Answer:
Total BF for the interior wall is 7.50BD
Explanation:
Given Data:
· Size of stud = 2” x 6”
· Height of Wall = 8 ft
· Top plates = 2
· Bottom Plate = 1
BF stands for board feet in lumber/wood terminology. It is the unit of volume.
1 BF (Board feet) = 1 ft x 1 ft x 1 inch
Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.
Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft
Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD
Total BF for the interior wall is 7.50BD
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
Answer:
≅ 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
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.
Therefore, = × 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 
Substituting our data from above; we have:
= × 
= 
= 110,990N
in N (newton) to KN (kilo-newton) will be:
= 
= 110.990 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.
Answer:
RAM, which stands for random access memory, and ROM, which stands for read-only memory, are both present in your computer. RAM is volatile memory that temporarily stores the files you are working on. ROM is non-volatile memory that permanently stores instructions for your computer.
Explanation:
