Answer:
hello your question lacks the required image attached to this answer is the image required
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:
6.37 inch
Explanation:
Thinking process:
We need to know the flow rate of the fluid through the cross sectional pipe. Let this rate be denoted by Q.
To determine the pressure drop in the pipe:
Using the Bernoulli equation for mass conservation:
![\frac{P1}{\rho } + \frac{v_{2} }{2g} +z_{1} = \frac{P2}{\rho } + \frac{v2^{2} }{2g} + z_{2} + f\frac{l}{D} \frac{v^{2} }{2g}](https://tex.z-dn.net/?f=%5Cfrac%7BP1%7D%7B%5Crho%20%7D%20%2B%20%5Cfrac%7Bv_%7B2%7D%20%7D%7B2g%7D%20%2Bz_%7B1%7D%20%20%3D%20%5Cfrac%7BP2%7D%7B%5Crho%20%7D%20%2B%20%5Cfrac%7Bv2%5E%7B2%7D%20%7D%7B2g%7D%20%2B%20z_%7B2%7D%20%2B%20f%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B2g%7D)
thus
![\frac{P1-P2}{\rho } = f\frac{l}{D} \frac{v^{2} }{2g}](https://tex.z-dn.net/?f=%5Cfrac%7BP1-P2%7D%7B%5Crho%20%7D%20%20%3D%20f%5Cfrac%7Bl%7D%7BD%7D%20%5Cfrac%7Bv%5E%7B2%7D%20%7D%7B2g%7D)
The largest pressure drop (P1-P2) will occur with the largest f, which occurs with the smallest Reynolds number, Re or the largest V.
Since the viscosity of the water increases with temperature decrease, we consider coldest case at T = 50⁰F
from the tables
Re= 2.01 × 10⁵
Hence, f = 0.018
Therefore, pressure drop, (P1-P2)/p = 2.70 ft
This occurs at ae presure change of 1.17 psi
Correlating with the chart, we find that the diameter will be D= 0.513
= <u>6.37 in Ans</u>
Answer with Explanation:
The general equation of simple harmonic motion is
![x(t)=Asin(\omega t+\phi)](https://tex.z-dn.net/?f=x%28t%29%3DAsin%28%5Comega%20t%2B%5Cphi%29)
where,
A is the amplitude of motion
is the angular frequency of the motion
is known as initial phase
part 1)
Now by definition of velocity we have
![v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bdx%7D%7Bdt%7D%5C%5C%5C%5C%5Ctherefore%20v%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%28Asin%28%5Comega%20t%2B%5Cphi%20%29%5C%5C%5C%5Cv%28t%29%3DA%5Comega%20cos%28%5Comega%20t%2B%5Cphi%20%29)
part 2)
Now by definition of acceleration we have
![a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bdv%7D%7Bdt%7D%5C%5C%5C%5C%5Ctherefore%20a%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%28A%5Comega%20cos%28%5Comega%20t%2B%5Cphi%20%29%5C%5C%5C%5Ca%28t%29%3D-A%5Comega%20%5E%7B2%7Dsin%28%5Comega%20t%2B%5Cphi%20%29)
part 3)
The angular frequency is related to Time period 'T' as
where
is the angular frequency of the motion of the particle.
Part 4) The acceleration and velocities are plotted below
since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is
and
respectively.
Answer:
The molecular weight will be "28.12 g/mol".
Explanation:
The given values are:
Pressure,
P = 10 atm
= ![10\times 101325 \ Pa](https://tex.z-dn.net/?f=10%5Ctimes%20101325%20%5C%20Pa)
=
Temperature,
T = 298 K
Mass,
m = 11.5 Kg
Volume,
V = 1000 r
= ![1 \ m^3](https://tex.z-dn.net/?f=1%20%5C%20m%5E3)
R = 8.3145 J/mol K
Now,
By using the ideal gas law, we get
⇒ ![PV=nRT](https://tex.z-dn.net/?f=PV%3DnRT)
o,
⇒ ![n=\frac{PV}{RT}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D)
By substituting the values, we get
![=\frac{1013250\times 1}{8.3145\times 298}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1013250%5Ctimes%201%7D%7B8.3145%5Ctimes%20298%7D)
![=408.94 \ moles](https://tex.z-dn.net/?f=%3D408.94%20%5C%20moles)
As we know,
⇒ ![Moles(n)=\frac{Mass(m)}{Molecular \ weight(MW)}](https://tex.z-dn.net/?f=Moles%28n%29%3D%5Cfrac%7BMass%28m%29%7D%7BMolecular%20%5C%20weight%28MW%29%7D)
or,
⇒
![=\frac{11.5}{408.94}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B11.5%7D%7B408.94%7D)
![=0.02812 \ Kg/mol](https://tex.z-dn.net/?f=%3D0.02812%20%5C%20Kg%2Fmol)
Answer: yes
Explanation: ontop of a tall building, you drop a small peace of metal covered in zinc. it is possible to be very dangerus because of gravity. some one walking on the side walk who gets hit in the head can get a concusion maybe even a brain injury.