Answer:
Hello your question is incomplete attached below is the missing part and answer
options :
Effect A
Effect B
Effect C
Effect D
Effect AB
Effect AC
Effect AD
Effect BC
Effect BD
Effect CD
Answer : 
 A  = significant 
  B  = significant
 C  = Non-significant
 D  = Non-significant
AB  = Non-significant 
AC  = significant
AD  = Non-significant
 BC  = Non-significant
  BD  = Non-significant
  CD = Non-significant
Explanation:
The dependent variable here is Time
Effect of A  = significant 
Effect of B  = significant
Effect of C  = Non-significant
Effect of D  = Non-significant
Effect of AB  = Non-significant 
Effect of AC  = significant
Effect of AD  = Non-significant
Effect of BC  = Non-significant
Effect of BD  = Non-significant
Effect of CD = Non-significant
 
        
             
        
        
        
Answer:
1.2727 stokes 
Explanation:
specific gravity of fluid A = 1.65 
Dynamic viscosity = 210 centipoise 
<u>Calculate the kinematic viscosity of Fluid A </u>
First step : determine the density of fluid A 
Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3
next : convert dynamic viscosity to kg/m-s
210 centipoise = 0.21 kg/m-s 
Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A 
                                             = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec
Convert to stokes = 1.2727 stokes 
 
        
             
        
        
        
Answer:
"Macro Instruction"
Explanation:
A macro definition is a rule or pattern that specifies how a certain input sequence should be mapped to a replacement output sequence according to a defined procedure. The mapping process that instantiates a macro use into a specific sequence is known as macro expansion.
It is a series of commands and actions that can be stored and run whenever you need to perform the task. You can record or build a macro and then run it to automatically repeat that series of steps or actions.
 
        
             
        
        
        
Answer:
Explanation:
ADT for an 2-D array:
struct array{
int arr[10];
}arrmain[10];
An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :
1) struct array{
int *p;
}arr[1000];
2) struct array{
int *p;
}arr[1000];
 
        
             
        
        
        
Answer:
 ≅ 111 KN
 ≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
 =
 =  × CρAv²
 × CρAv²
 where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, 
We can as well say: 

We can now replace  in the above equation.
 in the above equation.
Therefore,  =
 =  × CρAv²
 × CρAv²
The A which stands as the area of the jet is given by the formula:

We can now have a new equation after substituting our A into the previous equation as:
  =
 =  × Cρ
 × Cρ 
Substituting our data from above; we have:
 =
 =  ×
 × 
 =
 = 
 = 110,990N
 = 110,990N
 in N (newton) to KN (kilo-newton) will be:
  in N (newton) to KN (kilo-newton) will be:
  =
 = 
 = 110.990 KN
 = 110.990 KN
 ≅ 111 KN
 ≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.