Answer:
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answer : NOR1(q_) wave is complementary to NOR2(q)
Explanation:
Note ; NOR 2 will be addressed as q in the course of this solution while NOR 1 will be addressed as q_
Initial state is unknown i.e q = 0 and q_= 1
from the diagram the waveform reset and set
= from 0ns to 20ns reset=1 and set=0.from the truth table considering this given condition q=0 and q_bar=1 while
from 30ns to 50ns reset=0 and set=1.from the truth table considering this condition q=1 and q_bar=1.so from 35ns also note there is a delay of 5 ns for the NOR gate hence the NOR 2 will be higher ( 1 )
From 50ns to 65ns both set and reset is 0.so NOR2(q)=0.
From 65 to 75 set=1 and reset=0,so our NOR 2(q)=1 checking from the truth table
also from 75 to 90 set=1 and reset=1 , NOR2(q) is undefined "?" and is mentioned up to 95ns.
since q_ is a complement of q, then NOR1(q_) wave is complementary to NOR2(q)
Answer:
Technician A is wrong
Technician B is right
Explanation:
voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.
A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.
Solution :
Given :
The number of blows is given as :
0 - 6 inch = 4 blows
6 - 12 inch = 6 blows
12 - 18 inch = 6 blows
The vertical effective stress 
Now,
corrected N - value of overburden
effective stress at level of test
0 - 6 inch, 
= 9.86
6 - 12 inch, 
= 14.8
12 - 18 inch,
= 14.8
= 13.14
= 13
Answer:
The surface area of the primary settling tank is 0.0095 m^2.
The effective theoretical detention time is 0.05 s.
Explanation:
The surface area of the tank is calculated by dividing the volumetric flow rate by the overflow rate.
Volumetric flow rate = 0.570 m^3/s
Overflow rate = 60 m/s
Surface area = 0.570 m^3/s ÷ 60 m/s = 0.0095 m^2
Detention time is calculated by dividing the volume of the tank by the its volumetric flow rate
Volume of the tank = surface area × depth = 0.0095 m^2 × 3 m = 0.0285 m^3
Detention time = 0.0285 m^3 ÷ 0.570 m^3/s = 0.05 s
