The weight of the specimen in SSD condition is 373.3 cc
<u>Explanation</u>:
a) Apparent specific gravity = 
Where,
A = mass of oven dried test sample in air = 1034 g
B = saturated surface test sample in air = 1048.9 g
C = apparent mass of saturated test sample in water = 975.6 g
apparent specific gravity =
= 
Apparent specific gravity = 2.88
b) Bulk specific gravity 
= 2.76
c) Bulk specific gravity (SSD):
= 2.80
d) Absorption% :
Absorption = 1.44 %
e) Bulk Volume :
= 
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 3){
cout << "Too small";
cout << endl;
}
else if(bagOunces > 10){
cout << "Too large";
cout << endl;
}
else{
cout << (6 * bagOunces) << " seconds" << endl;
}
}
int main() {
PrintPopcornTime(7);
return 0;
}
Explanation:
Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.
Here’s some of them
6. J
7. I
10. O
13. F
14. E
15. N
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
Answer:
as slated in your solution, if delay time is 2.30 mins, hence 9 vehicle will be on queue as the improved service commenced.
Explanation:
4 vehicle per min, in 2 mins of the delay time 8 vehicles while in 0.3 min average of 1 vehicle join the queue. making 9 vehicle maximum
