When as-built drawings are received after a project is completed, they should contain which of the following before they are accepted?

A. Updated warranty information for the equipment maintenance program

B. RGI and change-order information for project costs

C. Owner-furnished equipment list for asset management

D. Field notes by the contractor during the construction project

Answer:

zero off your workpiece so you can work from a datum, usually the centre of your work on a lathe. change your tool to a drill and drill a hole to a size smaller than your thread diameter, change out your tool for a threaded tap and away you go.

I'm not sure which part they want but I'd say ensure your tool is set to the right height, you have the tool lines up where you want to cut and that you have calculated the speed you need to cut at safety. Drill a hole before you tap though.

If you have a CNC lathe you just set the programme to do the processes and tool change for you.

The first step to cutting internal threads on an engine lathe is to make calculations so that the thread will have proper dimensions.

The technique of thread cutting on the lathe results in a helical ridge with a consistent section on the workpiece.

To work from a datum, often the center of your work on a lathe, zero off your workpiece. Use a drill to create a hole that is less in diameter than the thread, then switch to a threaded tap and carry on.

One would advise making sure your tool is adjusted to the appropriate height, that it is lined up where you want to cut, and that you have determined the speed you must cut in order to be safe. Before you tap, drill a hole.

With a CNC lathe, you can simply program the machine to perform the processes and tool changes for you.

Therefore,  to do this, make a series of cuts with a threading toolkit that matches the needed thread form.

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False.In standard engineering practice, the number of digits past the decimal point in a final answer depends on the level of precision required for the specific engineering application.

Different engineering disciplines and industries may have different standards and guidelines regarding the number of significant figures or decimal places to be used in final answers.For example, in some engineering fields such as civil engineering or construction, it is common to round final answers to two decimal places.

This level of precision is typically sufficient for practical applications in these fields. However, in other engineering fields such as aerospace or microelectronics, where high precision is often required, final answers may be expressed with more than three decimal places.

Therefore, there is no universal rule that final answers in engineering must always be expressed in three digits past the decimal point. The level of precision should be determined based on the specific requirements and standards of the engineering discipline or industry

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Answer:

That is not a question...

This implies that the ratio of the inner and outer radii (r/rj) = 1. Hence, the use of the arithmetic mean area of the cylindrical wall with r = rj will not introduce more than 1% error in heat transfer calculation. However, the use of the arithmetic mean area always overestimates the heat transfer rate.

Given that the arithmetic mean area of the wall can be used to calculate the rate of heat transfer through a cylindrical wall of small thickness. We are required to determine the ratio of the inner and outer radii (r/rj) of the cylindrical wall for which the use of the arithmetic mean area does not introduce more than 1% error in heat transfer calculation.

The expression for the rate of heat transfer through a cylindrical wall of thickness 'dx' is given by dQ/dt = (2πL/kA) (T₁ − T₂), where 'L' is the length of the cylinder, 'k' is the thermal conductivity of the wall material, and 'A' is the area for heat transfer and is given by the arithmetic mean area of the wall as A = π(r² - rj²).

Let us assume that 'a' is the maximum allowable error, so we can express the acceptable limits of the area as (1 − a) A ≤ Am ≤ (1 + a) A, or A − aA ≤ Am ≤ A + aA, and π(r² − rj²) − aπ(r² − rj²) ≤ Am ≤ π(r² − rj²) + aπ(r² − rj²).

Since (r/rj) > 1, assume (r/rj) = α. The permissible range for the arithmetic mean area can be expressed as (1 − a) π(r² − rj²) ≤ Am ≤ (1 + a) π(r² − rj²), or (1 − a) π(r² − α²r²) ≤ Am ≤ (1 + a) π(r² − α²r²), or (1 − a)(1 − α²) ≤ Am/(π(r² − α²r²)) ≤ (1 + a)(1 − α²).

Since the arithmetic mean area does not introduce more than 1% error in heat transfer calculation, a = 0.01. Thus, (1 − 0.01)(1 − α²) ≤ Am/(π(r² − α²r²)) ≤ (1 + 0.01)(1 − α²). Therefore, (1 − α²) = 0.99(1 − α²), or 0.01(1 − α²) = 0.01, or (1 − α²) = 1. Therefore, α = 0.

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In order to find the temperature required to obtain 0.5% C at a distance of 0.5mm beneath the surface of a 0.2% C steel in 2 hours, given 1.1% C at the surface, we can make use of the Jominy end-quench test.

In the Jominy end-quench test, a standard size and shape of steel sample is heated to a high temperature and then quickly quenched at one end by spraying water on it. The water quenches the steel, causing it to cool rapidly from the high temperature. As the steel cools, it undergoes a transformation from austenite to a mixture of ferrite and pearlite.

As the steel cools, it undergoes a transformation from austenite to ferrite and pearlite. The distance from the quenched end to the point where this transformation is complete is a measure of the steel's hardenability. The hardenability of the steel depends on its composition and the cooling rate. A higher carbon content and a faster cooling rate will result in a higher hardenability.

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The differences which Rutherford would have observed if he used aluminum foil instead of gold foil are: More of the particles would have passed through the aluminum foil unobstructed, and less would have been deflected at a large angle.

Based on historical records and information, Ernest Rutherford performed a set of experiments in 1910 which has helped several scientists and researchers to better understand the structure of atoms by using gold foil.

In his experiment on structure of atoms, Rutherford directed a beam of small, positively charged particles referred to as alpha () particles at a very thin sheet of gold foil.

Basically, the differences which Ernest Rutherford would have observed if he had used aluminum foil instead of gold foil include the following:

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Complete Question:

What differences would Rutherford have observed if he used aluminum foil instead of gold foil?

Less of the particles would have passed through the aluminum foil unobstructed, and more would have been deflected at a large angle.

None of the particles would have passed through the aluminum foil, and all of them would have deflected at a large angle.

More of the particles would have passed through the aluminum foil unobstructed, and less would have been deflected at a large angle.

All of the particles would have passed through the aluminum foil unobstructed, and no particles would have deflected at a large angle.

Answer:

Cool.

Explanation:

The critical buckling load of the W8x24 column section is approximately 88.48 kip. To calculate the critical buckling load of a column section, we can use Euler's buckling formula.

Euler's formula provides an estimate of the critical load at which a column will buckle under an axial compressive load.

Euler's buckling formula is given by:

P_critical = (π^2 * E * I) / (l_effective)^2

Where:

P_critical is the critical buckling load

E is the modulus of elasticity of the material

I is the moment of inertia of the column section

l_effective is the effective length of the column

In this case, we have the following information:

Column section: W8x24

Length (l): 14 ft. (converted to inches: l = 14 * 12 = 168 inches)

Modulus of elasticity (E): 29,000 ksi

Yield stress: 50 ksi

To calculate the moment of inertia (I) for the W8x24 section, we can refer to standard reference tables or use the following data:

I = (b * h^3) / 12

Where:

b is the width of the section

h is the height of the section

For the W8x24 section, the dimensions are as follows:

b = 8 inches

h = 7.99 inches

Plugging in these values, we can calculate the moment of inertia:

I = (8 * 7.99^3) / 12 = 256.67 in^4

Now, we can calculate the effective length (l_effective) of the column. The effective length depends on the support conditions of the column. For simplicity, let's assume it is a pin-ended column, where the effective length is equal to the actual length (l).

l_effective = l = 168 inches

Finally, we can calculate the critical buckling load (P_critical) using Euler's buckling formula:

P_critical = (π^2 * E * I) / (l_effective)^2

          = (π^2 * 29000 * 256.67) / (168^2)

          ≈ 88.48 kip

Therefore, the critical buckling load of the W8x24 column section is approximately 88.48 kip.

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In the grocery shop when the pumkin with a mass of 14.5 lb on the spring scale, the spring elongates with the distance of 3.08 inches. Yes the spring elongates when the pumpkin with a mass of 14.5 lb.

Given that the spring in the scale operates and elongates 1 inch for each 4. 7 Ibf applied.

The gravitational acceleration is given as 32.2 ft/s2

The spring constant c can be used to indicate the force that causes a spring to elongate with times the elongation l,

                       Force = c*l

The pumpkin's weight, which can be calculated as the pumpkin's mass (m) times the acceleration of gravity, is what causes the force to be exerted (g),,

                       Force = Weight = mg = cl

With the correct conversion factor, the pumpkin mass, gravity's acceleration, and the spring constant values, together with the elongation problem, we obtain

               l = mg/c = (14.5)(32.2) / (4.7)  * (1/32.174) = 3.08 in

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Answer:

It is economical

Explanation:

Given data for determining

cost of product = £5 per kg

cost of solvent = £20 per /kg

additional solvent makeup = 10 kg/d

current loss of product = 0.7 kg/h

anticipated recovery of product = 80%

additional service ( utility cost ) = negligible

Determine if it is economical to install the recovery system

cost of product = £5 per kg

cost of solvent = £20 per /kg

additional solvent required = 10 kg / d

cost of additional solvent in a day  = £200 / day  ( £20 per kg * 10 kg / d )

product lost = 0.7 kg/h

product recovered = 0.7 * 80% = 0.56 kg/h

therefore product recovered in a day = 0.56 * 24 = 13.44 kg / day

Next :

calculate income generated from product recovery

13.44 kg/day *  £20 /kg =  £268.88 / day

Determine if it is economical

Income - cost

= £268.88 - £200

=  £68.88  ( hence it is economical )

The tensile stress in the rod is 11,299 psi.

We know that Tensile Stress is given by: Stress = Force/AreaIn this question, we have a rod of length 'l' and a circular cross-section of diameter 'd'. Let's calculate its area.Area of the cross-section of the rod = πd²/4= π(1.5 in)²/4= 1.77 in²Also, we know that Force applied (F) = p (Load applied) = 20,000 lbNow, we can find out the tensile stress using the formula mentioned above.Stress = F/A = 20,000 lb/1.77 in²= 11,299 psi.

We are given the values of load (p) and length (l) of the aluminum rod. We are also given the diameter of the circular cross-section of the rod.Using the formula of area of the cross-section of a circle, we find out the area of the cross-section of the rod. Then we use the formula of stress to find out the tensile stress in the rod.

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The available motor sizes for 2023 Ariya AC synchronous drive motor systems are:

40 kW.

62 kW.

160 kW.

A synchronous motor refers to an alternating current (AC) electric motor in which the rotational speed of the shaft is directly proportional (equal) to the frequency of the supply current, especially at steady state.

In Engineering, the available motor sizes for 2023 Nissan Ariya AC synchronous drive motor systems include the following:

40 kW.

62 kW.

160 kW.

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The displacement of end D with respect to end A is \(-0.488 \times 10^{-3}\;m\)

Given the following data:

To determine the displacement of end D with respect to end A:

First of all, we would do a sectional cut of the circular rods at point A and then determine the sum of forces acting on it:

\(\sum F=0\\\\ 10000+N_1=0\\\\N_1 = -10000\;N\)

Mathematically, the displacement of a circular rod is given by this formula:

\(d = \frac{NL}{AE}\)

Where:

Substituting the given parameters into the formula, we have;

\(d_1 = \frac{-10000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_1 = \frac{-4350}{0.0003142 \times 68.9 \times 10^9}\\\\d_1 =-0.201 \times 10^{-3}\;meter\)

For section cut BC:

\(\sum F=0\\\\ 10000-20000+N_2=0\\\\N_2 = 10000\;N\)

\(radius = outer\;radius^2 - inner\;radius^2\\\\radius = 0.04^2 -0.03^2=0.0007\;m\)

\(d_2 = \frac{10000 \times 0.435}{\frac{3.142 \times 0.0007^2}{4} \times 68.9 \times 10^9} \\\\d_2 = \frac{4350}{0.00054985 \times 68.9 \times 10^9}\\\\d_2 =0.115 \times 10^{-3}\;meter\)

For section cut D:

\(\sum F=0\\\\ -20000-N_3=0\\\\N_3 =- 20000\;N\)

\(d_3 = \frac{-20000 \times 0.435}{\frac{3.142 \times 0.02^2}{4} \times 68.9 \times 10^9} \\\\d_3 = \frac{-8700}{0.0003142 \times 68.9 \times 10^9}\\\\d_3 =-0.402 \times 10^{-3}\;meter\)

For the total displacement:

The total displacement is equal to the displacement of end D with respect to end A.

\(d_T = d_1 +d_2+d_3\\\\d_T = -0.201 \times 10^{-3} + 0.115 \times 10^{-3}-0.402 \times 10^{-3}\\\\d_T = -0.488 \times 10^{-3}\;m\)

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This question is incomplete, the complete question is;

A pool of contaminated water is lined with a 40 cm thick containment barrier. The contaminant in the pit has a concentration of 1.5 mol/L, while the groundwater circulating around the pit flows fast enough that the contaminate concentration remains 0. There is initially no contaminant in the barrier material at the time of installation. The governing second order, partial differential equation for diffusion of the contaminant through the barrier is:

dC/dt = D( d²C / dz²)

where c(z,t) represent the concentration of containment of any depth into the barrier at anytime and D is the diffusion coefficient (a constant) for the containment in the barrier material.

a) write all boundary and initial conditions needed to solve this equation for C(z, t)

b) Find the steady  state solution (infinite time) for C(z)

Answer:

a) At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b) C(z) = z² - 4.15z + 1.5

Explanation:

a)

The boundary and initial conditions are as follows

At t = 0, z= 0, c = 1.5 mol/L

at t =0, z = 0.4m, c = 0 mol/L

b)

The governing second order, partial differential equation for diffusion of the contaminant through the barrier is :

(dC/dt) = D*(d²C/dz²) ..............equ(1)

For steady state, above equation becomes,

(d²C/dz²) =0

Integrating above equation,

(dC/dz) = Z + C1  { where C1 is integration constant) }

again integrating above equation,

C = z² + C1*z + C2    ...................equ(2)

applying boundary condition : at t =0, z= 0, c = 1.5 mol/L, to above equation

 C = z² + C1*z + C2

1.5 = 0 + 0*0 + c2

C2 = 1.5

applying boundary condition : at t =0, z= 0.4m, c = 0 mol/L, to equation (2) ,

0 = 0.4² + C1*0.4 +  1.5

0 = 0.16 + 0.4C1 + 1.5

0.4C1 = - 1.66

C1 = -1.66/0.4

C1 = -4.15

So, the steady state solution for C(z) is:

C(z) = z² - 4.15z + 1.5

Answer:

Strap wrench (I think)

Explanation:

The strap doesnt have any rough edges so it shouldn't mark the workpiece (assuming that is what you meant when you said 'marring'). I would recommend to wait for some other answers though since i haven't studied this yet :)

Solution :

According to the question, the following datasheet contains eight items representing the colored points o the x-y plane.

We have to use this data as training set to run the k-nearest classification algorithm to decide most likely color for a new item with x = 3 and y = 3.

The distance between the points is actual distance on x-y plane, called as Eucledian distance.

We will make a data table, by calculating distance from (3, 3) of each point. By using formula :

Distance, \($d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$\)

x      y    Color       Distance from point (3, 3)

1       1     Red         \($\sqrt{(3-1)^2+(3-1)^2} = 2.82$\)

1      3    Green      \($\sqrt{(3-1)^2+(3-3)^2} = 2$\)              

2     5    Blue         \($\sqrt{(3-2)^2+(3-5)^2} = 2.23$\)      

3     5   Green       \($\sqrt{(3-3)^2+(3-5)^2} = 2$\)

4     1    Blue          \($\sqrt{(3-4)^2+(3-1)^2} = 2.23$\)

4     4   Red          \($\sqrt{(3-4)^2+(3-4)^2} = 1.41$\)

5     3   Blue        \($\sqrt{(3-5)^2+(3-3)^2} = 2$\)

5     4   Green     \($\sqrt{(3-5)^2+(3-4)^2} = 2.23$\)

Now, we will do sorting of colors with distance in ascending order.

We get, [ Red, Green, Green, Blue, Blue, Blue, Green, Red]  

Now if we run the algorithm with k = 1, then we pick only 1 color having the shortest distance that will be assigned to the given point.

Therefore the color is RED.

If we run the algorithm with k = 4, we will pick up \($4 \text{ colors}$\) with shortest distance which are \($\text{red, green, green, blue}$\). Since, now we know, Green has the greatest frequency among 4, hence the answer is Green.

Explanation:

rational

Step-by-step explanation:

The discriminant (d) of a quadratic equation ax^2 + bx + c = 0ax

2

+bx+c=0 is:

\boxed{\mathrm{d =} \ b^2 - 4ac}

d= b

2

−4ac

.

If:

• d > 0, then there are two real solutions

• d = 0, then there is a repeated real solution

• d < 0, then there is no real solution.

In this question, we are given the quadratic equation 3x^2 + 4x - 2 = 03x

2

+4x−2=0 . Therefore, the discriminant of the equation is:

b² - 4ac = (4)² - 4(3)(-2)

= 16 - (-24)rational

Step-by-step explanation:

The discriminant (d) of a quadratic equation ax^2 + bx + c = 0ax

2

+bx+c=0 is:

\boxed{\mathrm{d =} \ b^2 - 4ac}

d= b

2

−4ac

.

If:

• d > 0, then there are two real solutions

• d = 0, then there is a repeated real solution

• d < 0, then there is no real solution.

In this question, we are given the quadratic equation 3x^2 + 4x - 2 = 03x

2

+4x−2=0 . Therefore, the discriminant of the equation is:

b² - 4ac = (4)² - 4(3)(-2)

= 16 - (-24)

= 40

Since the discriminant, 40, is greater than zero, the quadratic equation has 2 rational solutions.

= 40

Since the discriminant, 40, is greater than zero, the quadratic equation has 2 rational solutions.

Polymers is the referred to the class of material which is generally considered to be the weakest at room temperature, offering the lowest elastic moduli and tensile strengths and is denoted as option B.

This refers to a type of compound which are long chained and are formed from smaller repeating chemical units known as monomers through the process which is referred to as polymerization

They are usually in the form of plastics or resins and have features such as lowest elastic moduli and tensile strengths thereby making it the most appropriate choice.

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The options incliude:

A.Metal

B.Polymer

C.Ceramics

D.Composites

Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

Answer:

The answer is B

Explanation: Hope this helps:)

The prevalence of developmental delays and nutritional risk factors can vary depending on various factors, including geographic location, socioeconomic conditions, and individual circumstances.

Nutrition plays a crucial role in the growth and development of infants and young children. Adequate nutrition is essential for their physical and cognitive development, and nutritional deficiencies or imbalances can potentially contribute to developmental delays.

Several nutritional risk factors can impact development in early childhood, including:

1. Malnutrition: Insufficient intake of essential nutrients such as protein, vitamins, and minerals can lead to malnutrition. Malnutrition can impair overall growth and development, including cognitive, motor, and socio-emotional development.

2. Iron Deficiency: Iron is necessary for proper brain development. Iron deficiency in infants and young children can result in cognitive impairments and developmental delays.

3. Vitamin and Mineral Deficiencies: Inadequate intake of key vitamins and minerals, such as vitamin D, vitamin B12, iodine, and zinc, can affect neurodevelopment and increase the risk of developmental delays.

4. Inadequate Caloric Intake: Insufficient calorie intake can lead to stunted growth and can impact cognitive development.

5. Feeding Difficulties: Infants and children with feeding difficulties, such as difficulty with breastfeeding or consuming solid foods, may experience inadequate nutrient intake, which can potentially contribute to developmental delays.

It is important to note that the presence of nutritional risk factors does not necessarily mean that an infant or child will have developmental delays. Developmental delays can have multiple causes and may not be solely attributed to nutritional factors. Genetic factors, environmental factors, and individual variations can also contribute to developmental delays.

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\(\huge{\orange}\fcolorbox{purple}{cyan}{\bf{\underline{\green{\color{pink}Answer}}}} \)

The code to

were written in JavaScript and are found in the attached images

The first code declares a variable called num and gives it an initial value of 0. It then enters a while loop that lauches a message box (using window.alert) to print the message "Keep going" as long as num remains less than 5.

If nothing is done within the loop to increment num towards the value 5, the loop will go on endlessly notifying the user to "Keep going".

So, an increment of 1 was added to the loop body to increment the variable num. This makes sure the loop terminates.

Here, a function was created that receives two arguments (n1  and n2), then returns the product (n1 * n2)

This last code segment creates an array using the new keyword. The new keyword is generally used in constructing objects.

In this case the object constructed is an array having three strings;

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The exit Mach number of the nozzle is 0.404.

Assumptions:

Solution:

From the problem, the exit-to-throat area ratio is 3.5, which means A2/A1 = 3.5. The total pressure in the air storage vessel is 2 atm, and the static pressure at the nozzle exit is 1 atm. The Mach number at the nozzle exit is required to be found.

The area ratio is given as:

Also, the pressure ratio across the shock wave is given as:

The isentropic relations can be used to determine the Mach number corresponding to the pressure ratio of 0.53, which is found to be:

M2 = 0.65

Now, using the mass flow rate equation, we can determine the Mach number at the nozzle exit as:

where rho is the density, A is the area, and V is the velocity of the fluid at the respective points.

We can write this equation in terms of Mach numbers as:

Substituting the given values, we get:

Solving for M1, we get:

Therefore, the exit Mach number of the nozzle is 0.404.

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Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

The reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor

Usually at no load, the Cosine of the phase angle between the stator current(I1) and the stator applied voltage(V1) is the power factor of an induction motor.

In conclusion, the reason for low power factor in no load test in induction motor is high rotor reactance at lower loading since the rotor reactance value gets decreased with increased loading on induction motor

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