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AleksandrR [38]
3 years ago
6

"At 195 miles long, and with 7,325 miles of coastline, the Chesapeake Bay is the largest and most complex estuary in the United

States.Though long and wide (30 miles wide at the Potomac River), the bay is very shallow, with an average depth of only 28 feet." The Bay's maximum depth is 174 ft. The hydraulic model of the Chesapeake Bay was built with a model length ratio Lr=1/1000
a. How wide was the model Bay [ft] at the Potomac River?
b. The Bay Bridge is 4.3 miles long; how long was it in the model [ft]?
c. If the hydraulic model occupied about 8 acres, approximately what is the real-world (prototype) area represented by the model [square miles]?
d. At this scale, what would be the average depth and maximum depth of the model Bay? [give answers in both ft and inches)
Engineering
1 answer:
Paraphin [41]3 years ago
4 0

Answer:

see explaination

Explanation:

Part a) Width of bay at Potomac River:

Given Data:

· Actual Width at Potomac River = 30 miles

· Bay Model Length Ratio Lr = 1/1000

In fluid mechanics models of real structures are prepared in simulation so that they can be analyzed accurately. A model is known to have simulation if model carries same geometric, kinematic and dynamic properties at a small scale.

Length of any part in model = Actual length x Lr

Hence,

Model Width of bay at Potomac River = 30 x 1/1000 = 0.03 miles

Since 1 mile = 5280 ft

Model Width of bay at Potomac River = 0.03 x 5280 = 158.4 ft

Part b) Model Length of bay bridge in model:

Given Data:

· Actual Length of bay bridge = 4.3 miles

· Bay Model Length Ratio Lr = 1/1000

Model Length = Actual Length x Lr = 4.3 x 1/1000 = 0.0043 miles

Since 1 mile = 5280 ft

Model Length in feet = 0.0043 x 5280 = 22.704 ft

Part c) Model Length of bay bridge in model:

Given Data:

· Model Area = 8 acre

· Bay Model Length Ratio Lr = 1/1000

Model Area = Actual Area x Lr x Lr

8 Model Area :: Actual Area =- (Lp)2 2 = 8,000,000 acre 1000

Since 1 square mile = 640 acre,

Actual Area in square miles = 8,000,000/640 = 12,500 square miles

Part d) Average and maximum depth of model:

Given Data:

· Actual Average depth = 28 ft

· Actual Maximum depth = 174 ft

· Bay Model Length Ratio Lr = 1/1000

Model average depth = Actual average depth x Lr = 28 x 1/1000 = 0.028 feet

Since 1 ft = 12 inch

Model average depth in inch = 0.028 x 12 = 0.336 in

Model maximum depth = Actual maximum depth x Lr = 174 x 1/1000 = 0.174 feet

Since 1 ft = 12 inch

Model maximum depth in inch = 0.174 x 12 = 2.088 in

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In the ______ phase of the organizational life cycle, the organization is usually very small and agile, focusing on new products
Vlad [161]

Answer:

Entrepreneurship

Explanation:

Different phases of  organizational life cycles can be observed depending on the stage of the company. These cycles are,

  • Entrepreneurship
  • Survival and Early Success
  • Sustained Success
  • Renewal (or Decline)

In the initial stage which is  Entrepreneurship Phase,  all of the founders take part in various activities, no formality is formed between founders and employees. Generally lots of ideas are present and company is actively searching to define correct market and products to focus on.

5 0
3 years ago
Four eight-ohm speakers are connected in parallel to an audio power amplifier. The amplifier can supply a maximum driver output
Vera_Pavlovna [14]

Answer:

112.5 watts

Explanation:

The output voltage (V) = 15 volts

The equivalent resistance for the speakers connected in parallel (R_T) is gotten by using the formula:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\\\\But\ R_1=R_2=R_3=R_4=8\ ohm\\\\\frac{1}{R_T} =\frac{1}{8} +\frac{1}{8} +\frac{1}{8} +\frac{1}{8} \\\\\frac{1}{R_T} =\frac{1+1+1+1}{8} \\\\\frac{1}{R_T} =\frac{4}{8} \\\\R_T=\frac{8}{4} \\\\R_T=2\ ohm

The current flowing through the amplifier (I) is:

I = V / R_T = 15 / 2 = 7.5 A

The audio power (P) when outputting this maximum voltage is given by:

P = V² / R_T = I²R_T. Therefore:

P = V² / R_T = 15² / R_T = 112.5 watts

6 0
3 years ago
Consider a step pn junction made of GaAs at T = 300 K. At zero bias, only 20% of the total depletion region width is in the p-si
Nat2105 [25]

Answer:

0.31 μm

Explanation:

this question wants us to Determine the depletion region width, xn​, in the n-side in unit of μm. using the information below.

density in the p-side = 5.68x10^16

density in the n-side = 1.42x10^16

\sqrt{\frac{2*12.7*8.85E-10}{1.6E-14}(\frac{1}{5.68E16}+\frac{1}{1.42E16} )(1.2)  }

= √(1.42x10⁵)(1.76056335x10⁻¹⁷ + 7.042253521x10⁻¹⁷)(1.2)

= √150.74x10⁻¹¹

= 3.882x10⁻⁵

approximately 0.39μm

xn = 0.39 x 0.8

= 0.31μm

0.31 um is the depletion region width. thank you!

3 0
3 years ago
Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
An engineer is designing a total hip implant. She intends to make the femoral stem out of titanium because it forms a good inter
makvit [3.9K]

Answer:

1. Yes.

2. Localized corrosion

Explanation:

Should she be worried about corrosion?

Yes, the engineer needs to be worried about corrosion as stainless steel has a lower resistance to corrosion, in other words, stainless steel corrodes faster than Titanium.

If so, what types of corrosion could take place?

The type of corrosion that takes place is called Localized corrosion. Localized corrosion occurs when a small part of a component experiences corrosion. In this case, the ball component of the femoral stem is made of stainless steel which will corrode faster than the other parts of the femoral stem which is made of Titanium.

3 0
3 years ago
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