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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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An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100
Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}

Where:

\sigma_{m} - Mean stress, measured in megapascals.

\sigma_{min} - Minimum stress, measured in megapascals.

\sigma_{max} - Maximum stress, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the mean stress is:

\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}

\sigma_{m} = 100\,MPa

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

\sigma_{a} = |\sigma_{max}-\sigma_{min}|

Where \sigma_{a} is the stress amplitude, measured in megapascals.

If we know that \sigma_{min} = -100\,MPa and \sigma_{max} = 300\,MPa, the stress amplitude is:

\sigma_{a} = |300\,MPa-(-100\,MPa)|

\sigma_{a} = 400\,MPa

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio (R) is the ratio of the stress amplitude to mean stress. That is:

R = \frac{\sigma_{a}}{\sigma_{m}}

If we know that \sigma_{m} = 100\,MPa and \sigma_{a} = 400\,MPa, the stress ratio is:

R = \frac{400\,MPa}{100\,MPa}

R = 4

The stress ratio of the aluminium part is 4.

3 0
3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
4 years ago
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator
MakcuM [25]

Answer:

Explanation:

The frequency components in the message signal are

f1 = 100Hz, f2 = 200Hz and f3 = 400Hz

When amplitude modulated with a carrier signal of frequency fc = 100kHz

Generates the following frequency components

Lower side band

100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\

Carrier frequency 100kHz

Upper side band

100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz

After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be

100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz

At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be

100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz

After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be

100k, 99.9k, 99.8k\ \ and \ \99.6kHz

At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be

100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz

3 0
3 years ago
Can someone explain it to me!
Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0
3 years ago
Dowry practice is considered to be a social problem . justify<br>​
Artemon [7]

Answer:

mark me brailinist

Explanation:

Dowry is a social evil in the society, that has caused unimaginable tortures and crimes towards women. The evil has taken lives of women from all stratas of society - be it poor, middle class or the rich. However it is the poor who succumb and fall prey to it, more due to their lack of awareness and education.

It is because of the dowry system, that daughters are not valued as much as the sons. In the society, many a times it has been seen that they are seen as a liability and are often subjected to subjugation and are given second hand treatment may it be in education or other amenities.

Today the government has come up with many laws and reforms, not only to eradicate the dowry system, but also to uplift the status of the girl child by bringing in many schemes.

It is now for the society at large to become aware and understand the situation. It up to all of us to take active steps in bringing about the necessary change and stop either giving or taking dowry. We must all know that we should start valuing our daughters first, so that others know their value once they grow up.

7 0
2 years ago
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