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.
Answer:
In the simplest terms, bioremediation is a waste management process using live organisms to neutralize or remove harmful pollutants from contaminated areas. Bioremediation is an environmental science that amplifies natural biological actions to remedy or remediate polluted groundwater and contaminated soil.
Explanation:
Answer:
a) v = +/- 0.323 m/s
b) x = -0.080134 m
c) v = +/- 1.004 m/s
Explanation:
Given:
a = - (0.1 + sin(x/b))
b = 0.8
v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.8))
- Separate variables:
v*dv = - (0.1 + sin(x/0.8)) . dx
-Integrate from v = 1 m/s @ x = 0:
0.5(v^2) = - (0.1x - 0.8cos(x/0.8)) - 0.8 + 0.5
0.5v^2 = 0.8cos(x/0.8) - 0.1x - 0.3
- Evaluate @ x = -1
0.5v^2 = 0.8 cos(-1/0.8) + 0.1 -0.3
v = sqrt (0.104516)
v = +/- 0.323 m/s
- v = v_max when a = 0:
-0.1 = sin(x/0.8)
x = -0.8*0.1002
x = -0.080134 m
- Hence,
v^2 = 1.6 cos(-0.080134/0.8) -0.6 -0.2*-0.080134
v = sqrt (0.504)
v = +/- 1.004 m/s
Answer:
See explanation below.
Explanation:
For this case the program needs to take the inputs as P,r and n and the output would be as A and printed on the system. The code is:
# Inputs
P = float(input("Enter the present value : "))
r = float(input("Enter your APR : "))
n = float(input("Enter the number of years : ") )
# Output
A = P*(1 +(r/100))**n
print("The future values is:", A)
And the result obtained is:
Enter the present value : 1000
Enter your APR : 0.95
Enter the number of years : 5
The future values is: 1048.4111145526908
