Answer:
Explanation:if you stretch the hose more tightly the speed of the pulse will reduce..
With reference to the NSPE Code of Ethics, which one of the following statements is true regarding the ethical obligations of the engineers involved in the VW Emissions Cheating Scandal.
a. The VW engineers involved conducted themselves honorably, responsibly, ethically, and lawfully so as to enhance the honor, reputation, and usefulness of the profession.
b. As faithful agents and trustees of Volkswagen the engineers involved could not ethically report the emissions cheating violations to public authorities.
c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
d. The VW engineers involved were not ethically obligated to report to their supervisors and to upper management the emissions cheating violations being implemented in the control system hardware and software.
Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.
One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.
The yield of a chemical process is being studied.The two most important variables are thought to be the pressure and the temperature.Three levels of each factor are selected, and a factorial experiment with two replicates is performed.The yield data follow:_____.
emperature 150 160 170 200 90.1 90.3 90.5 90.7 90.4 90.2 Pressure 215 90.5 90.6 90.8 90.9 90.7 90.6 230 89.9 90.1 90.4 90.1 90.2 90.4
a) Analyze the data and draw conclusions.Use α = 0.05.
b) Prepare appropriate residual plots and comment on the model’s adequacy.
c) Under what conditions would you operate this process?(a
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Which statement is not correct regarding the deformation of a circular shaft in torsion? a. Cross sections remain flat. b. Longitudinal lines remain straight. c. Circular sections remain circular. d. Radial lines on the sections remain straight.
Answer:
B
Explanation:
Torsion is application of torque to a shaft to turn it about its longitudinal axis. When torque is applied to a shaft the circle remains unchanged in a circular state, its cross section does not warp but remains flat with a straight radial lines but its longitudinal lines changes into an helix intersecting the circular shaft
What does the following program segment do? Declare Count As Integer Declare Sum As Integer Set Sum = 0 For (Count = 1; Count < 50; Count++) Set Sum = Sum + Count End For
1225
Explanation:
This segment helps initialize sum as 0. The for loop is used to increment with every execution and it is added to the sum. The loop runs 49 times and every time the count is added to the sum. In short it is the sum of first 49 natural numbers i.e 1+2+3+......+49.
In this lab, you add a loop and the statements that make up the loop body to a Java program that is provided. When completed, the program should calculate two totals: the number of left-handed people and the number of right-handed people in your class. Your loop should execute until the user enters the character X instead of L for left-handed or R for right-handed.The inputs for this program are as follows: R, R, R, L, L, L, R, L, R, R, L, XVariables have been declared for you, and the input and output statements have been written.
Answer:
use this to help www.code.org
Explanation:
this helped me alot
Air at 27oC and 50% relative humidity is cooled in a sensible cooling process to 18oC. The air is then heated to 45oC in a sensible heating process. Finally, the air experiences an adiabatic saturation process that increases the relative humidity back to 50%. Find the specific energy that is removed when the air is cooled to 18°C.
Answer:
[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]
Explanation:
First, it is required to find the absolute humidity of air at initial state:
[tex]\omega = \frac{0.622\cdot \phi \cdot P_{g}}{P-\phi \cdot P_{g}}[/tex]
The saturation pressure at [tex]T = 27^{\textdegree}C[/tex] is:
[tex]P_{g} = 3.601\,kPa[/tex]
Then,
[tex]\omega = \frac{0.622\cdot (0.5)\cdot (3.601\,kPa)}{101.325\,kPa-(0.5)\cdot (3.601\,kPa)}[/tex]
[tex]\omega = 0.0113\,\frac{kg\,H_{2}O}{kg\,air}[/tex]
A simple cooling process implies a cooling process with constant absolute humidity. The specific entalphies for humid air are:
Initial state:
[tex]h_{1} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (27^{\textdegree}C)+(0.0113)\cdot (2551.96\,\frac{kJ}{kg} )[/tex]
[tex]h_{1} = 55.972\,\frac{kJ}{kg}[/tex]
Final state:
[tex]h_{2} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (18^{\textdegree}C)+(0.0113)\cdot (2533.76\,\frac{kJ}{kg} )[/tex]
[tex]h_{2} = 46.722\,\frac{kJ}{kg}[/tex]
The specific energy that is removed is:
[tex]q_{out}= h_{1} - h_{2}[/tex]
[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]
g Create an array of five animals. Use a for loop to display the values stored in that array. Add two more animals to the end of that array. Sort the array and display the sorted array on the screen.
Answer:
The most common approach to accessing an array is to use a for loop:
var mammals = new Array("cat","dog","human","whale","seal");
var animalString = "";
for (var i = 0; i < mammals. length; i++) {
animalString += mammals[i] + " ";
}
alert(animalString);
Discussion
A for loop can be used to access every element of an array. The array begins at zero, and the array property length is used to set the loop end.
Though support for both indexOf and lastIndexOf has existed in browsers for some time, it’s only been formalized with the release of ECMAScript 5. Both methods take a search value, which is then compared to every element in the array. If the value is found, both return an index representing the array element. If the value is not found, –1 is returned. The indexOf method returns the first one found, the lastIndexOf returns the last one found:
var animals = new Array("dog","cat","seal","walrus","lion", "cat");
alert(animals.indexOf("cat")); // prints 1
alert(animals.lastIndexOf("cat")); // prints 5
Both methods can take a starting index, setting where the search is going to start:
var animals = new Array("dog","cat","seal","walrus","lion", "cat");
alert(animals.indexOf("cat",2)); // prints 5
alert(animals.lastIndexOf("cat",4)); /
Answer:
animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"]
for animal in animals:
print(animal)
x = animals.insert(5, "Lizard")
y = animals.insert(6, "Bat")
z = sorted(animals)
print(z)
Explanation:
The question can be solved using various back-end coding language like python, java, JavaScript etc. But I will be writing the code with python.
The first question said we should create an array or list of five animals.
animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"] → I created the five animals in a list and stored them in the variable animals.
for animal in animals: I used a for loop to iterate through the list print(animal) I used print statement to display the values stored in the array or list.
x = animals.insert(5, "Lizard") I added an animal, lizard to the end of the array
y = animals.insert(6, "Bat") I added another animal , Bat to the end of the array.
z = sorted(animals) I sorted the array according to alphabetical number.
print(z) I displayed the sorted array .
A cylindrical specimen of a brass alloy having a length of 60 mm (2.36 in.) must elongate only 10.8 mm (0.425 in.) when a tensile load of 50,000 N (11,240 lbf ) is applied. Under these circumstances what must be a radius of the specimen
Determining the radius of a brass cylinder based on its elongation under tensile load involves understanding stress, strain, and the material's properties. Without the elastic modulus of brass or additional details, an exact calculation can't be provided. Theoretically, one uses the relationship between stress, strain, and Young's modulus, and the formula for the area of a circle to find the radius.
Explanation:To determine the radius of a cylindrical specimen of brass that must elongate a specific amount under a given tensile load, one must approach the problem by considering the relationship between stress, strain, and the elastic modulus of the material. Given that the specimen is cylindrical, the cross-sectional area is crucial in these calculations, which is directly related to the radius of the cylinder.
The formula for stress (σ) is defined as the force (F) divided by the area (A), σ = F/A. Strain (ε) is the deformation (change in length) divided by the original length (L), ε = ΔL/L. However, without the elastic modulus of brass or a direct way to calculate the cross-sectional area from the provided data, finding the exact radius requires assuming or knowing additional properties of the material.
Without these specifics, a more detailed calculation cannot be accurately provided. In practice, one would use the known properties of brass and the relationship between stress, strain, and Young's modulus (Y) of the material (Y = stress/strain) to find the required dimensions. Typically, this involves rearranging the formulas to solve for the radius, given that the area (A) can be expressed in terms of the radius (r) for a circle (A = πr²).
Copper spheres of 20-mm diameter are quenched by being dropped into a tank of water that is maintained at 280 K . The spheres may be assumed to reach the terminal velocity on impact and to drop freely through the water. Estimate the terminal velocity by equating the drag and gravitational forces acting on the sphere. What is the approximate height of the water tank needed to cool the spheres from an initial temperature of 360 K to a center temperature of 320 K?
Answer:
The height of the water is 1.25 m
Explanation:
copper properties are:
Kc=385 W/mK
D=20x10^-3 m
gc=8960 kg/m^3
Cp=385 J/kg*K
R=10x10^-3 m
Water properties at 280 K
pw=1000 kg/m^3
Kw=0.582
v=0.1247x10^-6 m^2/s
The drag force is:
[tex]F_{D} =\frac{1}{2} Co*p_{w} A*V^{2}[/tex]
The bouyancy force is:
[tex]F_{B} =V*p_{w} *g[/tex]
The weight is:
[tex]W=V*p_{c} *g[/tex]
Laminar flow:
[tex]v_{T} =\frac{p_{c}-p_{w}*g*D^{2} }{18*u} =\frac{(8960-1000)*9.8*(20x10^{-3})^{2} }{18*0.00143} =1213.48 m/s[/tex]
Reynold number:
[tex]Re=\frac{1000*1213.48*20x10^{-3} }{0.00143} \\Re>>1[/tex]
Not flow region
For Newton flow region:
[tex]v_{T} =1.75\sqrt{(\frac{p_{c}-p_{w} }{p_{w} })gD }=1.75\sqrt{(\frac{8960-1000}{1000} )*9.8*20x10^{-3} } =2.186m/s[/tex]
[tex]Re=\frac{1000*2.186*20x10^{-3} }{0.00143} =30573.4[/tex]
[tex]Pr=\frac{\frac{u}{p} }{\frac{K}{pC_{p} } } =\frac{u*C_{p} }{k} =\frac{0.0014394198}{0.582} =10.31[/tex]
[tex]Nu=2+(0.4Re^{1/2} +0.06Re^{2/3} )Pr^{2/5} (u/us)^{1/4} \\Nu=2+(0.4*30573.4^{1/2}+0.06*30573.4^{2/3} )*10.31^{2/5} *(0.00143/0.00032)^{1/4} \\Nu=476.99[/tex]
[tex]Nu=\frac{h*d}{K_{w} } \\h=\frac{476.99*0.582}{20x10^{-3} } =13880.44W/m^{2} K[/tex]
[tex]\frac{T-T_{c} }{T_{w}-T_{c} } =e^{-t/T} \\T=\frac{m_{c}C_{p} }{hA_{c} } =\frac{8960*10x10^{-3}*385 }{13880.44*3} =0.828 s[/tex]
[tex]e^{-t/0.828} =\frac{320-280}{360-280} \\t=0.573\\heightofthewater=2.186*0.573=1.25m[/tex]
Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an environment with h=50 W/m2°C and k'=30°C, while the left face is exposed to h=75 W/m2°C and T[infinity]=50°C. What is the maximum allowable heat generation rate such that the maximum temperature in the solid does not exceed 300°C.
Answer:
2.46 * 10⁵ W/m³
Explanation:
See attached pictures for detailed explanation.
Answer:
[tex]q^.=2.46*10^5W/m^3[/tex]
Explanation:
[tex]Given\\k=2.5W/m\\h_{1} =75(left)\\h_{2} =50(right)\\T_{1} =50^oC\\T_{2} =30^oC[/tex]
so
[tex]T=-\frac{q^.x^2}{2k} +c_{1}x+ c_{2} \\T=T_{1} \\at \\x=-0.04\\T=T_{2} \\at\\x=+0.04[/tex]
[tex]dT/dx=-q^.x/k+c_{1} \\T=T_{max} =300\\at\\x=c_{1} \frac{k}{q^.} (1)[/tex]
[tex]h_{1}(T_{1infinity} -T_{1} )=-k\frac{dT}{dx} |_{x=0.04} (2)\\-k\frac{dT}{dx} |_{x=0.04} =h_{2} (T_{2}-T_{2infinity} (3)[/tex]
[tex]300=-\frac{q^.}{2k} [c_{1} \frac{k}{q} ]^2+c_{1} [c_{1} \frac{k}{q} ]+c_{2} (1)[/tex]
[tex]75[50+\frac{q^2}{2k} (0.04)^2+c_{1} (0.04)-c_{2} ]=-k[\frac{+q^2(0.04)}{2k} ](2)[/tex]
[tex]-k[\frac{-q^.(0.04)}{2k} ]=50[\frac{-q^.(0.04)}{2k} +c_{1} (0.04)+c_{2} -30](3)[/tex]
solving above 3 equations for 3 unknowns c1,c2,q
we get [tex]q^.=2.46*10^5W/m^3[/tex]
Consider a 20 * 105 m3 lake fed by a polluted stream having a flow rate of 4.2 m3/s and pollutant concentration equal to 25 mg/L and fed by a sewage outfall that discharges 0.5 m3/s of wastewater having a pollutant concentration of 275 mg/L. The stream and sewage wastes have a second order decay rate of 0.32 L/(mg-day). Assuming the pollutant is completely mixed in the lake and assuming no evaporation or other water losses or gains, find the steady-state pollutant concentration in the lake. Derive the equation from the mass balance on C in the lake for full credit. Note that the quadratic equation will be useful for solving this problem.
Answer:
Explanation:
The detailed steps and calculations is as shown in the attachment.
where s = flow rate of incoming stream
Cs = concentration of pollutant in stream
Qf = flow rate from factory
Cp = concentration of pollutant from factory
Qo = flow rate out of lake
Co = concentration of pollutant in the lake
V = volume of lake
k = reaction rate coefficient
Use the writeln() method of the document object to display the user agent in a
tag in the webpage. Hint: The userAgent property of the window.navigator object contains the user agent.
Answer:
Note that writeln() add a new line after each statement
var txt = "<p>User-agent header: " + navigator.userAgent + "</p>";
$("#agent").writeln(txt);
Then Anywhere in the body tag of the html file
create a div tag and include an id="agent"
A single axial load of magnitude P = 15 kips is applied at end C of thesteel rod ABC. Knowing that E = 30 × 106 psi, determine the diameterd of portion BC for which the deflection of point C will be 0.05 in.
Answer:
d = 0.868 in
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
This question involves solving for the diameter d at the portion BC of a steel rod under axial load. It involves utilizing Hooke's Law and expressing cross-sectional area in terms of diameter. The final calculation will depend on the length of the portion BC.
Explanation:The question appears to involve the concept of stress and strain in the area of mechanical engineering. In this problem, we have a steel rod ABC, where end C, experiencing a single axial load P = 15 kips, deflections to 0.05 inches. We are tasked to find the diameter, denoted as d, of portion BC.
To solve this, we could utilize Hooke's Law of elasticity that connects stress, strain, and the Young's modulus (E). The formula is represented as δ = PL/AE. In the equation, P is the load, L is the length, A is the cross-sectional area and E is Young's modulus. We are given P, E, and δ (deflection), leaving the cross-sectional area and length as variables to be found. Since we are looking for the diameter (d), we need to express the area (A) in terms of the diameter using the formula A = πd²/4.
The length L is dependent on the specific conditions of the problem and are not given directly. Depending on the conditions, L might need to be solved differently. First, solve the equation for diameter, afterwards, plug the known values in and evaluate, giving the diameter in inches of the steel rod at portion BC.
Learn more about Stress and Strain here:https://brainly.com/question/36010339
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Which one is dependent variable?
Develop the best possible linear regression model to predict the median value of the house based on its characteristics as well as the neighborhood characteristics
CRIM - per capita crime rate by town
ZN - proportion of residential land zoned for lots over 25,000 sq.ft.
INDUS - proportion of non-retail business acres per town.
CHAS - Charles River dummy variable (1 if tract bounds river; 0 otherwise)
NOX - nitric oxides concentration (parts per 10 million)
RM - average number of rooms per dwelling
AGE - proportion of owner-occupied units built prior to 1940
DIS - weighted distances to five Boston employment centres
RAD - index of accessibility to radial highways
TAX - full-value property-tax rate per $10,000
PTRATIO - pupil-teacher ratio by town
B - 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town
LSTAT - % lower status of the population
MEDV - Median value of owner-occupied homes in $1000's
Answer:
The dependent variable is MEDV - Median value of owner-occupied homes in $1000's
Explanation:
The median value of the house has to be predicted, based on its properties and neighborhood properties, this can be done by using a linear regression model.
The dependent variable in Machine Learning is the output variable that we want to predict.
Therefore, according to the question given "MEDV" is the dependent variable.
A developer is having a single-lane raceway constructed with a 180 km/h design speed. A curve on the raceway has a radius of 320 m, a central angle of 30 degrees, and PI stationing at 11 511.200. If the design coefficient of side friction is 0.20, determine the superelevation at the design speed (Hint: Consider normal component of the centripetal force). Also, compute the length of curve and stationing of the PC and PT.
Answer:
The answer is in the attachment.
Explanation:
A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N (2500 lb f ). Determine the minimum required wire diameter assuming a factor of safety of 2 and a yield strength of 1030 MPa (150,000 psi).
Answer:
5.24m
Explanation:
Data given
force, F= 11,100N
safety factor,N=2
yield strength, =1030MPa
To determine the minimum diameter, we first determine the working strength which is expressed as
[tex]working strength=\frac{yield strength}{safty factor}\\ Working strength =1030/2=515MPa\\[/tex]
Since also the working strength is define as the ration of the force to the area, we have
[tex]515=\frac{11100}{A}\\ A=21.55m^{2}[/tex]
hence the required diameter is given as
[tex]A=\pi d^2/4\\d=\sqrt{\frac{4*21.55 }{\pi }} \\d=5.24m[/tex]
In the casting of steel under certain mold conditions, the mold constant in Chvorinov's Rule is known to be 4.0 min/cm2, based on previous experience. The casting is a flat plate whose length = 35 cm, width = 10 cm, and thickness = 15 mm. Determine how long it will take for the casting to solidify.
Answer:
it will take for the casting to solidify 2.55 min
Explanation:
given data
mold constant = 4 min/cm²
length = 35 cm
width = 10 cm
thickness = 15 mm
solution
we use here Chvorinov's Rule that is
Chvorinov's Rule = mold constant × [tex](\frac{V}{A})^{1.9}[/tex] ..............1
put here value
Chvorinov's Rule = 4 × [tex](\frac{600}{760})^{1.9}[/tex]
Chvorinov's Rule = 2.55 min/in
so heer unit flow become [tex]min/in^{1.9}[/tex]
A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi). (a) Compute the maximum and minimum stress levels.
Answer:
[tex]\sigma_{max} = 275\,MPa[/tex], [tex]\sigma_{min} = - 175\,MPa[/tex]
Explanation:
Maximum stress:
[tex]\sigma_{max}=\overline \sigma + \sigma_{a}\\\sigma_{max}= 50\,MPa + 225\,MPa\\\sigma_{max} = 275\,MPa[/tex]
Minimum stress:
[tex]\sigma_{min}=\overline \sigma - \sigma_{a}\\\sigma_{min}= 50\,MPa - 225\,MPa\\\sigma_{min} = - 175\,MPa[/tex]
A fatigue test was conducted in which the mean stress was 50 MPa (7,250 psi) and the stress amplitude was 225 MPa (32,625 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Answer:(a) The maximum and minimum stress levels are 275MPa and -175MPa respectively.
(b) The stress ratio is 0.6
(c) The magnitude of the stress range is 450MPa
Explanation:(a )In fatigue, the mean stress ([tex]S_{m}[/tex]) is found by finding half of the sum of the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e
[tex]S_{m}[/tex] = [tex]\frac{S_{max} + S_{min}}{2}[/tex] ------------------------(i)
Also, the stress amplitude (also called the alternating stress), [tex]S_{a}[/tex], is found by finding half of the difference between the maximum stress ([tex]S_{max}[/tex]) and minimum stress ([tex]S_{min}[/tex]) levels. i.e
[tex]S_{a}[/tex] = [tex]\frac{S_{max} - S_{min}}{2}[/tex] ------------------------(ii)
From the question,
[tex]S_{m}[/tex] = 50 MPa (7250 psi)
[tex]S_{a}[/tex] = 225 MPa (32,625 psi)
Substitute these values into equations(i) and (ii) as follows;
50 = [tex]\frac{S_{max} + S_{min}}{2}[/tex]
=> 100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex] -------------------(iii)
225 = [tex]\frac{S_{max} - S_{min}}{2}[/tex]
=> 450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex] -------------------(iv)
Now, solve equations (iii) and (iv) simultaneously as follows;
(1) add the two equations;
100 = [tex]S_{max}[/tex] + [tex]S_{min}[/tex]
450 = [tex]S_{max}[/tex] - [tex]S_{min}[/tex]
________________
550 = 2[tex]S_{max}[/tex] --------------------------------(v)
_________________
(2) Divide both sides of equation (v) by 2 as follows;
[tex]\frac{550}{2}[/tex] = [tex]\frac{2S_{max} }{2}[/tex]
275 = [tex]S_{max}[/tex]
Therefore, the maximum stress level is 275MPa
(3) Substitute [tex]S_{max}[/tex] = 275 into equation (iv) as follows;
450 = 275 - [tex]S_{min}[/tex]
[tex]S_{min}[/tex] = 275 - 450
[tex]S_{min}[/tex] = -175
Therefore, the minimum stress level is -175MPa
In conclusion, the maximum and minimum stress levels are 275MPa and -175MPa respectively.
===============================================================
(b) The stress ratio ([tex]S_{r}[/tex]) is given by;
[tex]S_{r}[/tex] = [tex]\frac{S_{min} }{S_{max} }[/tex] ----------------------------(vi)
Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vi)
[tex]S_{r}[/tex] = [tex]\frac{-175}{275}[/tex]
[tex]S_{r}[/tex] = 0.6
Therefore, the stress ratio is 0.6
===============================================================
(c) The magnitude of the stress range ([tex]S_{R}[/tex]) is given by
[tex]S_{R}[/tex] = | [tex]S_{max}[/tex] - [tex]S_{min}[/tex] | ------------------------------(vii)
Insert the values of [tex]S_{max}[/tex] and [tex]S_{min}[/tex] into equation (vii)
[tex]S_{R}[/tex] = | 275 - (-175) |
[tex]S_{R}[/tex] = 450MPa
Therefore, the magnitude of the stress range is 450MPa
===============================================================
Note:1 MPa = 145.038psi
Therefore, the values of the maximum and minimum stress levels, the stress range can all be converted from MPa to psi (pounds per inch square) by multiplying the values by 145.038 as follows;
[tex]S_{max}[/tex] = 275MPa = 275 x 145.038psi = 39885.45psi
[tex]S_{min}[/tex] = -175MPa = -175 x 145.038psi = 25381.65psi
[tex]S_{R}[/tex] = 450MPa = 450 x 145.038psi = 65267.1psi
A pipe of inner radius R1, outer radius R2 and constant thermal conductivity K is maintained at an inner temperature T1 and outer temperature T2. For a length of pipe L find the rate at which the heat is lost and the temperature inside the pipe in the steady state.
Answer:
Heat lost from the cylindrical pipe is given by the formula,
[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]
Temperature distribution inside the cylinder is given by,
[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex]
Explanation:
Inner radius = [tex]R_{1}[/tex]
Outer radius = [tex]R_{2}[/tex]
Constant thermal conductivity = K
Inner temperature = [tex]T_{1}[/tex]
Outer temperature = [tex]T_{2}[/tex]
Length of the pipe = L
Heat lost from the cylindrical pipe is given by the formula,
[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]------------ (1)
Temperature distribution inside the cylinder is given by,
[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex] ------------ (2)
An earthen trapezoidal channel (n = 0.025) has a bottom width of 5.0 m, side slopes of 1.5 horizontal: 1 vertical and a uniform flow depth of 1.10 m. In an economic study to remedy excessive seepage from the canal two proposals, (a) to line the sides only and, (b) to line the bed only are considered. If the lining is of smooth concrete (n = 0.012), calculate the equivalent roughness in the above two cases for a bottom slope of 0.005
Answer:
A. 0.020
B. 0.018
Explanation: check the attached file
Answer:
a. n=0.020 b. n=0.018
Explanation:
a.
Case 1: To line the sides only
n=(Σn₁¹°⁵P₁)^2/3/P^2/3
n = equivalent roughness
n₁=corresponding roughness coefficients
P=length
At the bed: n₁=0.025 and P₁=5m
At the sides: n₂=0.012 and P₂= 2*1.1*√1+1.5²=2.2*1.8=3.96m
P = P₁+P₂=8.96m
Equivalent roughness, n = [5*(0.025)^1.5+3.96*(0.012)^1.5]^2/3/(8.96)^2/3
n= [(5*0.00395)+(3.96*0.0013)]^2/3/4.317
n=0.0249^2/3/4.317
n=0.0842/4.317
n=0.0195
n=0.020
b.
Case 2: To line the bed only
P₁=5m n₁=0.012
P₂=3.96 n₂=0.025
P=8.96
Equivalent roughness n= [5*(0.012)^1.5+3.96*(0.025)^1.5]^2/3/(8.96)^2/3
n=[(5*0.0013)+(3.96*0.00395)]^2/3/4.317
n=0.0221^2/3/4.317
n=0.078/4.317
n=0.018
P4.36. Real inductors have series resistance associated with the wire used to wind the coil. Suppose that we want to store energy in a 10-H inductor. Determine the limit on the series resistance so the energy remaining after one hour is at least 75 percent of the initial energy.
Answer:
The limit on the series resistance is R ≤ 400μΩ
Explanation:
Considering the circuit has a series of inductance and resistance. The current current in the current in the circuit in time is
[tex]i(t) = Iie^{\frac{R}{L} t}[/tex] (li = initial current)
So, the initial energy stored in the inductor is
[tex]Wi = \frac{1}{2} Li^{2}_{i}[/tex]
After 1 hour
[tex]w(3600) = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }[/tex]
Knowing it is equal to 75
[tex]w(3600) = 0.75Wi = 0.75 \frac{1}{2} Li^{2}_{i} = \frac{1}{2} Li_{i}e^{-\frac{R}{L} 3600 }\\[/tex]
This way we have,
R = [tex]-10 \frac{ln 0.75}{2 * 3600} = 400[/tex] μΩ
Than, the resistance is R ≤ 400μΩ
A two-stage, solid-propellant sounding rocket has the following properties:
First stage: m0 ¼ 249:5 kg ; mf ¼ 170:1 kg ; _ me ¼ 10:61 kg;sIsp ¼ 235 s
Second stage: m0 ¼ 113:4 kg ; mf ¼ 58:97 kg; _ me ¼ 4:053 kg;sIsp ¼ 235 s
Delay time between burnout of first stage and ignition of second stage: 3 s.
As a preliminary estimate, neglect drag and the variation of earth’s gravity with altitude to calculate the maximum height reached by the second stage after burnout.
Answer:
Explanation: see attachment below
public interface Frac { /** @return the denominator of this fraction */ int getDenom(); /** @return the numerator of this fraction */ int getNum(); } Which of the interfaces, if correctly implemented by a Fraction class, would be sufficient functionality for a user of the Fraction class to determine if two fractions are equivalent
The getDenom() and getNum() methods would be sufficient functionality for a user of the Fraction class to determine if two fractions are equivalent, provided that the class also includes methods to compare fractions based on their numerators and denominators.
The getDenom() and getNum() methods are essential for accessing the denominator and numerator of a fraction. With these methods, users can directly compare the numerators and denominators of two fractions to determine if they are equivalent.
By retrieving the numerator and denominator values separately, users can perform mathematical operations or comparisons to check for equivalence without needing additional methods specifically for equivalence testing.
The Complete Question
Explain the importance of the getDenom() and getNum() methods in the Fraction class for determining if two fractions are equivalent.
A ship tows a submerged cylinder, 1.5 m in diameter and 22 m long, at U = 5 m/s in fresh water at 20°C. Estimate the towing power in kW if the cylinder is (a) parallel, and (b) normal to the tow direction.
Based on the available information, the the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
Given:
- Diameter of the cylinder: 1.5 m
- Length of the cylinder: 22 m
- Towing speed (U): 5 m/s
- Water temperature: 20°C
(a) Cylinder parallel to the tow direction:
Step 1: Calculate the drag force for the parallel orientation.
The drag force for a submerged cylinder in parallel orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- ρ is the density of fresh water at 20°C, which is approximately 998 kg/m³.
- C_D is the drag coefficient for a cylinder in parallel orientation, which is approximately 0.82.
- A is the cross-sectional area of the cylinder, which is π × D × L = π × 1.5 m × 22 m = 103.87 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 0.82 × 103.87 m² × (5 m/s)² = 131,000 N
Step 2: Calculate the towing power for the parallel orientation.
Towing power = Drag force × Towing speed
Towing power = 131,000 N × 5 m/s = 655,000 W = 655 kW
(b) Cylinder normal to the tow direction:
Step 3: Calculate the drag force for the normal orientation.
The drag force for a submerged cylinder in normal orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- C_D is the drag coefficient for a cylinder in normal orientation, which is approximately 1.2.
- A is the cross-sectional area of the cylinder, which is π × D × D/4 = π × 1.5 m × 1.5 m/4 = 1.77 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 1.2 × 1.77 m² × (5 m/s)² = 823,350 N
Step 4: Calculate the towing power for the normal orientation.
Towing power = Drag force × Towing speed
Towing power = 823,350 N × 5 m/s = 4,116,750 W = 4,116.75 kW
Therefore, the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
A 1.5-m-long aluminum rod must not stretch more than 1 mm andthe normal stress must not exceed 40 MPa when the rod is subjectedto a 3-kN axial load. Knowing that E = 70 GPa, determine therequired diameter of the rod.
Using the maximum stress limit and elongation criteria, the required diameter of the aluminum rod to withstand a 3-kN axial load without exceeding a 1 mm stretch and keeping the normal stress below 40 MPa is calculated to be approximately 9.8 mm.
Explanation:Determining the Required Diameter of an Aluminum Rod
To ensure a 1.5-m-long aluminum rod does not stretch more than 1 mm (0.001 m) under a 3-kN (3000 N) axial load while keeping the normal stress below 40 MPa (40×106 N/m2), we first calculate the cross-sectional area required using the formula for stress (σ) which is σ = F/A, where F is the force applied and A is the cross-sectional area. Given that the maximum allowable stress σ is 40 MPa, we can reorganize the formula to solve for A, the required cross-sectional area of the rod. This gives us A = F/σ.
Substituting the given values, A = 3000 N / (40×106 N/m2) = 7.5×10-5 m2. To ensure the rod does not exceed the maximum stretch limit when this force is applied, we must also consider the modulus of elasticity (E) for aluminum, which is given as 70 GPa (70×109 N/m2). The formula for elongation (ΔL) under a force is ΔL = (FL)/(AE), where L is the original length of the rod. Given the requirements, the diameter can be calculated from the cross-sectional area (A = πd2/4), where d is the diameter of the rod.
From the area calculated earlier, we can determine the diameter is required to be sufficiently large to maintain stress and elongation within specified limits. Rearranging A = πd2/4 to solve for d, we find d to be approximately 9.8 mm, considering the area necessary to keep stress below 40 MPa while allowing for the specified elongation limit.
When the rope is at an angle of α = 30°, the 1-kg sphere A has a speed v0 = 0.6 m/s. The coefficient of restitution between A and the 2-kg wedge B is 0.8 and the length of rope l = 0.9 m. The spring constant has a value of 1500 N/m and θ = 20°. Determine (a) the velocities of A and B immediately after the impact, (b) the maximum deflection of the spring, assuming A does not strike B again before this poin
Answer:
Explanation: see the pictures attached
Write a loop that prints the first 128 ASCII values followed by the corresponding characters (see the section on characters in Chapter 2). Be aware that most of the ASCII values in the range "0..31" belong to special control characters with no standard print representation, so you might see strange symbols in the output for these values.
Explanation:
Please refer to the attached image
Python Code:
Please refer to the attached image
Output:
Please refer to the attached image
Following are the program to print the first 128 ASCII values:
Program:#include <iostream>//header file
using namespace std;
int main()//main method
{
int i=1;//defining integer variable
for(i=1;i<=128;i++) //defining loop that prints the first 128 ASCII values
{
char x=(char)i;//defining character variable that converts integer value into ASCII code value
printf("%d=%c\n",i ,x);//print converted ASCII code value
}
return 0;
}
Program Explanation:
Defining header file.Defining the main method.Inside the method, an integer variable "i" is declared which uses the for loop that counts 1 to 128 character values.Inside the loop, a character variable "x" is declared that converts integer values into a character, and use a print method that prints converter value.Output:
Please find the attached file.
Find out more information about the ASCII values here:
brainly.com/question/3115410
Find the median path loss under the Hata model assuming fc = 900 MHz, ht = 20m, hr = 5 m and d = 100m for a large urban city, a small urban city, a suburb, and a rural area. Explain qualitatively the path loss differences for these 4 environments.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
Please find the attached file for the calculation of the 4 environment solutions:
Given:
[tex]f_c = 900\ MHz\\\\ h_t = 20\ m\\\\ h_r = 5\ m\\\\ d = 100\ m\\[/tex]
To find:
environments=?
Solution:
Please find the attached file.
Learn more about the Qualitative:
brainly.com/question/276942
brainly.com/question/18011951
sed is a multipurpose tool that combines the work of several filters. sed performs noninteractive operations on a data stream. sed has a host of features that allow you ti select lines and run instructions on them.
True
False
Answer:
The answer is True.
Explanation:
SED is a command in UNIX for stream editors that parses and transforms text, using a simple, compact programming language. So the answer is TRUE that sed is a multipurpose tool that combines the work of several filters and performs noninteractive operations on a data stream. sed has a host of features that allow you ti select lines and run instructions on them.
What is the average distance in microns an electron can travel with a diffusion coefficient of 25 cm^2/s if the electron lifetime is 7.7 microseconds. Three significant digits and fixed point notation.
Answer: The average distance the electron can travel in microns is 1.387um/s
Explanation: The average distance the electron can travel is the distance an exited electron can travel before it joins together. It is also called the diffusion length of that electron.
It is gotten, using the formula below
Ld = √DLt
Ld = diffusion length
D = Diffusion coefficient
Lt = life time
Where
D = 25cm2/s
Lt = 7.7
CONVERT cm2/s to um2/s
1cm2/s = 100000000um2/s
Therefore D is
25cm2/s = 2500000000um2/s = 2.5e9um2/s
Ld = √(2.5e9 × 7.7) = 138744.37um/s
Ld = 1.387e5um/s
This is the average distance the excited electron can travel before it recombine