If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provideA. expansion joints
B. isolation joints
C. control joints
D. construction joints

Answer:

hello your question is incomplete attached is the complete question

A) Vc =  15 ( 1 - ) ,   ic =

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

= 8 k ohms || 24 k ohms = 6 k ohms

=  =  15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc =

hence Vc = 15 ( 1 - )

ic = = =

attached

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, = 18 kN/m³

maximum dry unit weight measured, = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

substitute the given values;

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

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

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Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s