Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at location 1 are 0.2 MPa and 800 K and at location 2 the pressure and temperature are 0.75 MPa and 700 K. The surroundings are at 300 K and the surface temperature of the device is 1000 K. Determine the rate that the device performs work on its surroundings if the rate of heat transfer from the surface of the device to the environment is 1 kW. Justify your answer. Note that the flow direction for the air is not specified so you need to consider all possibilities for the direction of the airflow. Assume that the air is an ideal gas, that R
Answers
Answer:
The answer is "+9.05 kw"
Explanation:
In the given question some information is missing which can be given in the following attachment.
The solution to this question can be defined as follows:
let assume that flow is from 1 to 2 then
Q= 1kw
m=0.1 kg/s
From the steady flow energy equation is:
[tex]m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\[/tex]
If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.
Related Questions
Consider two cycles: one is the ideal Carnot cycle and the other is the ideal Rankine cycle. Both cycles are saturated liquids at the inlet to the pump and saturated vapor at the inlet to the turbine. If the maximum temperature for the Carnot cycle is equal to the maximum temperature for the Rankine cycle, answer (and give a brief explanation) if the quantities below for the Rankine cycle are greater than, less than, or equal to those for the Carnot cycle. EXPLAIN each answer.
Answers
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem.
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2 (b) Show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs. (c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs
Answers
Answer:
Explanation:
a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2
We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX
If AB = 00 select [tex]I_o[/tex]
If AB = 01 select [tex]I_1[/tex]
If AB = 1_(B is don't care), select [tex]I_2[/tex]
However, the truth table is attached and shown in the first file below.
Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.
b) We are show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
For four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output
This question is a multiplexer which is a topic in digital circuit.
Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.
If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.
a)
Two 2-to-1 multiplexers to form a 3-to-1 MUX.
If AB = 00, select [tex]I_o[/tex]
If AB = 01, select [tex]I_1[/tex]
If AB = 1- (B is don't care) select I
The truth table for the above scenario is in the attached document below.
Figure 1 and 2 represents the solution to this question.
b).
Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.
In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.
Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.
In the attached diagram, figure 4 shows how it is structured.
We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.
Learn more about multiplexers here;
https://brainly.com/question/25953942
Explain why failure of this garden hose occurred near its end and why the tear occurred along its length. Use numerical values to explain your result. Assume the water pressure is 30 psistr
Answers
Answer:
hoop stresslongitudinal stressmaterial usedall this could led to the failure of the garden hose and the tear along the length
Explanation:
For the flow of water to occur in any equipment, water has to flow from a high pressure to a low pressure. considering the pipe, water is flowing at a constant pressure of 30 psi inside the pipe which is assumed to be higher than the allowable operating pressure of the pipe. but the greatest change in pressure will occur at the end of the hose because at that point the water is trying to leave the hose into the atmosphere, therefore the great change in pressure along the length of the hose closest to the end of the hose will cause a tear there. also the other factors that might lead to the failure of the garden hose includes :
hoop stress ( which acts along the circumference of the pipe):
αh = [tex]\frac{PD}{2T}[/tex] EQUATION 1
and Longitudinal stress ( acting along the length of the pipe )
αl = [tex]\frac{PD}{4T}[/tex] EQUATION 2
where p = water pressure inside the hose
d = diameter of hose, T = thickness of hose
we can as well attribute the failure of the hose to the material used in making the hose .
assume for a thin cylindrical pipe material used to be
[tex]\frac{D}{T}[/tex] ≥ 20
insert this value into equation 1
αh = [tex]\frac{20 *30}{2}[/tex] = 60/2 = 30 psi
the allowable hoop stress was developed by the material which could have also led to the failure of the garden hose
A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a maximum load of 250 kips, and the planned column support will consist of a 6 foot by 6 foot square footing placed 2 feet below the ground surface (to the bottom of the footing). Subsurface conditions consist of a 15-foot-thick layer of uniform silty sand (unit weight = 122 pcf, soil modulus = 160,000 psf) over stiff clay (unit weight = 118 pcf, soil modulus = 230,000 psf). Groundwater is deep.
a) Sketch the problem (freehand, not to scale) and state any necessary assumptions.
b) Calculate the immediate (elastic) settlement.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
he desk has a weight of 80 lb and a centerof gravity at G. Determine the initial acceleration of a desk when the man applies enough force F to overcomethe static friction at A and B. Also, find the vertical reactions on each of the two legs at A and at B. Thecoefficients of static and kinetic friction at A andB are μs= 0.45 andμk= 0.25, respectively.
Answers
Answer:
[tex]N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb[/tex]
Explanation:
Na and Nb are the vertical reactions on each of the two legs at A and at B
For the horizontal forces:
[tex]Fcos(30)-0.5N_a-0.5N_b=0\\0.5N_a+0.5N_b= Fcos(30)\\N_a+N_b= 2Fcos(30)[/tex]
For the vertical forces:
[tex]N_a+N_b-Fsin(30)-75=0\\N_a+N_b=Fsin(30)+80[/tex]
Therefore equating both equations:
[tex]2Fcos(30)=Fsin(30)+80\\F(2cos(30)-sin(30))=80\\F=\frac{80}{2cos(30)-sin(30)} =64.93N[/tex]
After the desk star to slide:
sum of all vertical force = ma , therefore:
[tex]N_a+N_b-64.93sin(30)-80=0\\N_a+N_b=64.93sin(30)+80[/tex]
sum of all horizontal force = ma
[tex]64.93cos(30)-0.2N_a-0.2N_b=\frac{80lb}{32.2ft/s^2}a\\ 0.2(N_a+N_b)=64.93cos(30)-\frac{80lb}{32.2ft/s^2}a\\N_a+N_b=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}a}{0.2}=324.65-12.42a[/tex]
equating both equations:
[tex]324.65-12.42a=64.93sin(30)+80\\12.42a=324.65-64.93sin(30)-80\\12.42a=212.185\\a=17.08ft/s^2[/tex]
From the moment equation:
[tex]4N_b-80(2)-64.93(3)=\frac{-80}{32.2} (17.08)(2)\\N_b=67.48lb[/tex]
[tex]N_a=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}(17.08)}{0.2}-67.48 = 45.04lb[/tex]
For each leg: [tex]N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb[/tex]
Air at 100°F, 1 atm, and 10% relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is 1765 ft3/min. Liquid water at 68°F enters the cooler and fully evaporates. Moist air exits the cooler at 70°F, 1 atm. There is no significant heat transfer between the device and its surroundings and kinetic and potential energy effects can be neglected. Determine the mass flow rate at which liquid enters, in lb(water)/min.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A thin‐walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.006 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 2.5 m/s. Assume the thermophysical properties of the exhaust gas are those of air
Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube.
Answers
Answer:
The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K
Explanation:
The detailed solution is attached as files below.
However, the steps followed are highlighted:
1) The average temperature was calculated as 380.5 K
2) The properties of air at 380.5 K was highlighted
3) The Prandti number was calculated. Pr = 0.693
4) The Reynold number was calculated, Re = 28716.77
5) The Nusselt umber was calculated, Nu = 75.94
6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.
The asymmetric roof truss is of the type used when a near normal angle of incidence of sunlight onto the south-facing surface ABC is desirable for solar energy purposes. The five vertical loads represent the effect of the weights of the truss and supported roofing materials. The 390-N load represents the effect of wind pressure. Determine the equivalent force-couple system at A. The couple is positive if counterclockwise, negative if clockwise. Also, compute the x-intercept of the line of action of the system resultant treated as a single force R.
Answers
Answer:
[tex]R= 337.75\ \bar i - 2275 \ \bar j \ \ N[/tex]
[tex]\sum M_A = -13650 \ N.m[/tex]
x = 6 m
Explanation:
From the diagram attached below :
Equivalent force:
[tex]R= -260 \bar j + 390 \ cos 30 \bar{ i} - 390 \ sin 30 \bar j - 520 \ \bar j- 520 \ \bar j- 520 \ \bar j- 260 \ \bar j[/tex]
[tex]R= 337.75\ \bar i - 2275 \ \bar j \ \ N[/tex]
The equivalent couple at point A is as follows:
[tex]\sum M_A = -390(3)-520(\frac{6}{2})- 520({6})- 520(6+\frac{6}{2}) -260(12)[/tex]
[tex]\sum M_A = -13650 \ N.m[/tex]
By applying the principles of momentum :
[tex]\sum M_A = Ry(x)[/tex]
[tex]- 13650 = - 2275 \ x[/tex]
x = [tex]\frac{13650}{2275}[/tex]
x = 6 m
Two identical bulbs are connected to a 12-volt battery in parallel. The voltage drop across the first bulb is 12 volts as measured with a voltmeter. What is the voltage drop across the other bulb?
Answers
Answer:
12 volts
Explanation:
The voltages across parallel-connected items are identical. (In fact, that's why you can measure the voltage by connecting the voltmeter in parallel with the circuit element.)
The voltage drop across each bulb is 12 volts.
The effective resistance of parallel resistors is always _____ than the lowest individual value.
a) more
b) less
c) no different than
Answers
Answer:
A
Explanation:
Answer:
the answer is a
Explanation:
it is a because thats what the answer is
is sampled at a rate of to produce the sampled vector and then quantized. Assume, as usual, the minimum voltage of the dynamic range is represented by all zeros and the maximum value with all ones. The numbers should increase in binary order from bottom to top. Find the bit combination used to store each sample when rounded to the nearest integer between and (clipping may occur). Note: A partially-correct answer will not be recognized. You must answer all three correctly on the same
Answers
Answer:
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
Explanation:
Assume that the number of bits is 8. The voltage range input is -8 to 7 volts. The range is thus 15V, and the resolution is 15/2^8 = 0.0586 volts. We will first add +8 to the input to convert it to a 0-15v signal. Then find the equivalent bit representation. For 7.8 volts, the binary signal will be all 1's, since the max input voltage for the ADC is 7 volts. For 4.95, we have 4.95+8 = 12.95 volts. Thus, N = 12.95/0.0586 = 221. The binary representation is 11011101. For -0.8, we have -0.8 + 8 = 7.2. Thus, N = 7.2/0.0586 = 123. The binary representation is 1111011.
Thus,
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
1. What is an op-amp? List the characteristics of an ideal op-amp
Answers
Answer:
An opamp is an operation amplifier. It takes an input signal and amplifies it on the output side.
An ideal opamp should have infinite impedance at its input, infinite gain on the output, and zero impedance on the output
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input is a unit step (1/s). (and show why it works); (2) Gives a settling time of 4 seconds; (3) has 10% overshoot. Use the standard 2nd order approximation. Plot the step response of the system and compare the standard approximation with the plot.
Answers
Answer:
Gc(s) = [tex]\frac{0.1s + 0.28727}{s}[/tex]
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial chemical concentration inside the tank was 0 mg/gal, the concentration of effluent coming in is 10 mg/gal. The volume of the tank is 10,000 gallons. The fluid coming in rate is equal to fluid going out is equal to 50 gal/min. Establish a dynamic model of how the concentration of the chemical inside the tank increases over time.
Answers
Answer:
0.05 mg / gallon
Explanation:
mass of chemecila coming in per minute = 50*10 = 500 mg/min
at a time t min , M = mass of chemical = 500*t mg
conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon
11.Three engines operate between reservoirs separated in temperature by 300 K. The reservoir temperatures are as follows: Engine a: , Engine b: , Engine c: , Rank the engines in order of their theoretically possible efficiency, from highest to lowest.
Answers
Answer:
Engine C > Engine B > Engine A
Explanation:
To calculate the theoretical efficiencies, we use the formula
1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir
And Th is the temperature of the hot reservoir
The ranking of the given engines in order of their theoretically possible efficiency, from highest to lowest is;
C > B > A
Theoretical Efficiency of EnginesThe formula for theoretical efficiency of engines is;
η = 1 - (Tc/Th)
Where;
Tc is the temperature of the cold reservoir.
Th is the temperature of the hot reservoir.
Thus the theoretical efficiencies are;
For Engine A;
η = 1 - (5700/6000)
η = 0.05
For Engine B;
η = 1 - (5500/5800)
η = 0.0517
For Engine C;
η = 1 - (5300/5600)
η = 0.0536
Looking at all the calculated theoretical efficiencies, Engine C has the highest followed by Engine B then Engine A.
The complete question is;
Three engines operate between reservoirs separated in temperature by 300 K. The reservoir temperatures are as follows:
Engine A: Th 6000 K, Tc 5700 K
Engine B: Th 5800 K, Tc 5500 K
Engine C: Th 5600 K, Tc 5300 K
Rank the engines in order of their theoretically possible efficiency, from highest to lowest.
Read more about Efficiency of Engines at; https://brainly.com/question/25819144
what is the Economic
Answers
(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each? (c) In bullet format compare and contrast the expected mechanical behavior of hypoeutectoid and hypereutectoid steels in terms of: (i) Yield strength (ii) Ductility (iii) Hardness (iv) Tensile strength (d) If you want to choose an alloy to make a knife or ax blade would you recommend a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your recommendation in 1-3 bullet points. (e) If you wanted a steel that was easy to machine to make a die to press powders or stamp a softer metal, would you choose a hypoeutectoid steel alloy or a hypereutectoid steel alloy? Explain your choice in 1-3 bullets.
Answers
Answer:
See explanation below
Explanation:
Hypo-eutectoid steel has less than 0,8% of C in its composition.
It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.
Ferrite has a higher tensile strength than cementite but cementite is harder.
Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:
Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel
Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel
Hyper-eutectoid steel is harder than Hyper-eutectoid steel
Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.
When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because
1. It is harder
2. It has low cost
3. It is lighter
When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because
1. It is ductile
2. It has high tensile strength
3. It is durable
Answer:
(a)
Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.
Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.
(b)
The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.
Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite. The carbon concentration for both ferrites is 0.022 wt% C.
(c)
(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.
(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.
(iii) Hardness: hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.
(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.
(d)
I would recommend hypereutectoid steel alloy to make a knife or ax blade
1- Hardness is required at the surface of the blades.
2- Ductility is not needed for such application.
3- Due to constant impact, the material will not easily yield to stress.
(e)
I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.
1- hypoeutectoid steel alloys have high machinability, hence better productivity
2- It will be used on softer metals, hence its fitness for the application
3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.
Explanation:
See all together above
Create an abstract class DiscountPolicy. It should have a single abstract method computeDiscount that will return the discount for the purchase of a given number of a single item. The method has two parameters, count and itemCost. 2. Derive a class BulkDiscount from DiscountPolicy, as described in the previous exercise. It should have a constructor that has two parameters, minimum and percent. It should define the method computeDiscount so that if the quantity purchased of an item is more than minimum, the discount is percent percent. 3. Derive a class BuyNItemsGetOneFree from DiscountPolicy, as described in Exercise 1. The class should have a constructor that has a single parameter n. In addition, the class should define the method computeDiscount so that every nth item is free. For example, the following table gives the discount for the purchase of various counts of an item that costs $10, when n is 3: count 1 2 3 4 5 6 7 Discount 0 0 10 10 10 20 20
4. Derive a class CombinedDiscount from DiscountPolicy, as described in Exercise 1. It should have a constructor that has two parameters of type DiscountPolicy. It should define the method computeDiscount to return the maximum value returned by computeDiscount for each of its two private discount policies. The two discount policies are described in Exercises 2 and 3. 5. Define DiscountPolicy as an interface instead of the abstract class described in Exercise 1.
Answers
Answer:
Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount
Explanation:
Solution
Code:
Main.java
public class Main {
public static void main(String[] args) {
BulkDiscount bd=new BulkDiscount(10,5);
BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);
CombinedDiscount cd=new CombinedDiscount(bd,bnd);
System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));
System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));
System.out.println("Combined discount :"+cd.computeDiscount(20, 20));
}
}
discountPolicy.java
public abstract class DiscountPolicy
{
public abstract double computeDiscount(int count, double itemCost);
}
BulkDiscount.java
public class BulkDiscount extends DiscountPolicy
{
private double percent;
private double minimum;
public BulkDiscount(int minimum, double percent)
{
this.minimum = minimum;
this.percent = percent;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if (count >= minimum)
{
return (percent/100)*(count*itemCost); //discount is total price * percentage discount
}
return 0;
}
}
BuyNItemsGetOneFree.java
public class BuyNItemsGetOneFree extends DiscountPolicy
{
private int itemNumberForFree;
public BuyNItemsGetOneFree(int n)
{
itemNumberForFree = n;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if(count > itemNumberForFree)
return (count/itemNumberForFree)*itemCost;
else
return 0;
}
}
CombinedDiscount.java
public class CombinedDiscount extends DiscountPolicy
{
private DiscountPolicy first, second;
public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)
{
first = firstDiscount;
second = secondDiscount;
}
at Override
public double computeDiscount(int count, double itemCost)
{
double firstDiscount=first.computeDiscount(count, itemCost);
double secondDiscount=second.computeDiscount(count, itemCost);
if(firstDiscount>secondDiscount){
return firstDiscount;
}else{
return secondDiscount;
}
}
}
An insulated, vertical piston–cylinder device initially contains 10 kg of water, 6 kg of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and 350°C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine (a) the final temperature in the cylinder and (b) the mass of the steam that has entered.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A wheel tractor is operating in it is fourth gear range with full rated revolution per minute. Tractors speed is 7.00 miles per hour. Ambien air temperature is 60 Fahrenheit, and operating attitude is sea level. This tractor is towing a fill material loaded pneumatic trailer while climbing with 5 % slop. Rolling resistance is 55lb/ton. Tractor is single axle and it is operating weight is 74,946 lb. The loaded trailer weighs 55,000 lb. The weight distribution is for the combined tractor- trailer unit is 53% to the drive axle and 47% to the rear axle.
The manufacturer, for the environmental conditions as given above, rates the tractive effort of the new tractor at 330 rimpull HP. What percentage of this manufacturers rated rimpul HP actually develop?
Answers
(20 points) (Assessment of Outcome 1): A plant has two identical standby generator units for emergency use. In the area of the generators, the normal noise level registers 82 dBA on the sound-level meter with the generators turned off. When one generator switches on, the SLM needle jumps to 85.8 dBA. What will the dBA reading be when the second generator also turns on (so that both generators are on)
Answers
Answer:
It wouldn't get any louder then maybe 3db more
Explanation:
There's even a equation if you wanted to check this out but, if they are the same generator same model and all and made the same precise noise it wouldn't increase more then 3db.
1. Deformation of a Cylindrical Pressure Vessel A thin–walled cylindrical pressure vessel of inner radius r and thickness t r is subjected to an internal pressure p. The pressure vessel is made from steel which has elastic material constants E = 200 GPa and ν = 0.30. If the vessel has a length ` = 3 m, an inner radius of r = 0.65 m, and a wall thickness t = 15 mm, determine the following items for a pressure p = 25 MPa. Assume that elongation is positive and contraction is negative. (a) Determine the hoop (circumferential) strain in the pressure vessel. (b) Determine the axial (longitudinal) strain in the pressure vessel. (c) Determine change in the circumference of the pressure vessel caused by the internal pressure. (d) Determine the change in the outer diameter of the pressure vessel caused by the internal pressure. (e) Determine the change in the length of the pressure vessel caused by the internal pressure.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A steady green traffic light means
Answers
Answer:
Its C. you may proceed, but only if the path is clear
Explanation:
I just gave Quiz and its correct
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
b) Compute the impulse of this event.
c) Obtain acceleration, velocity and displacement histories by using Newton’s second law of motion.
d) Compute the absorbed energy during collision.
Answers
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
Hot oil is to be cooled by water in a 1-shell-pass and 8-tube-passes heat exchanger. The tubes are thin-walled and are made of copper with an internal diameter of 1.4 cm. The length of each tube pass in the heat exchanger is 5 m, and the overall heat transfer coefficient is 310 W/m2.K. Water flows through the tubes at a rate of 0.2 kg/s, and the oil through the shell at a rate of 0.3 kg/is. The water and the oil enter at temperatures of 20 C and 150 C, respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.
Answers
Answer:
Rate of heat transfer is 66.8°C
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel, and concrete should always be installed and maintained __________ , __________ , and __________. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction. Please insert the answer into the spaces provided below.
Answers
Answer:
stability; resistance; rigidity.
Explanation:
Okay, let us fill in the gap in the question above. Please note that the capitalized words are the missing words to fill in the gap.
"Unless a structural element is intended to be battered, sloped, or cambered (or some other non-conforming position), wood, steel, and concrete should always be installed and maintained STABILITY , RESISTANCE , and RIGIDITY. These conditions are critical in order to ensure maximum stability, structural performance and user satisfaction."
This question has to deal with buildings or say structural (civil) engineering. The definition to the missing words are given below:
STABILITY: Stability occurs when we have the center of gravity coinciding with the base of the structure.
RESISTANCE : Resistance simply means the 'tension' that is is how much the structure can resist an applied force.
RIGIDITY : RIGIDITY can also be associated with resistance and it is the property of a structure to resist bending.
A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 5.9 × 10^(-11) m^2/s, and the diffusion flux is found to be 4.7 × 10^(-7) kg/m^2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.5 kg/m^3.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
Answers
Answer:
0.544×10–³
Explanation:
Please see the attached file for the solution
Water enters the tubes of a cold plate at 70°F with an average velocity of 40 ft/min and leaves at 105°F. The diameter of the tubes is 0.25 in. Assuming 14 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation and the remaining 86 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. The properties of water at room temperature are rho = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm·°F.
Answers
Answer:
The total amount of heat generated;Q' = 2067 Btu/h
Explanation:
We are given;
Water entering temperature;T1 = 70°F
Water leaving temperature;T2 = 105°F
average velocity of water;V = 40 ft/min
Diameter of tube;D = 0.25 in = 0.25/12 ft = 0.02083 ft
Water density;ρ = 62.1 lbm/ft³
cp = 1.00 Btu/lbm·°F.
Now, the mass flow rate of the water is calculated from;
m' = ρAV
Where ρ is density, A is area and V is velocity
Area = πD²/4 = π*0.02083²/4 = 0.00034077555 ft²
m' = 62.1 * 0.00034077555 * 40
m' = 0.8465 lbm/min
Converting to lbm/hr = 0.8465 * 60 = 50.79 lbm/hr
From energy balance equation, we have;
E_in = E_out
So,
Q_in,w + m'h1 = m'h2
Q_in,w = m'h2 - m'h1
Q_in,w = m'(h2 - h1)
Now, m'(h2 - h1) can be written as;
m'cp(T2 - T1).
Thus ;
Q_in,w = m'cp(T2 - T1)
Plugging in the relevant values, we have;
Q_in,w = (50.79*1)(105 - 70)
Q_in,w = 1777.65 Btu/h
We are told that remaining 86 percent of heat generaged is removed by the cooling water. Thus;
The total amount of heat generated could be defined as;
Q' = Q_in,w/0.86
Q' = 1777.65/0.86
Q' = 2067 Btu/h
As discussed in the text, one possible performance enhancement is to do a shift and add instead of an actual multiplication. Since 9 x 6, for example, can be written (2 x 2 x 2 + 1) x 6, we can calculate 9 x 6 by shift ing 6 to the left 3 times and then adding 6 to that result. Show the best way to calculate 0 x33 x 0 x 55 using shift s and adds/subtracts. Assume both inputs are 8-bit unsigned integers.
Answers
Answer:
The best way to calculate 0 x33 x 0 x 55 using shift s and adds/subtracts and assuming both inputs are 8-bit unsigned integers is attached below
You are a software developer for a Retail Point of Sale System Company. A client has made a request to upgrade the current system from command line interface to a graphic user interface. The client concerns on an effect to the current system with the change. The client does not want to lose any current data on the system. Provide recommendations and the change plan including the below requirements to the customer.
a. Create a requirement plan (what need to be done to support this upgrade)
b. Create a development plan (how to implement this upgrade with programming details explanation including technical details on this document)
c. Create testing plan (how to ensure that the new upgrade works without affecting current system and no data lost.)
Answers
Answer:
The recommendations to be provided for the customer and the change in plan before upgrade are: Check system requirements, Check the specifications, Problems that are existing and detected should be resolved before upgrade, Run a normal maintenance routines, Having backup plan
Explanation:
Solution
Command line interface is a way of working with a computer program where the user gives out commands to the program in a method called successive lines of text.
The Graphic user interface is a type of interface that enables users to interact with electronic devices through visual indicators. and graphical icons .
As a software developer, if a request comes to upgrade the system from CLI to GUI. Then, there are a number of necessary points which should be considered and also having a plan which is stated below:
First thing that should be carried is a feasibility study. We need to check whether the upgrade which is required by the client is feasible or not.
There can be a number of issues in the way like resource, technical constraints, budgetary constraints
So as a software developer, we need to examine each of them before Going further.
Check System Requirements: Think not only the system requirements, but also OS present configuration and any i software installed you use.
Check the specifications: The first thing is to check Specification. this is an important true requirement for upgrading, which is to guarantee your system meets the basic system requirements for running GUI
Existing problems should be detected and resolved: before upgrade. also, ensure that your system is not experiencing any major issues currently
Run Normal maintenance routines
Backup: To ensure the security of data, to prevent data loss and to avoid expensive recovery there is nothing that is not better than a backup.
The Golden rule or number one rule is Data backup.
Before upgrading any system, make sure that your personal, critical, ,irreplaceable data is safely backed up fully to an external device, like an external hard drive ,USB disk, DVD, etc.
Again, don't even try to attempt to upgrade before doing this important step. another way is creating a clone system or RAID (Redundant array of independent disks).
Then upgrade can begin. then in this particular order of upgrade,the remote secondary replica nodes first, then local secondary replica node is second , and the primary replica node third.
