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soldier1979 [14.2K]
3 years ago
6

You are to design two CONCEPTUALLY different synchronous state machines (Mealy and Moore) that perform the task described below.

You are allowed to use any type of flip-flop. However, I recommend that you use either JK or D flip-flops. Win Announcer There is a game called screwball. Every time a ball is put into play either player 1 (P1) scores a point or player 2 (P2) scores a point. A winner is declared when one player is ahead by 2 points. Starting with a score of 0:0 forP1 vs. P2, a game might progress as follows:0:1 1:1 2:1 3:1 (P1 Wins!)
Design a finite state machine that uses information about which player scores a point at each stage of a game to determine when a player wins and which player wins. Your design should include one reset input to initialize your win announcer (this may be a synchronous or an asynchronous input, as you desire). The outputs from your designs should indicate BOTH when a win has occurred and which player is the winner?
Engineering
1 answer:
allochka39001 [22]3 years ago
3 0
Answer:








Explanation:









I hope this helps!
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For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
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Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
2 years ago
Preheat and postheating are necessary when welding gray cast iron. *<br> True<br> False
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True will be your answer have a great day
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