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Vilka [71]
3 years ago
6

Two piezometers have been placed along the direction of flow in a confined aquifer that is 30.0 m thick. The piezometers are 280

m apart. The difference in piezometric head between the two is 1.4 m. The aquifer hydraulic conductivity is 50 m · day−1, and the porosity is 20%. Estimate the travel time for water to flow between the two piezometers.
Engineering
1 answer:
Dafna11 [192]3 years ago
8 0

Answer:

time = 224 days

Explanation:

given data

thick = 30 m

piezometers =  280 m

head between  two = 1.4 m

aquifer hydraulic conductivity = 50 m

porosity =  20%

solution

we get here average pole velocity that is get by using Darcy law that is

Va = \frac{k}{\eta } \times \frac{\Delta h}{L}   ................1

here Va is average pole velocity and k is hydraulic conductivity and \eta is porosity  

here v is = k \times  \frac{dh}{dl}   ...........2

v = 50 × \frac{1.4}{280}

v = 0.25 m/day

and here average linear velocity Va will be

Va = \frac{v}{\eta }  

Va = \frac{0.25}{0.2}  

Va = 1.25 m/day  

travel time for water will be

time = \frac{280}{1.25}  

time = 224 days

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a)- True

Explanation:

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If a driver must drive when their signal lights or taillights aren't operating properly, they may instead signal by using their
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4 years ago
A fluid with a relative density of 0.9 flows in a pipe which is 12 m long and lies at an angle of 60° to the horizontal At the t
Minchanka [31]

Answer:

Q=7.3\times 10^{-3} m^3/s

Explanation:

Given that

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\frac{V_1}{V_2}=\left(\dfrac{d_2}{d_1}\right)^2

\frac{V_1}{V_2}=\left(\dfrac{30}{85}\right)^2

V_2=8.02V_1

Z_2=12 sin60^{\circ}

\dfrac{1000\times 1000}{900\times 9.81}+\dfrac{V_1^2}{2\times 9.81}+0=\dfrac{860\times 1000}{900\times 9.81 }+\dfrac{V_2^2}{2\times 9.81}+12 sin60^{\circ}

So V_1=1.30m/s

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Q=A_1V_1

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4 0
3 years ago
Light from a helium-neon laser (λ = 633 nm) illuminates two slits spaced 0.50 mm apart. A viewing screen is 2.5 m behind the sli
serg [7]

Answer:

Explanation:

Given

wavelength \lambda =633\ nm

distance between two slits d=0.5\ mm

Screen is placed at a distance L=2.5\ m

Location of a (n+1)th bright fringe is given by

x_{n+1}=\frac{L}{2d}(n+1)\lambda

for nth bright fringe

x_n=\frac{L}{2d}(n)\lambda

Distance between two bright fringes

x_{n+1}-x_n=\frac{L}{2d}\cdot \lambda

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7 0
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In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala
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Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

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In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, |F_{electrostatic}| = |F_{gravitational}|

where

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4 years ago
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