Answer:
The correct option is A
Explanation:
Top down budgeting is a type of budget in which the senior management prepares a high level budget for the company. The senior management then allocates the amounts for the individual departments, which then use this number in preparing their own budget.
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]
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
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 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
Pin, Password, Passphrases, Tokens, smart cards, and biometric devices are all items that can be
used for Authentication. When one of these item listed above in conjunction with a second factor to validate authentication, it provides robust authentication of the individual by practicing which of the following?
A. Multi-party authentication
B. Two-factor authentication
C. Mandatory authentication
D. Discretionary authentication
Answer:
B. Two-factor authentication
Explanation:
As far as identity as been established, authentication must be carried out. There are various technologies and ways of implementing authentication, although most method falls under the same category.
The three major types of authentication.
a. Authentication through knowledge, what someone knows.
b. Authentication through possessions, what someone has.
c. Authentication by characteristic features, who a person is.
Logical controls that are in relations to these types of authentication are known as factors.
Two factor authentication has to do with the combination of 2 out of the 3 factors of authentication.
The general term used when more than one factor is adopted is Multi-party authentication
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
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 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 worker is asked to move 30 boxes from a desk onto a shelf within 3 minutes (assume there is enough space and no other lifting work within 8 hours). The shelf height is 55 inches, and the desk height is 30 inches. The initial horizontal distance from the box to the body is 7 inches. Assume that all the boxes are in the same size (8 inches edge, cube), each weigh 20 lb., and have well designed handles. Is there any lifting risk according NIOSH lifting guide
Answer:
LI = Lifting Index = 0.71
No lifting risk is involved
Explanation:
NIOSH Lifting Index
LI = Load Weight / Recommended Weight Limit
LI = L / RWL ............. Eq (A)
NIOSH Recommended Weight Limit equation is following,
RWL = LC * HM * VM * DM * AM * FM * CM ........... Eq (B)
Where,
LC = Load constant
HM = Horizontal multiplier
VM = Vertical multiplier
DM = Distance multiplier
AM = Asymmetric multiplier
FM = Frequency multiplier
CM = Coupling multiplier
Given data
V = 30 + (8/2) = 34 in
H = 7 in
D = 55 - 30 = 25 in
A = 0
F = 10 boxes/min
C = 1 = Good coupling
According to NIOSH lifting guide
LC = 51 lb
HM = 10/H
VM = 1 - {0.0075*(v-30)}
DM = 0.82 + (1.8/D)
AM = 1 - (0.0032*A)
FM = 0.45 (Table 5 from NIOSH lifting guide)
CM = 1 (Table 7 from NIOSH lifting guide)
Solution:
RWL = 51 * (10/7) * [1-{0.0075(34-30)}] * (0.82+ 1.8/25) * (1-0.0032*0) * 0.45 * 1
RWL = 51 * (10/7) * 0.97 * 0.892 * 1 * 0.45 * 1
RWL = 28.37 lb
Using equation A
LI = L / RWL
LI = 20 / 28.37
LI = 0.71
According to NIOSH lifting guide LI <= 1
So No lifting risk is involved
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.
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.
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]
If the wire has a diameter of 0.5 inin., determine how much it stretches when a distributed load of 140 lb/ftlb/ft acts on the beam. The material remains elastic. Express your answer to three significant figures and include appropriate units.
Answer:
δ_AB = 0.0333 in
Explanation:
Given:
- The complete question is as follows:
" The rigid beam is supported by a pin at C and an A−36
steel guy wire AB. If the wire has a diameter of 0.5 in.
determine how much it stretches when a distributed load of
w=140 lb / ft acts on the beam. The material remains elastic."
- Properties for A-36 steel guy wire:
Young's Modulus E = 29,000 ksi
Yield strength σ_y = 250 MPa
- The diameter of the wire d = 0.5 in
- The distributed load w = 140 lb/ft
Find:
Determine how much it stretches under distributed load
Solution:
- Compute the surface cross section area A of wire:
A = π*d^2 / 4
A = π*0.5^2 / 4
A = π / 16 in^2
- Apply equilibrium conditions on the rigid beam ( See Attachment ). Calculate the axial force in the steel guy wire F_AB
Sum of moments about point C = 0
-w*L*L/2 + F_AB*10*sin ( 30 ) = 0
F_AB = w*L*L/10*2*sin(30)
F_AB = 140*10*10/10*2*sin(30)
F_AB = 1400 lb
- The normal stress in wire σ_AB is given by:
σ_AB = F_AB / A
σ_AB = 1400*16 / 1000*π
σ_AB = 7.13014 ksi
- Assuming only elastic deformations the strain in wire ε_AB would be:
ε_AB = σ_AB / E
ε_AB = 7.13014 / (29*10^3)
ε_AB = 0.00024
- The change in length of the wire δ_AB can be determined from extension formula:
δ_AB = ε_AB*L_AB
δ_AB = 0.00024*120 / cos(30)
δ_AB = 0.0333 in
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 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.
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
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.
Assume that a phase winding of the synchronous machine of Problem 4.11 consists of one 5-turn, full-pitch coil per pole pair, with the coils connected in series to form the phase winding. If the machine is operating at rated speed and under the operating conditions of Problem 4.11, calculate the rms generated voltage per phase.
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:
The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-1000t V,l = 50e -IOOOt mAa) Fund the maximum value of the power delivered to the circuit.b) Find the total energy delivered to the element.
The maximum power delivered to the circuit is found to be 3.75W, occurring at the start (t = 0). Calculating the total energy requires integrating the power over time, involving exponentials reducing over time to approach a finite total energy delivered.
Explanation:The question involves calculating the maximum power and the total energy delivered to a circuit element, where the voltage V and current I are given as time-dependent expressions.
Maximum Power Delivered
Power is calculated using P = IV. Inserting the given expressions for V and I,
P(t) = V(t) × I(t) = (75 - 75e^{-1000t}) × 50e^{-1000t} mA.
To find the maximum power, we would differentiate P(t) with respect to t and set the derivative equal to zero. However, because the setup includes exponential decay functions, their maximum product occurs at t = 0 for these specific functions.
Pmax = 75V × 50mA = 3.75W.
Total Energy Delivered
The total energy delivered can be found by integrating the power over time:
Energy = ∫ P(t) dt.
Considering the specific forms of V and I provided, this becomes an integral of the product of two exponentials, leading to an expression that evaluates the total energy consumed by the circuit over time.
Given the nature of exponential decay in V and I, the energy delivered approaches a finite value as t approaches infinity.
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
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
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.
- Draw the internal representation for the following lisp list(s). • (cons '( (Blue) (Yellow () Red) () Orange) '() ) • (cons '(Red (Yellow Blue)) '(() Yellow Orange) ) • (cons '(()(Green Blue)) '(Red (Yellow ()) Blue) )
Answer:
Explanation: see attachment
) You are using a load cell to measure the applied load to a test in the Civil Engineering Structures Lab. What should you do if the measured load does not return to zero? How should you troubleshoot this to determine if this is a load cell or a mechanical problem?
Answer:
Insulation Resistance Tests
Explanation:
An insulation resistance test is carried out when there are unstable readings and random changes in the zero balance point of the load cell. It is done by measuring the resistance between the load cell body and all its connected wires, as follows:
First, disconnect the load cell from the summing box and indicator panel.
Connect all the input, output and sense (if equipped) wires together.
Measure the insulator resistance between the connected wires and the load cell body with a mega-ohmmeter.
Measure the insulation resistance between the connected wires and the cable shield.
Measure the insulation resistance between the load cell body and the cable shield.
The insulation resistance should match the value in the product’s load cell datasheet. A lower value shows an electrical leakage caused by moisture; this causes short circuits, giving unstable load cell outputs.
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).
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:
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g For this project you are required to perform Matrix operations (Addition, Subtraction and Multiplication). For each of the operations mentioned above you have to make a function in addition to two other functions for ‘Inputting’ and ‘Displaying’ the Matrices in Row Order (rows are populated left to right, in sequence). In total there will be 5 functions: 3 for the operations and 2 functions for inputting and displaying the data.
Answer:
C++ code is explained below
Explanation:
#include<iostream>
using namespace std;
//Function Declarations
void add();
void sub();
void mul();
//Main Code Displays Menu And Take User Input
int main()
{
int choice;
cout << "\nMenu";
cout << "\nChoice 1:addition";
cout << "\nChoice 2:subtraction";
cout << "\nChoice 3:multiplication";
cout << "\nChoice 0:exit";
cout << "\n\nEnter your choice: ";
cin >> choice;
cout << "\n";
switch(choice)
{
case 1: add();
break;
case 2: sub();
break;
case 3: mul();
break;
case 0: cout << "Exited";
exit(1);
default: cout << "Invalid";
}
main();
}
//Addition Of Matrix
void add()
{
int rows1,cols1,i,j,rows2,cols2;
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
//Taking First Matrix
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
//Printing 1st Matrix
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
//Taking Second Matrix
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
//Displaying second Matrix
cout << "\n";
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m2[i][j] << " ";
cout << "\n";
}
//Displaying Sum of m1 & m2
if(rows1 == rows2 && cols1 == cols2)
{
cout << "\n";
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j]+m2[i][j] << " ";
cout << "\n";
}
}
else
cout << "operation is not supported";
main();
}
void sub()
{
int rows1,cols1,i,j,k,rows2,cols2;
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\n";
//Displaying Subtraction of m1 & m2
if(rows1 == rows2 && cols1 == cols2)
{
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j]-m2[i][j] << " ";
cout << "\n";
}
}
else
cout << "operation is not supported";
main();
}
void mul()
{
int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];
cout << "\nmatrix1 # of rows: ";
cin >> rows1;
cout << "\nmatrix1 # of columns: ";
cin >> cols1;
int m1[rows1][cols1];
for(i=0;i<rows1;i++)
for(j=0;j<cols1;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m1[i][j];
cout << "\n";
}
cout << "\n";
for(i=0;i<rows1;i++)
{
for(j=0;j<cols1;j++)
cout << m1[i][j] << " ";
cout << "\n";
}
cout << "\nmatrix2 # of rows: ";
cin >> rows2;
cout << "\nmatrix2 # of columns: ";
cin >> cols2;
int m2[rows2][cols2];
for(i=0;i<rows2;i++)
for(j=0;j<cols2;j++)
{
cout << "\nEnter element (" << i << "," << j << "): ";
cin >> m2[i][j];
cout << "\n";
}
cout << "\n";
//Displaying Matrix 2
for(i=0;i<rows2;i++)
{
for(j=0;j<cols2;j++)
cout << m2[i][j] << " ";
cout << "\n";
}
if(cols1!=rows2)
cout << "operation is not supported";
else
{
//Initializing results as 0
for(i = 0; i < rows1; ++i)
for(j = 0; j < cols2; ++j)
mul[i][j]=0;
// Multiplying matrix m1 and m2 and storing in array mul.
for(i = 0; i < rows1; i++)
for(j = 0; j < cols2; j++)
for(k = 0; k < cols1; k++)
mul[i][j] += m1[i][k] * m2[k][j];
// Displaying the result.
cout << "\n";
for(i = 0; i < rows1; ++i)
for(j = 0; j < cols2; ++j)
{
cout << " " << mul[i][j];
if(j == cols2-1)
cout << endl;
}
}
main();
}
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)
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
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