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RUDIKE [14]
3 years ago
10

The annual storage in Broad River watershed is 0 cm/y. Annual precipitation is 100 cm/y and evapotranspiration is 50 cm/y. The s

andy loam soil in the watershed has an infiltration rate of 5 × 10-7 cm/s. The area of the watershed is 40 km2. Calculate the annual flow at the outlet of Broad River. What is the runoff coefficient for this watershed?

Engineering
2 answers:
PSYCHO15rus [73]3 years ago
7 0

Answer:

The annual flow at the outlet of Broad River is 13688480 m³

The runoff coefficient for the watershed is 0.342212

Explanation:

Here, we have

The annual storage in  the watershed = 0 cm/y

Annual precipitation = 100 cm/y

Annual evapotranspiration = 50 cm/y

infiltration rate = 5 × 10-7 cm/s

Watershed area = 40 km²

From the question, annual infiltration rate is given by

Annual infiltration rate = 5 × 10-7 cm/s × ‪31557600‬ s/year = 15.7788 cm/y

Therefore, annual flow =

Annual precipitation - Annual evapotranspiration - Annual infiltration rate - Annual storage in  the watershed

Annual flow = 100 cm/y - 50 cm/y  - 15.7788 cm/y - 0 cm/y = 34.2212 cm/y

The area of the watershed = 40 km²

Therefore, the volume of the annual flow is given by;

Height of annual flow × area of the watershed

Volume of annual flow = 34.2212 cm × 40 km²  =  1368.848 cm·km²

= 13688480 m³

The runoff coefficient = \frac{Amount \hspace{0.1cm} of  \hspace{0.1cm}runoff}{Amount  \hspace{0.1cm} of \hspace{0.1cm}  precipitation \hspace{0.1cm}  received} =

The amount of precipitation received is given by

100 cm × 40 km² = 4000 cm·km² = 40000000 m³

Runoff coefficient = \frac{13688480 m^3}{40000000 m^3} = 0.342212.

REY [17]3 years ago
3 0

Answer:

0.34232

Explanation:

See attachment

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<u>a)</u>

<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in series  

<u>List the Knowns: </u>

Capacitance of the first capacitor: C_{1}= 2цF = 2 x 10-6 F

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Voltage of battery: V = 9 V  

<u>Set Up the Problem:   </u>

Capacitance of a series combination:  

\frac{1}{C_{s} } =\frac{1}{C_{1} } +\frac{1}{C_{2} } +\frac{1}{C_{3} }+............

\frac{1}{C_{s} } =\frac{1}{2} +\frac{1}{ 7.4} \\C_{s} =\frac{2*7.4}{2+7.4}=1.575 *10^-6 F\\

Capacitance of a series combination is given by:

C_{s}=\frac{Q}{V}

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Energy stored in the series combination is:  

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Q=1.575*10^-6*9=1.42*10^-5C\\\\U_{c} =\frac{1}{2}*9^{2} *1.575*10^-6=6.38*10^-5J

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<u>Identify the unknown:  </u>

The charge and energy stored if the capacitors are connected in parallel  

<u>Set Up the Problem:  </u>

Capacitance of a parallel combination:

C_{p} =C_{1} +C_{2} +C_{3}

C_{p} =2+7.4=9.4*10^-6F

Capacitance of a parallel combination is given by

C_{p} =\frac{Q}{V}

Then the charge stored in the parallel combination is

Q=C_{p} V

Energy stored in the parallel combination is:  

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<u>Solve the Problem: </u><em>  </em>

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