Answer: −3.46kJ/K
Explanation:
From the question above, we have:
The mass of the block (m) = 45kg
The initial temperature of the block (T1) = 280∘C
The weight of the water (mw) = 100kg
The temperature of water (Tw) = 18∘C
Recall the energy balance equation,
ΔUI = −ΔUw
In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.
[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w
Here cp is the specific heat at constant pressure.
The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.
Now, we substitute the values in above equation,
[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w
[20.25(T2 − 280)] = −[418(T2 − 18)]
20.25T2 − 5,670 = −[418T2 − 7,524]
20.25T2 − 5,670 = −418T2 + 7,524
20.25T2 + 418T2 = 7,524 + 5,670
438.35T2 = 13,194
T2 = 30.1K
Recall, the expression to calculate the total entropy change is given as:
ΔStotal = ΔSI + ΔSw
ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w
Now, we substitute the values in above equation,
ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w
ΔStotal = −12.55 + 9.09
ΔStotal = −3.46kJ/K
Thus the total entropy change is −3.46kJ/K.
You need to display output for all of the values between the starting and ending values. First two values are temperatures in Fahrenheit. You need to display all of the values from the first temperature to the last temperature. You increment from one temperature to the next by the increment value (the third value you read in). You need to convert these temperatures to Celsius and Kelvin. You need to output the temperatures as Fahrenheit, Celsius, and Kelvin. The numbers should be 18 characters wide with 4 digits of precision and need to be in fixed format. Do not use tab characters (\t) to output the values.
Answer:
Check the explanation
Explanation:
#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double temp1,temp3,inc,cel;
int i=1;
while(i==1)
{
i=0;
cin>>temp1>>temp3>>inc;
if(temp3<temp1||inc<=0)
{
i=1;
cout<<"Starting temperature must be <= ending temperature and increment must be >0.0\n";
}
}
cout<<endl;
cout<<setw(18)<<"Fahrenheit"<<setw(18)<<"Celsius";
while(temp1<=temp3)
{
cel=(temp1-32)/1.8;
cout<<endl;
cout<<fixed<<setprecision(4)<<setw(18)<<temp1<<setw(18)<<cel;
temp1+=inc;
}
}
An actuator has a stem movement which at full travel is 40 mm. It is mounted on a process control valve with an equal percentage plug and which has a minimum flow rate of 0.2 m3/s and a maximum flow rate of 4.0 m3/s. What will be the flow rate when the stem movement is (a) 10 mm, (b) 20 mm
Answer:
a) for a stem of 10 mm the flow rate is 0.4229 m³/s
b) for a stem of 20 mm the flow rate is 0.8944 m³/s
Explanation:
The mathematical expression for the flow rate is:
[tex]Q=Q_{min} *(\frac{Q_{max} }{Q_{min} } )^{\frac{S-S_{min}}{S_{max}-S_{min} } }[/tex]
a) Here:
Qmin=0.2m³/s
Qmax=4m³/s
Smax-Smin=40mm
S-Smin=10mm
Substituting these values in the first equation:
[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{10}{40} } =0.4229m^{3} /s[/tex]
b) In this time, S-Smin=20mm
[tex]Q=0.2*(\frac{4}{0.2} )^{\frac{20}{40} } =0.8944m^{3} /s[/tex]
Let f(t) be an arbitrary signal with bandwidth Ω. Determine the minimum sampling frequencies ωs needed to sample the following analog signals without causing aliasing error. (a) f1(t) = f(t) sin(4000πt) (b) f2(t) = f(t) ∗ sin(4000πt) (c) f3(t) = f(t) ∗ f(sample the following analog signals without causing aliasing error
Answer:
See explaination
Explanation:
We can describr Aliasing as a false frequency which one get when ones sampling rate is less than twice the frequency of your measured signal.
please check attachment for the step by step solution of the given problem.
Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the Mach number at the exit is 0.3, determine the direction and the rate of heat transfer. For the same conditions at the inlet, determine the amount of heat that must be transferred to the system if the flow is to be sonic at the exit of the duct.
Answer:
a. 318.2k
b. 45.2kj
Explanation:
Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
See attachment for detailed analysis
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm (0.60 in. × 0.75 in.) is pulled in tension with 44,500 N (10,000 lbf) force, producing only elastic deformation. Calculate the resulting strain. Assume elastic modulus of Cu to be 110GPa. (Points: 5).
Answer:
The resulting strain is [tex]1.39\times 10^{-3}[/tex].
Explanation:
A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm
Force, F = 44,500 N
Th elastic modulus of Cu to be 110 GPa
The resulting strain is given by the formula as follows :
[tex]\epsilon=\dfrac{F}{AE}[/tex]
E is elastic modulus of Cu is are of cross section
[tex]\epsilon=\dfrac{44500}{15.2\times 19.1\times 10^{-6}\times 110\times 10^9}\\\\\epsilon=1.39\times 10^{-3}[/tex]
So, the resulting strain is [tex]1.39\times 10^{-3}[/tex].
// global variables int a = 5, b = 6, c = 7, d = 8; void sub2() { int a = 0, b = 3; //local variables { int b = 5; //local variable System.out.println("a=" + a); System.out.println("b=" + b); System.out.println("c=" + c); System.out.println("d=" + d); } } void sub1() { int a = 2, b = 4, c = 1; //local variables sub2(); } void main() { int a = 1, b = 2, c = 3, d = 4; //local variables sub1(); }\
Answer:
1)
Static scoping
A=0
B=5
C=3
D=4
2) For dynamic scoping
A=2
B=5
C=1
D=8
Fizeau’s method for measuring the speed of light using a rotating toothed wheel. The speed of rotation of the wheel controls what an observer sees. For example, if the light passing the opening at point A should return at the instant that tooth B had rotated into position to cover the return path, the light would not reach the observer. At a faster rate of rotation, the opening at point C could move into position to allow the reflected beam to reach the observer. Toothed Wheel Mirror A B C d Calculate the minimum angular speed of the wheel for light that passed through opening A to return through opening C to reach the observer. In an experiment to measure the speed of light using the apparatus of Fizeau, the distance between the toothed wheel and mirror was 10.72 km and the wheel had 671 notches. The experimentally determined value of c was 2.908 × 108 m/s . Answer in units of rad/s.
Answer:
The minimum angular speed, w is 126.94 rad/s
Explanation:
Given:
C = 2.908×10⁸ m/s
d = 10.72 km ⇒ 10.72×10³ m
There are 671 notches
⇒ Δθ = [tex]\frac{2\pi}{671}[/tex] ==> 9.359×10⁻³ rad
C = 2d / Δt ⇒ Δt = 2d/C
⇒
w = Δθ / Δt = CΔθ / 2d
substitute for given parameters
w = [2.908×10⁸×9.359×10⁻³] / [2×10.72×10³]
= 27.215972×10⁵ / 21.44×10³
= 1.2694×10²
w ⇒ 126.94 rad/s
(a) Design a half-subtractor circuit with inputs x and y and outputs Diff and B out . The circuit subtracts the bits x – y and places the difference in D and the borrow in B out .
(b) Design a full-subtractor circuit with three inputs x , y , B in and two outputs Diff and B out . The circuit subtracts x – y – B in , where B in is the input borrow, B out is the output borrow, and Diff is the difference.
Answer:
See the attachment
Explanation:
A half subtractor circuit is made up of on NOT gate, one XOR and one AND gate.
A full subtractor is made up of two half subractor with their outputs in an OR gate for Bout as shown in attachment
A four‐lane freeway (two lanes in each direction) operates at flow rate of 1700 during the peak hour. It has 11‐ft lanes, 4‐ft shoulders, and there are three ramps within three miles upstream of the segment midpoint and four ramps within three miles downstream of the segment midpoint. The freeway has only regular users. There are 8% heavy trucks, and it is on rolling terrain with a peak‐hour factor of 0.85. It is known that 12% of the AADT occurs in the peak hour and that the directional factor is 0.6.
What is the freeway’s AADT?
Answer:
Please see the attached file for the complete answer.
Explanation:
A photovoltaic panel of dimensions 6 m x 5 m is located on top of the roof of a house. Solar irradiation Gs = 900 W/m2 is incident on the panel. The panel has an absorptivity to solar irradiation, αs of 0.92. The freestream air temperature is Tinf and the surrounding temperature for radiation exchange with the sky is Tsurr. For this particular problem, it is given that Tinf= Tsurr. The convective heat transfer coefficient of air blowing over the panel is h W/m2-K.
Answer:
2.7 W/m^2K
Explanation:
Area of pane = 5 m x 6 m = 30 m^2
Solar irradiation Gs = 900 W/m2
Heat rate on panel = Gs x area = 900 x 30 = 27000 W
absorptivity to solar irradiation αs = 0.92
Therefore, absorbed heat is
0.92 x 27000 = 24840 W
For heat gain,
From E = §AT^4
Where § = stefan's constant = 5.7x10^-8 Wm^-2K^-1
T = temperature of panel
24840 = 5.7x10^-8 x 30 x T^4
24840 = 1.71x10^-6 x T^4
1.453x10^10 = T^4
T = 347.167 K
For net heat gain,
From E = §A(T^4 - T^4sur)
24840 = 5.7x10^-8 x 30 x (T^4 - T^4sur)
24840 = 1.71x10^-6 x (1.453x10^10 - T^4sur)
24840 = 24846.3 - 1.71x10^-6(T^4sur)
-6.3 = -1.71x10^-6(T^4sur)
3684210.526 = T^4sur
Tsur = 43.81 K
Also for convective heat,
E = Ah(T - Tsur)
24840 = 30h(347.167 - 43.81)
24840 = 30h x 303.357
81 = 30h
h = 2.7 W/m^2K
Using a forked rod, a smooth 3-lb particle is forced to move along around the horizontal path in the shape of a limaçon, r = (5+sin θ) ft. If θ =LaTeX: \frac{1}{8}t^21 8 t 2 rad, where t is in seconds, determine the force of the rod and the normal force of the slot on the particle at the instant t= 3sec. The fork and path contact the particle on only one side.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
The detailed solution is in the attached file
Air from a workspace enters the air conditioner unit at 30°C dry bulb and 25°C wet bulb temperatures. The air leaves the air conditioner and returns to the space at 25°C dry bulb and 6.5°C dew point temperature (ϕ2≠100%). If there is any, the condensate leaves the air conditioner at the same temperature of the air leaving the cooling coils (i.e., different from our previous assumption that condensate has same temperature of the cooling coils). The volume flow rate of the air returned to the workspace is 1000 m3 /min (Rair=0.287 kJ/kg∙K). Atmospheric pressure is 101.325 kPa. Compute
Answer:
See explaination
Explanation:
The volume flow rate Q Q QQ of a fluid is defined to be the volume of fluid that is passing through a given cross sectional area per unit time.
Kindly check attachment for the step by step solution of the given problem.
A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional stress of 15 kpsi and an alternating bending stress of 25 kpsi Find the factor of safety guarding against a static failure, and either the factor of safety guard- ing against a fatigue failure or the expected life of the part. For the fatigue analysis use:
(a) Modified Goodman criterion.
(b) Gerber criterion.
(C) ASME-elliptic criterion.
Answer:
(a) Modified Goodman criterion:
Factor of safety against fatigue failure = 1.0529
(b) Gerber criterion:
Factor of safety against fatigue failure = 1.31
(c) ASME-elliptic criterion:
Factor of safety against fatigue failure = 1.315
Explanation:
See the attached file for the calculation.
(a) Modified Goodman: SF ≈ 1.264, Static FS ≈ 2.058.
(b and c ) Gerber & ASME-elliptic: SF ≈ 1.783, Static FS ≈ 2.058.
To calculate the factor of safety against static and fatigue failures using different criteria, we need to determine the critical stress limits under the given loading conditions and compare them with the material properties.
Given:
- Torsional stress [tex](\( \tau \))[/tex] = 15 kpsi
- Alternating bending stress [tex](\( \sigma_a \))[/tex] = 25 kpsi
- Minimum endurance limit [tex](\( S_e \))[/tex] = 40 kpsi
- Ultimate tensile strength [tex](\( S \))[/tex] = 60 kpsi
- Endurance limit for reversed bending [tex](\( S_{-80} \))[/tex] = 80 kpsi
(a) Modified Goodman criterion:
The modified Goodman criterion accounts for both tensile and torsional stress, given by:
[tex]\[ \frac{1}{SF} = \frac{\frac{\sigma_a}{S} + \frac{\tau}{S_e}}{1} \][/tex]
Where [tex]\( SF \)[/tex] is the safety factor against fatigue failure.
Substitute the given values:
[tex]\[ \frac{1}{SF} = \frac{\frac{25}{60} + \frac{15}{40}}{1} \][/tex]
[tex]\[ \frac{1}{SF} = \frac{0.4167 + 0.375}{1} \][/tex]
[tex]\[ \frac{1}{SF} = 0.7917 \][/tex]
[tex]\[ SF = \frac{1}{0.7917} \][/tex]
[tex]\[ SF \approx 1.264 \][/tex]
The factor of safety against static failure [tex](\( FS_{static} \))[/tex] can be calculated by comparing the maximum applied stress with the ultimate tensile strength:
[tex]\[ FS_{static} = \frac{S}{\sigma_{max}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{\sigma_a^2 + \tau^2}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{25^2 + 15^2}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{625 + 225}} \][/tex]
[tex]\[ FS_{static} = \frac{60}{\sqrt{850}} \][/tex]
[tex]\[ FS_{static} \approx \frac{60}{29.1547} \][/tex]
[tex]\[ FS_{static} \approx 2.058 \][/tex]
(b) Gerber criterion:
The Gerber criterion considers the bending and torsional stresses, given by:
[tex]\[ \frac{1}{SF} = \sqrt{\frac{\sigma_a^2}{S^2} + \frac{\tau^2}{S_e^2}} \][/tex]
Substitute the given values:
[tex]\[ \frac{1}{SF} = \sqrt{\frac{25^2}{60^2} + \frac{15^2}{40^2}} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{\frac{625}{3600} + \frac{225}{1600}} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]
[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]
[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]
[tex]\[ SF \approx 1.783 \][/tex]
(c) ASME-elliptic criterion:
The ASME-elliptic criterion also considers bending and torsional stresses:
[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{\sigma_a}{S}\right)^2 + \left(\frac{\tau}{S_e}\right)^2} \][/tex]
Substitute the given values:
[tex]\[ \frac{1}{SF} = \sqrt{\left(\frac{25}{60}\right)^2 + \left(\frac{15}{40}\right)^2} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.1736 + 0.1406} \][/tex]
[tex]\[ \frac{1}{SF} = \sqrt{0.3142} \][/tex]
[tex]\[ \frac{1}{SF} \approx 0.5608 \][/tex]
[tex]\[ SF \approx \frac{1}{0.5608} \][/tex]
[tex]\[ SF \approx 1.783 \][/tex]
For all three criteria:
- Factor of safety against static failure [tex](\( FS_{static} \))[/tex] ≈ 2.058
- Safety factor against fatigue failure [tex](\( SF \))[/tex] ≈ 1.264 for the Modified Goodman criterion, and ≈ 1.783 for the Gerber and ASME-elliptic criteria.
Select all of the true statements.
A. Diodes are used in voltage regulators and limiters.
B. Different diode models are used to replace the real diode with a simpler version that approximates the i-v characteristics of the real diode.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Answer:
Diodes consists of two-terminal electronic parts which conducts current mainly in one direction. It has low resistance in one direction and high resistance in the other direction. These are correct statements about a Diode.
A. Diodes are used in voltage regulators and limiters.
C. In the ideal diode plus voltage source model, the forward bias region is characterized by VD
Explanation:
Diodes
Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combining the sub-colors red, green, and blue (rgb). Each sub-color's value can range from 0 to 255. Thus (255, 0, 0) is bright red, (130, 0, 130) is a medium purple, (0, 0, 0) is black, (255, 255, 255) is white, and (40, 40, 40) is a dark gray. (130, 50, 130) is a faded purple, due to the (50, 50, 50) gray part. (In other words, equal amounts of red, green, blue yield gray). Given values for red, green, and blue, remove the gray part. Ex: If the input is 130 50 130, the output is: 80 0 80 Find the smallest value, and then subtract it from all three values, thus removing the gray. Note: This page converts rgb values into colors.
Answer:
answer is given below
Explanation:
we write this program in C++ that is given below
and result is attach here
#include <iostream>
using namespace std;
int main() {
int r,g,b,small;
//input values
cin>>r>>g>>b;
//find the smallest value
if(r<g && r<b)
small=r;
else if(g<b)
small=g;
else
small=b;
//subtract smallest of the three values from rgb values hence removing gray
r=r-small;
g=g-small;
b=b-small;
cout<<r<<" "<<g<<" "<<b<<endl;
}
We can see here that the Python code is thus:
# Read input values for red, green, and blue
red, green, blue = map(int, input().split())
# Find the smallest value among red, green, and blue
smallest = min(red, green, blue)
# Subtract the smallest value from all three colors
red -= smallest
green -= smallest
blue -= smallest
# Output the modified values
print(red, green, blue)
Assuming the input values for red, green, and blue are 50, 30, and 40 respectively, the output will be:
10 0 10
This is because:
The smallest value among 50, 30, and 40 is 30.
Subtracting 30 from each color gives:
red = 50 - 30 = 20green = 30 - 30 = 0blue = 40 - 30 = 10The output displays the modified values: 20 0 10
To implement this functionality in Python, you can follow these steps:
Read the input values for red, green, and blue.Find the smallest value among red, green, and blue.Subtract the smallest value from all three colors.Output the modified values.Modify any of the previous labs which would have crashed when non-numeric data was entered by adding exception handling so that the non-numeric input no longer crashes the program. The program must handle the invalid input gracefully by informing the user of the bad input and re-querying until the user provides valid input. It must re-query for the specific input which was incorrectly entered.
Answer:
see explaination
Explanation:
import java.util.InputMismatchException;
import java.util.Scanner;
public class calculate {
static float a=0,b=0;
double cal()
{
if(a==0||b==0)
{
System.out.println("no values found in a or b");
start();
}
double x=(a*a)+(b*b);
double h=Math.sqrt(x);
a=0;
b=0;
return h;
}
float enter()
{
float val=0;
try
{
System.out.println("Enter side");
Scanner sc1 = new Scanner(System.in);
val = sc1.nextFloat();
return val;
}
catch(InputMismatchException e)
{
System.out.println("Enter correct value");
}
return val;
}
void start()
{
calculate c=new calculate();
while(true)
{
System.out.println("Enter Command");
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
switch(input)
{
case "A":
a=c.enter();
break;
case "B":
b=c.enter();
break;
case "C":
double res=c.cal();
System.out.println("Hypotenuse is : "+res);
break;
case "Q":
System.exit(0);
default:System.out.println("wrong command");
}
}
}
public static void main(String[] args) {
calculate c=new calculate();
c.start();
}
}
Determine the average power, complex power and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by: v(t) = 100cos(377t-30)v, i(t) = 2.5cos(377t-60)A
Answer:
Average power = 108.25W
Complex power = 125W
PF = 0.866
Explanation:
Check attachment for step by step instructions.
Answer:
Average power: 108.25 W
Complex power: 125 VA
Power factor: 0.866
Refer below for the explanation.
Explanation:
Refer to the picture for brief explanation.
A certain process requires 3.0 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90o elbows, determine the pipe diameter. Elevation differences are negligible.
Answer:
diameter of the pipe = 0.4932ft.
Explanation:
assuming d = 0.4932
Re = 3.16 x 10∧5/0.4932
= 6.4 x 10 ∧5
E/d = 0.0005/0.4932
= 0.0010 from moody chat t = 0.02
if 0.02 is beign substituted in equation 2 we will get the same required diameter of the pipe which is 0.4932ft.
check the attachment for better explanation.thanks
A technician connects a voltmeter in parallel across two points in a circuit. Technician A says this will provide a reading of the potential difference in volts. Technician B says that this will also show the amperage flowing in the circuit. Who is correct?
A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. Technician A is correct while Technician B is wrong.
What is a voltmeter?A voltmeter is a device that measures the difference in electric potential between two locations in an electric circuit. It is linked in parallel. It generally has a high resistance and draws very little current from the circuit.
Given the technician connects the voltmeter in parallel across two points in a circuit. Therefore, the technician A is correct the meter will provide a reading of the potential difference in volts.
Hence, Technician A is correct while Technician B is wrong.
Learn more about Voltmeter:
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Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow. Who is right?
Answer:
A
Explanation:
Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.
Who is technician?A technician is a worker in the technology industry who possesses the necessary knowledge, abilities, and skills as well as a practical comprehension of the theoretical underpinnings.
Expert technicians in a certain tool domain often have expert competency in technique and a moderate comprehension of theory. As a result, technicians in that field of technology are typically much more knowledgeable about technique than the average layperson and even general professionals.
For instance, although not as knowledgeable in acoustics as acoustical engineers, audio technicians are more adept at using sound equipment and are likely to know more about acoustics than other studio staff members, such as performers.
Therefore, Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit controls current flow Person A is right.
To learn more about technician, refer to the link:
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Consider a low speed open circuit subsonic wind tunnel with an inlet to throat area ratio of 12. The tunnel is turned on and the pressure difference between the inlet (the settling chamber) and the test section is read as a height difference of 10 cm on a U-tube mercury manometer. (The density of liquid mercury is 1.36 x 10-4 kg/m3 ). Calculate the velocity of the air in the test section.
Answer:
velocity = 147.57 m/s
Explanation:
given data
inlet to throat area ratio = 12
height difference Δh = 10 cm = 0.1 m
density of liquid mercury = 1.36 × [tex]10^{4}[/tex] kg/m³
solution
we get here weight of mercury that is express as
weight of mercury = density of liquid mercury × g .................1
weight of mercury = 1.36 × [tex]10^{4}[/tex] × 9.8
weight of mercury = 1.33 × [tex]10^{5}[/tex] N/m²
and
area ratio is
[tex]\frac{a1}{a2}[/tex] = 12
so velocity of air in the test section will be
velocity = [tex]\sqrt{\frac{2\times w \times \triangle h}{\rho (1-(area\ ration)^2)}}[/tex] .......................1
put here value and we get
velocity = [tex]\sqrt{\frac{2\times 1.33\times 10^5 \times 0.1}{1.23 (1-(\frac{1}{12})^2)}}[/tex]
velocity = 147.57 m/s
The velocity of air in the test section is; v = 147.802 m/s
Velocity of airWe are given;
inlet to throat area ratio; a1/a2 = 12height difference Δh = 10 cm = 0.1 mdensity of liquid mercury; ρ = 1.36 × 10⁴ kg/m³Thus, weight of mercury is;
W = ρg
W = 1.36 × 10⁴ × 9.8
W = 13.328 × 10⁴ N/m²
Formula for the velocity of the air in the test section is;
v = √[(2ρgΔh)/(ρ_air × (1 - a2/a1)]
Where ρ_air is density of air = 1.23 kg/m³
Thus;
v = √[(2 × 1.36 × 10⁴ × 9.8 × 0.1)/(1.23 × (1 - 1/12)]
Solving this gives us;
v = 147.802 m/s
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"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United States.Though long and wide (30 miles wide at the Potomac River), the bay is very shallow, with an average depth of only 28 feet." The Bay's maximum depth is 174 ft. The hydraulic model of the Chesapeake Bay was built with a model length ratio Lr=1/1000
a. How wide was the model Bay [ft] at the Potomac River?
b. The Bay Bridge is 4.3 miles long; how long was it in the model [ft]?
c. If the hydraulic model occupied about 8 acres, approximately what is the real-world (prototype) area represented by the model [square miles]?
d. At this scale, what would be the average depth and maximum depth of the model Bay? [give answers in both ft and inches)
Answer:
see explaination
Explanation:
Part a) Width of bay at Potomac River:
Given Data:
· Actual Width at Potomac River = 30 miles
· Bay Model Length Ratio Lr = 1/1000
In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.
Length of any part in model = Actual length x Lr
Hence,
Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles
Since 1 mile = 5280 ft
Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft
Part b) Model Length of bay bridge in model:
Given Data:
· Actual Length of bay bridge = 4.3 miles
· Bay Model Length Ratio Lr = 1/1000
Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles
Since 1 mile = 5280 ft
Model Length in feet = 0.0043 x 5280 = 22.704 ft
Part c) Model Length of bay bridge in model:
Given Data:
· Model Area = 8 acre
· Bay Model Length Ratio Lr = 1/1000
Model Area = Actual Area x Lr x Lr
8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000
Since 1 square mile = 640 acre,
Actual Area in square miles = 8,000,000/640 = 12,500 square miles
Part d) Average and maximum depth of model:
Given Data:
· Actual Average depth = 28 ft
· Actual Maximum depth = 174 ft
· Bay Model Length Ratio Lr = 1/1000
Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet
Since 1 ft = 12 inch
Model average depth in inch = 0.028 x 12 = 0.336 in
Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet
Since 1 ft = 12 inch
Model maximum depth in inch = 0.174 x 12 = 2.088 in
A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun. To account for the difference in time, every 4 years, a leap year takes place. A leap year is when a year has 366 days: An extra day, February 29th. The requirements for a given year to be a leap year are: The year must be divisible by 4 If the year is a century year (1700, 1800, etc.), the year must be evenly divisible by 400 Some example leap years are 1600, 1712, and 2016. Write a program that takes in a year and determines whether that year is a leap year. Ex: If the input is 1712, the output is: 1712 is a leap year. Ex: If the input is 1913, the output is: 1913 is not a leap year.
Answer:
Explanation:
def is_leap_year(year):
if(year % 400 == 0):
return True
elif year % 100 == 0:
return False
elif year%4 == 0:
return True
else:
return False
if __name__ == '__main__':
n = int(input())
if(is_leap_year(n)):
print(n,"- leap year")
else:
print(n, "- not a leap year")
check the attachment for the output
Answer:
def is_leap_year(year):
if(year % 400 == 0):
return True
elif year % 100 == 0:
return False
elif year%4 == 0:
return True
else:
return False
if __name__ == '__main__':
n = int(input())
if(is_leap_year(n)):
print(n,"is a leap year.")
else:
print(n, "- not a leap year")
Explanation:
A convergentâdivergent nozzle has an exit area to throat area ratio of 4. It is supplied with air from a large reservoir in which the pressure is kept at 500 kPa and it discharges into another large reservoir in which the pressure is kept at 10 kPa. Expansion waves form at the exit edges of the nozzle causing the discharge flow to be directed outward.
(a) Find the angle that the edge of the discharge flow makes to the axis of the nozzle.
Answer:
Angle of discharge make at the edge of tube=64.9 degrees.
During the recovery of a cold-worked material, which of the following statement(s) is (are) true?
O Some of the internal strain energy is relieved.
O All of the internal strain energy is relieved.
O There is some reduction in the number of dislocations.
O There is a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state.
O The electrical conductivity is recovered to its precold-worked state.
O The thermal conductivity is recovered to its precold-worked state.
O The metal becomes more ductile, as in its precold-worked state.
O Grains with high strains are replaced with new, unstrained grains.
Answer:
Some of the internal strain energy is relieved.
There is some reduction in the number of dislocations.
The electrical conductivity is recovered to its precold-worked state.
The thermal conductivity is recovered to its precold-worked state
Explanation:
The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.
A 15.0 in by 2.0 in work part is machined in a face milling operation using a 2.5 in diameter cutter with a single carbide insert. The machine is set for a feed of 0.010 in/tooth and a depth of 0.20 in. If a cutting speed of 400 ft/min is used, the tool lasts for 3 pieces. If a cutting speed of 200 ft/min it used, the tool lasts for 12 parts. Determine the Taylor tool life equation. (20 points)
Answer:
Explanation:
N1 =v/πD = 400(12)/2.5π = 611 rev/min fr = Nfnt = 611(0.010)(1) = 6.11 in/min A=D/2 = 2.50 / 2 = 1.25 Tm = (L+2A)/fr = (15 + 2(1.25))/6.11 = 2.863 min T1 = 3Tm = 3(2.863) = 8.589 min when v1 = 400 ft/min N2 = 200(12)/2.5π = 306 rev/min fr = Nfnt = 306(0.010)(1) = 3.06 in/min Tm = (15 + 2(1.25))/3.06 = 5.727 min T2 = 12Tm = 12(5.727) = 68.724 min when v2 = 200 ft/min n = ln (v1/v2)/ln(T2/T1) = ln (400/200)/ln (68.724/8.589) = 0.333 C = vTn = 400(8.589 )0.333 = 819
Answer:
C = 787.2
Explanation:
The Taylor tool life is referred to as the duration of the actual cutting time after which the tool can no longer be used.
To Quantify the end of a tool life we equate it to a limit on the maximum acceptable flank wear.
For the tool life equation, With the slope, n and intercept, c, Taylor derived the simple equation as
VTn = C where.
n is called, Taylor's tool life exponent. The values of both 'n' and 'c' depend mainly upon the tool-work materials and the cutting environment (cutting flake application)
Please kindly go through the attached file for a step by step solution to how the answer to your question is solved.
Given that the size of Elmer's property is 180 acres, and given that there are 20 volunteers, what would be a reasonable approach to surveying the land for houndstongue?
Answer:
Weed Mapping Application is the most reasonable approach for surveying.
Answer:
The most reasonable approach to survey the property would be a GPS survey using a dual-frequency GPS system.
Explanation:
Surveying makes use of coordinates to determine positions of objects, it is the first step towards developing a property. There are different types of survey which are specific to various needs and sizes of a property.
GPS SURVEY
GPS survey makes use of the global positioning system to determine positions of a property or object. It is quite accurate and also an efficient type of survey based on its flexibility.
In the illustration, the property size is 180 acres and 20 volunteers are expected to work on it. To carry out this survey, a station whose coordinate is known normally referred to the master station is used to get the coordinates of the remote stations. The volunteers are stationed at 20 remote stations and the surveyor at the master station makes use of dual-frequency GPS system to pick the coordinates of the remote stations. The dual-frequency GPS system would be used because it covers more distance and for a survey as massive as 180 acres this is reasonable.
For the survey to be carried out seamlessly by the 20 volunteers, a dual-frequency GPS system would be used for the GPS survey.
5. A straight round shaft is subjected to a torque of 5000 lb - in. Determine the required diameter, using steel with a tensile yield strength of 60 ksi and a safety factor of 2 based on initial yielding: (a) According to the maximum-normal-stress theory. (b) According to the maximum-shear-stress theory. (c) According to the maximum-distortion-energy theory. Discuss briefly the relative validity of the three predictions.
Answer:
a. 0.95 in
b. 1.19 in
c. 1.137 in
Explanation:
Express the factor of safety equation for maximum-normal-stress theory as:
S SF = Eau
Here, the factor of safety is SF, the yield strength is S„ and the maximum stress :I
Modify the above equation for shear stress acting on the solid rod as:
S. SF = To
Here, the combined shear stress on solid rod s
Calculate the combined shear stress for solid rod.
16T r2:1 = trd3
(1)
Here, the torque is T, and the diameter of the solid rod is d.
Substitute 5,000 'bin. for T.
— 16(5,000 lb • in ) v ird 80 000lb - in. _
rd3
60ksi 2 — 80,000lb -in. trd
Solve the above equation for d.
60 x 1031bAn? 80,00016 -in. ird3 — 2(80, 000 lb in.) d3 rrt60x103Ibfln?)
v3 d —[ 2(80,000lb-in.) rz-(60 x103112,An?) = 0.8488 in.3r3 =0.95 in.
check the attached files for clear cut details
investigation, determine when the Elastic Potential Energy is zero. Make sure you test your idea with several masses, all three springs and vary the stiffness of spring three. Write down how you determined the zero location(s) and explain why the position for zero makes sense.
Elastic Potential Energy is zero detailed description is given below.
Explanation:
It is the energy stored in stretched or compressed elastic materials. This also means that elastic potential energy is zero in objects that have not been stretched or compressed.To determine the gravitational potential energy of an object, a zero height position must first be arbitrarily assigned. Typically, the ground is considered to be a position of zero height. But this is merely an arbitrarily assigned position that most people agree upon. Since many of our labs are done on tabletops, it is often customary to assign the tabletop to be the zero height position. Again this is merely arbitrary. If the tabletop is the zero position, then the potential energy of an object is based upon its height relative to the tabletop. For example, a pendulum bob swinging to and from above the tabletop has a potential energy that can be measured based on its height above the tabletop. By measuring the mass of the bob and the height of the bob above the tabletop, the potential energy of the bob can be determined. Potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height. An object possesses elastic potential energy if it is at a position on an elastic medium other than the equilibrium position.
Since the gravitational potential energy of an object is directly proportional to its height above the zero position, a doubling of the height will result in a doubling of the gravitational potential energy. A tripling of the height will result in a tripling of the gravitational potential energy.
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Your program this week will have the same output as lab 10, except that instead of redirecting the filename for the input file, you will get the filename from the command-line, and use a file pointer to open the file. You will also dynamically allocate space in memory after reading in the first value from the file indicating the number of lean proteins that are in the file.
Answer:
Check the explanation
Explanation:
defs.h
#ifndef DEFS_H
#define DEFS_H
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
char item[20];
char quantity[10];
int calories;
float protein;
float carbs;
float fats;
} food;
int size;
void printArray(int size, food arr1[]);
#endif
arrayProcessing.c
#include "defs.h"
void printArray(int size, food arr1[])
{
int i = 0;
printf("\nFOOD ITEM\t\tQUANTITY\tCALS\tPRO\tCARBS\tFAT");
for (i = 0; i < size; i++)
{
printf("\n%i.%s", i + 1, arr1[i].item);
printf("\t\t%s", arr1[i].quantity);
printf("\t\t%i", arr1[i].calories);
printf("\t%.2f", arr1[i].protein);
printf("\t%.2f", arr1[i].carbs);
printf("\t%.2f\n", arr1[i].fats);
}
}
lab12.c
#include "defs.h"
int main(int argc, char *argv[])
{
int i = 0;
FILE *inFile;
inFile = fopen(argv[1], "r");
if(inFile==NULL){
fprintf(stderr, "File open error. Exiting program\n");
exit(1);
}
fscanf(inFile, "%i", &size);
food *arr = (food *)malloc(sizeof(food)*size);
for (i = 0; i < size; i++)
{
fscanf(inFile, "\n%[^\n]s", arr[i].item);
fscanf(inFile, "%s", arr[i].quantity);
fscanf(inFile, "%i", &arr[i].calories);
fscanf(inFile, "%f", &arr[i].protein);
fscanf(inFile, "%f", &arr[i].carbs);
fscanf(inFile, "%f", &arr[i].fats);
}
printArray(size, arr);
return 0;
}
Kindly check the Output below,