An engineer at Pied Piper has suggested that the current version of software that is being used cannot grow any further without being moved to a cloud environment. In order to expand, it will be necessary to adopt a methodology and set of tools that will allow for more efficient processes and deployments of the environments in which the software will run. Which of the following should Pied Piper start implementing

The unit shear along the south shear walls is 450 lb/ft.

The unit shear along the south shear walls of the warehouse one-story building is 450 lb/ft in the east-west direction. Unit shear refers to the shear force per unit length, in this case, per foot, acting on the shear walls. Shear walls play a crucial role in resisting lateral forces and maintaining the structural stability of a building.

In this context, the calculated unit shear of 450 lb/ft indicates the maximum shear force that each foot of the south shear walls can withstand without experiencing excessive deformation or failure. It is important to ensure that the shear walls are designed and constructed to withstand these shear forces adequately.

The adequate anchorage of the roof diaphragm to the shear wall is another important aspect mentioned in the question. Proper anchoring ensures a strong connection between the roof diaphragm and the shear wall, allowing the transfer of shear forces effectively and enhancing the overall structural performance.

By calculating the unit shear, engineers and designers can assess the capacity of the south shear walls to resist the applied forces and ensure the stability and safety of the warehouse building. This information is crucial for designing and evaluating the structural integrity of the building to meet the required safety standards and codes.

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Since the armature is wave wound, the magnetic flux per pole is 0.0274 Weber.

Given the following data:

To find the magnetic flux per pole:

Mathematically, the emf generated by a DC generator is given by the formula;

\(E = \frac{\theta ZN}{60}\) × \(\frac{P}{A}\)

Where:

First of all, we would determine the total number of armature conductors:

\(Z = 144\) × \(2\) × \(3\)

Z = 864

Substituting the given parameters into the formula, we have;

\(520 = \frac{\theta (864)(660)}{60}\) × \(\frac{4}{2}\)

\(520 = \theta (864)(11)\) × \(2\)

\(520 = 19008 \theta \\\\\Theta = \frac{520}{19008}\)

Magnetic flux = 0.0274 Weber.

Therefore, the magnetic flux per pole is 0.0274 Weber.

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Answer:

It does not take into consideration what the responsible party knew about the law or regulation they violated. Environmental criminal liability is triggered through some level of intent.

In regards to whether an employee can be held legally liable for environmental violations, this statement is False.

This means that if an employee notices that a company is responsible for environmental violations, and does nothing about it, they could be held legally responsible for the violation as well.

In conclusion, this is true.

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Answer:

I think its A

Explanation:

The normal reaction at A is equal to 190 lbs.

The x component of the force that the truck exerts on the pipe at B is 1140 lb-ft/s^2, and the y component is zero.

1) The normal reaction at point A, where the pipe is in contact with the truck, can be determined by analyzing the forces acting on the pipe in the vertical direction. Since the pipe is not accelerating vertically, the sum of the vertical forces must be equal to zero. The weight of the pipe (190 lbs) acts downwards, so the normal reaction at A acts upwards to balance out the weight. Therefore, the normal reaction at A is equal to 190 lbs.

2) To determine the x and y components of the force that the truck exerts on the pipe at point B, we need to break the force vector into its horizontal and vertical components. Since the pipe has a negligible diameter, we can assume that the force is acting at point B, where the pipe is in contact with the truck.

The force exerted by the truck on the pipe is equal to the mass of the pipe times its acceleration, which is:

force = mass x acceleration
force = 190 lbs x 6 ft/s^2
force = 1140 lb-ft/s^2

To find the horizontal and vertical components of this force, we can use trigonometry. Let θ be the angle between the force vector and the horizontal axis. Then:

horizontal component = force x cos(θ)
vertical component = force x sin(θ)

Since the force vector is in the direction of motion, which is horizontal, θ = 0. Therefore:

horizontal component = force x cos(0)
horizontal component = force

vertical component = force x sin(0)
vertical component = 0

So the x component of the force is 1140 lb-ft/s^2, and the y component is zero.

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Answer:

Pressure = 51.8 x 10³ Pa = 51.8 KPa

Explanation:

First, we calculate the total force exerted on the piston:

Total Force = Force Exerted by Spring + Weight of Piston

Total Force = 150 N + (mass)(g)

Total Force = 150 N + (3.2 kg)(9.8 m/s²)

Total Force = 150 N + 31.36 N

Total Force = 181.36 N

Now, for cross-sectional area of piston:

Area = (35 cm²)(1 x 10⁻⁴ m²/1 cm²)

Area = 0.0035 m²

Now, for pressure inside the cylinder:

Pressure = Total Force/Area

Pressure = 181.36 N/0.0035 m²

Pressure = 51.8 x 10³ Pa = 51.8 KPa

Highway safety is a multifaceted approach that involves a combination of engineering, enforcement, and education strategies. The design features contribute to the overall safety of expressways, but driver attentiveness, adherence to traffic laws, and responsible driving behavior are equally crucial for safe travel on these roadways.

Highway safety design features on expressways that help make them safer roadways to travel include:

1. Controlled Access: Expressways are designed with limited access points, such as ramps and interchanges, to regulate the flow of vehicles entering and exiting the roadway. This helps reduce the potential for collisions and congestion.

2. Median Barriers: Medians with physical barriers, such as concrete walls or cable barriers, are installed to separate opposing traffic. These barriers help prevent head-on collisions and minimize the severity of accidents.

3. Clear Zone Areas: Clear zones are areas adjacent to the roadway that are kept free from fixed objects, such as trees or poles. This provides a buffer zone and allows drivers to recover safely if they veer off the road.

4. High Visibility Markings: Expressways often have clear and highly visible lane markings, signage, and pavement markings. These aids help drivers navigate the roadway, change lanes safely, and understand upcoming exits and interchanges.

5. Lighting: Adequate lighting is installed along expressways to enhance visibility during nighttime conditions. Well-lit roadways improve driver awareness and reduce the risk of accidents.

6. Guardrails: Guardrails are installed along certain sections of the expressway, such as curves or areas with significant grade changes, to prevent vehicles from leaving the road and reduce the impact severity of collisions.

7. Emergency Pull-off Areas: Expressways may have designated emergency pull-off areas where drivers can safely stop in case of emergencies, breakdowns, or for accessing emergency services.

8. Traffic Control Devices: Intersections, interchanges, and merging areas on expressways are equipped with traffic control devices such as signals, yield signs, and merge signs to facilitate the safe movement of vehicles.

9. Advanced Warning Systems: Expressways may incorporate advanced warning systems, such as electronic message signs, variable speed limit signs, and intelligent transportation systems, to provide real-time information to drivers about hazards, congestion, or incidents ahead.

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The region in which the flow is both hydrodynamically and thermally developed is called the fully developed region. This region is characterized by a steady flow with no change in velocity profile along the flow direction, an irrotational flow with no swirls or vortices, and a fully developed boundary layer with constant temperature and velocity profiles.

In fluid mechanics, the fully developed flow region is the section of a pipe or channel where the flow is fully developed, meaning that the velocity profile and temperature distribution are constant across the pipe or channel cross-section. At the entrance to a pipe or channel, the flow may not be fully developed, and the velocity profile and temperature distribution may change over time until the fully developed region is reached. In this region, the flow is considered to be fully developed hydrodynamically and thermally.

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The correct option about word embeddings presented in the lecture is C) One way to learn word embeddings is by maximizing cosine similarity between words with related meaning.

Word embeddings are a way to represent words as vectors of real numbers in a high-dimensional space. The goal of word embeddings is to capture the semantic meaning of words, allowing for better natural language processing tasks such as language translation and sentiment analysis.

To learn word embeddings, one common approach is through the use of neural networks. Specifically, a neural network is trained to predict the context of a given word, such as the surrounding words in a sentence. The weights of the neural network are then used as the word embeddings.

One way to maximize the effectiveness of word embeddings is by optimizing the cosine similarity between words with related meaning. Cosine similarity measures the cosine of the angle between two vectors, which can be used to determine how similar two words are in meaning. By optimizing cosine similarity, word embeddings can better capture the relationships between words, such as synonyms and antonyms.

It is important to note that word embeddings are typically learned automatically, without the need for manual encoding by a natural language domain expert. This is because the neural network is able to learn the relationships between words based on the context in which they appear, without the need for explicit rules or definitions.

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To determine the amount of the mixture that can be made, we need to calculate the quantity of each ingredient based on the given ratio. Given that the ratio of cement, gravel, and sand is 3:5:7 respectively, we can calculate the quantities of gravel and sand by multiplying their respective ratios with the available quantity of cement.

For gravel:
Ratio of cement to gravel = 3:5
Quantity of gravel \(= (5 tons of cement) * (5/3) = 25/3 tons of gravel\)

For sand:
Ratio of cement to sand = 3:7
Quantity of sand = (5 tons of cement) * (7/3) = 35/3 tons of sand

Finally, to calculate the total amount of the mixture, we add the quantities of cement, gravel, and sand:
Total quantity of mixture = 5 tons of cement + 25/3 tons of gravel + 35/3 tons of sand

Converting the fractions to a common denominator, we have:
Total quantity of mixture = (15/3 tons of cement) + (25/3 tons of gravel) + (35/3 tons of sand)
Total quantity of mixture = 75/3 + 25/3 + 35/3 = 135/3 = 45 tons

Therefore, 45 tons of the mixture can be made using the given quantities of cement, gravel, and sand.

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Answer:

busway

Explanation:

Explanation:

I won't answer each of these, but will give you an explaination of how to solve for each.

You'll need to use Ohm's Law and Kirchhoff's Voltage Law.

Remember Ohm's Law as \(V = IR\).

Kirchoff's Voltage Law says that the sum of voltages for a given circuit "loop" must equal zero. In the circuit shown, this means that the voltage provided by the battery (E_T) equals the voltage drop across each of the three resisters in the loop.

\(E_T = E_1 + E_2 + E_3\)

A couple of other helpful notes:

These three resistors are in series which means that the current flowing through them is equal.

So it is easy to see that... \(V = IR \rightarrow E_1 = 4*2 = 8 V\)

Solve for the voltage across E_2 and E_3. The sum of the three voltages equals the voltage of the battery (E_T).

Technician a says kinetic friction changes kinetic energy into thermal energy. technician b says static friction holds the car in place when it is stopped. Both A and B are correct.

In static friction, there is found to be some frictional force that tends to resists force that is known to be applied to an object, and the object will be at the same sport at rest until the force of static friction is known to be overcome.

In kinetic friction, the frictional force is one that tend to resists the movement of an object.

Therefore, Technician a says kinetic friction changes kinetic energy into thermal energy and technician b says static friction holds the car in place when it is stopped. Both A and B are correct.

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The given statement "one of the keys to value engineering workshops is that they utilize time to decide solely to focus on improvement" is True.

What is Value Engineering? Value Engineering (VE) is a methodology for improving the worth of products, services, and procedures. It entails an organized and creative analysis that investigates the requirement for a product or service to achieve its purpose at the lowest cost without jeopardizing quality, dependability, or safety.

Value engineering workshops: Value engineering workshops or VE workshops are generally regarded as the greatest approach to generating the most money-saving ideas. These workshops may also help in the creation of excellent product designs that are value-engineered and offer a competitive edge to an organization. It is critical to utilize time in a VE workshop to focus solely on improvement. Engineers and other members of the team collaborate to establish new concepts, ideas, and designs in the workshop. The purpose is to make the processes and designs more efficient, resulting in cost savings and improved production capabilities.

So, the given statement is True.

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Answer:

I dont think you can blur text in google docs

The specific gravity of each of the tabularized items are given as attached.

To determine the specific gravity, we need to compare the density of each material to the density of water (which is 62.4 lb/ft^3). The specific gravity is defined as the ratio of the density of the material to the density of water.

Therefore, the specific gravity of Gold is 19.36, the specific gravity of Platinum is 21.47, the specific gravity of Silver is 10.48, the specific gravity of Sand is 1.52, the specific gravity of Freshly fallen snow is 0.50, the specific gravity of Tar is 1.19, the specific gravity of Hard Rubber is 1.00, and the specific gravity of Water is 1.00.

Note that Specific gravity is the ratio of the density of a substance to the density of a reference substance, typically water at a specified temperature and pressure. It is a dimensionless quantity that helps determine the buoyancy of a substance.

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Answer:

STANDARD DAVIATION OR SD

Explanation:

BECAUSE it is commonly in measuring of dispersion and it also the most roubst measure of variability

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.

Since the Earth contains a constant temperature range heat exchangers can be built underground without other added expenses yes they are expensive to put underground but the pay off is greater long term wise since the heat rises it can be said you could put a heatpump in a small area and it can over time increase the relative heat in that area of effect.

hope this helps look into, heat displacement thru underground heating to get a clearer picture

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Answer:

HVACR Industry Trade Groups

Explanation:

The value of the site is $10,472,000.

The total construction costs are:

$6,000/m2 * 10,000m2 + $400,000 = $64,000,000

The professional fees are:

$64,000,000 * 0.13 = $8,320,000

The total development costs are:

$64,000,000 + $8,320,000 = $72,320,000

The developer's return for risk and profit is:

$72,320,000 * 0.20 = $14,464,000

The total cost of the development is:

$72,320,000 + $14,464,000 = $86,784,000

The expected rent is:

$3,000/m2 * 8,000m2 * 12 months = $28,800,000

The initial yield is:

$28,800,000 * 0.07 = $2,016,000

The value of the site is:

$86,784,000 - $2,016,000 = $10,472,000

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Answer:

it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength.

Step-by-step:

Fluorescence spectroscopy is more sensitive than UV-Vis absorption spectroscopy because it involves the detection of light emitted by a sample after it has absorbed light of a specific wavelength. This emission of light is called fluorescence.

In UV-Vis absorption spectroscopy, the amount of light absorbed by a sample is measured as a function of wavelength. The amount of light absorbed is proportional to the concentration of the absorbing species in the sample, but the sensitivity of this technique is limited by the intensity of the incident light and the path length of the sample.

On the other hand, fluorescence spectroscopy measures the emission of light that occurs when the excited molecules return to their ground state. This emitted light is typically at a longer wavelength than the absorbed light, and it is much weaker than the incident light. However, the sensitivity of fluorescence spectroscopy is enhanced by the fact that the emitted light is measured at right angles to the excitation light, which reduces background noise from scattered light and improves the signal-to-noise ratio.

Additionally, fluorescence spectroscopy can be more selective than UV-Vis absorption spectroscopy because it can detect specific molecular species based on their unique fluorescence spectra. This selectivity is due to the fact that the fluorescence emission spectra of different molecules can be quite distinct, even for molecules with similar UV-Vis absorption spectra.

Overall, the increased sensitivity and selectivity of fluorescence spectroscopy make it a powerful technique for the detection and quantification of trace amounts of fluorescent molecules in complex samples.

Hope this helps!

Answer:

Fluorescence and UV-Vis absorption spectroscopy are two commonly used analytical techniques in the field of chemistry. Both techniques rely on the interaction of light with molecules to provide information about their electronic structure and chemical properties. However, fluorescence is generally considered to be more sensitive than UV-Vis absorption spectroscopy for several reasons.

Firstly, fluorescence is a more selective technique than UV-Vis absorption spectroscopy. When a molecule absorbs light, it undergoes a transition from its ground state to an excited state. This transition can occur via a number of different pathways, depending on the energy of the absorbed light and the electronic structure of the molecule. In contrast, fluorescence occurs when a molecule emits light after being excited by light of a specific wavelength. This means that fluorescence only occurs when certain conditions are met, such as the presence of specific functional groups or the correct excitation wavelength. As a result, fluorescence is generally more selective than UV-Vis absorption spectroscopy, which can detect any absorbing species in a sample.

Secondly, fluorescence is generally more sensitive than UV-Vis absorption spectroscopy because it produces a larger signal-to-noise ratio. In UV-Vis absorption spectroscopy, the signal is proportional to the concentration of absorbing species in the sample. However, the signal is also affected by other factors such as path length, instrument sensitivity, and background noise. In contrast, fluorescence produces a much stronger signal because it involves emission of light at a different wavelength than that used for excitation. This means that background noise and other interfering factors are less likely to affect the signal-to-noise ratio in fluorescence measurements.

Finally, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it can detect lower concentrations of analyte in a sample. This is because fluorescence is an amplification process – each absorbed photon can result in multiple emitted photons if the molecule undergoes multiple cycles of excitation and emission. This amplification effect means that even low concentrations of fluorescent analyte can produce a measurable signal, whereas in UV-Vis absorption spectroscopy, the signal becomes weaker as the concentration of analyte decreases.

In summary, fluorescence is more sensitive than UV-Vis absorption spectroscopy because it is a more selective technique, produces a larger signal-to-noise ratio, and can detect lower concentrations of analyte in a sample.

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We have a total of 6 nodes with 5 components. We must determine the temperature at each of these six nodes. Mesh refining is the process of increasing the number of elements.

What is temperature?

Temperature is a scientific term that describes how hot or cold something is.There are three types of temperature scales: those described in terms of the average translational kinetic energy per freely moving microscopic particle in a body, such as an atom, molecule, or electron, such as the SI scale; those which rely solely on purely macroscopic properties as well as thermodynamic principles, including such Kelvin's original definition; and those that are not defined by theoretical principles, and yet are defined by convenient empirical properties of particulate matter.

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Answer:

Explanation:

The 7 Habits of Highly Effective People, is a book written and first published in 1989. It is a business and self-help book that was written by Stephen Covey. The seven habits include

Being proactive

Starting anything with the end in mind

First things first

Always thinking towards a win-win situation

Seeking initially to understand, then going on to want to be understood

Synergize, and lastly

Growing

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

a) Tresca is a criterion that depends on the difference between the maximum and minimum principal stresses. The minimum wall thickness.  t = PD/2ty = 5,000(12)/2(36,000) = 0.0083 inches≈ 1/12 inches

According to Tresca, the yielding will begin when the maximum shear stress is equal to the shear strength. The formula for Tresca is given as;  

2ty = σ₁ − σ₂

Where σ₁ and σ₂ are the principal stresses;

ty is the yield strength of the material.

For thin-walled vessels, the wall thickness can be determined using the equation

t = PD/2ty

From the given information, the diameter (D) is equal to 12 inches,

the yield strength (ty) is equal to 36,000 psi, and the pressure (P) is equal to 5,000 psi,

Substitute these values in the above equation to obtain the minimum wall thickness.  

t = PD/2ty = 5,000(12)/2(36,000) = 0.0083 inches≈ 1/12 inches

b) Determination of the minimum wall thickness according to von Mises:

The von Mises yield criterion is a measure of the amount of distortion energy in the material. According to von Mises, the material will start yielding when the distortion energy exceeds the distortion energy at yield. The formula for von Mises is given as;

σv² = [(σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²] / 2

Where σ₁, σ₂, and σ₃ are the principal stresses.

The wall thickness for a thin-walled vessel can be calculated using the equation;

t = PD/4σv

Substitute the given values in the above equation to determine the minimum wall thickness.

t = PD/4σv = 5,000(12)/4(40,500)

= 0.0086 inches≈ 1/12 inches

c) Comparison of the answers from (a) and (b) and select which one to use as a designer:

As the results from Tresca and von Mises are almost the same and this is because both formulas result in the same yield criterion, therefore, as a designer we can use either of the two to determine the minimum thickness of the wall.

d) Discuss how the wall thickness would be altered to account for a safety factor:

A safety factor  is a ratio of the minimum strength needed to support the loading, divided by the actual strength of the object. The wall thickness is usually increased to account for the safety factor. The wall thickness would be multiplied by a factor of 1.5 to 2.0 to account for safety factors.

For example, if the minimum wall thickness is 0.0083 inches, then the new thickness for the safety factor can be calculated as follows;

t = 2t = 2(0.0083)

= 0.0166 inches≈ 1/16 inches

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Answer:

\(U=\pm 0.382N\)

Explanation:

From the question we are told that:

Resolution: 0.25 N

Range: 0 to 100 N

Linearity error: within 0.20 N over range

Hysteresis error: within 0.30 N over range

Generally the equation for Stage Uncertainty is mathematically given by

\(U=\sqrt{u_0^2+u_T^2}\)

Where

\(u_0=Zero\ order\ uncertainty\)

\(u_0=\pm 0.5*0.25\)

\(u_0=\pm=0.125\)

And

u_T=Total instrumental Uncertainty

\(u_T=\sqrt{l_e^2+h_e^2}\)

Where

l_e=Error of linearity

h_e=Error due to hysteresis

Hence

\(u_T=\sqrt{0.20^2+0.30^2}\)

\(u_T=\pm 0.36\)

Therefore

\(U=\sqrt{(0.125)^2+0.36^2}\)

\(U=\pm 0.382N\)