What solid is generated when the semicircle is rotated about the line?
2 answers:
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Answer:
246.3%
the complete question is found in the attached document
Step-by-step explanation:
1st step:
using 1936 data,
w1= 356% = 3.56(356/100) , H= 79 feet
specific gravity = 2.65
Sₓ= 100%= 1
initial void ratio(e₀)= (w1 x specific gravity)/Sₓ
=3.56 x 2.65/1 = 9.434
2nd step
using 1996 data
ΔH= 22ft
ΔH/H = Δe/(1 + e₀)
22/79 = Δe/(1+9.434)
0.278=Δe/10.434
Δe= 0.278 x 10.434
Δe= 2.905
Δe= e₀ - eₓ
eₓ= e₀-Δe
eₓ= 9.434 - 2.905
eₓ= 6.529
3rd step
calculating water content in 1996
eₓ =6.529, specific gravity= 2.65, Sₓ= 100%
W2 X 2.65 = 1 x 6.529
w2 = 6.529/2.65 = 2.463 = 246.3%
