Answer:
a.)
US Sieve no. % finer (C₅ )
4 100
10 95.61
20 82.98
40 61.50
60 42.08
100 20.19
200 6.3
Pan 0
b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4
c.) Cu = 3.33
d.) Cc = 1
Explanation:
As given ,
US Sieve no. Mass of soil retained (C₂ )
4 0
10 18.5
20 53.2
40 90.5
60 81.8
100 92.2
200 58.5
Pan 26.5
Now,
Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g
⇒ w = 421.2 g
As we know that ,
% Retained = C₃ = C₂×
∴ we get
US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)
4 0 0
10 4.39 4.39
20 12.63 17.02
40 21.48 38.50
60 19.42 57.92
100 21.89 79.81
200 13.89 93.70
Pan 6.30 100
Now,
% finer = C₅ = 100 - C₄
∴ we get
US Sieve no. Cummulative % retained (C₄) % finer (C₅ )
4 0 100
10 4.39 95.61
20 17.02 82.98
40 38.50 61.50
60 57.92 42.08
100 79.81 20.19
200 93.70 6.3
Pan 100 0
The grain-size distribution is :
b.)
From the diagram , we can see that
D10 = 0.12
D30 = 0.22
D60 = 0.12
c.)
Uniformity Coefficient = Cu = 
⇒ Cu = 
d.)
Coefficient of Graduation = Cc = 
⇒ Cc = 
= 1
Answer:
Explanation:
Civil engineers have become experts in creating sustainable and environmentally friendly buildings and systems. Multiplied over many communities, the energy and emissions savings can make a real difference in the environment. Other life-improving functions can also make communities better places to live.Jul 19, 2017
Answer:
Yes, this because the specific heat does not violate the third law of thermodynamics.
Explanation:
The third law of thermodynamic is usually used for a closed system to relate the thermodynamic properties of the system at equilibrium conditions. For this law, a body that exists at a temperature of 0 K will cease to move or stop moving. It can be inferred that heat capacity at (T = 0 K) is equivalent to 0. Therefore, the equation is valid for the given temperature range.
The design speed was used for the freeway exit ramp is 11 mph.
<h3>Design speed used in the exit ramp</h3>
The design speed used in the exit ramp is calculated as follows;
f = v²/15R - 0.01e
where;
v = ωr
v = (θ/t) r
θ = 90⁰ = 1.57 rad
v = (1.57 x 19.4)/2.5 s
v = 12.18 ft/s = 8.3 mph
<h3>Design speed</h3>
f = v²/15R - 0.01e
let the maximum superelevation, e = 1%
f = (8.3)²/(15 x 19.4) - 0.01
f = 0.22
0.22 is less than value of f which is 0.4
<h3>next iteration, try 10 mph</h3>
f = (10)²/(15 x 19.4) - 0.01
f = 0.33
0.33 is less than 0.4
<h3>next iteration, try 11 mph</h3>
f = (11)²/(15 x 19.4) - 0.01
f = 0.4
Thus, the design speed was used for the freeway exit ramp is 11 mph.
Learn more about design speed here: brainly.com/question/22279858
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