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alexdok [17]
3 years ago
5

What is the saturated zone

Physics
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

Saturated zone is area below the water table in which the soil is completely saturated with groundwater.

Explanation:

The saturated zone lies below the ground. It is mainly the lower zone of rock along with the water table where pore spaces are completely filled with water. Even the saturated zone is sometimes separated into 2 subzones: the phreatic zone and the capillary fringe.

The area where pores spaces are not saturated with water is also unsaturated zone. Localized saturated zones can occur within the unsaturated zone. The unsaturated zone lies above the groundwater table.

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14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari
Inessa [10]
In order to make things easier to describe and explain, let's call
the resistance of each bulb 'R', and the battery voltage 'V'.

a).  In series, the total resistance is 3R.
In parallel, the total resistance is R/3.
Changing from series to parallel, the total resistance of the circuit
decreases to 1/9 of its original value.

b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
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c).  In series, the power dissipated by the circuit is 

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In parallel, the power dissipated by the circuit is

                                   (V) · 3V/R  =  3V² / R .

Changing from series to parallel, the power dissipated by
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8 0
3 years ago
In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an
natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa

σ=5517.25 Pa

Strain in x direction

ε=σ/E

=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

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