The temperature at which the resistance is [tex]1 k\Omega[/tex] is [tex]89.4^{\circ}C[/tex]
Explanation:
For the thermistor in this problem, the relationship between temperature and resistance is linear.
We have:
[tex]R_1 = 5000 \Omega[/tex] when the temperature is [tex]T=25^{\circ}C[/tex]
[tex]R_2=340 \Omega[/tex] when the temperature is [tex]T=100^{\circ}C[/tex]
Assuming a straight-line relationship, we can find the slope of the line:
[tex]m=\frac{R_2-R_1}{T_2-T_1}=\frac{340-5000}{100-25}=-62.1 \Omega/^{\circ}C[/tex]
Now that we know the slope, we can extrapolate the temperature when the resistance is
[tex]R_3 = 1 k\Omega = 1000 \Omega[/tex]
In fact, we can write:
[tex]m=\frac{R_3-R_2}{T_3-T_2}[/tex]
And solving for [tex]T_3[/tex],
[tex]m(T_3-T_2)=R_3-R_2\\T_3 = T_2 +\frac{R_3-R_2}{m}=100 + \frac{1000-340}{-62.1}=89.4^{\circ}C[/tex]
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The temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.
Explanation:The relationship between the resistance of a thermistor and temperature can be modeled by a straight line equation, which is given by:
R = Ro(1 + αΔT)
where R is the resistance at a given temperature, Ro is the resistance at a reference temperature, α is the temperature coefficient of resistance, and ΔT is the change in temperature from the reference temperature.
To find the temperature at which the thermistor's resistance equals 1 kΩ, we can rearrange the equation:
ΔT = (R/Ro - 1) / α
Plugging in the given values:
Ro = 5 kΩ, R = 1 kΩ, α = (340 Ω - 5 kΩ) / (100 °C - 25 °C)
ΔT = (1 kΩ / 5 kΩ - 1) / [(340 Ω - 5 kΩ) / (100 °C - 25 °C)]
Simplifying the equation, we find that ΔT ≈ -14.12 °C.
Therefore, the temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.
Print a message telling a user to press the letterToQuit key numPresses times to quit. End with newline. Ex: If letterToQuit = 'q' and numPresses = 2, print:
Answer:
Vb.Net
msgbox ("Press "q" twice to quit", msgboxstyle.information)
if char.q = keypress and keypress.count = 2 then
End
End if
Explanation:
In a residential heat pump, Refrigerant-134a enters the condenser at 800 kPa and 50oC at a rate of 0.022 kg/s and leaves at 750 kPa subcooled by 3oC. The refrigerant enters the compressor at 200 kPa superheated by 4oC. Determine (a) the isentropic efficiency of the compressor, (b) the rate of heat supplied to the heated room, and (c) the COP of the heat pump. Also, (d) determine the COP and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle between the pressure limits of 200 kPa and 800 kPa.
Answer:
a) 0.813
b) 4.38 KW
c) COP = 5.13
d) 3.91 KW , COP = 6.17
Explanation:
Data obtained A-13 tables for R-134a:
[tex]h_{1} = 247.88 \frac{KJ}{kg} \\s_{1} = 0.9575 \frac{KJ}{kgK}\\h_{2s} = 279.45 \frac{KJ}{kg}\\h_{2} = 286.71 \frac{KJ}{kg}\\h_{3} = 87.83 \frac{KJ}{kg}[/tex]
The isentropic efficiency of the compressor is determined by :
[tex]n_{C} = \frac{h_{2s} - h_{1} }{h_{2} - h_{1} }\\= \frac{279.45 - 247.88 }{286.71 - 247.88}\\= 0.813[/tex]
The rate of heat supplied to the room is determined by heat balance:
[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(286.71 - 87.83)\\\\= 4.38KW[/tex]
Calculating COP
[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(286.71-246.88)}\\\\COP = 5.13[/tex]
Part D
Data Obtained:
[tex]h_{1} = 244.5 \frac{KJ}{kg} \\s_{1} = 0.93788 \frac{KJ}{kgK}\\h_{2} = 273.31 \frac{KJ}{kg}\\h_{3} = 95.48 \frac{KJ}{kg}[/tex]
The rate of heat supplied to the room is determined by heat balance:
[tex]Q = m(flow) * (h_{2} -h_{3})\\= (0.022)*(273.31 - 95.48)\\\\= 3.91KW[/tex]
Calculating COP
[tex]COP = \frac{Q_{H} }{W} \\COP = \frac{Q_{H} }{m(flow) * (h_{2} - h_{1}) }\\\\COP = \frac{4.38}{(0.022)*(273.31-244.5)}\\\\COP = 6.17[/tex]
It is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water entering at 20 °C?
Answer:
Yes, it is possible to maintain a pressure of 10 kPa in a condenser that is being cooled by river water that is entering at 20 °C because this temperature (20 °C) of the external cooling water is less than the saturation temperature of steam which is which is 45.81 °C, and heated by a boiler; as a result of this condition, coupled with the assumption that the turbine, pump, and interconnecting tube are adiabatic, and the condenser exchanges its heat with the external cooling river water, it possible to maintain a pressure of 10 kPa.
Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of the time and the unit cost of electricity is $0.09/kWh, the electricity cost of this refrigerator per month (30 days) is
A. $3.56
B. $5.18
C. $8.54
D. $9.28
E. $20.74
Answer:
B. $5.18
Explanation:
Cost of electricity per kWh = $0.09
Power consumption of refrigerator = 320W = 320/1000 = 0.32kW
In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours
Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh
Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18
The electricity cost of the refrigerator per month is approximately $0.65. The correct answer is none of the given options.
Explanation:To calculate the electricity cost of the refrigerator per month, we first need to find out how many kWh it consumes when it is running for one hour. We can do this by converting the power consumption of the refrigerator from watts to kilowatts:
320 W = 0.32 kW
Since the refrigerator runs only one quarter of the time, we can calculate the kWh consumed per month as follows:
0.32 kW * 1/4 * 24 hours/day * 30 days/month = 7.2 kWh/month
Now, we can calculate the cost of electricity using the unit cost of $0.09/kWh:
7.2 kWh/month * $0.09/kWh = $0.648/month
Therefore, the electricity cost of the refrigerator per month is approximately $0.65.
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Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) Cylindrical wall (b) Plane shell (c) Spherical shell (d) All of them (e) None of them
Answer:
Plane shell which is option b
Explanation:
The temperature in the case of a steady one-dimensional heat conduction through a plane wall is always a linear function of the thickness. for steady state dT/dt = 0
in such case, the temperature gradient dT/dx, the thermal conductivity are all linear function of x.
For a plane wall, the inner temperature is always less than the outside temperature.
What are the challenges posed by strategic information systems, and how should they be addressed?
Answer:
Explanation:
Implementing strategic system requires extensive organizational change coupled with a period of changing from one stage of socio-technical level to another. This changes are known as strategic transitions and are not easily achieved.
It must be noted that not all strategic systems are rewarding and can be very expensive to put together. It is easier to copy most information systems from other firms because strategic advantage can be most times unsustainable.
A sample of coarse aggregate has an oven dry weight of 1034.0 g and a moisture content of 4.0 %. The saturated surface dry weight is 1048.9g and the weight of the aggregate in water is 675.6 g. Determine using phase volume relationships: a) Apparent Specific Gravity (GA) b) Bulk Specific Gravity (GB) c) Bulk Specific Gravity SSD (GB (SSD)) d) Absorption, % e) Bulk Volume
Answer:
Apparent Specific Gravity = 2.88
bulk specific gravity = 2.76
Bulk Specific Gravity SSD = 2.80
absorption = 1.44%
bulk volume = 373.3
Explanation:
given data
oven dry weight A = 1034.0 g
moisture content = 4.0 %
saturated surface dry weight B = 1048.9 g
weight of the aggregate in water C = 675.6 g
solution
we get here Apparent Specific Gravity that is express as
Apparent Specific Gravity = [tex]\frac{A}{A-C}[/tex] ..........1
put here value
Apparent Specific Gravity = [tex]\frac{1034}{1034-675.6}[/tex]
Apparent Specific Gravity = 2.88
and
now we get bulk specific gravity that is
bulk specific gravity = [tex]\frac{A}{B-C}[/tex] ...................2
put here value
bulk specific gravity = [tex]\frac{1034}{1048.9-675.6}[/tex]
bulk specific gravity = 2.76
and
now we get Bulk Specific Gravity SSD
Bulk Specific Gravity SSD = [tex]\frac{B}{B-C}[/tex] ............3
Bulk Specific Gravity SSD = [tex]\frac{1048.9}{1048.9-675.6}[/tex]
Bulk Specific Gravity SSD = 2.80
and
now absorption will be here as
absorption = [tex]\frac{B-A}{A}[/tex] × 100% ................4
absorption = [tex]\frac{1048.9-1034}{1034}[/tex] × 100%
absorption = 1.44%
and
last we get bulk volume that is
bulk volume = [tex]\frac{weight\ displce\ water}{density\ water }[/tex]
bulk volume = [tex]\frac{1048.9-675.6}{1}[/tex]
bulk volume = 373.3
Assume we have already defined a variable of type String called password with the following line of code: password' can have any String value String password ???"; However, because of increased security mandated by UCSD's technical support team, we need to verify that passwo rd is secure enough. Specifically, assume that a given password must have all of the following properties to be considered "secure": It must be at least 7 characters long .It must have characters from at least 3 of the following 4 categories: Uppercase (A-Z), Lowercase (a-z), Digits (0-9), and Symbols TASK: If password is secure, print "secure", otherwise, print "insecure" HINT: You can assume that any char that is not a letter (A-Z, a-z) and is not a digit (0-9) is a symbol. EXAMPLE: "UCSDcse11" would be valid because it is at least 7 characters long (it's 9 characters long), it has at least one uppercase letter (U', 'C', 'S', and 'D'), it has at least one lowercase letter (c', 's', and 'e'), and it has at least one digit (1' and '1') EXAMPLE: "UCSanDiego" would be invalid because, while it is 10 characters long, it only has uppercase and lowercase letters, so it only has characters from 2 of the 4 categories listed Sample Input: UCSDcse11 Sample Output: secure
Answer:
The Java code is given below with appropriate comments for better understanding
Explanation:
import java.util.Scanner;
public class ValidatePassword {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter a password: ");
String password = input.nextLine();
int count = chkPswd(password);
if (count >= 3)
System.out.println("Secure");
else
System.out.println("Not Secure");
}
public static int chkPswd(String pwd) {
int count = 0;
if (checkForSpecial(pwd))
count++;
if (checkForUpperCasae(pwd))
count++;
if (checkForLowerCasae(pwd))
count++;
if (checkForDigit(pwd))
count++;
return count;
}
// checks if password has 8 characters
public static boolean checkCharCount(String pwd) {
return pwd.length() >= 7;
}
// checks if password has checkForUpperCasae
public static boolean checkForUpperCasae(String pwd) {
for (int i = 0; i < pwd.length(); i++)
if (Character.isLetter(pwd.charAt(i)) && Character.isUpperCase(pwd.charAt(i)))
return true;
return false;
}
public static boolean checkForLowerCasae(String pwd) {
for (int i = 0; i < pwd.length(); i++)
if (Character.isLetter(pwd.charAt(i)) && Character.isLowerCase(pwd.charAt(i)))
return true;
return false;
}
// checks if password contains digit
public static boolean checkForDigit(String pwd) {
for (int i = 0; i < pwd.length(); i++)
if (Character.isDigit(pwd.charAt(i)))
return true;
return false;
}
// checks if password has special char
public static boolean checkForSpecial(String pwd) {
String spl = "!@#$%^&*(";
for (int i = 0; i < pwd.length(); i++)
if (spl.contains(pwd.charAt(i) + ""))
return true;
return false;
}
}
A quack is a data structure combining properties of both stacks and queues. It can be viewed as a list of elements written left to right such that three operations are possible:
Answer:
Three operations possible in a quack is QUACKPUSH, QUACKPOP and QUACKPULL.
Explanation:
A quack is a data structure combining properties of both stacks and queues. It can be viewed as a list of elements written left to right such that three operations are possible, these include:
QUACKPUSH(x): add a new item x to the left end of the list; QUACKPOP(): remove and return the item on the left end of the list; QUACKPULL(): remove the item on the right end of the list.Elements in the quack are stored in stacks. The component stacks can be accessed only through the standard stack functions PUSH and POP.
The method of breaking digital messages into small fixed bundles, transmitting them along different communication paths, and reassembling them at their destination is called:
Answer:
Packet switching
Explanation:
Packet switching - it is referred to as the transmission process which involved sending of packets through communication path and last reassembling them again.
The sending of packets is done by broken the digital message into required pieces for efficient transfer.
in this process, every individual parcel is sent separately.
Print "Censored" if userInput contains the word "darn", else print userInput. End with newline. Ex: If userInput is "That darn cat.", then output is: Censored Ex: If userInput is "Dang, that was scary!", then output is: Dang, that was scary! Note: If the submitted code has an out-of-range access, the system will stop running the code after a few seconds, and report "Program end never reached." The system doesn't print the test case that caused the reported message.
Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply the same method for any programming language, pseudocode or flowchart.
Answer:
1 userInput = input("Please enter 'userInput':\n")
2 if "darn" in userInput.lower():
3 print("Censored")
4 else:
5 print(userInput\n)
Explanation line by line:
Line one asked for the user input, and store the value in a variable called userInput. Inside the input function, you put some text to indicate to the user to enter something and \n is the new line character. Line two checked if the word "darn" is in the user input, for this, we use the built-in function in. Because we don't know if the user inputs the word "darn" in upper case, lower case or a mix of both we use the method .lower() to change the user input to lowercase and make the validation (Python is case sensitive -> A is not equal to a). Line three prints the word "Censored" if and only if the word "darn" is in userInput.Line fourth allows going to line five if and only if the word "darn" is NOT in userInput.Finally, line five prints the word entered by the user if and only if the word "darn" in NOT in userInput.One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight last?
Answer:
time = 5.22 hr
Explanation:
Given data:
Energy of battery = 9400 J
Power consumed by three led bulb is 0.5 watt
we know Power is give as
[tex]Power = \frac{ energy}{time}[/tex]
plugging all value and solve for time
[tex]time = \frac{energy}{power}[/tex]
[tex]time = \frac{9400}{0.5}[/tex]
time = 18,800 sec
in hour
1 hour = 3600 sec
therefore in 18,800 sec
[tex]time = \frac{18800}{3600} = 5.22 hr[/tex]
Design a sequential circuit with two D flip-flops A and B, and one input x_in.
(a)When xin = 0, the state of the circuit remains the same. When x_in = 1, the circuit goes through the state transitions from 00 to 01, to 11, to 10, back to 00, and repeats.
(b)When xin = 0, the state of the circuit remains the same. When x_in =1, the circuit goes through the state transitions from 00 to 11, to 01, to 10, back to 00, and repeats.
Answer:
View Image
Explanation:
The question is basically asking you to build a 2-bit asynchronous counter.
What the counter does is it increase it's value by 01₂ every clock pulse. So at 0₂, nothing happens, but at 1₂ it'll count up by 1. It then reset to 00₂ when it overflows.
The design for it is pretty much universal so I kinda did this from memory.
a.) A count-up counter (from 00-11) is simply made by connecting Q' to D, and the output of the previous DFF to the clock of the next one.
b.) A count-down counter (from 11-00) is simply made by using the same circuit as the count-up counter, but you connect Q' to the clock instead of Q.
How far do you jog each morning? You prefer to jog in different locations each day and do not have a pedometer to measure your distance. Create an application to determine the distance jogged given the average number of strides ran during the fist minute, average number ran during the last minute, and the total minutes jogging. Design a modularized solution (with methods) to display the distance traveled. Pedometers measure the distance you run. However, you can also do a good estimate of the distance if you know your foot stride, how many strides you complete per minute, and the number of minutes you job. Foot stride is the distance covered by one average step length. Since everyone has a different foot size, strides differ. Manny people average 3 feet per setup when jogging. For this application, assume the foot stride is 2.5 feet. There are 5,280 feet in a mile. To establish how many strides per minutes, allow the user to input the number of strides made during the first minute jogging and the number of strides made string the last minutes of jogging. Use the average of those values to represent the strides per minute. Allow the user to input the total time spent jogging in hours and minutes. Write code that will display to distance traveled in miles.
Answer:
The program is given below with appropriate comments for better understanding
Explanation:
#Program
# foot stride = 2.5 feet
# 1 mile = 5280 feet
no_stride_first_min = int(input('Enter the number strides made durng the first minute of jogging: '))
no_stride_last_min = int(input('Enter the number strides made durng the last minute of jogging: '))
avg_stride_one_min = (no_stride_first_min + no_stride_last_min)/2 # calculates the average stride per minute
jogging_duration = float(input('Enter the total time spent jogging in hours and minute: '))
jogging_duration_hours = int(jogging_duration) # gets the hour
jogging_duration_min = jogging_duration - int(jogging_duration) # gets the minute
tot_jogging_duration_min = jogging_duration_hours*60 + jogging_duration_min # calculates total time in minutes
dist_feet = (avg_stride_one_min*2.5)*tot_jogging_duration_min # calculates the total distance in feet
dist_miles = dist_feet/5280 # calculates the total distance in mile
print('Distance traveled in miles = {0:.2f} miles'.format(dist_miles))
The gage pressure in a liquid at a depth of 3 m is read to be 48 kPa. Determine the gage pressure in the same liquid at a depth of 9 m.
The gage pressure in the same liquid at a depth of 9 m is__________ kPa.
Answer: 144kpa
Explanation:
pressure density =p
The Gage pressure (P1)=48kpa
The height (h1)=3m
The Gage pressure (P2)=?
The height (h2)=9m
P=p * g * h
P1/P2=p * g * h1/p * g * h2
Cancelling out similar terms:
Therefore, P1/P2=h1/h2
P2=P1*h2/h1
Hence, P2=48*9/3=144kpa.
Electrical failure can lead to ________failure, which in turn can lead to software failure.
Answer:
Electrical failure can lead to _hardware__failure, which in turn can lead to software failure
The net potential energy between two adjacent ions, EN, may be represented by EN = -A/r + B/rn Where A, B, and n are constants whose values depend on the particular ionic system. Calculate the bonding energy E0 in terms of the parameters A, B, and n using the following procedure: 1. Differentiate EN with respect to r, and then set the resulting expression equal to zero, since the curve of EN versus r is a minimum at E0. 2. Solve for r in terms of A, B, and n, which yields r0, the equilibrium interionic spacing. 3. Determine the expression for E0 by substitution of r0 into the above equation for EN. What is the equation that represents the expression for E0?
The equation that represents the expression for the bonding energy [tex](E_0\))[/tex] in terms of the parameters A, B, and n is: [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].
Step 1: Differentiate EN with Respect to r
Given:
[tex]\[ E_N = -\frac{A}{r} + \frac{B}{r^n} \][/tex]
Differentiate [tex]\(E_N\)[/tex] with respect to r:
[tex]\[ \frac{dE_N}{dr} = \frac{A}{r^2} - \frac{nB}{r^{n+1}} \][/tex]
Set the derivative equal to zero to find the minimum (equilibrium) point:
[tex]\[ \frac{A}{r^2} - \frac{nB}{r^{n+1}} = 0 \][/tex]
Solve for r as
[tex]\[ \frac{A}{r^2} = \frac{nB}{r^{n+1}} \][/tex]
[tex]\[ r^n = \frac{nB}{A} \][/tex]
[tex]\[ r = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]
Step 2: Solve for r in terms of A, B, and n
The equilibrium interionic spacing [tex]\(r_0\)[/tex] is the value of r at the minimum point,
[tex]\[ r_0 = \left(\frac{nB}{A}\right)^{\frac{1}{n}} \][/tex]
Step 3: Determine the Expression for E0
Substitute [tex]\(r_0\)[/tex] into the expression for [tex]\(E_N\):[/tex]
[tex]\[ E_0 = -\frac{A}{r_0} + \frac{B}{r_0^n} \][/tex]
[tex]\[ E_0 = -\frac{A}{\left(\frac{nB}{A}\right)^{\frac{1}{n}}} + \frac{B}{\left(\frac{nB}{A}\right)^{\frac{n}{n}}} \][/tex]
Simplify:
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{(nB)^{\frac{n}{n}}} \][/tex]
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{B}{nB} \][/tex]
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]
Combine the terms:
[tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex]
So, the equation is [tex]\[ E_0 = -\frac{A^{\frac{n-1}{n}}}{(nB)^{\frac{1}{n}}} + \frac{1}{n} \][/tex].
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Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balance. The current balance must always be non-negativevalue. The variable types must be selected wisely. A sample run is below. The user’s response is in boldface (C by discovery)BANK ACCOUT PROGRAM!----------------------------------Enter the old balance: 1234.50Enter the transactions now.Enter an F for the transaction type when you are finished.Transaction Type (D=deposit, W=withdrawal, F=finished): DAmount: 568.34Transaction Type (D=deposit, W=withdrawal, F=finished): WAmount: 25.68Transaction Type (D=deposit, W=withdrawal, F=finished): WAmount: 167.40Transaction Type (D=deposit, W=withdrawal, F=finished): FYour ending balance is $1609.76Program is ending
Answer:
Explanation:
Sample output:
BANK ACCOUT PROGRAM!
----------------------------------
Enter the old balance: 1234.50
Enter the transactions now.
Enter an F for the transaction type when you are finished.
Transaction Type (D=deposit, W=withdrawal, F=finished): D
Amount: 568.34
Transaction Type (D=deposit, W=withdrawal, F=finished): W
Amount: 25.68
Transaction Type (D=deposit, W=withdrawal, F=finished): W
Amount: 167.40
Transaction Type (D=deposit, W=withdrawal, F=finished): F
Your ending balance is $1609.76
Program is ending
Code to copy:
// include the necessary header files.
#include<stdio.h>
// Definition of the function
float withdraw(float account_balance, float withdraw_amount)
{
// Calculate the balace amount.
float balance_amount = account_balance - withdraw_amount;
// Check whether the withdraw amount
// is greater than 0 or not.
if (withdraw_amount > 0 && balance_amount >= 0)
{
// Assign value.
account_balance = balance_amount;
}
// return account_balance
return account_balance;
}
// Definition of the function deposit.
float deposit(float account_balance, float deposit_amount)
{
// Check whether the deposit amount is greater than zero
if (deposit_amount > 0)
{
// Update account balance.
account_balance = account_balance + deposit_amount;
}
// return account balance.
return account_balance;
}
int main()
{
// Declare the variables.
float account_balance;
float deposit_amount;
float withdrawl_amount;
char input;
// display the statement on console.
printf("BANK ACCOUT PROGRAM!\n");
printf("----------------------------------\n");
// prompt the user to enter the old balance.
printf("Enter the old balance: ");
// Input balance
scanf("%f", &account_balance);
// Display the statement on console.
printf("Enter the transactions now.\n");
printf("Enter an F for the transaction type when you are finished.\n");
// Start the do while loop
do
{
// prompt the user to enter transaction type.
printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");
// Input type.
scanf(" %c", &input);
// Check if the input is D
if (input == 'D')
{
// Prompt the user to input amount.
printf("Amount: ");
// input amount.
scanf("%f", &deposit_amount);
// Call to the function.
account_balance=deposit(account_balance,deposit_amount);
}
// Check if the input is W
if (input == 'W')
{
printf("Amount: ");
scanf("%f", &withdrawl_amount);
// Call to the function.
account_balance = withdraw(account_balance,withdrawl_amount);
}
// Check if the input is F
if (input == 'F')
{
// Dispplay the amount.
printf("Your ending balance is $%.2f\n", account_balance);
printf("Program is ending\n");
}
// End the while loop
} while(input != 'F');
return 0;
}
the picture uploaded below shows the program screenshot.
cheers, i hope this helps.
A multilane highway (two lanes in each direction) is on level terrain. The free-flow speed has been measured at 45 mi/h. The peak-hour directional traffic flow is 1300 vehicles with 6% large trucks and buses and 2% recreational vehicles (f_p = 0.95).If the peak-hour factor is 0.85, determine the highway's level of service.
The level of service of a highway is determined by the volume-to-capacity ratio. Using the provided traffic information, we can calculate the highway's capacity and determine its level of service, which ranges from A to F.
Explanation:The level of service of a highway is determined by the volume-to-capacity ratio (V/C ratio).
Given peak-hour directional traffic flow, vehicle types, peak-hour factor, and free-flow speed, we can calculate the capacity of the highway and determine the level of service.
In this case, we would need to calculate the volume of traffic compared to the highway's capacity to determine the level of service, which can range from A to F.
A cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
Answer:
1.505
Explanation:
cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.
stress is force per unit area
stress=P/A
A = πd^2/4.
uncertainty of axial force P= +/-.11
s=+/-.20, strength
d=+/-.04 diameter
fail load/max allowed
minimum design=fail load/max allowed
minimum design =s/(P/A)
sA/P
A=([tex]\pi[/tex].96d^2)/4, so Amin=
[tex]0.96^{2}[/tex] (because the diameter at minimum is (1-0.04=0.96)
minimum design=Pmax/(sminxAmin)
1.11/(.80*.96^2)=
1.505
Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to building foundations.
Answer:PHYSICALLY- sand is spherical in shape with large particles up to 0.18g
Silt also contains more of spherical particles with weights around 0.0029g.
Clay contain plate like particles,which are flat or layered. It's particles are generally lower that silt and sand below 0.0029g.
BEHAVIORS Sand particles are coarse and very porous,water can easily penetrate with no particles cohesion,does not retain water, difficult to expand,it is IDEAL FOR BUILDINGS.
Silt particles are also porous with some coarseness and no particles cohesion. retains water and can expand. NOT IDEAL FOR BUILDINGS.
Clay contain particles that are not coarse,they have high cohesiveness,they are not porous as it is difficult for water to penetrate. NOT IDEAL FOR BUILDINGS.
Explanation: Sand particles are coarse,very porous,contains spherical particles with large particles sizes and IDEAL FOR BUILDINGS
Silt particles are also porous, with some coarseness and with little or no particles cohesion, NOT IDEAL FOR BUILDINGS.
Clay soils are not porous water can not easily flow thought it,it is hard when dry and soft when wet. IT IS NOT A GOOD OPTION FOR BUILDINGS.
Ensure at least ___ distance around fire sprinkler heads, safety showers, eyewash units, and heating and cooling units to ensure proper operation.
90 inches
Explanation:
According to OSHA requirement, the distance around safety showers and eyewash should be between 82-96 inches off the flow. This will allow for maximum diameter of spray.
Learn More
Safety distance around safety showers:https://brainly.com/question/11123362
Keywords: distance, fire sprinkler head, safety showers, eyewash units,heating and cooling units
#LearnwithBrainly
A unit cell has a lattice constant a, where a is the length of a single side of the unit cell. Draw a cube and derive the length of the diagonal from one top corner of the unit cell to the bottom far corner (ie. The largest distance between corners in a 3-D cube).
Answer:
The largest distance between corners in a 3-D cube = a root 3
Explanation : The detailed and step by step explanation is given in the attached file below.
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 13.2 ft/sft/s at point A and 17.6 ft/sft/s at point C. The cart takes 3.00 ss to go from point A to point C, and the cart takes 1.90 ss to go from point B to point C. What is the car's speed at point B?
Answer:
15.99 ft/s
Explanation:
From Newton's equation of motion, we have
v = u + at
v = Final speed
u = initial speed
a = acceleration
t = time
now
for the points A and C
v = 17.6 ft/s
u = 13.2 ft/s
t = 3 s
thus,
17.6 = 13.2 + a(3)
or
3a = 17.6 - 13.2
3a = 4.4
or
a = 1.467 m/s²
Thus,
For Points A and B
v = speed at B i.e v'
u = 13.2 ft/s
a = 1.467 ft/s²
t = 1.90 s
therefore,
v' = 13.2 + (1.467 × 1.90 )
v' = 13.2 + 2.7867
v' = 15.9867 ≈ 15.99 ft/s
Write Python expressions using s1, s2, and s3 and operators and * that evaluate to: (a) 'ant bat cod'
To form the string 'ant bat cod', you can concatenate the strings s1, s2, and s3 using the + operator or repeat each string based on the desired repetition using the * operator.
Explanation:Python expressions using s1, s2, and s3 and operators and * that evaluate to: (a) 'ant bat cod'.
s1 + ' ' + s2 + ' ' + s3s1 * 1 + ' ' + s2 * 2 + ' ' + s3 * 3Consider a steady flow Carnot cycle with water as the working fluid. The maximum and minimum temperatures in the cycle are 350 and 50 C. The quality of water is 0.891 at the beginning of the heat rejection process and 0.1 at the end. Determine:
(a) the thermal efficiency (how much percent).
(b) the pressure at the turbine inlet, and
(c) the network output
a) Etath,C= %
b) P2=MPa
c) wnet = kJ/kg
Answer:
a) Etath = 48.2 %
b) P2 = 1.195 MPa
c) wnet = 1749.14 KJ/kg
Explanation:
Given:
T1 = T2 = 50 , TL = 50 + 273 = 323 K
T3 = T4 = 350 , TH = 350 + 273 = 623 K
x3 =0.891
x4 = 0.1
Part a
Thermal efficiency (Etath) of carnot cycle is:
Etath = 1 - (TL / TH )
Etath = 1 - (323) / (623) = 48.2 %
Part b
Note:
From A - 4 Table
T1 = 50 C @sat
sf = 0.7038 KJ/kg.K
sfg = 7.3710 KJ/Kg.K
s2 = s3 = sf + x3 * (sfg)
s2 = s3 = 0.7038 + 0.891*7.3710 = 7.271361 KJ/kg.K
Thus,
@ T2 = 350 C ; sg = 5.2114 KJ/kg.K
s2 = 7.271361 KJ/kg.K
s2 > sg@T = 350 C, hence super-heated region.
Use Table A-6
T2 = 350 C
s2 = 7.271361 KJ/kg.K
P2 = 1.195 MPa (using interpolation)
part c
wnet can be calculated by finding the enclosed area on a T-s diagram:
s4= sf + x4*sfg = 0.7038 + 0.1*7.3710 = 1.4409 KJ/kg.K
wnet = (TH - TL)*(s3 - s4)
wnet = (623 - 223) * (7.271361 - 1.4409) = 1749.14 KJ/kg
a 100mh inductor is placed parallel with a 100 ohm resistor in the circuit , the circuit has a source voltage of 30 vac and a frequancy of 200 Hz what is the current through the inductor
Answer:
current through the inductor = 0.24 A
Explanation:
given data
inductor = 100mh
resistor = 100 ohm
voltage = 30 vac
frequancy = 200 Hz
to find out
current through the inductor
solution
current through the inductor will be when inductor is place parallel with resistor
across resistor
XL = i2πl
XL = i2π×200×100×[tex]10^{-3}[/tex] = i126.66
so
current through the inductor = [tex]\frac{voltage}{XL}[/tex]
current through the inductor = [tex]\frac{30}{125.66}[/tex]
current through the inductor = 0.24 A
A harmonic oscillator with spring constant, k, and mass, m, loses 3 quanta of energy, leading to the emission of a photon.
a. What is the energyon in terms of k adm
b. If the oscillator is a bonded atom with k = 15 N/m and m = 4 × 10-26 kg, what is the frequency (Hz) of the emitted photon? (Note: the energy of a photon is Ephoton= hf)
c. In which region of the electromagnetic spectrum (x-ray, visible, microwave, etc.) does this photon belong?
The answer explains the energy of a quantum harmonic oscillator, calculates the frequency of an emitted photon, and identifies the region of the electromagnetic spectrum it belongs to.
Explanation:a. Energy: The energy of a quantum harmonic oscillator can be represented as En = (n+1/2)h(sqrt(k/M)), where n = 0,1,2... and h represents Planck's constant.
b. Frequency Calculation: Using the given values of k = 15 N/m and m = 4 x 10^-26 kg, you can calculate the frequency of the emitted photon using the formula w = sqrt(k/M)/(2pi).
c. Electromagnetic Spectrum: To determine the region of the electromagnetic spectrum the photon belongs to, compare the frequency calculated to the known ranges of various regions like x-ray, visible, and microwave.
A) Fix any errors to get the following program to run in your environment. B) Document each line of code with comments and describe any changes you had to make to the original code to get it to work. C) Write a summary of what your final version of the program does.
You may also add white space or reorder the code to suit your own style as long as the intended function does not change.
#include
using namespace std;
void m(int, int []);
void p(const int list[], int arraySize)
{
list[0] = 100;
}
int main()
{
int x = 1;
int y[10];
y[0] = 1;
m(x, y);
cout << "x is " << x << endl;
cout << "y[0] is " << y[0] << endl;
return 0;
}
void m(int number, int numbers[])
{
number = 1001;
numbers[0] = 5555;
}
Answer:
Answer is explained below
Explanation:
Part A -:
Error statement -:
/*
prog.cpp: In function ‘void p(const int*, int)’:
prog.cpp:7:15: error: assignment of read-only location ‘* list’
list[0] = 100;
*/
There is one error in the code in following part
void p( const int list[], int arraySize)
{
list[0] = 100;
}
you are passing list as constant but changing it inside the function that is not allowed. if you pass any argument as const then you can't change it inside the function. so remove const from function argument solve the error.
Part B -:
change made
void p( int list[], int arraySize)
{
list[0] = 100;
}
Executable fully commented code -:
#include <iostream> // importing the iostream library
using namespace std;
void m(int, int []); // Function declearation of m
void p( int list[], int arraySize) // definition of Function p
{
list[0] = 100; // making value of first element of list as 100
}
int main()
{
int x = 1; // initilizing x with 1
int y[10]; // y is a array of 10 elements
y[0] = 1; // first element of y array is 1
m(x, y); // call m function
// printing the desired result
cout << "x is " << x << endl;
cout << "y[0] is " << y[0] << endl;
return 0;
}
void m(int number, int numbers[]) // Function definition of m
{
number = 1001; // value of number is 1001 locally
numbers[0] = 5555; // making value of first element of numbers array 5555
}
Part C :-
In program we initilize x with value 1 and create an array y of 10 elements.
we initilize the y[0] with 1\
then we call function m. In function m ,first argument is value of x and second argument is the pointer to the first element of array y.
so value of x is changed locally in function m and change is not reflected in main function.
but value of y[0] is changed to 5555 because of pass by refrence.
So we are getting the following result :-
x is 1
y[0] is 5555
The lattice constant of a simple cubic lattice is a0.
(a) Sketch the following planes:
(i) (110),
(ii) (111),
(iii) (220), and (iv) (321).
(b) Sketch the following directions:
(i) [110],
(ii) [111],
(iii) [220], and (iv) [321]
Answer:
A)The sketches for the required planes were drawn in the first attachment.
B)The sketches for the required directions were drawn in the second attachment.
To draw a plane in a simple cubic lattice, you have to follow these instructions:
1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment)
2- The coordinates of that plane are written as: π:(1/a₀ 1/b₀ 1/c₀) (if one of the coordinates is 0, for example (1 1 0), c₀ is ∞, therefore that plane never cross the direction c).
3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.
4- Join the points.
To draw a direction in a simple cubic lattice, you have to follow these instructions:
1- Identify the points a₀, b₀, and c₀ in the cubic cell.
2- Draw the direction as a vector-like (a₀ b₀ c₀).