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4 years ago

Is this a

Engineering

1 answer:

Oduvanchick [21]4 years ago

4 0

Answer:

cam and follower

Explanation:

I believe its that because I took the class 2 years ago

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Simplify the following expressions, then implement them using digital logic gates. (a) f = A + AB + AC (b) f = AB + AC + BC (c)

Tema [17]

<u>Explanation</u>:

(a)

$f=A+A B+A C\\=A[1+B+C]=A \quad[1+x=1]\\F=A$

No gate is required to implement this function

(b)

$\begin{aligned}&\ f=A B+\bar{A} C+B C\\&\therefore f=A B+A C\end{aligned}$ $\begin{array}{l}(A B+\bar{A} C+B E=A B+\bar{A} C \\B C \text { is redendant })\end{array}$

Note: Refer the first image.

(c)

$\begin{aligned}f &=\overline{A+B}+A \bar{B}+B \bar{C} \\&=(\bar{A} \bar{B})+A \bar{B}+B \bar{C} \\&=\bar{B}[A+\bar{A}]+B \bar{C} \\& F=\bar{B}+B \bar{C} =\bar{B}+\bar{C}\end{aligned}$

Note: Refer the second image

(d)

$\begin{aligned}f=& A B \bar{c}+\overline{A+\bar{c}} \\=& A B \bar{c}+\bar{A} \bar{c}=\bar{A} B \bar{c}+\bar{A} c \\f=& \bar{A}[c+B \bar{c}] . \\& f=\bar{A} B+\bar{A} c=\bar{A}(B+c)\end{aligned}$

Note: Refer the third image

(e)

$\begin{aligned}f=& A \bar{B}+\bar{B} C+A \bar{B} \\&=\bar{B}[A+\bar{A}+c] \\&=\bar{B}[1+C]\end{aligned}$

$f=\frac{}{B}$

(f)

$\begin{aligned}f &=A B C+A B D+A B C \\&=A B[C+C]+A B D \\&=A B+A B D \\&=B[A+A D] \\&=B[A+D] \\\therefore & A=B[A+D]\end{aligned}$

Note: Refer the fourth image

6 0

3 years ago

Please. my brain isn’t working right now

Alex787 [66]

Answer:

10kQ is to 36......... D2 _ D1 D4

7 0

3 years ago

The percentage modulation of AM changes from 50% to 70%. Originally at 50% modulation, carrier power was 70 W. Now, determine th

adoni [48]

Answer:

What is percentage modulation in AM?

The percent modulation is defined as the ratio of the actual frequency deviation produced by the modulating signal to the maximum allowable frequency deviation.

3 0

3 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

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4 years ago

"A fluid at a pressure of 7 atm with a specific volume of 0.11 m3/kg is constrained in a cylinder behind a piston. It is allowed

AlekseyPX

Answer:

Work done by the fluid in the piston=164.5kJ/kg

Specific gas constant= 0.263 kJ/kg K

Molecular weight of gas= 31.54 kmol

7 0

4 years ago