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3 years ago

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Rain falls on a 1346 acre urban watershed at an intensity of 1.75 in/hr for a duration of 1 hour. The catchment land use is 20%

parkland, 30% residential, and 50% multiple detached residential area. The time to concentration is 1 hr. Determine the peak runoff. Use runoff coefficients published in APWA 5600.

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer:

Detailed solution is given below:

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What's the best way to find the load capacity of a crane? Select the best option. Call the manufacturer Ask co-workers Look at t

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Answer: You need to lift a load of 15 tons (30,000 pounds) a distance of 25 feet. The distance is measured from the center pin of the crane to the center of the load. Once you determine the distance, look on that line for the largest capacity; that will indicate how many feet of boom must be extended

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3 years ago

Complete the following sentence. The skills and content of several subject areas were combined to form a new field known as a me

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The answer is "discipline" , I was doing the same thing a minute ago. I hope this anwser is helpful! ^^

Explanation:

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3 years ago

Read 2 more answers

A heat engine operates between a source at 477°C and a sink at 27°C. If heat is supplied to the heat engine at a steady rate of

lara [203]

Answer:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Explanation:

For this case we can use the fact that the maximum thermal efficiency for a heat engine between two temperatures are given by the Carnot efficiency:

$e = 1 -frac{T_C}{T_H}$

We have on this case after convert the temperatures in kelvin this:

$T_C = 27+273.15 = 300.15 K$

$T_H = 477+273.15 = 750.15 K$

And replacing in the Carnot efficiency we got:

$e= 1- \frac{300.15}{750.15}= 0.59988 = 59.98 \%$

And the maximum power output on this case would be defined as:

$W_{max}= e* Q_H = 0.59988 * 65000 \frac{KJ}{min}= 38992.2 \frac{KJ}{min}$

Where $Q_H$ represent the heat associated to the deposit with higher temperature.

4 0

3 years ago

While discussing run-flat tires: Technician A says that some are self-sealing tires and are designed to quickly and permanently

musickatia [10]

Answer:

The correct option is d ( Neither A nor B)

Explanation:

Technician A made 2 mistakes in his statement.Firstly the tire is self supporting not self sealing.

Secondly, this tire does not provide permanent sealing of punctured area option a is incorrect.

This self-supporting tire after being affected with complete air leakage can temporarily bear the load of the car and avoid rolling over a distance of 80 km at a maximum speed of 55 mph. Here is what technician B suggested incorrectly as the tire after being.Here the technician B suggested incorrectly as the tire after being affected with puncture can not travel at any speed so option B is wrong

Since option a and b are incorrect and c is invalid.

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3 years ago

A 50 (ohm) lossless transmission line is terminated in a load with impedance Z= (30-j50) ohm. the wavelength is 8cm. Determine:

Serjik [45]

Answer:

Reflection Coefficient = $0.57e^{-i79.8}$

SWR=3.65

Position of $V_{max} =3.11cm$

position of $i_{max} =1.11cm$

Explanation:

To determine the above answers, let outline the useful formulas

refection coefficient $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$ .

where terminal impednce = (30-i50)Ω

characteristics impedance= 50Ω

Secondly, the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

Now let us substitute values and solve,

a. $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$

$p=\frac{(30-i50)-50}{(30-i50)-50} \\$

$p=\frac{-20-i50}{80-i50} \\$

multiplying the numerator and denominator by the conjugate of the denominator. we have

$p=\frac{-20-i50}{80-i50}*\frac{80+i50}{80+i50}\\$

by carrying out careful operation, we arrived at

$p=\frac{900-i5000}{8900} \\p=0.1011-i0.56179\\$

To express in polar form i.e $re^{i alpha}$

$r=\sqrt{0.1011^{2}+0.56179^{2}} \\r=0.57\\$

to get the angle

alpha= $tan^{-1} \frac{0.56179}{0.1011} \\alpha=-79.8\\$

hence the Reflection Coefficient,<em>p</em> = $0.57e^{-i79.8}$

b. we now determine the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

$swr=\frac{1+0.57}{1-0.57} =3.65\\$

c. to determine the position of the maximum voltage nearest to the load,

we use the equation

$Position of V_{max}=\frac{\alpha λ}{4\pi}+\frac{λ}{2}\\$

were $λ$ is the wavelength of 8cm

lets convert α to rad by multiplying by π/180

$Position of V_{max}=\frac{-79.8 *8cm*\pi}{4\pi*180 } +\frac{8cm}{2} \\$

$Position of V_{max}=-0.89+4.0=3.11cm\\$ .

d. also were we have minimum voltage,there the maximum current will exist, to find this position nearest to the load

$Position of v_{min}=Position of V_{max}-\frac{λ}{4} \\\\Position of v_{min}=3.11cm-\frac{8cm}{4}=1.11cm$ .

since the voltage minimum occure at 1.11cm. we can conclude that the current maximum also occur at this point i.e 1.11cm

3 0

3 years ago