Answer:

**Answer:**

0.25 J/K

**Explanation:**

**Given data** in given question

heat (Q) = 100 J

temperature (T) = 400 K

**to find out **

the change in entropy of the given system

**Solution **

we use the entropy change equation here i.e

**ΔS = ΔQ / T ...................a**

Now we put the value of heat (Q) and Temperature (T) in equation a

ΔS is the entropy change, Q is heat and T is the temperature,

so that

ΔS = 100/400 J/K

**ΔS = 0.25 J/K **

An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft when inflated to a pressure of 0.60 psi. A longitudinal seam runs the entire length of the structure. The seam fails in tension when the load is 600 pounds per inch of seam. What is the factor of safety with respect to longitudinal seam failure?

(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinusodal force given by p(t) = P sin ωt. All positions are measured from equilibrium. Use m_1=1.5 kg, m_2=2 kg, k=7 N/m, b=3.2 (N∙s)/m, P=15 N, =12 rad/sec. Hint: first create the state space model for the system. Then use SS2TF to make the two transfer functions and then the two Bode plots (include with submission). Use the plots to find the steady-state equations.

Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

A ___________ is defined as a change in shape of the part between the damaged and undamaged area hat is smooth and continuous. When the part is straightened, it is returned to proper shape and state without any areas of permanent deformation

(25) Consider the mechanical system below. Obtain the steady-state outputs x_1 (t) and x_2 (t) when the input p(t) is the sinusodal force given by p(t) = P sin ωt. All positions are measured from equilibrium. Use m_1=1.5 kg, m_2=2 kg, k=7 N/m, b=3.2 (N∙s)/m, P=15 N, =12 rad/sec. Hint: first create the state space model for the system. Then use SS2TF to make the two transfer functions and then the two Bode plots (include with submission). Use the plots to find the steady-state equations.

Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?

Reduce the following lambda-calculus term to the normalform. Show all intermediate steps, with one beta reduction at a time. In the reduction, assume that you are supplied with extra rules thatallow you to reduce the multiplication of two natural numbers into thecorresponding result.(λf.λx.f(f x))(λy.y≠3) 2

A ___________ is defined as a change in shape of the part between the damaged and undamaged area hat is smooth and continuous. When the part is straightened, it is returned to proper shape and state without any areas of permanent deformation

**Answer:**

3 m/s²

**Explanation:**

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45 m/s

t= 15 s

a= 45 /15

a= 3 m/s²

**Answer:**

Time=2.72 seconds

Front wheel reactions= 1393 lb

Rear wheel reactions= 857 lb

**Explanation:**

The free body diagram is assumed to be the one attached here

The mass, m of the car is

where W is weight and g is acceleration due to gravity

Taking g as then

Considering equilibrium in x-axis

At point A using the law of equilibrium, the sum of moments is 0 hence

The normal reaction at B is therefore

Consider equilibrium in y-axis

To find time that the car takes to a speed of 10 ft/s

Using kinematic equation

V=u+at

10=0+3.68t

b. The contract specifications require an average asphalt content of 5.5% +/- 0.5% every day. Plot the daily average asphalt content. Show upper and lower control limits.

c. Do all of these samples meet the contract specifications? Explain your answer.

d. What trend do you observe based on the data? What could cause this trend?"

**Answer:**

hello your question is incomplete attached below is the complete question

A) overall mean = 5.535, standard deviation ≈ 0.3239

B ) upper limit = 5.85, lower limit = 5.0

C) Not all the samples meet the contract specifications

D) fluctuation ( unstable Asphalt content )

**Explanation:**

B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday

The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85

The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0

attached below is the required plot

C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :

15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20

D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed

9514 1404 393

**Answer:**

746.7 ft²

**Explanation:**

You can add them up, or you can take advantage of multiplication to make the repeated addition simpler.

(112.5 ft² +136.4 ft²) +(112.5 ft² +136.4 ft²) +(112.5 ft² +136.4 ft²)

= (3)((112.5 ft² +136.4 ft²) = 3(248.9 ft²) = **746.7 ft²**

**The total area of the decks on the 3 homes is 746.7 ft²**.

**Answer:**

//Annual calendar

#include <iostream>

#include <string>

#include <iomanip>

void month(int numDays, int day)

{

int i;

string weekDays[] = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};

// Header print

cout << "\n----------------------\n";

for(i=0; i<7; i++)

{

cout << left << setw(1) << weekDays[i];

cout << left << setw(1) << "|";

}

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

int firstDay = day-1;

//Space print

for(int i=1; i< firstDay; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

int cellCnt = 0;

// Iteration of days

for(int i=1; i<=numDays; i++)

{

//Output days

cout << left << setw(1) << "|" << setw(2) << i;

cellCnt += 1;

// New line

if ((i + firstDay-1) % 7 == 0)

{

cout << left << setw(1) << "|";

cout << "\n----------------------\n";

cellCnt = 0;

}

}

// Empty cell print

if (cellCnt != 0)

{

// For printing spaces

for(int i=1; i<7-cellCnt+2; i++)

cout << left << setw(1) << "|" << setw(2) << " ";

cout << "\n----------------------\n";

}

}

int main()

{

int i, day=1;

int yearly[12][2] = {{1,31},{2,28},{3,31},{4,30},{5,31},{6,30},{7,31},{8,31},{9,30},{10,31},{11,30},{12,31}};

string months[] = {"January",

"February",

"March",

"April",

"May",

"June",

"July",

"August",

"September",

"October",

"November",

"December"};

for(i=0; i<12; i++)

{

//Monthly printing

cout << "\n Month: " << months[i] << "\n";

month(yearly[i][1], day);

if(day==7)

{

day = 1;

}

else

{

day = day + 1;

}

cout << "\n";

}

return 0;

}

//end

Cut that photo by

1. Left click your mouse on the photo

2. Click cut

Then enter the file where you want to transfer and press

1. ctrl+v

**Answer:**

you can go to your file and then select the phpto and hold on a little bit and choose the delete option