Answer:
c < 7
Explanation:
The cardinal indicates the number or quantity of the elements of a set, be this finite or infinite quantity.
Given, A = {1, 2, 42, 57, 99, 538, 677}
B = {1, 5, 6, 7, {2, 3} }
C = {1, {7}, {8, 9} }
Let's say the cardinality of the subset is: c
Inequality that represents the cardinality of any proper subset of A is: c < 7
Hope this helps!
An inequality that represents the cardinality of any proper subset of A is c<7.
Given, A = {1, 2, 42, 57, 99, 538, 677}
B = {1, 5, 6, 7, {2, 3} }
C = {1, {7}, {8, 9} }
Inequalities are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.
Let's say the cardinality of the subset is: c
Inequality that represents the cardinality of any proper subset of A is: c < 7
Therefore, an inequality that represents the cardinality of any proper subset of A is c<7.
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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.
Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.
Answer:
W = 1.7593 * 10 ^ (-7) KJ
Explanation:
The work done by the bubble is given:
[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]
Answer: W = 1.7593 * 10 ^ (-7) KJ
A glass window pane, 1 m wide, 1.5 m high, and 5 mm thick, has a thermal conductivity of kg = 1.4 W/(m∙K). On a cold winter day, the indoor and outdoor temperatures are 15 °C and −25 °C respectively. (a) For a single-pane window at steady state, what is the rate of heat transfer through the glass? (10 pts) (b) To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by a dead-air space. The thermal conductivity of air is ka = 0.024 W/(m∙K). If the spacing between the two glasses is 10 mm. Calculate the temperatures of the glass surfaces in contact with the dead-air at the steady state. (10 pts) (c) Calculate the heat loss through the double-pane window in (b). (5 pts)
Answer:
a) Rate of heat transfer = 16.8KW
b) Temperature of glass surface = 15degree Celsius
c) Heat loss through frame = 141.34W
Explanation:
The concept used to approach this question is the Fourier's law of head conduction postulated by Joseph Fourier. it states that the rate of heat flow through a single homogeneous solid is directly proportional to the area and to the direction o heat flow and to the change in temperature with respect to the path length. Mathematically,
Q = -KA dt/dx
The detailed and step by step calculation is attached below
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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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.
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]
Consider 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
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.
My Notes How many grams of perchloric acid, HClO4, are contained in 39.1 g of 74.9 wt% aqueous perchloric acid?
How many grams of water are in the same solution?
Answer:
a)29.9 b) 9.81
Explanation:
Wt% = mass of solute / mass of solvent × 100
0.749 = mass of solute / mass of solvent
a) Mass of perchloric acid = 0.749 × 39.1 = 29.29
b) Mass of water = 39.1 - 29.29 = 9.81
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
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
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 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
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).
If a barrel of oil weighs 1.5 kN, calculate the specific weight, density, and specific gravity of the oil. The barrel weighs 110 N
Answer
given,
oil barrel weight = 1.5 k N = 1500 N
weight of the barrel = 110 N
Assuming volume of barrel = 0.159 m³
weight of oil = 1500-110
= 1390 N
[tex]specific\ weight = \dfrac{weight}{volume}[/tex]
[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]
= 8742.14 N/m³
[tex]mass = \dfrac{weight}{g}[/tex]
[tex]mass = \dfrac{1390}{9.8}[/tex]
= 141.84 kg
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]density = \dfrac{141.84}{0.159}[/tex]
= 892.05 kg/m³
[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]
= 0.892
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.
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.
Determine the percent increase in the nominal moment capacity of the section in Problem 2 when including compression steel at top equal to 0.5 the area of the tension steel at the bottom.
Explanation:
Please kindly share your problem two with us as to know the actual problem we are dealing with, the question looks incomplete
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 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
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 a point in a structural member that is subjected to plane stress. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sx = 175 MPa, Sy = 90 MPa, and Sxy = 75 MPa.
Answer:
C = 132.5 MPa
R = 86.20 MPa
Explanation:
Given
σx = 175 MPa
σy = 90 MPa
τxy = 75 MPa
For the given state of stress at a point in a structural member, determine the center C and the radius R of Mohr’s circle.
We apply the following equation for the center C
C = (σx + σy) / 2
C = (175 MPa + 90 MPa) / 2
C = 132.5 MPa
The Radius can be obtained as follows
R = √(((σx - σy) / 2)² + (τxy)²)
R = √(((175 MPa - 90 MPa) / 2)² + (75 MPa)²)
R = 86.20 MPa
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.
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 * 3A 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.
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;
}
}
(1 point) Consider the initial value problem 2????y′=4y, y(−1)=−2. Find the value of the constant ???? and the exponent ???? so that y=???????????? is the solution of this initial value problem. y= help (formulas) Determine the largest interval of the form ????<????<???? on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)
Answer:
Here is the missing part of the question ;
find the value of the constant C and the exponent r so that y = Ctr is the solution of this initial value problem. Determine the largest interval of the form a<t<b
Explanation:
The step by step explanation is as given in the attachment. You will notice I used β as the exponential in place of r.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.
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.
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]