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:
8 to 10 times
Explanation:
For dry road
u= 15 mph ( 1 mph = 0.44 m/s)
u= 6.7 m/s
Let take coefficient of friction( μ) of dry road is 0.7
So the de acceleration a = μ g
a= 0.7 x 10 m/s ² ( g=10 m/s ²)
a= 7 m/s ²
We know that
v= u - a t
Final speed ,v=0
0 = 6.7 - 7 x t
t= 0.95 s
For snow road
μ = 0.4
de acceleration a = μ g
a = 0.4 x 10 = 4 m/s ²
u= 30 mph= 13.41 m/s
v= u - a t
Final speed ,v=0
0 = 30 - 4 x t'
t'=7.5 s
t'=7.8 t
We can say that it will take 8 to 10 times more time as compare to dry road for stopping the vehicle.
8 to 10 times
Answer:
Machinist
Explanation:
A skilled worker with the ability to operate computer numerically controlled (CNC) machines is qualified to work in a machinist position.
A machinist is a person who is properly skilled and consists of advanced knowledge regarding the functions of a CNC machine. He can use different mechanisms and complex numerical functions of the machine to carry out different tasks. Any person who lacks the official learning of mechanisms cannot operate such machines effectively.
Answer:
W= 8120 KJ
Explanation:
Given that
Process is isothermal ,it means that temperature of the gas will remain constant.
T₁=T₂ = 400 K
The change in the entropy given ΔS = 20.3 KJ/K
Lets take heat transfer is Q ,then entropy change can be written as
Now by putting the values
Q= 20.3 x 400 KJ
Q= 8120 KJ
The heat transfer ,Q= 8120 KJ
From first law of thermodynamics
Q = ΔU + W
ΔU =Change in the internal energy ,W=Work
Q=Heat transfer
For ideal gas ΔU = m Cv ΔT]
At constant temperature process ,ΔT= 0
That is why ΔU = 0
Q = ΔU + W
Q = 0+ W
Q=W= 8120 KJ
Work ,W= 8120 KJ
Answer:
connecting two independent clauses
