Answer:
Technician B says that a ratchet is used to loosen fasteners that are very tight.
Explanation:
A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.
Answer:
A ratchet wrench is usually used to loosen and tighten parts like steering linkages, tie rod end clamps and muffler clamps. Basically when nut is used on long thread a ratchet wrench is being used.
hence the technician B (option b) is correct.
Explanation:
A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2 mA is required when the device is biased in the saturation region at VGS = 3 V. Determine the required channel width of the device.
Answer:
[tex]W= 3.22 \mu m[/tex]
Explanation:
the transistor In saturation drain current region is given by:
[tex]i_D}=K_a(V_{GS}-V_{IN})^2[/tex]
Making [tex]K_a[/tex] the subject of the formula; we have:
[tex]K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}[/tex]
where;
[tex]i_D = 1.2m[/tex]
[tex]V_{GS}= 3.0V[/tex]
[tex]V_{TN} = 0.6 V[/tex]
[tex]K_a=\frac {1.2m} {(3.0 - 0.6)^2}[/tex]
[tex]K_a = 208.3 \mu A/V^2[/tex]
Also;
[tex]k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}[/tex]
where:
[tex]\mu n (\frac{cm^2}{V-s} ) = 600[/tex]
[tex]\epsilon _{ox}=3.9*8.85*10^{-14}[/tex]
[tex]{t_{ox}(cm)=200*10^{-8}[/tex]
substituting our values; we have:
[tex]k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}[/tex]
[tex]k'_n}=103.545 \mu A/V^2[/tex]
Finally, the width can be calculated by using the formula:
[tex]W= \frac{2LK_n}{k'n}[/tex]
where;
L = [tex]0.8 \mu m[/tex]
[tex]W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}[/tex]
[tex]W= 3.22 \mu m[/tex]
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]
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 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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The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic turbines of this dam to produce 100 MW of power if the turbines are 100 percent efficient?
Answer:
m' = 4948.38 kg/s
Explanation:
For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion
GIVEN THAT
Power = 100 MW
rate of Potential energy = (m')*g*h
100*10^6 = (m')*9.81*206
m' = 4948.38 kg/s
Answer:
49.484 m³ / s
Explanation:
Volume flow rate = Power in W / (efficiency × density × height × acceleration due to gravity)
Volume flow rate = 100 × 10⁶ / ( 1 × 1000 kg/m³ × 206 m × 9.81 m/s²)
V = 49.484 m³ / s
3. Suppose that a class named Bicycle contains a private nonstatic integer named height, a public nonstatic String named model, and a public static integer named wheels. Which of the following are legal statements in a class named BicycleDemo that has instantiated an object as Bicycle myBike new Bicycle C);? f. Bicycle. model Hurricane a. myBike height 26; b. my Bike model Cyclone g. Bicycle. int 3 3; c. myBike Wheels 3 d. my Bike .model 108; i. Bicycle wheels 2 e. Bicycle height 24; j. Bicycle yourBike myBike
Answer:
The solution to the given problem is provided below.
Explanation:
a.) myBike.height = 26; Not Legal statement
b.) myBike.model = “Cyclone”: Legal statement
c.) myBike.wheels = 3; Legal statement
d.) myBike.model = 108; Not legal statement
e.) Bicycle.height = 24; Not Legal statement
f.) Bicycle.model = “Hurricane”; Not legal statement
g.) Bicycle.int = 3; Not Legal statement
h.) Bicycle.model = 108; Not Legal Statement
i.) Bicycle.wheels = 2; Legal Statement
j.) Bicycle yourBike = myBike; Legal Statement
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.
. A steam turbine operates between 500°C and 3.5 MPa to 200°C and 0.3 MPa. If the turbine generates 750 kW and the heat loss is 100 kW, what is the flow rate of steam through the turbine?
Answer:
1.757 kg/s
Explanation:
According to the First Law of Thermodynamics, the physical model for a turbine working at steady state is:
[tex]-\dot Q_{out} - \dot W_{out} + \dot m \cdot (h_{in}-h_{out})=0[/tex]
The flow rate of steam is:
[tex]\dot m = \frac{\dot Q_{out}+\dot W_{out}}{h_{in}-h_{out}}[/tex]
Water enters and exits as superheated steam. After looking for useful data in a property table for superheated steam, specific enthalpies at inlet and outlet are presented below:
[tex]h_{in} = 3451.7 \frac{kJ}{kg} \\h_{out} = 2967.9 \frac{kJ}{kg} \\[/tex]
Finally, the flow rate is calculated:
[tex]\dot m = \frac{100 kW + 750 kW}{3451.7 \frac{kJ}{kg} - 2967.9 \frac{kJ}{kg}}\\\dot m =1.757 \frac{kg}{s}[/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
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 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:
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]
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.
Silicon carbide nanowires of diameter 15 nm can be grown onto a solid silicon carbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Silicon carbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array of nanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronics packages in which an extremely small, 10 μm × 10 μm electronics device is sandwiched between two 100-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at 20°C. A heat transfer coefficient of 1.0 × 105 W/m2·K exists on the top and bottom of the unfinned package and on all surfaces of the exposed silicon carbide fins, which are each 300 nm long. Each nano-heat sink includes a 50 × 50 array of nanofins. Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperature is maintained at 95°C for (a) the unfinned and (b) the finned packages
Answer:
Please find attached file for complete answer.
Explanation:
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 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)
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
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
Checkpoint 4.72 Write a statement that uses a conditional expression that determines if the credits variable is less than 0. If the condition is true, assign the value 0 to the credits variable, otherwise the value of the credits variable should remain unchanged.
Answer:
See step to step explanations for answer.
Explanation:
This is the expression to check if credits is less than 0 assuming variable name is credits:
if(credits<0)
{
credits = 0;
}
Q 4.69:
this is the expression checking the name is Swordfish or not
if(name=="Swordfish")
{
cout<<"We have a match";
}
Q 4.67
This is the expression to check the character assuming that character is ch
if(ch>='1' && ch<='9')
{
cout<<"Digit detected";
}
Q 4.59:
This is the statement to check if the number is between 0 and 500 inclusive . Assuming the variable is n
if(n>=0 && n<=500)
{
cout<<"The number is valid";
}
To address the programming question about conditional expressions, a ternary operator can be used to set the variable 'credits' to 0 if its value is less than 0, and otherwise keep the current value.
Explanation:The student's question pertains to the use of a conditional expression in a programming context. A conditional expression evaluates a condition and based on its truth value, executes one of two expressions. The statement required should check if the credits variable is less than 0 and if so, assign the value 0 to credits; otherwise, it should leave the value of credits as is.
An example in a generic programming language would be:
credits = 0 if credits < 0 else creditsThis line of code is known as a ternary operator or conditional assignment. It reads as 'Set credits to 0 if credits is less than 0; otherwise, keep the value of credits.'
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
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]
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
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
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
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]
The portable lighting equipment for a mine is located 100 meters from its dc supply source. The mine lights use a total of 5 kW and operate at 120 V dc. Determine the required cross-sectional area of the copper wires used to connect the source to the mine lights if we require that the power lost in the copper wires be less than or equal to 5 percent of the power required by the mine lights.
To calculate the required cross-sectional area of copper wires needed to connect the DC supply source to the mine lights with less than 5% power loss, calculate the permissible power loss, and then use the relationship between power loss, current, and wire resistance. Considering the resistivity of copper and the round trip length of the wire, the formula A = ρL/(δP/I²) determines the necessary wire thickness.
Explanation:To determine the required cross-sectional area of copper wires for the lighting equipment, we need to ensure the power loss (δP) is no more than 5% of the lights' power usage (P), which is 5 kW (5000 W). As the power loss in wires is given by δP = I²R, where I is the current and R is the resistance of the wire, we must first calculate the current using P = IV, giving I = P/V = 5000W/120V = 41.67 A. The power loss allowed is thus 5% of 5000 W, equating to 250 W. Given δP, we can find R using δP = I²R, which gives R = δP/I².
The resistance of a copper wire is also given by R = ρL/A, where ρ is the resistivity of copper (1.68 x 10^-8 Ωm), L is the length of the wire (200 meters round trip), and A is the cross-sectional area we need to find. Equating the two expressions for R and solving for A gives A = ρL/(δP/I²). Substituting the given and calculated values yields the required cross-sectional area. Finally, as resistance depends on the entire length of the circuit, remember to double the distance to account for both the outgoing and return paths.
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
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
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.
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).