Answer:
The differences are listed below
Explanation:
The differences between consolidation and compaction are as follows:
In compaction the mechanical pressure is used to compress the soil. In consolidation, there is an application of stead pressure.
In compaction, there is a dynamic load by rapid mechanical methods like tamping, rolling, etc. In consolidation, there is static and sustained pressure applied for a long time.
In compaction, the soil volume is reduced by removing air from the void. In consolidation, the soil volume is reduced by squeezing out water from the pores.
Compaction is used for sandy soil, consolidation on the other hand, is used for clay soil.
It is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information
<h3>What are online information?</h3>
Online informations are information which are available on the internet such as search engines, social handles and other websites
In conclusion, it is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information
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Answer: Fiber Optic Network Fiber-optic networks have been used for decades to transmit large volumes of traffic across the country. The economics of fiber networks have only recently allowed for connecting the fiber directly to the home, creating a fiber-to-the-home (FTTH) network.
Explanation:
If there is a random child, you should ask where they should be and where their parents went. if it your child, they prolly want food or sum.
The amount of settlement that would occur at the end of 1.5 year and 5 year are 7.3 cm and 13.14 cm respectively.
<h3>How to determine the amount of settlement?</h3>
For a layer of 3.8 m thickness, we were given the following parameters:
U = 50% = 0.5.
Sc = 7.3 cm.
For Sf, we have:
Sf = Sc/U
Sf = 7.3/0.5
Sf = 14.6
Therefore, Sf for a layer of 38 m thickness is given by:
Sf = 14.6 × 38/3.8
Sf = 146 cm.
At 50%, the time for a layer of 3.8 m thickness is: 
= 1.5 year.
At 50%, the time for a layer of 38 m thickness is:
= 1.5 × (38/3.8)²
= 150 years.
For the thickness of 38 m, U₂ is given by:
![\frac{U_1^2}{U_2^2} =\frac{(T_v)_1}{(T_v)_2} = \frac{t_1}{t_2} \\\\U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{1.5}{150} ]\\\\U_2^2 = 0.25 \times 0.01\\\\U_2=\sqrt{0.0025} \\\\U_2=0.05](https://tex.z-dn.net/?f=%5Cfrac%7BU_1%5E2%7D%7BU_2%5E2%7D%20%3D%5Cfrac%7B%28T_v%29_1%7D%7B%28T_v%29_2%7D%20%3D%20%5Cfrac%7Bt_1%7D%7Bt_2%7D%20%5C%5C%5C%5CU_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B1.5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.01%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0025%7D%20%5C%5C%5C%5CU_2%3D0.05)
The new settlement after 1.5 year is:
Sc = U₂Sf
Sc = 0.05 × 146
Sc = 7.3 cm.
For time, t₂ = 5 year:
![U_2^2 = U_1^2 \times [\frac{t_2}{t_1} ]\\\\U_2^2 = 0.5^2 \times [\frac{5}{150} ]\\\\U_2^2 = 0.25 \times 0.03\\\\U_2=\sqrt{0.0075} \\\\U_2=0.09](https://tex.z-dn.net/?f=U_2%5E2%20%3D%20U_1%5E2%20%5Ctimes%20%5B%5Cfrac%7Bt_2%7D%7Bt_1%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.5%5E2%20%5Ctimes%20%5B%5Cfrac%7B5%7D%7B150%7D%20%5D%5C%5C%5C%5CU_2%5E2%20%3D%200.25%20%20%5Ctimes%200.03%5C%5C%5C%5CU_2%3D%5Csqrt%7B0.0075%7D%20%5C%5C%5C%5CU_2%3D0.09)
The new settlement after 5 year is:
Sc = U₂Sf
Sc = 0.09 × 146
Sc = 13.14 cm.
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