Answer:
P= 5.5 bar
Explanation:
Given that
L= 4000 m
d= 0.2 m
Friction factor(F) = 0.01
speed V= 2 m/s
Head = 5 m
Head loss due to friction



So the total head(H) = 5 + 40.77 + 10.3 =56.07
Where 10.3 m is the atmospheric head.
We know that
P=ρ g H
So total Pressure
P= 1000 x 9.81 x 56.07 Pa

P= 5.5 bar
Answer:
The answer is option
C . the JFET has a PN junction
Explanation:
Not only is option C in the question a dissimilarity between the MOSFET and the JFET we can go on with some more dissimilarities.
1.MOSFET stands for Metal Oxide Silicon Field Effect Transistor or Metal Oxide Semiconductor Field Effect Transistor.
(JFET) stands for junction gate field-effect transistor (JFET)
2. JFET is a three-terminal semiconductor device, whereas MOFET a four-terminal semiconductor device.
3. In terms of areas of application of JFETs are used in low noise applications while MOSFETs, are used for high noise applications
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)