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Tresset [83]
2 years ago
13

What is the average linear (seepage) velocity of water in an aquifer with a hydraulic conductivity of 6.9 x 10-4 m/s and porosit

y of 30% if the hydraulic gradient is 0.0014? a. 0.28 m/d b. 3.2 m/d c. 0.08 m/d d. 0.003 m/d
Engineering
1 answer:
jeka942 years ago
4 0

Answer:

a. 0.28

Explanation:

Given that

porosity =30%

hydraulic gradient = 0.0014

hydraulic conductivity = 6.9 x 10⁻4 m/s

We know that average linear velocity given as

v=\dfrac{K}{n_e}\dfrac{dh}{dl}

v=\dfrac{6.9\times 10^{-4}}{0.3}\times0.0014\ m/s

v=3.22\times 10^{-6}\ m/s

The velocity in m/d      ( 1 m/s =86400 m/d)

v= 0.27 m/d

So the nearest answer is 'a'.

a. 0.28

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Reception of signals from a radio facility, located off the airway being flown, may be inadequate at the designated mea to ident
lakkis [162]

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

<h3>What is altitude?</h3>

Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.

The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.

The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.

Therefore, the correct answer is 22 NM of a VOR.

To learn more about altitudes refer to:

brainly.com/question/1159693

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3 0
1 year ago
Which reference source may be consulted to answer questions regarding the Professional Engineers Act?
9966 [12]

Answer: c. The Professional Engineers Act and Board Rules

Explanation:

The reference source may be consulted to answer questions regarding the Professional Engineers Act is the The Professional Engineers Act and Board Rules.

The Professional Engineers Act and Board Rules is an Act that was established in order to regulate the qualifications for professional engineered, register them and also make sure that their conducts and behavior are looked into.

6 0
3 years ago
A 1-w, 350-ω resistor is connected to 24 v. Is this resistor operating within its power rating?
seraphim [82]

Answer:

No.

Explanation:

P_r = Power rating = 1 W

R = Resistance = 350\ \Omega

V = Voltage = 24\ \text{V}

Power is given by

P=\dfrac{V^2}{R}\\\Rightarrow P=\dfrac{24^2}{350}\\\Rightarrow P=1.65\ \text{W}

1.65\ \text{W}>1\ \text{W}

So

P>P_r

Hence, the resistor is not operating within its power rating.

8 0
2 years ago
the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an
Mashcka [7]

Answer:

(a) The velocity of the collar = 1.624er-15.54eo in/s

(b) the acceleration of the collar=-49.94er-9.74eo in/s²

(c)the acceleration of collative rod =-3.284er in/s²

Explanation:

Check attachment for calculation

4 0
2 years ago
Rsidential Solar Solution:a. A type of photovoltaic solar panel manufactured in China receives a rating of 250W. The rating proc
aksik [14]

Answer:

a) \eta = 13.455\%, b) E_{day} = 812.716\,kJ, c) C_{month. total} = 19.505\, USD, d) t = 40.588\,years

Explanation:

a) The area of the solar panel is:

A = (20\,ft^{2})\cdot (\frac{0.3048\,m}{1\,ft} )^{2}

A = 1.858\,m^{2}

The energy potential is determined herein:

\dot E_{o} = (1000\,\frac{W}{m^{2}} )\cdot (1.858\,m^{2})

\dot E_{o} = 1858\,W

The efficiency of the solar panel is:

\eta = \frac{\dot E}{\dot E_{o}}\times 100\%

\eta = \frac{250\,W}{1858\,W}\times 100\%

\eta = 13.455\%

b) The energy generated by the solar panel is presented below:

E_{day} = (0.135)\cdot (150\,\frac{W}{m^{2}} )\cdot (20\,ft^{2})\cdot \left(\frac{0.3048\,m}{1\,ft} \right)^{2}\cdot (6\,h)\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{1\,kJ}{1000\,J} )

E_{day} = 812.716\,kJ

c) The energy generated per month and per panel is:

E_{month} = 30\cdot E_{day}

E_{month} = 30 \cdot (812.716\,kJ)\cdot \left(\frac{1\,kWh}{3600\,kJ}  \right)

E_{month} = 6.773\,kWh

Monthly energy savings due to the use of 18 panels are:

C_{month, total} = 18\cdot E_{month}\cdot c

C_{month, total} = 18\cdot (6.773\,kWh)\cdot (\frac{0.16\,USD}{1\,kWh} )

C_{month. total} = 19.505\, USD

d) The payback of the solar energy system is:

t = \frac{9500\,USD}{12\cdot (19.505\,USD)}

t = 40.588\,years

6 0
3 years ago
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