Please help me with this question

Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

Given data:

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

Describe the procedure you would follow to obtain this material.

assuming 1040 steel experiences cracking at 40%CW

attached below is a detailed procedure of obtaining the material

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

The pressure given is P = 10N/cm²

The cross sectional area is A = 50cm²

Formula ,

P = Force / Area

F = P X A

F = ( 10N/cm² ) X ( 50cm² )

F= 500N

From cars to us carrying bags everything we use the concept of pressure.. The work that requires high pressure over a finite and small area is known as the  the hydraulic press.

The hydraulic press depends on Pascal's principle:  which says that the pressure through closed system is constant.

The hydraulic press has rapidly improved with time is providing great precision, control and effective functioning.

Hence extra load will the output piston IS 500N

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Answer:

Q = [ mCp ( ΔT) ] \(_{cooling water }\)

(ΔT)\(_{cooling water}\) and  Q  is given

\(m_{cooling water}\)  = \(\frac{Q}{Cp[ T_{out} - T_{in} ] }\)

next the rate of condensation of the steam

Q = [ m\(h_{fg}\) ]\(_{steam}\)

  \(m_{steam} = \frac{Q}{h_{fg} }\)

Total resistance of the condenser is

R = \(\frac{Q}{change in T_{cooling water } }\)

Explanation:

How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined

Q = [ mCp ( ΔT) ] \(_{cooling water }\)

(ΔT)\(_{cooling water}\) and  Q  is given

\(m_{cooling water}\)  = \(\frac{Q}{Cp[ T_{out} - T_{in} ] }\)

next the rate of condensation of the steam

Q = [ m\(h_{fg}\) ]\(_{steam}\)

  \(m_{steam} = \frac{Q}{h_{fg} }\)

Total resistance of the condenser is

R = \(\frac{Q}{change in T_{cooling water } }\)

The advice that will be given to the client based on the information is that the samples have arsenic levels within the limits.

Soil, also known as earth or dirt, is a complex mixture of organic matter, minerals, gases, liquids, and organisms that support life. Soil is the loose surface material that covers the majority of the land. It is made up of both inorganic and organic particles. Soil provides structural support to agricultural plants as well as a source of water and nutrients. The chemical and physical properties of soils vary greatly.

All of the samples collected have arsenic levels that are within the limits. The concentration of arsenic in samples 2 and 3 is well within the limit. The average value of 21 mg/kg has shown no adverse effects on residents and animals living in already-established residential areas.

Garden vegetable consumption has also not been linked to higher urinary arsenic levels. As a result, there will be no effects of such concentration for residential purposes.

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When in regular operation, ash and other debris should be removed daily from the combustion chamber of a wood-burning heater.

A wood stove radiates heat as it warms up via the stove's top and walls. The nearby region is warmed by the radiant heat, which can also warm other areas of the house thanks to the natural airflow in the building.

Fans using electric or convection power can help move this heat around to warm a bigger space. Some wood stoves use a convection chamber that wraps around the firebox to combine radiant and convection heat into a single appliance. Cool air is drawn into this convection chamber where it is heated and then circulated once again around the space.

Ash and other debris should be taken out of a wood-burning heater's combustion chamber every day when it is in regular use.

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Answer:

Normal operating temperatures of a converter are 500-800°F, and up to 1200°F when the vehicle is under heavy load.To melt the catalytic converter's substrate, the temperature inside the converter would have to exceed 2000°F.

Explanation:

Answer:

Email etiquette is important

Explanation:

It is important to understand how to use correct email etiquette because it helps you communicate more clearly. It also makes you seem a bit more professional too. For example depending in who you're emailing like say you're emailing your teacher for help then here's how it'd go:

Dear(teacher name, capitalize, never use first name unless they allow it)

Hello (teacher name), my name is (first and last name) from your (number class) and I was wondering if you could please help me out with (situation, be clear on what you need help with otherwise it won't get through to them)? If you could that would be greatly appreciated!

Sincerely,

(your name first and last)

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:

59°C

Explanation:

Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)

and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)

Therefore the minimum and maximum heat capacities are:

Cmin = Cc = 5803.2(W/K)

Cmax = Ch = 11634.3(W/K)

The capacity ratio is:

Cr = Cmin / Cmax = 0.499 = 0.5

The maximum possible heat transfer rate is:

Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)

And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99

Given that from the appropriate graph in the handouts we can read  = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)

Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C

The C function float sum(float a[], int n) takes an array of floating-point values and an integer that tells how many floating-point values are in the array, and returns the sum of the floating-point values in the array.

The given C function takes as its arguments an array of floating-point values and an integer that tells how many floating-point values are in the array. The function should return as its value the sum of the floating-point values in the array.

The following is the C function code with comments:

```cfloat sum(float a[], int n){// This is a float function named "sum" that takes an array of floating-point values (a[]) and an integer value (n) as its arguments. float sumAcc = 0.0; // Declare a float variable named "sumAcc" and initialize it with 0.0. int i; // Declare an integer variable named "i". // The for loop is used to iterate through each element of the array and add them to the "sumAcc" variable.for (i = 0; i < n; i++){sumAcc += a[i];}return sumAcc; // The function returns the final value of "sumAcc".}```

Hence, the correct answer is: float sum(float a[], int n).

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Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

   = 20.4kg · 9.81 \(\frac{N}{kg}\) · 1.50m

   = 35.316 \(\frac{N}{m}\)

   = 35.316 W

Answer: 8.33333333 or 6.1989778

Explanation:

Answer:

True

Explanation:

Principles of plasma arc cutting, Uses a constricting nozzle to create, concentrate,  and direct the high-velocity plasma. Plasma gas is always used

in plasma arc cutting When shielding gas is also used, the process is called dual flow plasma arc cutting. I hope this helps.

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

Hey there ..

I think the answer is as they put brake ..

I am not sure .. just check the image also provided above .. ..

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Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

A flexible budget prepared (before) the period begins allows management to make adjustments to increase profits or decrease losses.

A budget can be defined as a financial plan that is typically used for the estimation of revenue and expenditures of an individual, business organization or government for a specified period of time, often one (1) year.

In Economics, there are different types of budget and this include the following:

In conclusion, management can make several adjustments to increase profits or decrease losses when a flexible budget is prepared before the period begins.

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Answer:

filter drier

Explanation:

Any refrigeration or air conditioning system needs filter driers as a crucial part. In a refrigeration or air conditioning system, a filter drier serves two crucial purposes: first, it adsorbs system impurities like water, which can produce acids, and second, it provides physical filtering. To guarantee optimal and cost-effective dryer design, each component must be evaluated. The most crucial role of a dryer is to be able to drain water from a refrigeration system. Water can originate from a variety of places, including motor windings, equipment leaks, retained air from poor drainage, and others.

An airplane flying across the sky experience drag force determined by the factors including the speed of flight, coefficient of skin friction and the reference surface area

The boundary layer thickness is approximately 0.233 cm

The total skin friction drag, is approximately 265 N

Reason:

First part:

Given parameters are;

Chord length, L = 3 m

Velocity of the plane, V = 200 m/s

Density of the air, ρ = 1.225 kg/m³

Viscosity of the air, μ = 1.81 × 10⁻⁵ kg/(m·s)

The Reynolds number is given as follows;

\(R_{eL} = \dfrac{\rho \times V \times L}{\mu}\)

Therefore;

\(R_{eL} = \dfrac{1.255 \times 200 \times 3}{1.81 \times 10^{-5}} = 4.16022099448 \times 10^7 \approx 4.16 \times 10^7\)

Boundary layer thickness, \(\delta_L\), for laminar flow, is given as follows;

\(\dfrac{ \delta_L }{L}=\dfrac{5.0}{\sqrt{R_{eL} } }\)

\({ \delta_L }=\dfrac{5.0 \times L}{\sqrt{R_{eL} } }\)

Which gives;

\({ \delta_L }=\dfrac{5.0 \times 3}{\sqrt{4.16 \times 10^{7}} } \approx 2.33 \times 10^{-3 }\)

The boundary layer thickness, \(\delta_L\) ≈ 2.33 × 10⁻³ m = 0.233 cm

Second Part

The total skin friction is given as follows;

\(Dynamic \ pressure, q = \dfrac{1}{2} \cdot \rho \cdot V^2\)

Therefore;

\(q = \dfrac{1}{2} \times 1.225 \times 200^2 = 24,500\)

The dynamic pressure, q = 24,500 N/m²

Skin friction drag coefficient, \(C_D\), is given as follows;

\(C_D = \dfrac{1.328}{\sqrt{R_{eL} } }\)

Therefore;

\(C_D = \dfrac{1.328}{\sqrt{4.16 \times 10^7 } } \approx 2.06 \times 10^{-4}\)

Skin friction drag, \(D_f\), is given as follows;

\(D_f\) = q × \(C_D\) × A

Where;

A = The reference area

∴ \(D_f\) = 24,500 N/m² × 2.06 × 10⁻⁴ × 3 m × 17.5 m = 264.9675 N ≈ 265 N

The total skin friction drag, \(D_f\) ≈ 265 N

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Using the Newton's second law of motion we will see that the acceleration is  0.0875 m/s^2.

The acceleration of a vehicle is determined by the force acting on it and its mass, as described by Newton's second law of motion:

F = m * a

where F is the force acting on the object, m is the mass of the object, and a is the resulting acceleration.

In this case, the force acting on the vehicle is 0.07 N, and the mass of the vehicle is 0.8 kg. Substituting these values into the equation above, we get:

0.07 N = 0.8 kg * a

Solving for a, we get:

a = 0.07 N / 0.8 kg

a = 0.0875 m/s^2

Therefore, the acceleration of the 0.8 kg vehicle powered by 0.07 N of force is 0.0875 m/s^2.

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Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter \(d_{o}\) = 12.8 mm

Final diameter \(d_{f}\) = 10.7

Engineering stress  \(\alpha _{E}\) = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(\(d_{o}\) )²  ; Ag = π/4(\(d_{f}\) )²

% = π/4 [ ( (\(d_{f}\) )² - (\(d_{o}\) )²) / ( π/4  (\(d_{o}\) )²) ]

= ( (\(d_{f}\) )² - (\(d_{o}\) )²) / (\(d_{o}\) )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \(\alpha _{T}\) = \(\alpha _{E}\) ( 1 +  \(E_{E}\) )

\(E_{E}\)  is engineering strain

\(E_{E}\)  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

\(E_{E}\) = 0.431

so we substitute the value of \(E_{E}\)  into our initial equation;

True stress  \(\alpha _{T}\) = 460 ( 1 +  0.431)

True stress  \(\alpha _{T}\) = 460 (1.431)

True stress  \(\alpha _{T}\) = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

Technician A is correct. Transistors are semiconductors that are used to amplify or switch electronic signals and electrical power. Technician B's statement is more applicable to diodes, which function as one-way valves for current flow.

Technician A is correct. Transistors are indeed semiconductors, which are materials that can conduct electricity under certain conditions but not others. They are commonly used in electronic devices as amplifiers or switches. Technician B's statement is not entirely accurate. While transistors can control the flow of current, they do not function as one-way valves. Instead, they rely on the properties of semiconductors to regulate the flow of electrons.

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Answer:

21 m

Explanation:

Since F₁ = 100 N and acts at an angle of 20° to the horizontal, it has horizontal component F₁' = 100cos20° = 93.97 N and vertical component F₁" = 100sin20° = 34.2 N.

Also, F₂ = 75 N and acts at an angle of -35° to the horizontal, it has horizontal component F₂' = 75cos(-35°) = 75cos35° = 61.44 N and vertical component F₂" = 75sin(-35°) = -75sin35° = -43.02 N

The resultant horizontal force F₃' = F₁' + F₂' = 93.97 N + 61.44 N = 155.41 N

The resultant vertical force F₃" = F₁" + F₂" = 34.2 N - 43.02 N = -8.82 N

If f is the frictional force on the block, the net horizontal force on the block is F = F₃' - f.

Since f = μN where μ = coefficient of kinetic friction = 0.4 and N = normal force on the block.

For the block to be in contact with the surface, the vertical forces on the block must balance.

Since the normal force, N must equal the resultant vertical force F₃" and the weight, W = mg of the object for a zero net vertical force,

N = mg + F₃" (since both the weight and the resultant vertical force act downwards)

N = mg + F₃"

Since m = mass of block = 35 kg and g = acceleration due to gravity = 9.8 m/s² and F₃" = 8.82 N

So,

N = mg + F₃"

N = 35 kg × 9.8 m/s² + 8.82 N

N = 343 N + 8.82 N

N = 351.82 N

So, the net horizontal force F = F₃' - f.

F = 155.41 N - 0.4 × 351.82 N

F = 155.41 N - 140.728 N

F = 14.682 N

Since F = ma, where a = acceleration of block,

a = F/m = 14.682 N/35 kg = 0.42 m/s²

To find the distance the block moved, x we use the equation

x = ut + 1/2at² where u = initial speed of block = 0 m/s, t = time = 10 s and a = acceleration of block = 0.42 m/s²

Substituting the values of the variables into the equation, we have

x = ut + 1/2at²

x = 0 m/s × 10 s + 1/2 × 0.42 m/s² × (10 s)²

x = 0 m + 1/2 × 0.42 m/s² × 100 s²

x = 0.21 m/s² × 100 s²

x = 21 m

So, the distance moved by the block is 21 m.

Using APA reference style, the references is attached below from Feldman, R. S. (2013) to American Psychology Association (2020)

APA reference style is a style of citing sources used by the American Psychological Association. It is primarily used in the social sciences and includes guidelines for citing sources both within the text of a document and in the reference list at the end of the document. The guidelines provide a consistent way of citing sources to make it easier for readers to identify and locate the sources cited in a document.

In the given problem, if we need to write the reference using APA style, it will take this format below

Reference Page:

Feldman, R. S. (2013). Psychology and your life (2nd ed.). New York, NY: McGraw Hill Companies.

Frontline. (n.d.). League of denial. Retrieved from https://www.pbs.org/wgbh/frontline/film/league-of-denial/

Friends. (2004). The one after Joey and Rachel kiss [Television series episode]. In D. Crane, & M. Kauffman (Executive producers), Friends. Burbank, CA: Warner Bros. Television.

Silver Linings Playbook. (2012). [Motion Picture]. United States: The Weinstein Company.

U.S. Energy Information Administration. (2013). Annual energy outlook 2013 with projections to 2040. Retrieved from http://www.eia.gov/forecasts/aeo/er/pdf/0383er(2013).pdf

Snickers. (2013, April 2). You’re not you when you’re hungry [Video file]. Retrieved from https://youtu.be/2rF_FRCd_LA

American Psychological Association. (2020). Publication manual of the American Psychological Association (7th ed.). doi:10.1037/0000165-000

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To find how many ways, you must simply multiply 10 tasks and 5 exercises

    Ways: = 10 * 5 = 50 ways

Hope that helps!

Answer:

Emails and phone calls are both common forms of communication that are used in professional and personal settings. There are several similarities between email and phone calls:

1. Both are asynchronous forms of communication: Unlike instant messaging or face-to-face conversations, both emails and phone calls allow the sender or recipient to respond at their convenience. They don't require immediate attention or an instant response.

2. Both are written forms of communication: While phone calls rely on spoken words, emails are written. As a result, both can be used to convey detailed information and allow the sender to carefully consider their words before sending.

3. Both are forms of direct communication: Emails and phone calls both allow for direct communication between two parties. This can be beneficial for discussing sensitive information or resolving issues quickly.

4. Both can be used for formal and informal communication: Emails and phone calls can be used in both personal and professional contexts. They are both flexible forms of communication that can be adapted to fit different situations.

5. Both require attention to tone and etiquette: Just like with phone calls, emails require attention to tone and proper etiquette. Both forms of communication should be approached professionally and respectfully to ensure effective communication.

In conclusion, while there are differences between emails and phone calls, there are also similarities that make them useful communication tools. Both allow for direct, asynchronous communication and can be adapted to fit different situations.

Explanation:

Answer:

Parasite drag and induced drag.

Answer:

The issues related to the privacy are:

1. Informational privacy

2. Discrimination factors

3. Biased grouping on the basis of Data mining

4. Lack of consent

5. Morally wrong

6. Illegal distribution of information risks

7. Possibility of threat to life

Let's look at some major concerns:

1. Informational privacy : The concept of privacy of the personal information is totally nullified when the information is being used for a purpose other than the intended one for which it was given. This unethical use of information even for general purposes is not correct and is a matter of concern. It is more like using the sensitive data of others for personal benefit which is purely objectionable and raises security issues. Sometimes the data is also shared with the potential employers which might have certain impacts we are unaware of.

2. Data mining issues : The process of using a certain information to arrive and understand the trend and outcomes is called data mining. In this case, the consumer's data undergoes grouping and might get placed in the wrong group rather than the actual one. Also, there can be a case of biasing towards the groups which are not be focused on, or are not a part of the intended audience. This leads to the discrimination factors if we see it from a social point of view.

3. Lack of consent : Use of information without the consent or awareness of the consumers raises concern over the business ethics followed by the company. No one deserves the right to misuse information for his personal benefits without any of its information to the consumer. It is morally wrong and againt the work ethics. Moreover, it raises trust issues between the two involved, and hence is socially unacceptable.Explanation:

The amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.

Heat is defined as the process of kinetic energy being transferred from an energy source to a medium or an item. These three mechanisms radiation, conduction, and convection can all transmit energy. These three modes of heat transfer are typically combined, and it often happens on its own. The heat energy can be transformed into other types of energy, such as mechanical and kinetic energy. Thermal (heat) energy is a type of energy that is produced by the movement of particles and is a component of all matter.

The process rejects a certain amount of heat, which is

-q = w + Δu

q = - w - Δu

q = - 185.5 kJ/kg - 1948.9 kJ/kg

q = 2133.4 kJ/kg

Thus, the amount of heat rejected during the process in kJ/kg is 2133.4 kJ/kg.

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