If you are pouring a large concrete slab and wish to avoid random cracks caused by shrinkage, you would likely provide Control joints. The correct answer would be C.
Control joints are used to prevent random cracks from forming in large concrete slabs caused by shrinkage. These joints are placed at strategic locations in the slab to allow for the concrete to expand and contract without cracking. Expansion joints, on the other hand, are used to separate concrete from other structures or materials, and isolation joints are used to separate different sections of concrete.
Construction joints are used to connect two different pours of concrete. Therefore, the best option for preventing random cracks caused by shrinkage would be to use control joints.
Learn more about Control joints:
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b) sketch vc and ic.
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
and exit cross-sectional areas. The inlet cross-sectional area is
6 cm26cm
2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy
effects and modeling air as an ideal gas with constant cp=1.008 kJ/kg⋅Kc
p =1.008kJ/kg⋅K, determine
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm2
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm
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vehicles are abreast of each other, the police car starts to pursue the automobile at
a constant acceleration of 1.96 m/s². 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. (Hint: The police will not go against the law.)
a) Find the total time required for the police car to overtake the automobile.
b) Find the total distance travelled by the police car while overtaking the
automobile.
c) Find the speed of the police car at the time it overtakes the automobile.
d) Find the speed of the automobile at the time it was overtaken by the police car.
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