Answer:
a. [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
b.833.3(0.006-x)+112
c. 117 deg C
Explanation:
Consider the base plate of an 800-W household iron with a thickness of L 5 0.6 cm, base area of A 5 160 cm2, and thermal conductivity of k 5 60 W/m·K. The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be 112°C. Disregarding any heat loss through the upper part of the iron,
Assumption
Heat conduction is steady state and unidimensional 2. thermal conductivity is constant. Heat supplied is not in the plate
4. we disregard heat loss
Heat flux=heat/area
[tex]\alpha[/tex]/A=800W/160*10^-4
with direction to the surface been in the x direction,
the mathematical expression will be
[tex]\frac{d^2T}{dx^2}[/tex]=0..............1
and [tex]\frac{-kdT(0)}{dx} =q_{0}[/tex]=5000W/m^2
from fourier law, for conductivity
T(L)=T2=112C
b. integrating equation 1 twice we have\dT/dx=c1
T(x)=C1x+C2
C1 and C2 are arbitrary constant
at x=0 the boundary conditions become
-kC1=qo
C1=-(qo/k)
at x=L
=T(L)=C1L+C2=T2
C2=T2-cL1
C2=T2+qoL/k
Juxtaposing C1 and C2 into the general equation , we have
T(x)=-qo/k+T2+qoL/k=qo(L-k)/k+T2
50000*(0.006-x)/60+112
833.3(0.006-x)+112
c. inner surface plate temperature is
T(0)=833.33(0.006-0)+112 ( using the derivation in answer b)
117 deg C
Primary U.S. interstate highways are numbered 1-99. Odd numbers (like the 5 or 95) go north/south, and evens (like the 10 or 90) go east/west. Auxiliary highways are numbered 100-999, and service the primary highway indicated by the rightmost two digits. Thus, the 405 services the 5, and the 290 services the 90 Given a highway number, indicate whether it is a primary or auxiliary highway. If auxiliary, indicate what primary highway it serves. Also indicate if the (primary) highway runs north/south or east/west. Ex: If the input is: 90 the output is: The 90 is primary, going east/west. Ex: If the input is: 290 the output is: The 290 is auxiliary, serving the 90, going east/west Ex: If the input is G here to search The 290 is auxiliary, serving the 90, going eas Ex: If the input is: 0 or any number not between 1 and 999, the output is: 0 is not a valid interstate highway number. See Wikipedia for more info on highway numbering LAB 4.16.1: LAB: Interstate highway numbers ACTIVITY LabProgr- 1 import java.util.Scanner; 2 3 public class LabProgram public static void main(Stringl] args) t Scanner scnr new Scanner(System.in); int highwayNumber; int primaryNumber; 4 5 6 7 highwayNumber scnr.nextInt(); 1e Type your code
Answer:
The Java code is given below with appropriate variable names and tags for better understanding
Explanation:
import java.util.Scanner;
public class LabProgram{
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int highwayNumber;
int primaryNumber;
highwayNumber = scnr.nextInt();
if(highwayNumber<1 || highwayNumber>999)
System.out.println(highwayNumber+" is not a valid interstate highway number.");
else{
if(highwayNumber>=1 && highwayNumber<=99){
System.out.print("The "+highwayNumber+" is primary, going ");
if(highwayNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
else{
primaryNumber = highwayNumber%100;
System.out.print("The "+highwayNumber+" is auxillary, serving the "+primaryNumber+", going ");
if(primaryNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
}
}
}
The program is an illustration of conditional statements.
Conditional statements are statements whose execution is dependent on its truth value.
The missing code segment in Java where comments are used to explain each line is as follows:
import java.util.Scanner;
public class Main {
public static void main(String [] args){
Scanner scnr = new Scanner(System.in);
int highwayNumber; int primaryNumber;
highwayNumber = scnr.nextInt();
//This checks if the highwayNumber is not between 1 and 999 (inclusive)
if (highwayNumber <1 || highwayNumber > 999){
//If yes, the highwayNumber is invalid
System.out.print(highwayNumber+" is not a valid interstate highwayNumber number.");
}
//If otherwise
else{
//This checks if highwayNumber is less than 100
if (highwayNumber< 100){
if (highwayNumber%2 == 0){
//Even highwayNumber are primary going east/west
System.out.print("I-"+highwayNumber+" is primary, going east/west.");
}
else{
//Odd highwayNumber are primary going north/south
System.out.print("I-"+highwayNumber+" is primary, going north/south.");
}
}
//Otherwise
else{
if ((highwayNumber%100) % 2 == 0){
//Even highwayNumber are auxiliary going east/west
System.out.print("I-"+highwayNumber+" is auxiliary, going east/west.");
}
else{
//Even highwayNumber are auxiliary going north/south
System.out.print("I-"+highwayNumber+" is auxiliary, going north/south.");
}
}
}
}
}
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A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m3/s. Just upstream from the discharge point, the stream has an ultimate BOD of 3.0 mg/L The deoxygenationconstant kd is estimated at 0.22/day.(a) Assuming complete and instantaneous mixing. find the ultimate BOD of the mixture of waste and river just downstreamfrom the outfall.(b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find ata point 10.000 m downstream?
Answer:
a) 6.4 mg/l
b) 5.6 mg/l
Explanation:
Given data:
effluent Discharge Q_w = 1.0 m^3.s
Ultimate BOD L_w = 40 mg/l
Discharge of stream Q_r = 10 m^3.s
Stream ultimate BOD L_r = 3 mg/l
a) Ultimate BOD of mixture[tex] = \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}[/tex]
[tex] = \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l[/tex]
b) utlimate BOD at 10,000 m downstream
[tex]t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times \frac{day}{24 hr}[/tex]
putting [tex]Q_r + Q_w = 1+ 10 = 11 m^3/s[/tex]
t = 0.578 days
we know
[tex]L_t = L_o e^{-kt}[/tex]
[tex]L_t = 6.4 \times e^{-0.22 \times 0.578}[/tex]
[tex]L_t = 5.6 mg/l[/tex]
Determine the tension developed in cord DE required for equilibrium of the lamp. Express your answer to three significant figures and include the appropriate units.
Answer:
Fde = 1080 N (3 sig fig)
Explanation:
Taking point E
Sum of forces at E in x direction:
Fde cos (30) - Fcd = 0 ..... Eq 1
Sum of forces at E in y direction:
Fde sin (30) - Wf = 0 ..... Eq 2
Wf = m*g = 55 * 9.81 = 539.55 N
Using Eq 2 to evaluate Fde
Fde = Wf / sin (30) = 539.55 / sin (30) = 1079.1 N
Answer: Fde = 1080 N (3 sig fig)
Write a function called pyramid(height) that acceptsa parameter ""height"". It then prints a pyramid of that height
Answer:
I am writing a function in C++
Explanation:
C++ program#include <iostream>
using namespace std;
int pyramid(int height) // function pyramid with parameter height
{int distance; //variable for spaces
for(int i = 1, j = 0; i <= height; ++i, j= 0) //handle the rows
{
for(distance= 1; distance<= height-i; ++distance)
// handles the spaces & columns
{ cout <<" "; } //prints spaces between stars
while(j!= 2*i-1) //handles shape and spaces
{ cout << "* "; // printing stars
++j; }
cout << '\n'; } } // for the next line
int main()
{
int height; //declare height variable
cout <<"Enter height of pyramid "; //asks user to enter height of pyramid
cin>>height; //stores value of height
pyramid(height); //calls pyramid function
}
Previously , you wrote a program named Hurricane that classified hurricanes into five categories using the Saffir-Simpson Hurricane Scale. Now, create a modified version named Hurricane Modularized that passes a user’s input wind speed to a method that returns the hurricane category.
Answer:
Answer of the Java class explained below with appropriate comments
Explanation:
using System;
class MainClass {
public static void Main (string[] args) {
//entering wind speed of the hurricane
Console.WriteLine ("Enter wind speed in mph: ");
int windSpeed = Convert.ToInt32(Console.ReadLine());
//checking for the category with the credited wind speed
int category = GetCategory(windSpeed);
Console.WriteLine ("Category: " + category);
}
public static int GetCategory(int wind) {
//function for classifying the wind speeds
if(wind >= 157) {
return 5;
}
if(wind >= 130) {
return 4;
}
if(wind >= 111) {
return 3;
}
if(wind >= 96) {
return 2 ;
}
return 1;
}
}
If E = 94.2 mJ of energy is transferred when Q = 1.66 C of charge flows through a circuit element, what is the voltage across the circuit element?
Answer:
V = 56.8 mV
Explanation:
When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:
P = V*I
By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:
E/t = V*q/t ⇒ E = V*Q
Solving for V:
V = E/Q = 94.2 mJ /1.66 C = 56.8 mV
Using phasors, the value of 37 sin 50t + 30 cos(50t – 45°) is _________ cos(50t+(_____°)). Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 80 degrees
Answer: 62 cos(50t - 70°)
Explanation:
First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)
Therefore, 37sin50t = 37cos(50t - 90°)
Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].
So,
37cos(50t - 90°)
= 37∠-90°
= 37[cos(-90°)+jsin(-90°)
=37[0+j(-1)]
= -j37
Similarly,
30 cos(50t – 45°)
=30∠-45
=30[cos(-45)+jsin(-45)
=30[0.707-j0.707]
=21.21 - j21.21
37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21
Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.
Here a=21.21 and b=58.21
magnitude = √(21.21)²+(58.21)²=61.95≅62
For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.
∅=inversetan(-58.21/21.21)
= -69.9°≅-70°
So the final answer is 62cos(50t-70°)
The value of [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] is [tex]\(61.97 \cos(50t - 70^\circ)\)[/tex].
To solve [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] using phasors, we follow these steps:
Step-by-Step Calculation:
1. Express each term as a phasor:
- The term [tex]\(37 \sin 50t\)[/tex] :
- Convert to cosine form: [tex]\(37 \sin 50t = 37 \cos(50t - 90^\circ)\)[/tex]
- Phasor form: [tex]\(37 \angle -90^\circ\)[/tex]
- The term [tex]\(30 \cos(50t - 45^\circ)\)[/tex] :
- Phasor form: [tex]\(30 \angle -45^\circ\)[/tex]
2. Convert the phasors to rectangular form:
- [tex]\(37 \angle -90^\circ\)[/tex] :
[tex]\[ 37 \cos(-90^\circ) + j 37 \sin(-90^\circ) = 0 - j 37 = -j 37 \][/tex]
- [tex]\(30 \angle -45^\circ\)[/tex]:
[tex]\[ 30 \cos(-45^\circ) + j 30 \sin(-45^\circ) = 30 \left(\frac{\sqrt{2}}{2}\right) - j 30 \left(\frac{\sqrt{2}}{2}\right) = 21.2132 - j 21.2132 \][/tex]
3. Add the rectangular components:
[tex]\[ -j 37 + (21.2132 - j 21.2132) \][/tex][tex]\[ = 21.2132 - j (37 + 21.2132) \][/tex]
[tex]\[ = 21.2132 - j 58.2132 \][/tex]
4. Convert the result back to polar form:
- Magnitude:
[tex]\[ R = \sqrt{21.2132^2 + 58.2132^2} = \sqrt{449.54 + 3388.78} = \sqrt{3838.32} = 61.97 \][/tex]
- Angle:
[tex]\[ \theta = \tan^{-1}\left(\frac{-58.2132}{21.2132}\right) = \tan^{-1}(-2.743) \approx -70^\circ \][/tex]
5. Express the final result:
Using the positive magnitude, the expression in cosine form is:
[tex]\[ 37 \sin 50t + 30 \cos(50t - 45^\circ) = 61.97 \cos(50t - 70^\circ) \][/tex]
Liquid toluene is flowing through a pipe at a rate of 175 m 3 / h 175m3/h . What is the mass flow rate of this stream in kg/min? What is the molar flow rate in mol/s? In fact, the answer to Part (a) is only an approximation that is almost certain to be slightly in error. What did you have to assume to obtain the answer?
Answer:
(a) Mass flow rate is 2528.75 kg/min
(b) Molar flow rate is 458.1 mol/s
(c) Toulene is in-compressible at standard temperature and pressure
Explanation:
The volume flow rate of liquid toulene is given:
Volume Flow Rate = 175 m^3/h
Density of Liquid Toulene = 867 kg/m^3
Molar mass of Toulene = 92 g/mol = 0.092 kg/mol)
(a)
Thus, the mass flow rate of toulene will be:
Mass Flow Rate = (Volume Flow Rate)(Density)
Mass Flow Rate = (175 m^3/h)(867 kg/m^3)
Mass Flow Rate = (151725 kg/h)(1 h/60 min)
Mass Flow Rate = 2528.75 kg/min
(b)
Now, for molar flow rate we use formula:
Molar Flow Rate = (Mass Flow Rate)/(Molar Mass)
Molar Flow Rate = (2528.75 kg/min)/(0.092 kg/mol)
Molar Flow Rate = (27486.41 mol/min)(1 min/60sec)
Molar Flow Rate = 458.1 mol/s
(b)
The assumption is that Toulene is in-compressible at standard temperature and pressure. So, that its density can be taken constant.
The loss of power a signal suffers as it travels from the transmitting computer to a receivingcomputer is: a.white noiseb.spikingc.attenuation d.intermodulation noisee.echo
Answer:
c.attenuation
Explanation:
In telecommunication, it is called attenuation of a signal, be it acoustic, electrical or optical, to the loss of power suffered by it when passing through any transmission medium.
Attenuation is usually not expressed as a difference in powers but in logarithmic units such as decibels, which are more comfortable to use when calculating.
The attenuation is expressed in decibels (db) and the power is measured with this formula:
α = 10 * log (P1 / P2)
where P is the power both initial and final.
The article "Display of Health Risk Behaviors on MySpace by Adolescents"† described a study in which researchers looked at a random sample of 500 publicly accessible MySpace web profiles posted by 18-year-olds. The content of each profile was analyzed. One of the conclusions reported was that displaying sport or hobby involvement was associated with decreased references to risky behavior (sexual references or references to substance abuse or violence).
a. It's not entirely reasonable to generalize the conclusion to all 18-year-olds with publicly accessible MySpace profiles due to the study's limited sample size and potential biases.
b. b. No, it's not reasonable to generalize the conclusion to all 18-year-old MySpace users because not all users have publicly accessible profiles.
c. While somewhat reasonable, generalizing to MySpace users with publicly accessible profiles requires caution, considering potential individual differences within this subgroup.
The study only looked at a random sample of 500 profiles, which might not be fully representative of all 18-year-olds on MySpace. Generalizing to all 18-year-old MySpace users is not reasonable as not all users have publicly accessible profiles, which could skew the findings.
Additionally, factors such as cultural differences, regional variations, and individual preferences could influence whether adolescents choose to display sport or hobby involvement on their profiles.
See text
The article "Display of Health Risk Behaviors on MySpace by Adolescents" (Archives of Pediatrics and Adolescent Medicine [2009]:
) described a study in which researchers looked at a random sample of 500 publicly accessible MySpace web profiles posted by 18-year-olds. The content of each profile was analyzed. One of the conclusions reported was that displaying sport or hobby involvement was associated with decreased references to risky behavior (sexual references or references to substance abuse or violence).
a. Is it reasonable to generalize the stated conclusion to all 18-year-olds with a publicly accessible MySpace web profile? What aspect of the study supports your answer?
b. Not all MySpace users have a publicly accessible profile. Is it reasonable to generalize the stated conclusion to all 18-year-old MySpace users? Explain.
c. Is it reasonable to generalize the stated conclusion to all MySpace users with a publicly accessible profile? Explain.
A certain battery has terminals labeled a and b. The battery voltage is vab = 12 V. a) To increase the chemical energy stored in the battery by 900J, how much charge must move through the battery? b) Should electrons move from a to b or from b to a?
Answer:
(a) The charges of 75 C must flow through battery.
(b) The electrons must flow from a to b.
Explanation:
(a)
The relationship between the energy and voltage is as follows:
Energy = (Voltage)(Current)(Time)
but, (current)(time) = charge
therefore,
Energy = (Voltage)(Charge)
Charge = (Energy)/(Voltage)
Charge = (900 J)/(12 V)
Charge = 75 C
(b)
Since, Vab = Va - Vb = 12 V
Hence, the equation suggest that Va is at higher potential then Vb.
As, the current flows from higher to lower potential.
Thus, electrons must flow from a to b
A thermistor is a temperature‐sensing element composed of a semiconductor material, which exhibits a large change in resistance proportional to a small change in temperature. A particular thermistor has a resistance of 5 kΩ at 25°C. Its resistance is 340 Ω at 100°C. Assuming a straight‐line relationship between these two values, at what temperature will the thermistor's resistance equal 1 kΩ?
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.
For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in
barometers containing one of the following liquids a) mercury, b) water, c) ethyl alcohol.
Calculate the heights including the effect of vapour pressure and compare the results with those obtained
neglecting vapour pressure. Do these results support the widespread use of a particular fluid for
barometers? Explain the reason.
Answer:
a) 0.759 m , 0.759 m
b) 10.1 m , 10.3 m
c) 12.3 m , 13 m
Explanation:
Vapour Pressure included:
[tex]P_{atm,vp} = y*h + P_{vp} \\\\h = \frac{P_{atm,vp}-P_{vp}}{y} ... Eq1[/tex]
mercury
[tex]h_{Hg,vp} = \frac{101*10^3-1.6*10^(-1)}{133*10^3} \\\\h_{Hg,vp} = 0.759m[/tex]
[tex]h_{Hg} = \frac{101*10^3}{133*10^3} \\\\h_{Hg} = 0.759m[/tex]
water
[tex]h_{H2O,vp} = \frac{101*10^3-1.77*10^3}{9.8*10^3} \\\\h_{H2O,vp} = 10.1m[/tex]
[tex]h_{H2O} = \frac{101*10^3}{9.8*10^3} \\\\h_{H20} = 10.3m[/tex]
Ethyl Alcohol
[tex]h_{EA,vp} = \frac{101*10^3-5.9*10^3}{7.74*10^3} \\\\h_{EA,vp} = 12.3m[/tex]
[tex]h_{EA} = \frac{101*10^3-5}{7.74*10^3} \\\\h_{EA} = 13m[/tex]
For mercury barometers the effects of vapour pressure are negligible and required height for mercury barometer is reasonable as compared to that of water and ethyl alcohol.
the net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared. Steam enters the turbine in both cases at 5 MPa as a saturated vapor and the condenser pressure is 50 kPa. in the Rankine cycle, the condenser exit state is saturated liquid and in the Carnot cycle, the boiler inlet state is saturated liquid. Draw the T-s diagrams for both cycles.
Answer:
a) Rankine
Net work output = 719.1 KJ/kg
Thermal Eff = 0.294
a) Carnot
Net work output = 563.2 KJ/kg
Thermal Eff = 0.294
For T-s diagrams see attachments
Explanation:
Part a Rankine Cycle
The obtained data from water property tables:
[tex]P_{L,sat liquid} = 50 KPa \\v_{1} = 0.00103m^3/kg\\\\ h_{1} = 340.54KJ/kg\\\\P_{H} = 5000KPa\\h_{2} = h_{1} + v_{1} *(P_{H} - P_{L} )\\h_{2} =350.54 + (0.00103)*(5000 - 50)\\\\h_{2} = 345.64KJ/kg\\\\P_{H,satsteam} = 5000KPa\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\s_{3} = s_{4} = 5.9739KJ/kgK\\P_{L}= 50KPa\\\\h_{4}= 2070KJ/kg\\\\[/tex]
Heat transferred from boiler
[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-345.64\\\\q_{b} =2448.56KJ/kg\\\\[/tex]
Heat transferred from condenser
[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-340.54\\\\q_{b} =1729.46KJ/kg\\\\[/tex]
Thermal Efficiency
[tex]u_{R} = 1- \frac{q_{c}}{q_{b}}\\\\u_{R} = 1 - \frac{1729.46}{2448.56}\\\\u_{R} =0.294[/tex]
Net work output
[tex]w_{R} = q_{b}-q_{c}\\w_{R} = 2448.56-1729.46\\\\w_{R}=719.1KJ/kg[/tex]
Part b Carnot Cycle
The obtained data from water property tables:
[tex]P_{H,sat-steam} = 5000KPa\\T_{3} = 263.94 C\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\T_{2,sat-liquid} = T_{3} = 263.94C\\s_{2} = 2.920KJ/kgK\\\\h_{2} = 1150KJ/kg\\\\P_{L} = 50KPa\\s_{1}=s_{2} = 2.920KJ/kgK\\\\h_{1} = 989KJ/kg\\\\s_{3} = s_{4} = 5.9737KJ/kgK\\P_{L} = 50KPa\\\\h_{4} = 2070KJ/kg[/tex]
Heat transferred from boiler
[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-1150\\\\q_{b} =1644.2KJ/kg\\\\[/tex]
Heat transferred from condenser
[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-989\\\\q_{b} =1081KJ/kg\\\\[/tex]
Thermal Efficiency
[tex]u_{C} = 1- \frac{q_{c}}{q_{b}}\\\\u_{C} = 1 - \frac{1081}{1644.2}\\\\u_{C} =0.343[/tex]
Net work output
[tex]w_{C} = q_{b}-q_{c}\\w_{C} = 1644.2-1081\\\\w_{C}=563.2KJ/kg[/tex]
How much energy in joules does a 12 volt battery in your car use to move 2.5 coulombs through the electrical circuit?
Answer: 30joules
Explanation
voltage of battery =12volts
2.5coulombs is the same as 2.5ampere-seconds
Energy required =12V*2.5A-S
=30watts-seconds
Hence the required energy is equal to 30joules.
In studying of traffic flow at a highway toll booth over the course of 60 minutes, it is determined that the arrival and departure rates are deterministic, but not uniform. The arrival rate is found to vary according to the function A(t) = 1.8 + 0.25t - 0.0030t^2. The departure rate function is D(t) = 1.4 + 0.11t. In both of these functions, t is in minutes after the beginning of the observation and A(t) and D(t) are in vehicles per minute. At what time does the maximum queue length occur?
a. 2.7 min
b. 9.4 min
c. 49.4 min
d. 60.0 min
The time that the maximum queue length occurs would be c. 49.4 min
To find the time at which the maximum queue length occurs, we need to determine when the arrival rate equals the departure rate.
The queue length increases when the arrival rate exceeds the departure rate and decreases when the departure rate exceeds the arrival rate. The maximum queue length occurs when the arrival rate equals the departure rate.
Substituting the given functions into the equation, we get:
[tex]\[ 1.8 + 0.25t - 0.0030t^2 = 1.4 + 0.11t \][/tex]
Rearranging the terms, we get a quadratic equation:
[tex]\[ -0.0030t^2 + 0.25t - 0.11t + 1.8 - 1.4 = 0 \][/tex]
[tex]\[ -0.0030t^2 + 0.14t + 0.4 = 0 \][/tex]
Now, we can solve this quadratic equation to find the value(s) of t at which the maximum queue length occurs. We can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where:
a = -0.0030
b = 0.14
c = 0.4
Calculate the values of t using this formula. Then, we'll choose the appropriate value based on the physical meaning of the problem.
Using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{(0.14)^2 - 4(-0.0030)(0.4)}}{2(-0.0030)} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{0.0196 + 0.0048}}{-0.0060} \][/tex]
[tex]\[ t = \frac{-0.14 \pm 0.156}{-0.0060} \][/tex]
Now, we have two possible values for t :
1. [tex]\( t_1 = \frac{-0.14 + 0.156}{-0.0060} \)[/tex]
2. [tex]\( t_2 = \frac{-0.14 - 0.156}{-0.0060} \)[/tex]
Calculate these values:
1. [tex]\( t_1 = \frac{0.016}{-0.0060} = -2.67 \)[/tex]
2. [tex]\( t_2 = \frac{-0.296}{-0.0060} = 49.4[/tex]
Since t represents time, it cannot be negative. The maximum queue length occurs at approximately t = 49.4 minutes after the beginning of the observation.
Create a conditional expression that evaluates to string "negative" if userVal is less than 0, and "non-negative" otherwise. Ex: If userVal is -9, output is: -9 is negative.
Answer:
(userVal < 0) ? "negative" : "non-negative" ;
Explanation:
The above code has been written using Java's ternary operator (? : ).
Java uses this operator to write conditional expressions that are similar to the regular if ... else statements.
This expression has three parts,
i. the conditional statement: this is the expression before the ? mark. In this case, (userVal < 0). This expression contains the condition to be tested for and it returns true or false.
ii. the second part is the statement after the ? mark. In this case, "negative". This expression will be executed if the conditional statement returns true.
iii. the third part is the statement after the : mark. In this case, "non-negative". This expression will be executed if the conditional statement returns false.
So in this case, if userVal < 0, the string "negative" will be evaluated. Otherwise, "non-negative".
What are nonexecuting statements that you can use to document or add notes to assist in the use of the program?
Answer:
Program comments
Give two circumstances in which in situ treatment would be better, and two in which ex situ treatment would be better. Both using chemical oxidation as the remediation technique.
Answer:
in situ treatment would be better if; (1) a large volume of soil is to be treated at once, and (2) a permeable sandy soil (un-compacted) is to be treated.
ex situ treatment would be better would be better if; (1) a wider range of contaminants and soil types are to be treated, and (2) there is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.
Explanation:
in situ treatment would be better if;
a large volume of soil is to be treated at oncea permeable sandy soil (un-compacted) is to be treatedex situ treatment would be better would be better if;
a wider range of contaminants and soil types are to be treatedthere is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.Recycled materials content is environmental information that is typically: a. self-declared by the manufacturer. b. independently certified by third-party entities. c. listed in the building code. d. part of an environmental label. e. None of the above
Answer:
a. Self declared by the manufacturer
Explanation:
Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are
post-consumer .
A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft lanes with a 4-ft shoulder on the right side and 2-ft shoulder on the left. The peak-hour factor is 0.84 and the traffic stream has a heavy vehicle adjustment factor of 0.847. If the analysis flow rate is 1250 pc/h/ln, what is the peak-hour volume? 976 veh/h 1345 veh/h 1389 veh/h 1779 veh/h
Answer:
v = 1779 veh/h
Explanation:
We will calculate the peak hour volume by using the following formula:-
Vp = v / ( PHF)(N)(Fg)(Fhv)
here,
1. v is the peak hour volume
2. vp is the analysis flow rate
3. PHF is the peak hour factor
4. N is the number of lanes
5. Fg is the grade adjustment factor which is 0.99 for rolling terrain > 1200 pc/h/ln
6. Fhv is the heavy vehicle adjustment factor
vp = v / (PHF) (N) (Fg) (Fhv)
1250 = v / (0.84)(2)(1)(0.847)
v = 1779 veh/h
Chlorate (ClO3 − ) is a toxic anion in polluted water from the use of high energy materials, herbicide, and drinking water disinfection with chlorine chemicals. In a reaction for the catalytic reduction of ClO3 − into the non-toxic chloride (Cl− ), the reaction rate constant was measured as 9.98 hr−1 , then how many minutes does the reaction take to reduce 90% of the chlorate in the water?
Answer:
Time taken = 5.41mins
Explanation:
A step by step calculations with detailed explanation has been attached below.
An automotive Battery is rated at120 A-h. This means that under certain test conditions, it canoutput 1 A at 12 V for 120 hours.
(a) How much total energy is storedin battery?
(b) If the headlights are left on overnight (8h),how much energy will still be stored in the battery in the morning(Assume a 150 w total power rating for both headlightstogether)
Answer:
Part A:
In W-h:
Energy Stored=1440 W-h
In Joules:
[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]
Part B:
In W-h:
Energy left=240 W-h
In Joules:
Energy left= 8.64*10^5 J
Explanation:
Part A:
We are given rating 120A-h and voltage 12 V
Energy Stored= Rating*Voltage (Gives us units W-h)
Energy Stored=120A-h*12V
Energy Stored=1440 W-h
Converting it into joules (watt=joules/sec)
Energy Stored=[tex]1440 Joules * \frac{3600sec}{h}h[/tex]
Energy Stored=5184000 Joules
[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]
Part B:
Energy used by lights for 8h=150*8
Energy used by lights for 8h=1200W-h
Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h
Energy left=1440-1200
Energy left=240 W-h
Energy left=[tex]240 Joules * \frac{3600sec}{h}h[/tex]
Energy left=864000 Joules
Energy left= 8.64*10^5 J
The total energy stored in the battery is 1440 Wh. If the headlights are left on overnight (8h), the energy remaining in the battery in the morning is 1320 Wh.
Explanation:(a) The total energy stored in the battery can be calculated by multiplying the battery's capacity (120 A-h) by the voltage (12 V). So, the total energy stored in the battery is 1440 Wh.
(b) To calculate the energy consumed by the headlights in 8 hours, we need to find the power (P) using the formula P = V × I, where V is the voltage (12 V) and I is the current (the total power rating divided by the voltage). Then, we can find the energy consumed by multiplying the power by the time (8 hours). We can subtract this energy consumption from the total energy stored in the battery to find the energy remaining in the morning. So, the energy remaining in the battery is 1440 Wh - 12 kWh = 1320 Wh.
Learn more about Energy stored in an automotive battery here:https://brainly.com/question/34222184
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A scrubber on a coal-fired power plant is useful to remove
a. sulfur dioxide.
b. carbon dioxide.
c. particulate matter.
Answer:
sulfur dioxide
Explanation:
The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack. Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.
Answer:
sulfur dioxide.
Explanation:
dentify a semiconducting material and provide the value of its band gap) that could be used in: (a) (1 point) red LED (b) (1 point) UV LED c) (1 point) infrared detector (d) (1 point) blue LED (e) (1 point) green LED
Answer:
(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV ≈ 2eV
(b) Gallium Nitride (GaN). Band Gap = 3.4eV
(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV
(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV
(e) Gallium Phosphide (GaP). Band Gap = 2.24eV
Explanation:
LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.
The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).
Which of the following elements of the CIA triad refers to maintaining and assuring the accuracy of data over its life-cycle?
Confidentiality
Integrity
Availability
Authentication
Answer:
Integrity: involves maintaining and assuring the accuracy of data over its life-cycle
Explanation:
Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.
Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.
Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.
Authentication:This is a security control that is used to protect the system with regard to the CIA properties.
A signal whose timing is completely influenced by the traffic volumes, when detected, on all of the approaches operates in the following mode: a. Pretimed b. Semi-actuated c. Fully-actuated
Answer: c. Fully actuated
Explanation: fully actuated signals are completely influenced by volumes if traffic and employs sensors at all approaches for its detection.
A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at a mass flow of 0.5 kg/s. The pipe is 5 m long, has an inner radius of 50 cm and pipe wall thickness of 50 mm. The pipe has a thermal conductivity of 20 W/m-K, and the inner pipe surface is at a uniform temperature of 110 ºC. The convection heat transfer coefficient of the air surrounding the pipe is 100W/m2 -K. The temperature of the water at inlet of pipe is 130 ºC and the constant pressure specific heat of hot water is 4000 J/kg-ºC. If the temperature of the air surrounding the pipe is 20 ºC, determine the exit temperature of the water at the end of the pipe.
The calculation of the exit temperature of water in the heated pipe involves using the energy balance equation, considering the heat lost through the pipe walls by convection, and then finding the change in thermal energy of the water via the heat transfer equation to solve for the exit temperature.
Explanation:Exit Temperature of Water in a Heated Pipe
To determine the exit temperature of water at the end of an over-pressurized heated pipe, we must consider the energy balance for the water flowing through the pipe. Based on the first law of thermodynamics, the change in thermal energy of the water will be equal to the heat lost through the pipe walls by convection:
Q = mc_p
(Exit Temperature - Inlet Temperature)
In this case, Q will be negative, since the water is losing heat to the surrounding air. The heat transfer from the pipe to the air is given by:
Q = hA(T_surface - T_air)
The area A for heat transfer is the external surface area of the pipe (
2
classes Math.PI
* radius * length of the pipe). Since we have the heat transfer coefficient h, the surface temperature of the pipe T_surface, and the air temperature T_air, we can calculate Q. Then we can use the mass flow rate m and the constant pressure specific heat c_p to find the exit temperature of the water.
To solve, we first calculate Q, then rearrange the first equation to solve for the Exit Temperature.
A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.
a. If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2 and is constant with height.
b. If the stored energy is used to accelerate the battery with 100 percent efficiency, what velocity is attained?
c. Gasoline contains about rho1 = 4.5×107 J/kg. Compare this with the energy content per unit mass for the fully charged battery (rho2).
ans: rho2/ rho1=?
Answer: attached below. rate brainliest if correct please
The viscosity of a fluid is to be measured by an viscometer constructed of two 5 ft long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2lbf *ft.
A) Determine the viscosity of the fluid.
Answer:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]
Explanation:
For this case we need to remember first thet the torque T is defined as:
[tex] T = FR[/tex]
Where T represent the torque, F the force acting in the inner cylinder and R the radius for the inner cylinder.
For the inner cylinder the force acting can be expressed as:
[tex] F = \mu A \frac{v}{l}[/tex]
Where [tex] \mu[/tex] represent the viscosity of the fluid, A the area of the inner cylinder, v represent the tangential velocity and l the thickness of fluid between the two cylinders.
And the tangential velocity for this case can be esxpressed as [tex] v = wR[/tex]
The info given is:
[tex] l = 0.035 in *\frac{1ft}{12in}=0.00292 ft[/tex]
[tex] R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft[/tex]
[tex] L = 5 ft[/tex] from the info given
N= 250 rpm represent the reveolutions per minute
[tex] T = 1.2 lbf ft[/tex] represent the torque given
We can find the surface area for the cylinder with this formula:
[tex] A= 2\pi R L[/tex]
And if we replace we got:
[tex] A= 2\pi 0.25 ft *5 ft= 7.85 ft^2[/tex]
Now we can find the tangential velocity like this:
[tex] v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}[/tex]
Now we can set up the following equation for the torque:
[tex] T = FR[/tex]
[tex] T = \mu A \frac{v}{l} R[/tex]
And we can find the value for the viscosity [tex]\mu[/tex] like this:
[tex] \mu = \frac{T}{A \frac{v}{l} R}[/tex]
And if we replace we got:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]