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WARRIOR [948]
2 years ago
14

How deep is a 6ft hole?​

Engineering
2 answers:
Pavlova-9 [17]2 years ago
3 0

Answer:

I know this sounds quite deep but it is as deep as a grave

Explanation:

It's reality

denis23 [38]2 years ago
3 0

Answer:

6ft or 72 inches

Explanation:

The hole is most likely 6ft or also known as 72 inches.

Hope this helps :-)

Have a great rest of your day or night!

<3 simplysun

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Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per
Tomtit [17]

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Explanation :

The given  information to be listed can are Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per Gallon.

Check the attached document for the solution.

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Whats the best used for cable -stayed bridge
nalin [4]

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a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.

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3 years ago
a stem and leaf display describes two-digit integers between 20 and 80. for one one of the classes displayed, the row appears as
allochka39001 [22]

Answer:

  52, 50, 54, 54, 56

Explanation:

The "stem" in this scenario is the tens digit of the number. Each "leaf" is the ones digit of a distinct number with the given tens digit.

  5 | 20446 represents the numbers 52, 50, 54, 54, 56

8 0
3 years ago
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
Fofino [41]

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
3 years ago
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fomenos

Answer:

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2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0
3 years ago
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