Design a 4 bit binary weighted resistor D/A converter for the following specifications Use LM741 op-amp. R = 10 k, Vref=2.5 V. Full scale output 5V. 3. i. Which is the fastest A/D converter? Give reason.

The correct SI Form of the following combinations of Units are given as follows:

A) kN/μs = GN/s

B) Mg/mN; = Gg/N

C) MN/(kg.ms)  = GN/(kg.s)

The International System of Units, abbreviated SI in all languages and often pleonastically as the SI system, is the current version of the metric system and the world's most extensively used measuring system.

The System of Units, often known as the metric system, is frequently shortened as SI, which originates from the original French word, Système international d'unités.

A) kN/μs  = (10) ³N/ (10) ⁻6s

= (10)⁹ N/s

= GN/s

B) Mg/mN = (10⁶)g/10⁻³/N

= Gg/N

C) MN/ (kg.ms) = 10⁶N/kg * (10⁻³)s

= 10⁹ (N/kg · s)

= GN/Kg · S)

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Full Question:

Represent Each Of The Following Combinations Of Units In The Correct Si Form Using An Appropriate Prefix:

A) kN/μs

B) Mg/mN; and

MN/(kg.ms)

Answer:

Alternating

Explanation:

Answer:

When possible, use public transportation, walk, or ride a bike

Explanation:

Answer:

water; water; Oxygen enters the waterways and causes an imbalance of plant life to animal life.

Explanation:

Answer:

  (√6)/3 ≈ 0.8165

Explanation:

The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...

  \(\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}\)

Answer:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time

Explanation:

Fast Fourier ( FFT ) algorithms speed up computation of Fourier coefficients by simply reducing the the computing time of a traditional direct form  Fourier series. it  achieves this by breaking complex DFTS into smaller DFTS to reduce its complexity and in turn reduce its computing time. an example of such FFT is Cooley-Tukey algorithm

Answer:

The correct answer is He forgot to include indirect costs such as advertising, legal fees, and repairs.

Explanation:

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

The relationship between density, pipe average velocity, and cross-sectional area perpendicular to the velocity can be used to calculate mass flow rate.

The quantity of mass flowing across a cross-section per unit of time is known as the mass flow rate. The volume of the fluid moving through a cross-section per unit time is referred to as a volume flow rate. By multiplying the fluid's density, the average pipe velocity, and the cross-sectional area of the pipe normal to the velocity, one can compute the mass flow rate, which is a measurement of the amount of mass flowing through a particular cross-section per unit time. Contrarily, volume flow rate is defined as the amount of fluid moving across a specific cross-section in a unit of time. The mass flow rate into a control volume is proportional to the velocity component parallel to the cross-sectional area for fluid flow in a pipe, when the velocity is constant across the pipe's cross-section.

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The first step to performing hardware maintenance is Turn off the computer and remove its power source.

What is hardware maintenance?
Hardware maintenance is the process of monitoring, inspecting, and repairing hardware components to ensure that they remain in good working order. It involves the regular inspection of computers, servers, printers, and other devices to identify and address any hardware-related issues in a timely manner. This type of maintenance is important for the longevity of hardware components and for the smooth functioning of a network. Regular hardware maintenance can help prevent costly repairs, system crashes, and data loss. It also helps to maximize the performance of the hardware, which can lead to increased productivity and efficiency. Common maintenance tasks include cleaning and dusting computer components, checking for any signs of damage or wear and tear, replacing defective parts, and running diagnostic tests.

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When it comes to the intrinsic silicon becoming an insulator, the temperature range where this occurs is 1420 to 1720 Kelvin (1147 to 1447 °C).

Intrinsic silicon is an extremely pure form of silicon. Because the atoms in the crystalline lattice are of the same type, it is referred to as an intrinsic semiconductor.

This is distinct from extrinsic silicon, which has had impurities implanted to change its electrical characteristics. Because it lacks a majority carrier, it is called an intrinsic semiconductor (hole or electron).

Pure silicon crystal has no impurities, resulting in a uniform lattice, no dopant atoms, and no excess carriers in the crystal's bulk.The electrical conductivity of silicon varies with temperature, with its conductivity decreasing as temperature increases.

In contrast, the intrinsic silicon becomes an insulator at temperatures above 1720 Kelvin (1447°C).

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The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10. She estimates that she is going to earn grade points depending on the number of periods spent on each course.

Tyler Galpin, a friend of Wednesday Addams, arrives in town and calls Wednesday Addams for a date. Assessing her situation, Wednesday decides that all she really needs is a total of 16 grade points gained from any of the courses to graduate.

She wants to allocate her time so that she spends the fewest number of study periods necessary to guarantee her receiving at least 16 grade points. We need to formulate this decision problem as an integer programming model and solve using OPL.

So she gets 4 grade points if she spends one period studying an easy course and 2 grade points if she spends one period studying a difficult course. Number of periods studied Grade points from Easy course Difficult course04                 0             0        14                 4             2        24                 4             2        37                 7             4        48                 8             6        58                 8             9        

 So, the given problem can be formulated as follows: Minimize \(Z = x11 + x12 + x13 + x14\)

Subject to\(4 x11 + 4 x12 + 7 x13 + 8 x14 ≥ 16\)

(Easy courses)\(2 x11 + 2 x12 + 4 x13 + 6 x14 ≥ 16\)

(Difficult courses\()Y ≥ 1x11, x12, x13, x14,\)

Y are integers\(xij ≥ 0 (i = 1,2,3,4; j = 1, 2, …, 10)\)

Below is the OPL code:

int easy\([1..4]=[4,4,7,8];\)

int hard\([1..4]=[2,2,4,6];\)

dvar int \(x[1..4][1..10] in 0..10\);

dvar int y in 1..4;

minimize sum\((i in 1..4, j in 1..10)\)

The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10.

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The dynamic viscosity of the oil in poise is 13.625 pois

The kinematic viscosity of the oil in strokes is 14.34

F viscous is given as n*  ΔFr / Δy

where n = F * Δy / A * ΔVn

We have to define the terms of the formula

Δy = 12.5 x 10⁻³

ΔVr = 2.5m /s

A = 60 x 60 cm² = 0.36m

F = 98.1 n

We have to put the values in the formula

98.1 n *  12.5 x 10⁻³ /  0.36m * 2.5m /s

n = 1.3625  ns / m²

The kinematic viscosity of the oil in strokes if the specific gravity of the oil is 0.95

y = n / e

n =  1.3625  

e = 0.95 x 10³

y = 1.3625  / 0.95 x 10³

= 1.434 x 10⁻³

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Answer:

the equivalent resistance of the circuit decreases and the total current of the circuit increases. 

Explanation:

hope this helps

Answer:

"Physical" would be the right approach.

Explanation:

So the answer here is just the perfect one.

Answer: Option B

Explanation:

In order to enhance net radiation exchange between a high temperature surface and a low temperature surface in an enclosure, one should: paint both surfaces so that they are close to a blackbody surface but don't polish both surfaces so that they are highly reflective

The list of common malfunctions given in the question belongs to the 'impulse tubing' sensor-related instrument.

Impulse is referred to the sensing process of medium pressure, and tubing that directly connects this sensing process tapping to the transmitters is called impulse tubing. Impulse tubing is used to transmit the pressure signal from a process to a transmitter. Impulse tubing can be affected by several common malfunctions such as leakage, plugged or partially plugged tubing, improperly sized tubing, heat tracing too hot or too cold, excessive vibration, and mechanical damage.

"

Correct question is:

The following list of common malfunctions are for which sensor related instrument?

List of common malfunctions is as follows:

"

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Robert Harshaw, Events Coordinator for STEM Mentors, organizes an annual conference showcasing education software for High School STEM teachers. Survey responses in the Excel table named "Survey" are analyzed to generate a report on conference feedback.

Robert Harshaw, an Events Coordinator for STEM Mentors, is responsible for organizing a conference held annually in July.

The conference, sponsored by STEM Mentors, focuses on showcasing educational software for High School STEM teachers and offers informative speakers and workshops on technology in science and math education.

After each conference, Robert compiles the results from a survey to gather feedback and insights for future conferences.

To generate a report on the conference response, you need to analyze the answers provided in the Excel table named "Survey" on the "Survey Results" worksheet.

The responses for the first four questions range from "very satisfied" to "very dissatisfied," represented by the letters a through d, while the survey questions themselves can be found on the "Survey Questions" worksheet.

1. Open the Excel file containing the conference survey results.

2. Navigate to the "Survey Results" worksheet.

3. Locate the table named "Survey" which contains the survey responses.

4. Review the responses for the first four questions, which are represented by the letters a through d.

5. Refer to the "Survey Questions" worksheet to understand the text of each survey question.

6. Analyze the responses to identify patterns, trends, and overall satisfaction levels.

7. Summarize the findings in a clear and concise report.

8. Include relevant statistics, such as the percentage of respondents for each response option.

9. Consider including any notable feedback or suggestions provided by the respondents.

10. Use graphs or charts to visualize the data if necessary.

11. Proofread and edit the report for clarity and accuracy.

12. Submit the report to Robert Harshaw for review and further action.

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Answer:

The dimension of the  square cross section is = 30mm * 30mm

Explanation:

Before proceeding with the calculations convert MPa to Kpsi

Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi

the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut

at 10^6 cycles" for 111.65 Kpsi  = f ( fatigue strength factor) = 0.83

To calculate the endurance limit  use  Se' = 0.5 Sut      since Sut < 1400 MPa

Se'( endurance limit ) = 0.5 * 770 = 385 Mpa

The surface condition modification factor

Ka = 57.7 ( Sut )^-0.718

Ka = 57.7 ( 770 ) ^-0.718

Ka = 0.488

Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one

The endurance limit at the critical location of a machine part can be expressed as :

Se = Ka*Kb*Se'

Se = 0.488 * 0.85 * 385 = 160 MPa

Next:

Calculating the constants to find the number of cycles

α = \(\frac{(fSut)^2}{Se}\)

α =\(\frac{(0.83*770)^2}{160}\)  =  2553 MPa

b = \(-\frac{1}{3} log(\frac{fSut}{Se} )\)

b = \(-\frac{1}{3} log (\frac{0.83*770}{160} )\)  = -0.2005

Next :

calculating the fatigue strength using the relation

Sf = αN^b

N = number of cycles

Sf = 2553 ( 10^4) ^ -0.2005

Sf = 403 MPa

Calculate the maximum moment of the beam

M = 2000 * 0.6 = 1200 N-m

calculating the maximum stress in the beam

∝ = ∝max = \(\frac{Mc}{I}\)

Where c = b/2 and   I = b(b^3) / 12

hence ∝max = \(\frac{6M}{b^3}\)  =  6(1200) / b^3   =  7200/ b^3   Pa

note: b is in (meters)

The expression for the factor of safety is written as

n = \(\frac{Sf}{\alpha max }\)

Sf = 403, n = 1.5 and ∝max = 7200 / b^3

= 1.5 = \(\frac{403(10^6 Pa/Mpa)}{7200 / B^3}\)   making b subject of the formula in other to get the value of b

b = 0.0299 m * 10^3 mm/m

b = 29.9 mm therefore b ≈ 30 mm

since  the size factor  assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2

the dimension = 30 mm by 30 mm

Neither technician is correct it is because vehicles that are capable of using 83% ethanol are called flex-fuel vehicles and such vehicles are able to achieve 100 horsepower.

Flex fuel is an alternative fuel prepared with a combination of ethanol and gasoline. Flux fuel vehicles are vehicles that have internal combustion engines made to operate on more than one type of fuel - gasoline and any blend of gasoline and ethanol up to 83%. Ethanol contains a higher octane number than gasoline so flex-fuel vehicles that use up to 83% ethanol to run have increased power and performance (100 horsepower).

So both technicians a and b are incorrect as per discussion vehicles that are able to operate on 83% ethanol are referred to as flex-fuel vehicles and those vehicles are able to provide 100 horsepower.

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Answer:

A

Explanation:

Eclipsing stars have orbital planes that are parallel to the sky's plane. Since the length of ingress and the length of the eclipse rely on the disparity in the diameters of the stars

The orbital inclination over which eclipses can occur depends on the relative diameters of the stars, but in general, only a tiny percentage of the known binary systems will be observed to experience eclipses. The varying light has two functions. Since the length of ingress and the length of the eclipse rely on the disparity in the diameters of the stars, it enables the estimation of the respective radii of the stars. That is, if Δt1 is the total time between first and last contact, and Δt2 is the period of totality, assuming that the eclipse is annular or total, then.

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The basic requirement of measurements is to have a standard or reference point against which to compare the quantity being measured. This standard or reference point should be well-defined and stable, and the measurement process should be repeatable and consistent. Additionally, it is important to ensure that the measurement equipment is calibrated and in good working condition.

Answer:

The most basic requirement for measurements is the presence of a standard or reference point against which the quantity being measured can be compared. The measurement process should be repeatable and consistent, and the standard or reference point should be well-defined and stable. Furthermore, ensure that the measurement equipment is calibrated and in good working order.

Explanation:

Answer:

mark me brainliest

Explanation:

Superposition theorem is one of those strokes of genius that takes a complex subject and simplifies it in a way that makes perfect sense. A theorem like Millman’s certainly works well, but it is not quite obvious why it works so well. Superposition, on the other hand, is obvious.

Option (d) ClassNotFoundException is not a runtime exception in Java. ClassNotFoundException is a checked exception in Java, which means that it must be either handled or declared in the method signature using the throws keyword.

Runtime exceptions are a type of exception that can occur at any point during program execution and are not checked at compile time. ClassNotFoundException, on the other hand, is a checked exception that occurs when the JVM tries to load a class at runtime, but the class cannot be found or accessed. This can occur when the class file is missing, the class is in a different package, or there is a problem with the classpath. Checked exceptions are typically used for errors that can be anticipated and handled by the program, unlike runtime exceptions which usually indicate a bug or error in the program logic. The other options - a. NullPointerException, b. ArraylnderOutOfBoundException, and c. ArithmeticException - are all runtime exceptions in Java.

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Option a is correct

Channel width will decrease as you move from headwaters to the mouth.

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A career in engineering is interesting and fun. It involves a lifetime of continuous learning to adapt to changes in society and the natural world. It often involves working in multi-disciplinary, multi-cultural, multi-site teams.

Answer:

Force is a kinetic quantity. Force = m x a. Hence, the mass of the body has an effect on force calculation.

This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.

Answer:

the required time to achieve the same concentration at a 4.9 is 83.733 hrs

Explanation:

Given the data in the question;

treatment time t₁ = 11.3 hours

Carbon concentration = 0.444 wt%

thickness at surface x₁ = 1.8 mm = 0.0018 m

thickness at identical steel x₂ = 4.9 mm = 0.0049 m

Now, Using Fick's second law inform of diffusion

\(x^2\) / Dt = constant

where D is constant

then

\(x^2\) / t = constant

\(x^2_1\) / t₁ = \(x^2_2\) / t₂

\(x^2_1\) t₂ = t₁\(x^2_2\)

t₂ = t₁\(x^2_2\) / \(x^2_1\)

t₂ = (\(x^2_2\) / \(x^2_1\))t₁

t₂ = \((\) \(x_2\) / \(x_1\) \()^2\) × t₁

so we substitute

t₂ = \((\) 0.0049  / 0.0018  \()^2\) × 11.3 hrs

t₂ = 7.41 × 11.3 hrs

t₂ = 83.733 hrs

Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs