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:
LAOD = 6669.86 N
Explanation:
Given data:
width
thickness 
crack length 2c = 0.5 mm at centre of specimen
stress intensity factor = k will be
we know that
[c =0.5/2 = 2.5*10^{-4}]
K = 0.1724 Mpa m^{1/2} for 1000 load
if
then load will be
LAOD = 6669.86 N
Answer:
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Explanation:
Answer:
Time of concentration, 
⇒ 1280 min
Peak runoff rate, 
⇒4.185 ff³/s
Explanation:
See detailed explanation
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.
