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shepuryov [24]
3 years ago
13

Realize the function f(a, b, c, d, e) = Σ m(6, 7, 9, 11, 12, 13, 16, 17, 18, 20, 21, 23, 25, 28)using a 16-to-1 MUX with control

inputs b, c, d, and e. Each data input should be0, 1, a, or a′.(Hint: Start with a minterm expansion of F and combineminterms to eliminate a and a′ wherepossible.)(3pts)

Engineering
1 answer:
NNADVOKAT [17]3 years ago
7 0

Answer:

See explaination

Explanation:

please see attachment for the detailed diagram used in solving the given problem.

It is attached as an attachment.

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Which tool ensures that a fastener has the proper amount of tightness
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A torque wrench tool is a tool that ensures that a fastener has the proper amount of tightness.

<h3>What is the torque wrench used for?</h3>

The torque wrench tool is used to ensure screws and bolts are properly tightened. When performing home repairs and maintenance of equipment it is quite important that a torque wrench is used in other to prevent a scenario where a fastener (screws and bolts) does not become loose leading to equipment failure or damage. Because of its many advantages, this tool is often found in the possession of construction workers.

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is my guess but yeah

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3 years ago
A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c
Eduardwww [97]

Answer:

h=1.99998\ W/m^2.C

k=33.333\ W/m.C

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0       Eq (1)

Where:

k is thermal Conductivity

\dot e_{gen} is uniform thermal generation

T(x) = a(L^2-x^2)+b

\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a

Putt in Eq (1):

-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C

Energy balance is given by:

Q_{convection}=Q_{conduction}

h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L     Eq  (2)

T(x) = a(L^2-x^2)+b

Putting x=L

T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC

\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2

From Eq (2)

h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C

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3 years ago
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