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
Complete Question
The complete question is shown on the first uploaded image
Answer:
The of the lander would be 75.66 pounds
Explanation:
From the question we are given that the
Weight of the Philae on Earth is = 220 pound-mass
gravity of Mars is = 3.71 m/
Weight of the Philae on Mars is = ?
Now Mass is quantity of matter in an object so it is always constant for a particular object
Generally Weight = Mass × Gravity :
Hence
Weight of the Philae on Mars is =
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d =
a)(100)
a=5.28 Å
Distance = =5.28 Å
b)(110)
Distance = = 3.73 Å
c)(111)
Distance= = 3.048 Å
Answer:
t = 6179.1 s = 102.9 min = 1.7 h
Explanation:
The energy provided by the resistance heater must be equal to the energy required to boil the water:
E = ΔQ
ηPt = mH
where.
η = efficiency = 84.5 % = 0.845
P = Power = 2.61 KW = 2610 W
t = time = ?
m = mass of water = 6.03 kg
H = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)
<u>t = 6179.1 s = 102.9 min = 1.7 h</u>
Answer:
The level of service of of compound grade freeway is LOSB.
Explanation:
Find the provided attachments for explanation
