In this lab, you will be creating a class that implements the Rule of Three (A Destructor, A Copy Constructor, and a Copy Assignment Operator). You are to create a program that prompts users to enter in contact information, dynamically create each object, then print the information of each contact to the screen. Some code has already been provided for you. To receive full credit make sure to implement the following:
Default Constructor - set Contact id to -1
Overloaded Constructor - used to set the Contact name, phoneNumber and id (Should take in 3 parameters)
Destructor
Copy Constructor
Copy Assignment Operator
Any other useful functions (getters/setters)
Main.cpp
#include
#include
#include "Contact.h"
using namespace std;
int main() {
const int NUM_OF_CONTACTS = 3;
vector contacts;
for (int i = 0; i < NUM_OF_CONTACTS; i++) {
string name, phoneNumber;
cout << "Enter a name: ";
cin >> name;
cout << "Enter a phoneNumber; ";
cin >> phoneNumber;
// TODO: Use i, name, and phone number to dynamically create a Contact object on the heap
// HINT: Use the Overloaded Constructor here!
// TODO: Add the Contact * to the vector...
}
cout << "\n\n----- Contacts ----- \n\n";
// TODO: Loop through the vector of contacts and print out each contact info
// TODO: Make sure to call the destructor of each Contact object by looping through the vector and using the delete keyword
return 0;
}
Contact.h
#ifndef CONTACT_H
#define CONTACT_H
#include
#include
using std::string;
using std::cout;
class Contact {
public:
Contact();
Contact(int id, string name, string phoneNumber);
~Contact();
Contact(const Contact& copy);
Contact& operator=(const Contact& copy);
private:
int *id = nullptr;
string *name = nullptr;
string *phoneNumber = nullptr;
};
#endif
Contact.cpp
#include "Contact.h"
Contact::Contact() {
this->id = new int(-1);
this->name = new string("No Name");
this->phoneNumber = new string("No Phone Number");
}
Contact::Contact(int id, string name, string phoneNumber) {
// TODO: Implement Overloaded Constructor
// Remember to initialize pointers on the heap!
}
Contact::~Contact() {
// TODO: Implement Destructor
}
Contact::Contact(const Contact ©) {
// TODO: Implement Copy Constructor
}
Contact &Contact::operator=(const Contact ©) {
// TODO: Implement Copy Assignment Operator
return *this;
}
Answers
Answer:
Explanation:
============== main.cpp =======================
#include <iostream>
using namespace std;
#include <iostream>
#include <vector>
#include "Contact.h"
using namespace std;
int main() {
const int NUM_OF_CONTACTS = 3;
vector<Contact *> contacts;
for (int i = 0; i < NUM_OF_CONTACTS; i++) {
string name, phoneNumber;
cout << "Enter a name: ";
cin >> name;
cout << "Enter a phoneNumber: ";
cin >> phoneNumber;
// Use i, name, and phone number to dynamically create a Contact object on the heap
// HINT: Use the Overloaded Constructor here!
Contact * newContact = new Contact(i+1,name,phoneNumber);
// Add the Contact * to the vector...
contacts.push_back(newContact);
}
cout << "\n\n----- Contacts ----- \n\n";
// Loop through the vector of contacts and print out each contact info
std::vector<Contact *>::iterator itrContact;
for(itrContact = contacts.begin(); itrContact !=contacts.end(); itrContact++)
{
cout<< " Id : " << (*itrContact)->getId();
cout<< " Name : " << (*itrContact)->getName();
cout<< " Phone-Number : " << (*itrContact)->getPhoneNumber() <<endl;
}
// Make sure to call the destructor of each Contact object by looping through the vector and using the delete keyword
for(int i = 0; i < NUM_OF_CONTACTS; i++)
{
delete contacts[i];
}
return 0;
}
================== contact.h =====================
#ifndef CONTACT_H
#define CONTACT_H
#include <string>
#include <iostream>
using std::string;
using std::cout;
class Contact {
public:
Contact();
Contact(int id, string name, string phoneNumber);
~Contact();
Contact(const Contact& copy);
Contact& operator=(const Contact& copy);
// getters
int getId() const;
string getName() const;
string getPhoneNumber() const;
//setters
void setId(int id);
void setName(string name);
void setPhoneNumber(string phoneNumber);
private:
int *id = nullptr;
string *name = nullptr;
string *phoneNumber = nullptr;
};
#endif
================= contact.cpp ========================
#include "Contact.h"
Contact::Contact() {
this->id = new int(-1);
this->name = new string("No Name");
this->phoneNumber = new string("No Phone Number");
}
Contact::Contact(int id, string name, string phoneNumber) {
// Implement Overloaded Constructor
// Remember to initialize pointers on the heap!
this->id = new int(id);
this->name = new string(name);
this->phoneNumber = new string(phoneNumber);
}
Contact::~Contact() {
// Implement Destructor
if(this->id != nullptr)
delete this->id;
if(this->name != nullptr)
delete this->name;
if(this->phoneNumber != nullptr)
delete this->phoneNumber;
}
Contact::Contact(const Contact ©) {
// Implement Copy Constructor
this->id = new int(copy.getId());
this->name = new string(copy.getName());
this->phoneNumber = new string(copy.getPhoneNumber());
}
Contact &Contact::operator=(const Contact ©) {
// Implement Copy Assignment Operator
if(this == ©) // Checks for self Assignment
return *this;
this->id = new int(copy.getId());
this->name = new string(copy.getName());
this->phoneNumber = new string(copy.getPhoneNumber());
return *this;
}
// getters
int Contact::getId() const
{
return *(this->id);
}
string Contact::getName() const
{
return *(this->name);
}
string Contact::getPhoneNumber() const
{
return *(this->phoneNumber);
}
//setters
void Contact::setId(int id)
{
*(this->id) = id;
}
void Contact::setName(string name)
{
*(this->name) = name;
}
void Contact::setPhoneNumber(string phoneNumber)
{
*(this->phoneNumber) = phoneNumber;
}
====================Output =========================
is attached below
Related Questions
A convergentâdivergent nozzle has an exit area to throat area ratio of 4. It is supplied with air from a large reservoir in which the pressure is kept at 500 kPa and it discharges into another large reservoir in which the pressure is kept at 10 kPa. Expansion waves form at the exit edges of the nozzle causing the discharge flow to be directed outward.
(a) Find the angle that the edge of the discharge flow makes to the axis of the nozzle.
Answers
Answer:
Angle of discharge make at the edge of tube=64.9 degrees.
Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air conditioner and returns to the space at 25°C dry bulb and 6.5°C dew point temperature (ϕ2≠100%). If there is any, the condensate leaves the air conditioner at the same temperature of the air leaving the cooling coils (i.e., different from our previous assumption that condensate has same temperature of the cooling coils). The volume flow rate of the air returned to the workspace is 1000 m3 /min (Rair=0.287 kJ/kg∙K). Atmospheric pressure is 101.325 kPa. Compute
Answers
Answer:
See explaination
Explanation:
The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.
Kindly check attachment for the step by step solution of the given problem.
"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United States.Though long and wide (30 miles wide at the Potomac River), the bay is very shallow, with an average depth of only 28 feet." The Bay's maximum depth is 174 ft. The hydraulic model of the Chesapeake Bay was built with a model length ratio Lr=1/1000
a. How wide was the model Bay [ft] at the Potomac River?
b. The Bay Bridge is 4.3 miles long; how long was it in the model [ft]?
c. If the hydraulic model occupied about 8 acres, approximately what is the real-world (prototype) area represented by the model [square miles]?
d. At this scale, what would be the average depth and maximum depth of the model Bay? [give answers in both ft and inches)
Answers
Answer:
see explaination
Explanation:
Part a) Width of bay at Potomac River:
Given Data:
· Actual Width at Potomac River = 30 miles
· Bay Model Length Ratio Lr = 1/1000
In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.
Length of any part in model = Actual length x Lr
Hence,
Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles
Since 1 mile = 5280 ft
Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft
Part b) Model Length of bay bridge in model:
Given Data:
· Actual Length of bay bridge = 4.3 miles
· Bay Model Length Ratio Lr = 1/1000
Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles
Since 1 mile = 5280 ft
Model Length in feet = 0.0043 x 5280 = 22.704 ft
Part c) Model Length of bay bridge in model:
Given Data:
· Model Area = 8 acre
· Bay Model Length Ratio Lr = 1/1000
Model Area = Actual Area x Lr x Lr
8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000
Since 1 square mile = 640 acre,
Actual Area in square miles = 8,000,000/640 = 12,500 square miles
Part d) Average and maximum depth of model:
Given Data:
· Actual Average depth = 28 ft
· Actual Maximum depth = 174 ft
· Bay Model Length Ratio Lr = 1/1000
Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet
Since 1 ft = 12 inch
Model average depth in inch = 0.028 x 12 = 0.336 in
Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet
Since 1 ft = 12 inch
Model maximum depth in inch = 0.174 x 12 = 2.088 in
How many D-cell batteries would it take to power a human for 1 day?
All numbers must be entered as 5000 or 5e3 or 5.0e3 and not with commas as in "5,000" and not as fractions as in "3/4" and not as percentages as in "70%".
Estimate the recomended daily food intake (in Food Calories).
Estimate the daily energy (in joule) a human needs.
Estimate the voltage (in volt) of one D cell battery.
Estimate the charge (in amp-hours) of one D cell battery.
Estimate the energy of one D cell battery (in watt-hour).
Estimate the number of D-cell batteries it takes to power a human for 1 day.
Answers
Answer:
it would take approximately 232 to 258 D cell batteries to power a human for 1 day.
Explanation:
Estimate the recommended daily food intake (in Food Calories).
An average adult man requires between 2000 to 3000 calories per day.
Estimate the daily energy (in joule) a human needs.
As we know 1 food calorie is equal to 4.184 Joules of energy
2000*4.184 to 3000*4.184
8368 to 12552 Joules
But for engineering calculations 1 food calorie is equal to 1000 engineering calories, so
8368*1000 to 12552*1000
8368000 to 12552000 Joules
Estimate the voltage (in volt) of one D cell battery.
The voltage of a D cell battery is around 1.5 Volts
Estimate the charge (in amp-hours) of one D cell battery.
The amp-hours of a D cell battery varies with the manufacturing company, the typical amp-hours are in the range of 6 to 10 amp-hours.
Estimate the energy of one D cell battery (in watt-hour).
Energy in watt-hour is given by
voltage*amp-hour
1.5*6 to 1.5*10
9 to 15 watt-hour
Estimate the number of D-cell batteries it takes to power a human for 1 day.
First let us calculate the energy in a D cell battery,
1 watt-hour is equal to 3600 Joules
9*3600 to 15*3600
32400 to 54000 Joules
The number of D cell batteries required is found by dividing the energy need of a human by the energy stored in a D cell battery.
12552000/54000 to 8368000/32400
232 to 258 batteries
Therefore, it would take approximately 232 to 258 D cell batteries to power a human for 1 day.
A three-point bending test is performed on a glass specimen having a rectangular cross section of height d = 5.4 mm (0.21 in.) and width b = 12 mm (0.47 in.); the distance between support points is 43 mm (1.69 in.).(a) Compute the flexural strength if the load at fracture is 292 N (66 lbf). (b) The point of maximum deflection, ?y, occurs at the center of the specimen and is described by where E is the modulus of elasticity and I is the cross-sectional moment of inertia. Compute ?y at a load of 266 N (60 lbf). Assume that the elastic modulus for the glass is 60 GPa. What is the flexural strength of this sample in MPa if the load in part (a) is applied? What is I, the moment of inertia, in m4? What is the deflection, ?y, in mm, at the load given in the problem?
Answers
Answer:
5.21e-2mm
Explanation:
Please see attachment
The flexural strength of the glass specimen under the given load is approximately 84.4 MPa. The cross-sectional moment of inertia is about 0.655 *10^-12 m^4. When subjected to a load of 266N, the deflection at the center of the specimen is approximately 0.87 mm.
Explanation:The subject matter of the question refers to concepts from material science and mechanical engineering, specifically pertaining to the calculation of flexural strength, the moment of inertia, and deflection.
Flexural strength or 'Modulus of rupture' is calculated with the formula σf = 3FL / 2bd^2, where F is the fracture load, L the distance between support points, b the specimen's width, and d its height. Substituting the given values, we get σf = (3*292 * 43*10^-3) / (2 * 12 *10^-3 * (5.4 *10^-3)^2), which gives us about 84.4 MPa.
The moment of inertia for a rectangular cross-section is given by I = b*d^3 / 12. Substituting the given values, we get I = 12*10^-3 * (5.4 *10^-3)^3 / 12, which gives approximately 0.655 *10^-12 m^4.
Deflection, typically denoted by the Greek letter δ (or in this case, Υ), is computed with the formula Υ = FL^3 / 48EI. Given that the modulus of elasticity (E) for glass is 60 GPa or 60*10^9 Pa and we already calculated I, we substitute all values to get Υ = (266 * (43*10^-3)^3) / (48 * 60*10^9 * 0.655*10^-12) which gives us approximately 0.87 mm.
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11.4 Compute the volume percent of graphite VGr in a 3.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.
Answers
Answer:
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
Explanation:
Consider 100 grams 3.5% C cast iron.
Mass of carbon in 100 grams of cast iron = [tex]m_1=100 g\times \frac{3.5 }{100}=3.5 g[/tex]
Mass of iron in cast iron =[tex]m_2[/tex] = 100 g - 3.5 g = 96.5 g
Density of graphite = [tex]d_1=2.3g/cm^3[/tex]
Volume of the graphite = [tex]v_1[/tex]
[tex]v_1=\frac{m_1}{d_1}=\frac{3.5 g}{2.3g/cm^3}=1.52 cm^3[/tex]
Density of iron= [tex]d_2=7.9 g/cm^3[/tex]
Volume of the iron = [tex]v_2[/tex]
[tex]v_2=\frac{m_2}{d_2}=\frac{96.5 g}{7.9 g/cm^3}=12.22 cm^3[/tex]
Total volume of the cast iron ,V=[tex]v_1+v_2=1.52 cm^3+12.22 cm^3=13.74 cm^3[/tex]
Volume percent of the graphite :
[tex]=\frac{v_1}{V}\times 100[/tex]
[tex]=\frac{1.52 cm^3}{13.74 cm^3}\times 100=11.06\%\approx 11\%[/tex]
The volume percent of graphite in a 3.5 wt% C cast iron is 11%.
The volume percent of graphite VGr in a 3.5 wt% C cast iron is;
V_gr = 11.08%
We are given;
Density of Ferrite; ρₐ = 7.9 g/cm³
Density of Graphite; [tex]\rho _{g}[/tex] = 2.3 g/cm³
Weight percent of C cast iron; C₀ = 3.5%
Now the formula for the mass fraction of ferrite is;
Wₐ = [tex]\frac{C_{g} - C_{0}}{C_{g} - C_{a}}[/tex]
If we consider 100 g of 3.5 wt% C cast iron, then [tex]C_{g}[/tex] = 100 and Cₐ = 0
Thus;
Wₐ = [tex]\frac{100 - 3.5}{100 - 0}[/tex]
Wₐ = 0.965
Thus, mass fraction of graphite is;
[tex]W _{g}[/tex] = 1 - Wₐ
[tex]W _{g}[/tex] = 1 - 0.965
[tex]W _{g}[/tex] = 0.035
Formula for the volume percent of graphite is;
[tex]V_{gr} = \frac{\frac{W_{g}}{\rho_{g}}}{\frac{W_{a}}{\rho_{a}} + \frac{W_{g}}{\rho_{g}}} * 100%[/tex]
Plugging in the relevant values gives;
V_gr = [(0.035/2.3)/((0.965/7.9) + (0.035/2.3))]
V_gr = 0.01521739/(0.1221518987 + 0.01521739)
V_gr = 11.08%
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Between 1975 and 1985, the volume of all iron and steel in a given automobile model decreased from 0.165 m3m3 to 0.118 m3m3 . In the same time frame, the volume of all aluminum alloys increased from 0.012 m3m3 to 0.023 m3m3 . Using the densities of pure FeFe and AlAl, estimate the mass reduction resulting from this trend in materials substitution.
Answers
Answer: 340.19kg
Explanation:
The reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. It is referred to as a quantity which allows the two-body problem to be solved as if it were a one-body problem. You should note, however, that the mass determining the gravitational force is not reduced. In the computation or calculation, one mass can be replaced with the reduced mass, if this is compensated by replacing the other mass with the sum of both masses.
It has the dimensions of mass, and SI unit kg.
Given two bodies, first with mass m1 and the other with mass m2, the equivalent one-body problem, with the position of one body with respect to the other referred to as the unknown, is that of a single body of mass.
The density of Iron (Fe) = 7.87Mg/m^3
The volume of Iron decreased from= (0.165 - 0.118) m^3 =0.047m^3
Therefore,
The weight of Iron decreased= (0.047m^3) x (7.87Mg/m^3) = 0.36989Mg
The density of aluminium= 2.70Mg/m^3
The volume of aluminum increased= (0.023 - 0.012)m^3= 0.011m^3
The weight of aluminium is= (0.011m^3) × (2.70Mg/m^3) = 0.0297Mg
Therefore,
The mass reduction is:
The weight of Iron decreased minus the weight of Aluminium:
0.36989Mg - 0.0297Mg= 0.34019Mg
0.34019Mg × 1000
= 340.19Kg
Therefore, the mass reduction resulting from the trend is= 340.19kg
A 15.0 in by 2.0 in work part is machined in a face milling operation using a 2.5 in diameter cutter with a single carbide insert. The machine is set for a feed of 0.010 in/tooth and a depth of 0.20 in. If a cutting speed of 400 ft/min is used, the tool lasts for 3 pieces. If a cutting speed of 200 ft/min it used, the tool lasts for 12 parts. Determine the Taylor tool life equation. (20 points)
Answers
Answer:
Explanation:
N1 =v/πD = 400(12)/2.5π = 611 rev/min fr = Nfnt = 611(0.010)(1) = 6.11 in/min A=D/2 = 2.50 / 2 = 1.25 Tm = (L+2A)/fr = (15 + 2(1.25))/6.11 = 2.863 min T1 = 3Tm = 3(2.863) = 8.589 min when v1 = 400 ft/min N2 = 200(12)/2.5π = 306 rev/min fr = Nfnt = 306(0.010)(1) = 3.06 in/min Tm = (15 + 2(1.25))/3.06 = 5.727 min T2 = 12Tm = 12(5.727) = 68.724 min when v2 = 200 ft/min n = ln (v1/v2)/ln(T2/T1) = ln (400/200)/ln (68.724/8.589) = 0.333 C = vTn = 400(8.589 )0.333 = 819
Answer:
C = 787.2
Explanation:
The Taylor tool life is referred to as the duration of the actual cutting time after which the tool can no longer be used.
To Quantify the end of a tool life we equate it to a limit on the maximum acceptable flank wear.
For the tool life equation, With the slope, n and intercept, c, Taylor derived the simple equation as
VTn = C where.
n is called, Taylor's tool life exponent. The values of both 'n' and 'c' depend mainly upon the tool-work materials and the cutting environment (cutting flake application)
Please kindly go through the attached file for a step by step solution to how the answer to your question is solved.
Using a forked rod, a smooth 3-lb particle is forced to move along around the horizontal path in the shape of a limaçon, r = (5+sin θ) ft. If θ =LaTeX: \frac{1}{8}t^21 8 t 2 rad, where t is in seconds, determine the force of the rod and the normal force of the slot on the particle at the instant t= 3sec. The fork and path contact the particle on only one side.
Answers
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The detailed solution is in the attached file
A rectifier charges a battery bank in a substation. The bank rated dc voltage is 48 V. The required charging current is 25 A. The available ac supply is 120 V. The internal resistance of the battery is 2.5 Ω. (a) Analyze the operating conditions of the charger. Plot the ac and dc voltage and current, and determine the feasibility of delay angle control. (b) Calculate the delay angle needed to maintain the 25 A charging current. (c) Calculate the power and power factor at the ac side.
Answers
Answer:
See explaination
Explanation:
A constant voltage source is called a DC Voltage with a voltage that varies periodically with time is called an AC voltage. Voltage is measured in volts, with one volt being defined as the electrical pressure required to force an electrical current of one ampere through a resistance of one Ohm.
Please check the attached file.
A non-profit foundation is hosting a fundraising dinner. A few days ahead of the event you need to notify the caterer the quantity of meals you want served. The cost to cater each meal is $25. Each guest donates $100 to the non-profit foundation to attend the dinner event. Any leftover meals not eaten have no residual value. If a donor arrives at the event wanting to donate and no meals are left (i.e., you under ordered), the disappointed, hungry donor will i) not donate $100 at this event, and ii) be less likely to donate in the future, costing the foundation $75 (goodwill cost).a) (4 pts) What is the underage cost? What is the overage cost? What is the target service level? Cu = 150 Co = 25 Target Service Level = .8571 or 85.71%b) (4 pts) This is the first year of your event, so the only historical data you have is from the caterer who says that from her past experience with other foundations, the number of donors who will attend is well-approximated by a normal distribution with a mean of 100 and a standard deviation of 9.How many meals should you order? What is your effective service level? (NOTE: It might help you to know that for a standard normal distribution: z.8824=1.19, z.8571=1.07, z.90=1.28, z.95=1.64) Order 110 meals Effective Service Level = 85.71%For the remainder of the questions: It is now 10 years later (assume all revenue and costs are the same as in the original question), and you believe the historical data from your past fundraisers will more accurately predict the amount of meals you should order. The following table contains historical donor attendance for your various past fundraisers:Number of Donors. 104 105 106 107% of fundraisers 63% 20% 10% 7%c) Using the historical data, now how many meals should you order? Justify. Q* = 106 mealsd) What is your effective service level? Service Level = 93%e) The president of your foundation comes to you and says that he wants you to ensure that you won’t run out of meals at least 95% for all future fundraisers. How many dinners do you need to buy in order to meet that service level? 107 Meals
Answers
Answer:
Whether you’re fundraising for a personal cause or for a nonprofit organization, there are countless reasons why you or your organization may need to raise money.
We’ve broken our fundraising ideas into several categories. Use this list to jump between sections:
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Explanation:
Koch traded Machine 1 for Machine 2 when the fair market value of both machines was $60,000. Koch originally purchased Machine 1 for $76,900, and Machine 1's adjusted basis was $40,950 at the time of the exchange. Machine 2's seller purchased it for $64,050 and Machine 2's adjusted basis was $55,950 at the time of the exchange. What is Koch's adjusted basis in machine 2 after the exchange?
Answers
Answer:
Koch's adjusted basis in machine 2 after the exchange is $60,000
Explanation:
given data
fair market value = $60,000
originally purchased Machine 1 = $76,900
Machine 1 adjusted basis = $40,950
Machine 2 seller purchase = $64,050
Machine 2 adjusted basis = $55,950
solution
As he exchanged machine for another at $60,000
and this exchanged in fair market
so adjusted basis = $50,000
Adjusted basis is the price of the item that affects the factors that are considered price. These factors usually include taxes, depreciation value, and other costs of acquiring and maintaining a given item. Adjusted basis is important so the right amount to sell
Adjusted basis increases when a person deducts expenses from factor taxes and operating statements
so Koch's adjusted basis in machine 2 after the exchange is $60,000
Use the indirect pattern when you need to soften or delay bad news until after an explanation is given. Understanding the four components of the indirect pattern will help you craft messages that convey empathy, present reasons, cushion bad news, and close pleasantly. What buffering technique are you using if you show in your opening that you care and are concerned
Answers
Answer:
The general method for indirect strategy is Start with a neutral buffer.
Explanation:
The general method for indirect strategy goes like this:
-Start with a neutral buffer, you should never start with good news because it will give the reader false hope that more good news will come. So a neutral “buffer” or a show of appreciation for your business is a good way to start. You are not apologizing for the bad news that is coming, you are simply preparing the reader for it.
-The next part is where you give reasons. There are many studies on the effectiveness of reasons in communication: people like to know why things are the way they are. By offering reasons, you will make the bad news easier to accept and once you have prepared the reader, you give the bad news.
The idea is also not to give infinite turns to the subject in general, since the information and the objective of the message can be lost. It is important not to spend too much time on this: the buffer should be short, so as not to make the moment tedious and lose the attention of the public before reaching the main topic.
-Finally, in your conclusion, divert attention from the bad news. Don't talk about it anymore, be nice, focus your final efforts on future opportunities and recover goodwill.
Identical wire loops are dipped into liquid a and liquid b, so that a film of liquid forms across the loops (like the bubble solution on a child's bubble blowing wand). the width of each loop is increased slowly and the forces fa and fb needed to make the loops 5% wider are measured.
(a) fa will be greater than fb
(b) fa will be less than fb
(c) fa will be equal to fb
(d) it is impossible to predict whether fa or fb will be greater without more information.
Answers
Without information about the liquids' surface tensions, it is impossible to predict whether Fa will be greater, less, or equal to Fb. Answer (d) is correct.
Explanation:When analyzing forces on a wire loop in magnetic fields, it's essential to consider various factors including current, magnetic field intensity, and loop dimensions. A uniform magnetic field will exert a force on each segment of the loop carrying a current; this applies to rectangular loops in a magnetic field as noted in the given scenario. The question seems to focus on the force needed to change the width of loops coated with a liquid film and could draw on concepts of surface tension and possibly elasticity if considering the loop's properties themselves. However, the provided text focuses on magnetic effects on current-carrying wires.
For the case of the film across the loop, the necessary force to stretch it will heavily depend on the surface tension of the liquid. Therefore, without information on the liquids' surface tensions or other properties, it is impossible to predict whether Fa will be greater than, less than, or equal to Fb. Answer (d) it is impossible to predict whether Fa or Fb will be greater without more information is the most suitable given the circumstances.
Use the following information to complete the concrete mixture designs for the following question.
The following materials are available for a concrete mixture design:
ASTM C 150 Type I/II Cement
Relative density of 3.15
Coarse Aggregate: well-graded 3⁄4 in. maximum-size aggregate (MSA), crushed limestone, angular
Oven-dry specific gravity: 2.6
Absorption: 0.72%
Oven-dry rodded bulk density: 102 lb/ft3 Moisture content of coarse stockpile: 0.6%
Fine Aggregate: well-graded natural sand Oven-dry specific gravity: 2.5 Absorption: 1.3%
Moisture content of fine stockpile: 3.6% Fineness Modulus: 2.2
An air entraining admixture will provide enough air for the design mixture at a dosage rate of 3 oz/ft3
QUESTION:
Concrete is required for an 8 in thick exterior concrete slab in central Texas. A specified compressive strength, f’c , of 5000 psi is required at 28 days using an ASTM C 150 Type II portland cement. The design calls for a minimum of 2 in. of concrete cover over the reinforcing steel. The minimum distance between reinforcing bars is 3 in. A slump of 3 inches should be the target. The sidewalk is located in an environment that does not require air entraining. The concrete is required to have low permeability when exposed to water and moderate sulfates. The concrete will be exposed water soluble sulfates in soil at 0.14%. No statistical data on previous mixes are available.
Determine:
a. The theoretical mixture proportions of all concrete constituents (fine aggregate, coarse aggregate, water, cement, chemical admixtures) based on 1 yd3 of concrete on an oven dry basis for aggregates.
b. The as-batched mixture proportions based on the moisture contents provided in the materials available section.
c. How much of each material would be required to make a 20 ft long by 15 ft slab from the concrete mixture described above?
Answers
Answer:
a. 147lb/dt^3
b. 148lb/dt^3
c. 7318.52lb
Explanation:
Please see attachments
A Permanent Electrical Safety Device (PESD) becomes a real safety device only after it is installed as part of a Lockout/Tagout procedure. Until this is done it is nothing but another electrical component.A. TrueB. False
Answers
Answer:
True
Explanation:
Permanent electrical safety devices (PESD) acts as a layer of protection between the electrical worker and the hazardous voltage.
Permanent electrical safety devices (PESDs) are deployed to reduce the risks in isolating electrical energy.
Electrical safety can be improved if a worker determines a zero electrical energy state irrespective of any voltage exposure to themselves.
The given statement is true
During the recovery of a cold-worked material, which of the following statement(s) is (are) true?
O Some of the internal strain energy is relieved.
O All of the internal strain energy is relieved.
O There is some reduction in the number of dislocations.
O There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state.
O The electrical conductivity is recovered to its precold-worked state.
O The thermal conductivity is recovered to its precold-worked state.
O The metal becomes more ductile, as in its precold-worked state.
O Grains with high strains are replaced with new, unstrained grains.
Answers
Answer:
Some of the internal strain energy is relieved.
There is some reduction in the number of dislocations.
The electrical conductivity is recovered to its precold-worked state.
The thermal conductivity is recovered to its precold-worked state
Explanation:
The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.
1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat transfer with the surroundings, what is the final temperature?
Answers
Answer:
final temperature T = 24.84ºC
Explanation:
given data
copper volume = 1 L
temperature t1 = 500ºC
oil volume = 200 L
temperature t2 = 20ºC
solution
Density of copper [tex]\rho[/tex] cu = 8940 Kg/m³
Density of light oil [tex]\rho[/tex] oil = 889 Kg/m³
Specific heat capacity of copper Cv = 0.384 KJ/Kg.K
Specific heat capacity of light oil Cv = 1.880 KJ/kg.K
so fist we get here mass of oil and copper that is
mass = density × volume ................1
mass of copper = 8940 × 1 × [tex]10^{-3}[/tex] = 8.94 kg
mass of oil = 889 × 200 × [tex]10^{-3}[/tex] = 177.8 kg
so we apply here now energy balance equation that is
[tex]M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}[/tex] = 0
put here value and we get T2
[tex]8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)[/tex] = 0
solve it we get
T = 24.84ºC
Water at 20 bar, 400oC enters a turbine operating at steady state and exits at 1.5 bar. Neglect heat transfer, kinetic energy and potential energy effects. Someone states that the vapor quality at the turbine exit is 98%. (a) (35%) Find the entropy generation based on this statement. (b) (5%) Is this statement possible?
Answers
Answer:
a) [tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex], b) No.
Explanation:
The turbine is modelled after the Second Law of Thermodynamics, which states:
[tex]s_{in} - s_{out} + s_{gen} = 0[/tex]
The entropy generation per unit mass is:
[tex]s_{gen} = s_{out} - s_{in}[/tex]
The specific entropy for steam at entrance and exits are obtained from property tables:
Inlet (Superheated Steam)
[tex]s_{in} = 7.1292\,\frac{kJ}{kg\cdot K}[/tex]
Outlet (Liquid-Vapor Mixture)
[tex]s_{out} = 7.1073\,\frac{kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = 7.1073\,\frac{kJ}{kg\cdot K} - 7.1292\,\frac {kJ}{kg\cdot K}[/tex]
[tex]s_{gen} = -0.0219\,\frac{kJ}{kg\cdot K}[/tex]
b) It is not possible, as it contradicts the Kelvin-Planck and Claussius Statements, of which is inferred that entropy generation can only be zero or positive.
A 280-km-long pipeline connects two pumping stations. If 0.56m3/s are to be pumped through a 0.62-m-diameter line, the discharge station is 250m lower in elevation than the upstream station, and the discharge pressure is to be maintained at300,000Pa,determinethepowerrequiredtopumptheoil.The oilhas akinematic viscosity of4.510"6m2/s anda density of 810kg/m3.Thepipeisconstructedofcommercialsteel.Theinlet pressure may be taken as atmospheric.
Answers
The question asks for pump pressure calculations in a pipeline, requiring applications of fluid dynamics principles, such as Bernoulli's equation, and considering factors like elevation changes, viscosity, and pipe characteristics.
Explanation:The student's question relates to the field of fluid dynamics, specifically the calculation of required pump pressure in a pipeline system using principles such as Bernoulli's equation and concepts like pressure, flow rate, pipe diameter, density, viscosity, and elevation change. To solve this type of problem, you typically apply the Bernoulli equation, along with any additional head losses due to viscosity, known as the head loss due to friction, which can be calculated using the Darcy-Weisbach equation or other empirical formulas. You must then calculate the total head needed, convert this to pressure, taking into account the fluid's density, gravitational acceleration, and add any other relevant pressures, such as atmospheric pressure if the pipeline opens to the atmosphere.
A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 kpsi Find the factor of safety guarding against a static failure, and either the factor of safety guard- ing against a fatigue failure or the expected life of the part. For the fatigue analysis use:
(a) Modified Goodman criterion.
(b) Gerber criterion.
(C) ASME-elliptic criterion.
Answers
Answer:
(a) Modified Goodman criterion:
Factor of safety against fatigue failure = 1.0529
(b) Gerber criterion:
Factor of safety against fatigue failure = 1.31
(c) ASME-elliptic criterion:
Factor of safety against fatigue failure = 1.315
Explanation:
See the attached file for the calculation.
(a) Modified Goodman: SF ≈ 1.264, Static FS ≈ 2.058.
(b and c ) Gerber & ASME-elliptic: SF ≈ 1.783, Static FS ≈ 2.058.
To calculate the factor of safety against static and fatigue failures using different criteria, we need to determine the critical stress limits under the given loading conditions and compare them with the material properties.
Given:
- Torsional stress [tex](\( \tau \))[/tex] = 15 kpsi
- Alternating bending stress [tex](\( \sigma_a \))[/tex] = 25 kpsi
- Minimum endurance limit [tex](\( S_e \))[/tex] = 40 kpsi
- Ultimate tensile strength [tex](\( S \))[/tex] = 60 kpsi
- Endurance limit for reversed bending [tex](\( S_{-80} \))[/tex] = 80 kpsi
(a) Modified Goodman criterion:
The modified Goodman criterion accounts for both tensile and torsional stress, given by:
[tex]\[ \frac{1}{SF} = \frac{\frac{\sigma_a}{S} + \frac{\tau}{S_e}}{1} \][/tex]
Where [tex]\( SF \)[/tex] is the safety factor against fatigue failure.
Substitute the given values:
[tex]\[ \frac{1}{SF} = \frac{\frac{25}{60} + \frac{15}{40}}{1} \][/tex]
[tex]\[ \frac{1}{SF} = \frac{0.4167 + 0.375}{1} \][/tex]
[tex]\[ \frac{1}{SF} = 0.7917 \][/tex]
[tex]\[ SF = \frac{1}{0.7917} \][/tex]
[tex]\[ SF \approx 1.264 \][/tex]
The factor of safety against static failure [tex](\( FS_{static} \))[/tex] can be calculated by comparing the maximum applied stress with the ultimate tensile strength:
[tex]\[ FS_{static} = \frac{S}{\sigma_{max}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{\sigma_a^2 + \tau^2}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{25^2 + 15^2}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{625 + 225}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{850}} \][/tex]
[tex]\[ FS_{static} \approx \frac{60}{29.1547} \][/tex]
[tex]\[ FS_{static} \approx 2.058 \][/tex]
(b) Gerber criterion:
The Gerber criterion considers the bending and torsional stresses, given by:
[tex]\[ \frac{1}{SF} = \sqrt{\frac{\sigma_a^2}{S^2} + \frac{\tau^2}{S_e^2}} \][/tex]
Substitute the given values:
[tex]\[ \frac{1}{SF} = \sqrt{\frac{25^2}{60^2} + \frac{15^2}{40^2}} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{\frac{625}{3600} + \frac{225}{1600}} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]
[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]
[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]
[tex]\[ SF \approx 1.783 \][/tex]
(c) ASME-elliptic criterion:
The ASME-elliptic criterion also considers bending and torsional stresses:
[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{\sigma_a}{S}\right)^2 + \left(\frac{\tau}{S_e}\right)^2} \][/tex]
Substitute the given values:
[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{25}{60}\right)^2 + \left(\frac{15}{40}\right)^2} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]
[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]
[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]
[tex]\[ SF \approx 1.783 \][/tex]
For all three criteria:
- Factor of safety against static failure [tex](\( FS_{static} \))[/tex] ≈ 2.058
- Safety factor against fatigue failure [tex](\( SF \))[/tex] ≈ 1.264 for the Modified Goodman criterion, and ≈ 1.783 for the Gerber and ASME-elliptic criteria.
An actuator has a stem movement which at full travel is 40 mm. It is mounted on a process control valve with an equal percentage plug and which has a minimum flow rate of 0.2 m3/s and a maximum flow rate of 4.0 m3/s. What will be the flow rate when the stem movement is (a) 10 mm, (b) 20 mm
Answers
Answer:
a) for a stem of 10 mm the flow rate is 0.4229 m³/s
b) for a stem of 20 mm the flow rate is 0.8944 m³/s
Explanation:
The mathematical expression for the flow rate is:
[tex]Q=Q_{min} *(\frac{Q_{max} }{Q_{min} } )^{\frac{S-S_{min}}{S_{max}-S_{min} } }[/tex]
a) Here:
Qmin=0.2m³/s
Qmax=4m³/s
Smax-Smin=40mm
S-Smin=10mm
Substituting these values in the first equation:
[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{10}{40} } =0.4229m^{3} /s[/tex]
b) In this time, S-Smin=20mm
[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{20}{40} } =0.8944m^{3} /s[/tex]
Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color's value can range from 0 to 255. Thus (255, 0, 0) is bright red, (130, 0, 130) is a medium purple, (0, 0, 0) is black, (255, 255, 255) is white, and (40, 40, 40) is a dark gray. (130, 50, 130) is a faded purple, due to the (50, 50, 50) gray part. (In other words, equal amounts of red, green, blue yield gray). Given values for red, green, and blue, remove the gray part. Ex: If the input is 130 50 130, the output is: 80 0 80 Find the smallest value, and then subtract it from all three values, thus removing the gray. Note: This page converts rgb values into colors.
Answers
Answer:
answer is given below
Explanation:
we write this program in C++ that is given below
and result is attach here
#include <iostream>
using namespace std;
int main() {
int r,g,b,small;
//input values
cin>>r>>g>>b;
//find the smallest value
if(r<g && r<b)
small=r;
else if(g<b)
small=g;
else
small=b;
//subtract smallest of the three values from rgb values hence removing gray
r=r-small;
g=g-small;
b=b-small;
cout<<r<<" "<<g<<" "<<b<<endl;
}
We can see here that the Python code is thus:
# Read input values for red, green, and blue
red, green, blue = map(int, input().split())
# Find the smallest value among red, green, and blue
smallest = min(red, green, blue)
# Subtract the smallest value from all three colors
red -= smallest
green -= smallest
blue -= smallest
# Output the modified values
print(red, green, blue)
Assuming the input values for red, green, and blue are 50, 30, and 40 respectively, the output will be:
10 0 10
This is because:
The smallest value among 50, 30, and 40 is 30.
Subtracting 30 from each color gives:
red = 50 - 30 = 20green = 30 - 30 = 0blue = 40 - 30 = 10The output displays the modified values: 20 0 10
To implement this functionality in Python, you can follow these steps:
Read the input values for red, green, and blue.Find the smallest value among red, green, and blue.Subtract the smallest value from all three colors.Output the modified values.A student is working with three sealed containers filled with water. The first container is filled with ice. The second is filled with liquid water, and the third is filled with water vapor. Each container has a plunger on top. What will happen to the volumes of each container when the student presses the plungers
Answers
Explanation:
The three containers each contains water in different states.
Solid state of matter is considered as not compressible because the molecules are already as closely packed as they can be.
The liquid sate of matter has a very minute to no compression ability at all as the molecules are relatively close to each other. Compression is difficult to achieve in the liquid state.
In the gaseous state of matter, the molecules have broken free of one another, and are fairly spaced one from another. This means that gases can be easily compressed.
Pressing down on the plunger, the container containing ice can't be compressed at all so it's volume stays the same.
For the container filled with water, only a minute compression can be achieved with great difficulty hence, the volume reduces by an insignificant amount.
For the container filled with vapour, compression can be easily achieved and the volume reduces significantly.
Only the volume of the third container will decrease.
What happens when the student plungers at the top of the container?There are three containers and each has water in three different states. Solid-state of the matter is ice as they are closely packed.
The liquid state is minute and has molecules that are relatively close to other, Compression is thus difficult to achieve.
The last state has water in gases matter and can be compressed easily.
Thus the volume of the third container will go down.
Find out more information about liquid water.
brainly.com/question/2608851.
Hydroelectric power plants convert the gravitational potential energy of falling water into electrical power, typically by allowing the water to flow through a pipe called a penstock to rotate a generator located below it. Let the bottom of the penstock be the origin of a Cartesian coordinate system and the point at which the gravitational potential energy is zero.
a) Consider a penstock that is vertical and has a height of h = 61 m. How long, t in seconds, does it take water to fall from the top of the penstock to the bottom? Assume the water starts at rest.
Answers
Answer:
3.527 seconds
Explanation:
The height of the falling water, assuming no friction or air resistance, is given by ...
h(t) = -(1/2)gt² +61
where g is the standard gravity value, 9.80665 m/s².
The the time required for h(t) = 0 is ...
1/2gt² = 61
t² = 2·61/g
t = √(2·61/9.80665) ≈ 3.527 . . . . seconds
A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed of 320 km/h in air at 15 C. The student decided to build a 1/50 scale model and test it in a water tunnel at the same temperature.
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
Answers
Answer:
b. 1232.08 km/hr
c. 1.02 kn
Explanation:
a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.
See attachment for the remaining.
Modify any of the previous labs which would have crashed when non-numeric data was entered by adding exception handling so that the non-numeric input no longer crashes the program. The program must handle the invalid input gracefully by informing the user of the bad input and re-querying until the user provides valid input. It must re-query for the specific input which was incorrectly entered.
Answers
Answer:
see explaination
Explanation:
import java.util.InputMismatchException;
import java.util.Scanner;
public class calculate {
static float a=0,b=0;
double cal()
{
if(a==0||b==0)
{
System.out.println("no values found in a or b");
start();
}
double x=(a*a)+(b*b);
double h=Math.sqrt(x);
a=0;
b=0;
return h;
}
float enter()
{
float val=0;
try
{
System.out.println("Enter side");
Scanner sc1 = new Scanner(System.in);
val = sc1.nextFloat();
return val;
}
catch(InputMismatchException e)
{
System.out.println("Enter correct value");
}
return val;
}
void start()
{
calculate c=new calculate();
while(true)
{
System.out.println("Enter Command");
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
switch(input)
{
case "A":
a=c.enter();
break;
case "B":
b=c.enter();
break;
case "C":
double res=c.cal();
System.out.println("Hypotenuse is : "+res);
break;
case "Q":
System.exit(0);
default:System.out.println("wrong command");
}
}
}
public static void main(String[] args) {
calculate c=new calculate();
c.start();
}
}
All areas of the country use a BAC of 0.100.10 g/dL as the legal intoxication level. Is it possible that the mean BAC of all drivers involved in fatal accidents who are found to have positive BAC values is less than the legal intoxication level? A. No, it is not possible. B. Yes, but it is not likely. C. Yes, and it is highly probable.
Answers
Answer:
C. Yes, and it is highly probable.
Explanation:
Null hypothesis, [tex]H_0[/tex] : ц 0.09
Alternative hypothesis, [tex]H_1[/tex]: ц > 0.09
Test statistic is,
x Z = x - ц / s
[tex]\frac{0.15 - 0.09 }{0.06}[/tex]
[tex]\frac{0.06}{0.06}[/tex]
=1
The p-value is,
p = P(Z > z)
=1—P(Z ≤1)
= 1— 0.841345
= 0.158655
(From normal tables)
The p-value is greater than the significance level, so we fail to reject the null hypothesis. The correct option is, C
Yes, and it is highly probable.
an electric circuit includes a voltage source and two resistances (50 and 75) in parallel. determine the voltage source required to provide 1.6 A of current through the 75 ohm resistance
Answers
Answer:
The voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.
Explanation:
Given;
Resistance, R₁ = 50Ω
Resistance, R₂ = 75Ω
Total resistance, R = (R₁R₂)/(R₁ + R₂)
Total resistance, R = (50 x 75)/(125)
Total resistance, R = 30 Ω
According to ohms law, sum of current in a parallel circuit is given as
I = I₁ + I₂
[tex]I = \frac{V}{R_1} + \frac{V}{R_2}[/tex]
Voltage across each resistor is the same
[tex]1.6 = \frac{V}{R_2}[/tex]
V = 1.6 x R₂
V = 1.6 x 75
V = 120 V
Therefore, the voltage source required to provide 1.6 A of current through the 75 ohm resistance is 120 V.
This voltage is also the same for 50 ohms resistance but the current will be 2.4 A.
Answer:
120 volts
Explanation:
Since the two resistances are connected in parallel across the voltage source, the effective resistance of the circuit can be obtained by using the formula.
[tex]\frac{1}{R_{eq}}=\frac{1}{R_{1}}+ \frac{1}{R_{2}}[/tex]
given that [tex]R_{1} and R_{2}[/tex] are 50 and 75 ohms respectively, we have the equivalent resistance as:
[tex]\frac{1}{R_{eq}}=\frac{1}{50}+ \frac{1}{75}=\frac{1}{30}[/tex]
hence,
[tex]R_{eq}= 30\Omega[/tex]
From Ohm's law, voltage = current X resistance.
given that the current through the 75 ohm resistor is 1.6 A
[tex]V= I\times R[/tex]
[tex]V= 1.6 \times 75\Omega[/tex]
voltage = 120 Volts.
Because the resistors are connected in parallel, it means that they are connected to the same voltage source.
Hence, the voltage source for the 75 Ohm resistance = 120 volts. This is same for the 50 Ohm resistor.
An insulated air nozzle with an inlet pressure of 10 bar operates with a mass flow rate of 1.2 kg s−1 . The inlet temperature is 600 K and the outlet velocity is 100 m s−1 . The inlet diameter is 0.5 m and the outlet diameter is 0.05 m. Determine the inlet velocity and the outlet temperature. What is the change in enthalpy across the nozzle? Assume air is an ideal gas. [0.95 m s−1 , 595.2 K, −5 kJ kg−1 ]
Answers
Answer:
Given mass flow rate =1.2 kg/s
mass flow rate=density*A*V
Area=pi(douter^2-dinner^2)/4
Area=0.194m^2
The velocity is given by
velocity=2.876 m/s