Answer:
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Explanation:
The heat treatment procedure is simply the heating of a metal to a high temperature and cooling the metal back. during this process the metal will undergo certain mechanical changes
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
We are given a CSP with only binary constraints. Assume we run backtracking search with arc consistency as follows. Initially, when presented with the CSP, one round of arc consistency is enforced. This first round of arc consistency will typically result in variables having pruned domains. Then we start a backtracking search using the pruned domains. In this backtracking search we use filtering through enforcing arc consistency after every assignment in the search.
Which of the following are true about this algorithm?
a) If after a run of arc consistency during the backtracking searchwe end up with the filtered domains of allof the not yetassigned variables being empty, this means the CSP has nosolution.
b) If after a run of arc consistency during the backtracking searchwe end up with the filtered domain of oneof the not yetassigned variables being empty, this means the CSP has nosolution.
c) None of the above.
A private plane pilot is what kind of individual transportation position? professional level mid-level entry-level EPA-certified
Answer:
A private plane pilot is a mid-level position.
Explanation:
The six pilot certifications in the US are as follows:
Sport PilotRecreational PilotPrivate Plane PilotCommercial Plane PilotFlight InstructorAirline Transport PilotFrom this listing, it is evident that the Private Plane Pilot is in the mid of the line so it is a mid-level position.
A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system, indicate whether the following statements are true or false. (Note the the volume MUST CHANGE upon the addition of water.)
A. The concentration of HCN will increase.
B. The concentration of CN" will decrease.
C. The equilibrium concentration of Hy0 will remain the same 4
D. The pH will remain the same.
E. The ratio of [HCN]/[CN] will decrease.
You will be hiking to a lake with some of your friends by following the trails indicated on a map at the trailhead. The map says that you will travel 1.7 mi directly north, then 2.7 mi in a direction 36° east of north, then finally 1.7 mi in a direction 15° north of east. At the end of this hike, how far will you be from where you started, and what direction will you be from your starting point?
Explanation:
Which of the following are the same as 1545.5347
Select one:
a. 1.545534e: 3
b. 1545534.0e 4
c0.15455340-5
d. 154553.4e 3
Explanation:
D the answer is D and anyone can answer this Question
Based on the provided options, the scientific expression that is the same as 1545.5347 is 154553.4e3. Therefore, the correct option is option D.
A scientific expression, often known as scientific notation, is a method of representing extremely big or extremely small integers. It is often used to simplify the representation of such numbers in scientific and mathematical operations. A number is expressed in scientific notation as a product of two parts: a coefficient and a power of ten.
The coefficient is a value between 1 and 10, and the power of 10 determines how far the decimal point should be moved. The "e3" signifies multiplying the integer by 10 raised to the power of 3, which is similar to moving the decimal point three places to the right in this formula. As a result, 154553.4e3 equals 154553.4103, which simplifies to 154553400.
Therefore, the correct option is option D.
To know more about scientific expression, here:
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Which location sharing service offers items for users as a gaming component and also allows them to collectively link their check ins to publish a trip
Answer:
Gowalla
Explanation:
Gowalla is a Social networking service operating as a gaming component in which it includes the registration of users' cities. Such information is then saved and utilized during the fun activities on the platform such as meeting up with people nearby and discuss interesting things relating to them.
Hence, considering there are no available options, GOWALLA is the location-sharing service that offers items for users as a gaming component and also allows them to collectively link their check-ins to publish a trip
please help with my economics problem
Answer:
You first get a new job, and make a new company and then by amazon to traumatize Jeff Bezos after his divorce
Explanation:
La probabilidad de que un nuevo producto tenga éxito es de 0.85. Si se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto, ¿cuál es la probabilidad de que por lo menos 8 adquieran el nuevo producto?
Answer:
La probabilidad pedida es [tex]0.820196[/tex]
Explanation:
Sabemos que la probabilidad de que un nuevo producto tenga éxito es de 0.85. Sabemos también que se eligen 10 personas al azar y se les pregunta si comprarían el nuevo producto. Para responder a la pregunta, primero definiremos la siguiente variable aleatoria :
[tex]X:[/tex] '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''
Ahora bien, si suponemos que la probabilidad de que el nuevo producto tenga éxito se mantiene constante [tex](p=0.85)[/tex] y además suponemos que hay independencia entre cada una de las personas al azar a las que se les preguntó ⇒ Podemos modelar a [tex]X[/tex] como una variable aleatoria Binomial. Esto se escribe :
[tex]X[/tex] ~ [tex]Bi(n,p)[/tex] en donde [tex]''n''[/tex] es el número de personas entrevistadas y [tex]''p''[/tex] es la probabilidad de éxito (una persona adquiriendo el producto) en cada caso.
Utilizando los datos ⇒ [tex]X[/tex] ~ [tex]Bi(10,0.85)[/tex]
La función de probabilidad de la variable aleatoria binomial es :
[tex]p_{X}(x)=P(X=x)=\left(\begin{array}{c}n&x\end{array}\right)p^{x}(1-p)^{n-x}[/tex] con [tex]x=0,1,2,...,n[/tex]
Si reemplazamos los datos de la pregunta en la función de probabilidad obtenemos :
[tex]P(X=x)=\left(\begin{array}{c}10&x\end{array}\right)(0.85)^{x}(0.15)^{10-x}[/tex] con [tex]x=0,1,2,...,10[/tex]
Nos piden la probabilidad de que por lo menos 8 personas adquieran el nuevo producto, esto es :
[tex]P(X\geq 8)=P(X=8)+P(X=9)+P(X=10)[/tex]
Calculando [tex]P(X=8), P(X=9)[/tex] y [tex]P(X=10)[/tex] por separado y sumando, obtenemos que [tex]P(X\geq 8)=0.820196[/tex]
Por definición de variable aleatoria binomial, la probabilidad de que por lo menos 8 adquieran el nuevo producto es 0.8202.
En primer lugar, debes saber que la Probabilidad es la mayor o menor posibilidad de que ocurra un determinado suceso.
En otras palabras, la probabilidad establece una relación entre el número de sucesos favorables y el número total de sucesos posibles.
Por otro lado, una variable aleatoria es una función que asocia un número real, perfectamente definido, a cada punto muestral.
Una variable aleatoria es discreta si los números a los que da lugar son números enteros.
Finalmente, una distribución binomial es una distribución de probabilidad discreta que describe el número de éxitos al realizar n experimentos independientes entre sí, acerca de una variable aleatoria.
La expresión para calcular la distribución binomial es:
P(X=x)=( [tex]n\\ x[/tex] ) pˣ qⁿ⁻ˣ
Donde:
n = Número de ensayos/experimentos x = Número de éxitos p = Probabilidad de éxito q = Probabilidad de fracaso (1-p)Recordar que:
( [tex]n\\ x[/tex] )= [tex]\frac{n!}{x!(n-x)!}[/tex]
donde el signo de exclamación representa el símbolo de factorial, es decir el producto de todos los números enteros positivos desde 1 hasta n.
En este caso, sabes que la probabilidad de que un nuevo producto tenga éxito es de 0.85 y se eligen 10 personas al azar. Entonces se define la siguiente variable aleatoria:
X= '' Número de personas que adquirirán el nuevo producto de 10 personas a las que se les preguntó ''
Suponiendo que hay independencia entre cada una de las personas al azar a las que se les preguntó, es posible usar una variable aleatoria Binomial. Siendo n el número de personas entrevistadas (en este caso, 10 personas) y p la probabilidad de éxito (una persona adquiriendo el producto, en este caso 0.85) en cada caso, la expresión a utilizar queda expresada como:
P(X=x)=( [tex]10\\ x[/tex] ) 0.85ˣ (1-0.85)¹⁰⁻ˣ
P(X=x)= [tex]\frac{10!}{x!(10-x)!}[/tex] 0.85ˣ (0.15)¹⁰⁻ˣ
Se desea calcular la probabilidad de que por lo menos 8 adquieran el nuevo producto. Siendo 10 las personas elegidas al azar, puede ser que sean 8,9 o 10 las personas que adquieran un nuevo producto. Entonces la probabilidad deseada es calculada como:
[tex]P(X\geq 8)= P(X=8) + P(X=9) + (P=10)[/tex]
Entonces:
[tex]P(X\geq 8)=[/tex] [tex]\frac{10!}{8!(10-8)!}[/tex] 0.85⁸ (0.15)¹⁰⁻⁸ + [tex]\frac{10!}{9!(10-9)!}[/tex] 0.85⁹ (0.15)¹⁰⁻⁹ + [tex]\frac{10!}{10!(10-10)!}[/tex] 0.85¹⁰ (0.15)¹⁰⁻¹⁰
Resolviendo:
[tex]P(X\geq 8)=[/tex] [tex]\frac{10!}{8!2!}[/tex] 0.85⁸ (0.15)² + [tex]\frac{10!}{9!1!}[/tex] 0.85⁹ (0.15)¹ + [tex]\frac{10!}{10!0!}[/tex] 0.85¹⁰ (0.15)⁰
[tex]P(X\geq 8)=[/tex] 45×0.85⁸×(0.15)² + 10×0.85⁹×(0.15)¹ + 1×0.85¹⁰×(0.15)⁰
[tex]P(X\geq 8)=[/tex] 0.2759 + 0.3474 + 0.1969
[tex]P(X\geq 8)=[/tex] 0.8202
Finalmente, la probabilidad de que por lo menos 8 adquieran el nuevo producto es 0.8202.
Aprender más sobre probabilidad y distribución binomial:
https://brainly.com/question/18735028?referrer=searchResultshttps://brainly.com/question/25008895?referrer=searchResultshttps://brainly.com/question/22304471.If aligned and continuous carbon fibers with a diameter of 6.90 micron are embedded within an epoxy, such that the bond strength across the fiber-epoxy interface is 17 MPa, and the shear yield strength of the epoxy is 68 MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is 3960 MPa.
Answer:
the required minimum fiber length is 0.80365 mm
Explanation:
Given the data in the question;
Diameter D = 6.90 microns = 6.90 × 10⁻⁶ m
Bond strength ζ = 17 MPa
Shear yield strength ζ[tex]_y[/tex] = 68 Mpa
tensile strength of carbon fibers [tex]6t_{fiber[/tex] = 3960 MPa.
To determine the minimum fiber length we make use of the following relation;
L = ([tex]6t_{fiber[/tex] × D) / 2ζ
we substitute our given values into the equation;
L = ( 3960 × 6.90 × 10⁻⁶) / (2 × 17 )
L = 0.027324 / 34
L = 0.000803647 m
L = 0.000803647 × (1000) mm
L = 0.80365 mm
Therefore, the required minimum fiber length is 0.80365 mm
What happens to the speed of light if the IOR increases?
A tank has a gauge pressure of 552 psi. The cover of an inspection port on the tank has a surface area of 18 square inches. What is the total force the cover is experiencing.
Answer:
44197.55 N
Explanation:
From the question,
Pressure of the pressure guage (P) = Total force experienced by the cover (F)/Area of the cover (A)
P = F/A................ Equation 1
make F the subeject of the equation
F = P×A............... Equation 2
Given: P = 552 psi = (552×6894.76) = 3805907.52 N/m², A = 18 square inches = (18×0.00064516) = 0.01161288 m²
Substitute these values into equation 2
F = ( 3805907.52×0.01161288)
F = 44197.55 N
_____ can be defined as the rate at which work is done or the amount of work done based on a period of time. (2 Points) voltage power resistant current
Answer: Power
Explanation:
The rate at which work is done or the amount of work done based on a period of time is referred to as power.
Power can also be defined as the amount of energy that is being transferred per unit time. The unit of power is one joule per second or simply called the watt.
3.Which of the following drawings are matched with the project specifications to form the bulk of the contract document?
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% level of signigicance, is that th mean shear strenfth of spot welds is at least 3.1 MPa. The engineer randomly selcts 15 Welds and measures the shear strength is 3.07 MPa with a sample standard deciation of 0.069 MPa. Which of the following statment is true?
a) The null hypothesis should not be rejected.
b) The null hypothesis should be rejected.
c) The alternate hypothesis should be rejected.
d) The null and alternate hypotheses are equally likely.
Answer:
b) The null hypothesis should be rejected.
Explanation:
The null hypothesis is that the mean shear strength of spot welds is at least
3.1 MPa
H0: u ≥3.1 MPa against the claim Ha: u< 3.1 MPa
The alternate hypothesis is that the mean shear strength of spot welds is less than 3.1 MPa.
This is one tailed test
The critical region Z(0.05) < ± 1.645
The Sample mean= x`= 3.07
The number of welds= n= 15
Standard Deviation= s= 0.069
Applying z test
z= x`-u/s/√n
z= 3.07-3.1/0.069/√15
z= -0.03/0.0178
z= -1.68
As the calculated z= -1.68 falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa
Illinois furniture , Inc produces all types of coffee furniture the executive secretary is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130 there are 480 minutes available during the day and the average daily demand has been 50 chairs there are eight tasks
This question is incomplete, the complete question is;
Illinois furniture , Inc produces all types of office furniture. The "Executive Secretary" is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130. There are 480 minutes available during the day and the average daily demand has been 50 chairs. There are eight tasks.
TASK PERFORMANCE TIME( MIN ) TASK MUST FOLLOW TASK LISTED
A 4 ---------
B 7 ----------
C 6 A,B
D 5 C
E 6 D
F 7 E
G 8 E
H 8 F,G
1) What is the cycle time for operation?
2) What is the theoretical minimum number of workstation?
Answer:
1) the cycle time for operation is 9.6 min
2) the theoretical minimum number of workstation is 5
Explanation:
Given the data in the question;
production time per day = 480 minutes
average daily demand = 50
Given the data in the question;
1) cycle time for operation
this is simply referred to as the total time for the process from start to finish.
cycle time = production time per day / units demand per day
we substitute
cycle time = 480 min / 50
cycle time = 9.6 min
Therefore, the cycle time for operation is 9.6 min
2) theoretical minimum number of workstation.
theoretical minimum number of workstation = total task time / cycle time
Total task time = ( 4 + 7 + 6 + 5 + 6 + 7 + 8 + 6 ) = 49 min
∴ theoretical minimum number of workstation = 49 min / 9.6 min
theoretical minimum number of workstation = 5.104 ≈ 5
Therefore, the theoretical minimum number of workstation is 5
Route Choice The cost of roadway improvements to the developer is a function of the amount of traffic being generated by the theater as well as the routes that these vehicles use in getting to/from the theater. Thus, you need to determine traffic demand on the available routes between the housing area (apartment complexes) and the proposed theater site. There are two potential routes from this housing area to the proposed theater site. The total flow on these two routes (origin-to-destination demand plus other traffic) is 5500 vehicles. These routes have the following speed and length characteristics:
Route Free Flow Speed(mi/h) Length(mi)
1 45 5.5
2 40 3.5
3 35 3.75
It is known that the individual route travel times increase (in units of minutes) according to the following functions (with x in units of 1000 vehicles per hour):
Route Travel time increase as a function of traffic volume
1 0.5x1
2 x2
3 0.25
Required:
Determine user equilibrium flows and travel times if the total flow on these routes (origin-to-destination demand plus other traffic) is 5500 veh/h.
Design a septic system with gravity flow subsurface soil adsorption for a three-bedroom house with two baths and a basement that is currently occupied by a family of four. The building site is depicted in topographic map A and in the cross sectional transects A3-C3, A2-C2, A1-C1. Based on an initial site investigation, soil percolation tests were conducted in areas C1 and A3 to assess their suitability as locations for the absorption bed. The pertinent data from these tests is given below.
Area Cl Water Level Drop in Inches per 30 Minute Interval Percolation Rate (min/in Hole 1 13.56.57.361.24.221.221.211.2 65 .92.75.71 2.2 1.781.1.78 63 65 3 2.441.46 1.02 7 4 51.85 1.5 .92.87.8 1.76.781.6 1.441.36.28 1.26 1.26 .78 .78.78 Average: Area A3 Water Level Drop in Inches per 30 Minute Interval Percolation Rate (min/in Hole 1 3 5 1.340.940.900.88 0.840.78 0.76 0.75 2 1.45 1.18 0.98 0.92 0.90.88 0.880.86 3 0.980.88 0.840.81 0.820.800.790.77 1.140.960.90.870.850.84 0.82 0.83 1.211.02 0.920.880.85 0.830.82 0.81 4 Average:
Your solution to this exercise should include the following and should be in accord with the State of Virginia "Sewage Handling and Disposal Regulations" (A summary of the appropriate regulations is available within the course notes):________.
a) The design of an appropriately sized septic tank. This design should include a fully labeled sketch that notes all appurtenances
b) The design for a complete gravity, subsurface trench drainfield system. Depict your system on the layout sketch for the site and note all setback distances. Discuss the reasons behind your choice of a location for the drainfield,
c) If you were to utilize a bed-based design, how would the surficial area of your drainfield change?
Consider a normal population distribution with the value of known.
Answer:
it is alba kk
Explanation:
rbeacuse wrong
5. Which of these materials in a shop contain metals and toxins and can pollute the environment? A) Antifreeze B) Solvents C) Batteries D) All of the above
I say it's D) All of the above
Answer:
D
Explanation:
All of the above
14. The top plate of the bearing partition
I
a. laps the plate of the exterior wall.
b. is a single member.
c. butts the top plate of the exterior wall.
d. is applied after the ceiling joists are
installed.
Answer:
d. is applied after the ceiling joists are
installed.
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor principal stresses.
Answer:
The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.
Explanation:
The major and minor principal stresses are given as follows:
[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]
[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}[/tex]
Here
[tex]\sigma_x[/tex] is the normal stress which is 1750 psf[tex]\sigma_y[/tex] is 0[tex]\tau_{xy}[/tex] is the shear stress which is 800 psfSo the formula becomes
[tex]\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}[/tex]
Similarly, the minimum normal stress is given as
[tex]\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}[/tex]
The maximum shear stress is given as
[tex]\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}[/tex]
Write a program that can take a tree as input and travers it in 3 different format (pre-order, in-order and post-order). Write a report, you can follow the attached file. You can modify the report as you want.
Answer:
Let us see different corner cases.
Complexity function T(n) — for all problem where tree traversal is involved — can be defined as:
T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of root and n-k-1 on the other side.
Let’s do an analysis of boundary conditions
Case 1: Skewed tree (One of the subtrees is empty and other subtree is non-empty )
k is 0 in this case.
T(n) = T(0) + T(n-1) + c
T(n) = 2T(0) + T(n-2) + 2c
T(n) = 3T(0) + T(n-3) + 3c
T(n) = 4T(0) + T(n-4) + 4c
el protozoos es del reino protista?
Answer: Si (Yes)
Explanation:
Answer:
Protozoario o protozoo es un organismo unicelular y eucariota (con núcleo celular definido) perteneciente al Reino protista. Los protozoarios se encuentran junto con los protófitos o algas simples, generalmente acuáticas, dentro del Reino protista o también denominado Reino protoctista.
Explanation:
You are designing a package for 200 g of snack food that is sensitive to oxygen, and fails when it absorbs 120 ppm of oxygen (by weight). Marketing tells you it wants the snack to be in a plastic pouch measuring 6 inches by 6 inches (ignore seems), so it will have a total surface area available for permeation of 72 in(6" x 6" x 2 sides). You need to recommend an appropriate plastic material for this product, to provide a minimum of 70 days shelf life. Follow these steps:
a. Calculate the allowable oxygen gain, in cm at STP. (5 pts)
b. At this point, you do not know the material you will use, so you do not know the permeability coefficient or the thickness. The better the barrier the plastic you choose, the thinner the material can be to provide the appropriate barrier. Rather than simple trial and error, a sensible approach is to solve for the ratio of P/L that is required. We can solve the basic permeability equation for this ratio: P = 9 At Ap L Use the information you have to determine the required value for P/L, expressing your answer in cm/(100 in? d atm). (5 pts)
c. Use the information in the textbook (chapters 4 and 14 or in another reliable source; provide reference if you have used chapter 4, 14 or any other source) on oxygen permeability coefficients for various polymers to select a polymer that would be suitable, and calculate the required thickness. (Be sure this is reasonable; for example, if the required thickness is more than 20 mils, you need to choose a different polymer!) Note that chapter 4 presents these values in the units you used in (b) while chapter 14 presents values with different units, so unit conversion would be required. In your answer, state the material you have chosen, its oxygen permeability coefficient, and the minimum thickness you recommend. (Be sure to express the thickness with no more than one decimal place.) Obviously, there is more than one solution to this problem, but you only need one. (10 pts)
hmmmmmmmm i already put the photo as attachment its
Answer:
letse see
Explanation:
The substance xenon has the following properties:
normal melting point: 161.3 K
normal boiling point: 165.0 K
triple point: 0.37 atm, 152.0 K
critical point: 57.6 atm, 289.7 K
A sample of xenon at a pressure of 1.00 atm and a temperature of 204.0 K is cooled at constant pressure to a temperature of 163.7 K.Which of the following are true?
a. One or more phase changes will occur.
b. The final state of the substance is a liquid.
c. The final state of the substance is a solid.
d. The sample is initially a gas.
e. The liquid initially present will vaporize.
Answer:
the liquid woulriekwvhrnsshsnekwb ndrhwmoadi
In a production facility, 3-cm-thick large brass plates (k 5 110 W/m·K, r 5 8530 kg/m3, cp 5 380 J/kg·K, and a 5 33.9 3 1026 m2/s) that are initially at a uniform temperature of 25°C are heated by passing them through an oven maintained at 700°C. The plates remain in the oven for a period of 10 min. Taking the convection heat transfer coefficient to be h 5 80 W/m2·K, determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be
Answer:
the surface temperature of the plates when they come out of the oven is approximately 445 °C
Explanation:
Given the data in the question;
thickness t = 3 cm = 0.03 m
so half of the thickness L = 0.015 m
thermal conductivity of brass k = 110 W/m°C
Density p = 8530 kg/m³
specific heat [tex]C_p[/tex] = 380 J/kg°C
thermal diffusivity of brass ∝ = 33.9 × 10⁻⁶ m²/s
Temperature of oven T₀₀ = 700°C
initial temperature T[tex]_i[/tex] = 25°C
time t = 10 min = 600 s
convection heat transfer coefficient h = 80 W/m².K
Since the plate is large compared to its thickness, the heat conduction is one dimensional. heat transfer coefficient and thermal properties are constant over the entire surface.
So, using analytical one-term approximation method, the Fourier number > 0.2.
now, we determine the Biot number for the process
we know that; Biot number Bi = hL / k
so we substitute
Bi = hL / k
Bi = (80 × 0.015) / 110 = 1.2 / 110 = 0.0109
Now, we get the constants λ₁ and A₁ corresponding to Biot Number ( 0.0109 )
The interpolation method used to find the
λ₁ = 0.1039 and A₁ = 1.0018
so
The Fourier number т = ∝t/L²
we substitute
Fourier number т = ( (33.9 × 10⁻⁶)(600) ) / (0.015)²
т = 0.02034 / 0.000225
т = 90.4
As we can see; 90.4 > 0.2
So, analytical one-term approximation can be used.
∴ Temperature at the surface will be;
θ(L,t)[tex]_{wall[/tex] = (T(x,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀) ----- let this be equation
θ(L,t)[tex]_{wall[/tex] = [A₁e^(-λ₁²т)]cos( λ₁L / L )
so we substitute
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- (0.1039)²× 90.4 )] cos( 0.1039 × 0.015 / 0.015 )
θ(L,t)[tex]_{wall[/tex] = [1.0018e^(- 0.975886984 )] cos( 0.1039 )
θ(L,t)[tex]_{wall[/tex] = [1.0018 × 0.376857938] × 0.999998
θ(L,t)[tex]_{wall[/tex] = 0.3775
so we substitute into equation 1
θ(L,t)[tex]_{wall[/tex] = (T(L,t) - T₀₀) / (T[tex]_i[/tex] - T₀₀)
0.3775 = ( T(L,t) - 700 ) / ( 25 - 700 )
0.3775 = ( T(L,t) - 700 ) / ( - 675 )
0.3775 × ( - 675 ) = ( T(L,t) - 700 )
- 254.8125 = T(L,t) - 700
T(L,t) = 700 - 254.8125
T(L,t) = 445.1875 °C ≈ 445 °C
Therefore, the surface temperature of the plates when they come out of the oven is approximately 445 °C
Cast irons containing graphite are formed under a metastable eutectic reaction. 2) ___ High carbon steels are easily welded. 3) ___ Hardenability refers to the ease with which a steel can be quenched to form pearlite. 4) ___ In the AISI system for designating steels, the first two numbers refer to the major alloying elements of the steel. 5) ___ Quenching hardens most steels while tempering increases the toughness.
Answer:
Stating which of the above is TRUE or FALSE;
1) True
2) False
3) False
4) True
5) True
Explanation:
1)True ___ Cast irons containing graphite are formed under a metastable eutectic reaction.
2) False___ High carbon steels are easily welded.
3) False ___ Hardenability refers to the ease with which a steel can be quenched to form pearlite.
4) True ___ In the AISI system for designating steels, the first two numbers refer to the major alloying elements of the steel.
5) True ___ Quenching hardens most steels while tempering increases the toughness.
Two consecutive, first order reactions (with reaction rate constant k1 and k2) take place in a perfectly mixed, isothermal continuous reactor (CSTR) A (k1) → B (k2) → C Volumetric flow rates (F) and densities are constant. The volume of the tank (V) is constant. The reactor operate at steady state and at constant temperature. The inlet stream to the reactor contains only A with CA,in = 10 kmol/m3. If k1 = 2 min-1, k2 = 3 min-1, and τ = V/F.= 0.5 min, determine the concentration of C in the stream leaving the reactor.
Answer:
3 kmol/m^3
Explanation:
Determine the concentration of C in the stream leaving the reactor
Given that the CTSR reaction ; A (k1) → B (k2) → C
K1 = 2 min^-1 , K2 = 3 min^-1 , time constant ; τ = V/F.= 0.5 min also n1 = n2
attached below is the detailed solution
concentration of C leaving the reactor= 3 kmol/mol^3
Given ; Ca = 5 kmol/m^3 , Cb = 2 kmol/m^3 ( from the attached calculations ) Cc = 3 kmol/m^3
banana with an average mass of 0.15 kg and average specific heat of 3.35 kJ/kg · °C is cooled from 20°C to 5°C. The amount of heat transferred from the banana is
a.
62.1 kJ
b.
7.5 kJ
c.
None of these
d.
6.5 kJ
e.
0.85 kJ
f.
17.7 kJ
Answer:
The amount of heat transferred from the banana is (-)7.54 KJ
Explanation:
As we know,
[tex]Q = m*c*\Delta T[/tex]
Q = Amount of heat transferred
m = mass of banana
[tex]T_2 = 5[/tex] degree Celsius
[tex]T_1 = 20[/tex] degree Celsius
The amount of heat transferred from the banana =
[tex]0.15 * 3.35 * (5 -20)\\-7.54[/tex]KJ (negative sign represents reduction in heat energy)