If 100 J of heat is added to a system so that the final temperature of the system is 400 K, what is the change in entropy of the system? a)- 0.25 J/K b)- 2.5 J/K c)- 1 J/K d)- 4 J/K

Answers

Answer 1
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


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Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Answers

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

The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

Answers

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  

M=\frac {W}{g} where W is weight and g is acceleration due to gravity

Taking g as 32.2 ft/s^(2) then  

M=\frac {4500}{32.2}=139.75 lbm

Considering equilibrium in x-axis

Ma_G-f=0

Ma_G-(\mu_g* 2N_B)=0

139.75* a_G-(0.3* 2* N_B)=0

0.6N_B=139.75a_g

N_B=232.92a_g

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

-2N_B(6)+4500(2)=-Ma_G(2.5)

-12N_B+9000=-139.75a_G* 2.5

-12(232.92a_G)+900=-349.375a_G

a_g\approx 3.68 ft/s^(2)

The normal reaction at B is therefore

N_B=232.92a_G=232.92* 3.68\approx 857 lb

Consider equilibrium in y-axis

4500-2N_A-2N_B=0

N_A+N_B=2250

N_A+857=2250

N_A=1393 lb

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

Using kinematic equation

V=u+at

10=0+3.68t

t=\frac {10}{3.68}\approx 2.72 s

Given asphalt content test data: a. Calculate the overall mean and standard deviation for the entire test period.
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?"

Answers

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

There are three homes being built, each with an identical deck on the back. Each deck is comprised of two separate areas. One area is 112.5 square feet, while the other is136,4 square feet. What is the total square footage of the decks for all three homos? Your answer should be to the nearest tonth of a square

Answers

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.

Write a C++ program to display yearly calendar. You need to use the array defined below in your program. // the first number is the month and second number is the last day of the month. into yearly[12][2] =

Answers

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

Every time I take a photo, that photo has to be stored in a file somewhere within "My Files" correct?How would I be able to take that photo out of the file it was stored?

Answers

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