock period of at least 225 ps is required for the correct execution of the beq instruction.

4.7.5:

The I-type instruction refers to the load and store instructions (ld and sd). Since we have already determined their latencies in previous questions:

I-type Instruction Latency = Latency of ld or sd = 345 ps

4.7.6:

The minimum clock period for this CPU would be equal to the highest latency among all the instructions. From the previous calculations, the highest latency is 345 ps.

Therefore, the minimum clock period for this CPU should be at least 345 ps.

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The battery is supplying electrical energy to the circuit at a constant rate, which is determined by its voltage and internal resistance.

The circuit is completed, the battery starts to release its stored energy in the form of electrical current, which flows through the wires and components of the circuit. The rate at which this energy is released depends on the resistance of the circuit, which determines how much current can flow at a given voltage.

If the resistance of the circuit is low, then the battery will supply a high current and therefore a high rate of energy output. Conversely, if the resistance is high, then the battery will supply a low current and a low rate of energy output.

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Answer:

Amplifier gain is equal to 10.67.

Explanation:

Voltage gain of an amplifier can be expressed as this equation:

A_v = (V_output) / (V_input)

Where A_v is the gain, V_output is the output voltage, and V_input is the input voltage.

For this we will solve:

A_v = 128 mV / 12 mV

A_v = 10.67

So the amplifier's gain is 10.67.

Cheers.

a. The equal-frequency partitioning for the given sales price records would be as follows: Bin 1: 5, 10, 11, 13  Bin 2: 15, 35, 50, 55  Bin 3: 72, 92, 204, 215

(a) Equal-Frequency (Equidepth) Partitioning:

Equal-frequency partitioning divides the data into bins of equal frequency. In this case, we have 12 records, so we need to partition them into three bins.

Now, we can assign the records to the bins based on their order. Starting from the lowest value, we assign four records to each bin until all records are assigned. If there are any remaining records, we distribute them evenly across the bins.

Using this method, the equal-frequency partitioning for the given sales price records would be as follows:

Bin 1: 5, 10, 11, 13

Bin 2: 15, 35, 50, 55

Bin 3: 72, 92, 204, 215

(b) Equal-Width Partitioning:

Equal-width partitioning divides the data into bins of equal width or range. In this case, we need to determine the range of the data and then divide it into three equal-width bins.

The range of the data is the difference between the maximum and minimum values, which is 215 - 5 = 210.

Each bin will have a width of 210 / 3 = 70. Starting from the minimum value, we assign the records to the bins based on their value falling within the corresponding width range.

Using this method, the equal-width partitioning for the given sales price records would be as follows:

Bin 1: 5, 10, 11, 13, 15, 35

Bin 2: 50, 55, 72, 92

Bin 3: 204, 215

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Reynolds number: This is a dimensionless parameter used to help in predicting flow patterns in different fluid flow systems.

It is important in fluid mechanics and is given by the formula as shown below:

Re= ρVD/μ

Where

Re is the Reynolds number

V is the velocity of the fluid

D is the diameter of the fluidρ is the density of the fluid

μ is the dynamic viscosity of the fluid

Calculation of volumetric flow rate: Volumetric flow rate can be defined as the volume of fluid that passes through a given cross-sectional area per unit of time. It is given by the formula;

Qv= A×V

Where by;

Qv is the volumetric flow rate

V is the velocity of the fluid

A is the cross-sectional area of the fluid

Qv for the trachea is given by;

Qv= π(\(0.009^2\))(80/100)

Qv= 0.0202 \(m^3\)/sQv

for each small bronchus is given by;

Qv= π(0\(.00065^2\))(15/100)

Qv= 8.3634 x \(10^{-7} m^3\)/s

Calculation of mass flow rate:Mass flow rate is the rate at which mass passes through a given cross-sectional area per unit of time. It is given by the formula as shown below;

Qm= ρ×A×V

Whereby;

Qm is the mass flow rate

A is the cross-sectional area of the fluid

V is the velocity of the fluidρ is the density of the fluid

Qm for the trachea is given by;

Qm= 1.2041×0.0202

Qm= 0.0244 kg/s

for each small bronchus is given by;

Qm= 1.2041×8.3634×\(10^{-7\)

Qm= 1.0066 x \(10^{-6\) kg/s

Calculation of molar flow rate:

Molar flow rate is defined as the rate at which the number of molecules of a substance passes through a given cross-sectional area per unit time. It is given by the formula as shown below;

Q= C×Qv

Whereby;

Q is the molar flow rate

C is the concentration of the substance

Qv is the volumetric flow rate

Q for the trachea is given by;

Q= (1/0.029)×0.0202

Q= 0.6979 mol/s

Q for each small bronchus is given by;

Q= (1/0.029)×8.3634×\(10^{-7\)

Q= 2.8756 x \(10^{-5\) mol/s

Calculation of Reynolds number: Reynolds number for the trachea is given by;

Re= (1.2041×0.0202×18/1000)/ (1.845×\(10^{-5\))

Re= 2194.167

Reynolds number for each small bronchus is given by;

Re= (1.2041×8.3634×\(10^{-7\)×1.3/1000)/ (1.845×\(10^{-5\))

Re= 7.041

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You typically increase the decimal places from 1 to 2 for a A. Currency field.

When working with fields in a database, it is important to understand the various data types and how they should be formatted. One aspect of formatting involves the number of decimal places displayed for numeric fields. In this case, the question is asking about which field typically requires an increase in decimal places from 1 to 2. The options given are Currency, AutoNumber, Long Text, and Money. The field that typically requires an increase in decimal places from 1 to 2 is A. Currency. This is because currency values often require greater precision in order to accurately reflect the monetary amount being represented. To summarize, when working with a Currency field in a database, it is common to increase the decimal places from 1 to 2 in order to display a more precise monetary value.

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Your Answer would be A I believe.

An access control system controls access to area restricted area or facility using physical and computer-related devices, equipment along with authentication and authorization mechanisms.

Answer:

yrt a

Explanation:

A contractor has a job that should be completed in 100 days. At present, he has 80 men on the job and it is estimated that they will finish the work in 130 days. Of the 80 men, 50 are each paid ₱120.00 a day, 25 at ₱180.00 a day, and 5 at ₱250.00 a day. For each day beyond the original 100 days, a contractor has to pay ₱500.00.

liquidated damages.(a) How many more men should the contractor add so that he would complete the work on time?In the first case, we see that the contractor already has 80 men and they are working for 130 days to complete the job. So, we can use the following formula to determine the additional number of workers required to finish the work in 100 days.

b) If of the additional men, 2 are paid ₱180.00 a day, and the rest at ₱120.00 a day, would the contractor save money by employing more men and not paying the fine Let’s assume that the contractor adds 440 workers, of which 2 are paid ₱180.00 a day and the rest are paid ₱120.00 a day.

The total cost of the new workers is, therefore, ₱9600.00 + ₱4500.00 + ₱49800.00 = ₱63,900.00.The cost of liquidated damages would be calculated as follows:  $$LD = (130-100) \cdot 500 = ₱15,000.00$$.

Therefore, the contractor would save money if he employs more men and not pays the fine. The contractor’s savings would be:$$Savings = LD - Additional cost$$$$= 15000.00 - 63900.00 $$$$= -48900.00$$

Thus, we can see that the contractor would save ₱48,900.00 by employing more men and not paying the fine.

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Answer:

engineers are people who fix or invent items that will support people’s lives. For fun, engineers may construct unique items that is hard to invent.

Explanation:

If you need more info, pls comment.

Answer:

An engineer uses science, technology and math to solve problems. It designes machinery, build skyscrapers and oversee public works.

Types; 2 Auto body repair; 3 Auto glass repair; 4 See also; 5 References ... Independent automobile repair shops in the US may also achieve certification through manufacturer sponsored ...

Answer:

The Scrum Master is a key role in the Scrum framework for agile project management. The Scrum Master's primary role is to ensure that the Scrum team follows the Scrum process and values, and to remove any impediments that may prevent the team from achieving their goals.

Some of the specific responsibilities of a Scrum Master include:

Facilitating the Scrum process: The Scrum Master is responsible for facilitating all Scrum ceremonies, including daily stand-ups, sprint planning, sprint review, and sprint retrospective meetings. They ensure that all participants understand the purpose and goals of each ceremony, and that the ceremonies are conducted effectively.

Protecting the team: The Scrum Master acts as a shield for the Scrum team, protecting them from external distractions and interruptions that could prevent them from achieving their sprint goals. They also ensure that the team has a safe and productive working environment.

Removing impediments: The Scrum Master identifies and removes any obstacles that may impede the team's progress. They work closely with team members to understand the root cause of any impediments and help the team find creative solutions to overcome them.

Coaching the team: The Scrum Master is responsible for coaching the team on the Scrum process and values, as well as agile principles and practices. They help the team continuously improve their processes and practices to increase efficiency and productivity.

Ensuring transparency: The Scrum Master ensures that there is transparency and visibility into the team's progress, by maintaining the Scrum board, updating burndown charts, and communicating progress to stakeholders.

Overall, the Scrum Master is a servant-leader who supports the Scrum team in achieving their goals and helps them continuously improve their processes and practices.

Hope this helps!

It is true that program-data dependence refers to the coupling of data stored in files and software programs that use this data such that changes in programs require changes to the data.

Program-data dependence refers to the coupling of data stored in files and the specific programs required to update and maintain those files such that changes in programs require changes to the data. Every traditional computer program has to describe the location and nature of the data with which it works.

Data Independence is defined as a property of DBMS that helps you to change the Database schema at one level of a database system without requiring to change the schema at the next higher level. Data independence helps you to keep data separated from all programs that make use of it.

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This scenario can be described as an experimental study. An experiment involves manipulating variables to observe the effects on an outcome.

In this case, the highway department manipulates the type of concrete used (Type A and Type B) and observes the time taken for cracks to appear in each section. The explanation for this conclusion is that by actively controlling the type of concrete used, the department can compare the performance and durability of both types, allowing them to make informed decisions about which concrete is more suitable for future use. In an observational study, no variables would be intentionally manipulated; rather, the researcher would simply observe and gather data without any intervention.

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Heat energy is the known to be a product of the movement of tiny particles called atoms, molecules or ions. The true statement is by Technician A.

Internal combustion engine is dependent on the heat of combustion so as to make torque to move the vehicle and power the system.

A lot of heat made during combustion is not often used productively and therefore need to be removed to avoid overheating of the engine.

The heat energy that is not used for is wasted in three ways: They are:

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A PFAS anchorage point for a 6-foot lanyard must be able to hold a minimum static load of 5,000 pounds.

PFAS stands for Personal Fall Arrest System, which is a type of fall protection equipment that is designed to protect workers from falls.
The 5,000-pound minimum static load requirement is based on the assumption that the PFAS will experience a shock load of 2,500 pounds in the event of a fall.

This shock load is due to the force of the fall and the weight of the worker, and it puts a significant amount of stress on the anchorage point. The 5,000-pound minimum static load ensures that the anchorage point can withstand this shock load without failing.
It's important to note that the 5,000-pound minimum static load requirement is a general guideline, and the actual load capacity of the anchorage point will depend on a variety of factors, including the type of anchorage point, the material it's made from, and the conditions of use.

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A 2 stage piston compressor has an intercooler where compressed air is cooled while passing to the second stage.

Answer:

0.314 kg

Explanation:

\(\frac{8}{1000}\) x 4 x 9.81

Answer:

.3139  joules

Explanation:

Weight in the metric system is   ma

 you have 4 fins that have MASS of 8 grams each

    or   32 grams   which is .032 kg

WEIGHT  would be   .032 kg * 9.81 m/s^2 = .31392 j

Answer:

Umm.... because the spit that comes up of your mouth when you talk mixes with the air

Explanation:

Answer:

Option B  Lower than

Explanation:

Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below.  On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure.  This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure

Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

The most frequent task is generally oil changes.

Answer:

20

Explanation:

Technician A says that primary vibration is created by slight differences in the inertia of the pistons between top dead center and bottom dead center. Technician B says that secondary vibration is a strong low-frequency vibration caused by the movement of the piston traveling up and down the cylinder. Who is correct? O A. Neither Technician A nor B OB. Technician B O C. Both Technicians A and B D. Technician A​