A loose, uncompacted sand fill of 6 ft depth has a relative density of 40%. Laboratory tests indicated that the minimum and maximum void ratios of the sand are 0.46 and 0.9 respectively. The specific gravity of solids of the sand is 2.65.a) What is the dry unit weight of the sand

1). A loose, uncompacted sand fill 6 ft in depth has a relative density of 40%.

Laboratory tests indicated that the minimum and maximum void ratios of the sand are 0.46 and 0.90, respectively. The specific gravity of solids of the sand is 2.65.

a). What is the dry unit weight of the sand?.

b). If the sand is compacted to a relative density of 75%, what is the decrease in thickness of the 6-ft fill?

The answers to the question are

a. 15.37 kN/m³

b. 5.457 ft

Step-by-step explanation:

Relative density =

Where = maximum void ratio

= minimum void ratio

e = void ratio property

0.4 = (0.9 - e)/(0.9 - 0.46)

e =  0.9 - 0.4×0.44

= 0.724

Dry unit weight is given by

= 10×(2.65)/(1+0.724)

15.37 kN/m³

b. Relative density =

0.75 = (0.9 - e')/(0.9- 0.46)

or e' = 0.57

Hence the dry unit weight is given by

New dry unit weight wt =

= 10×2.65÷(1+0.57) = 16.88 kN/³m

15.36/16.9 = V₂/V₁

 0.9095 = h₂/h₁

h₂ = 0.9095 ×6 = 5.457 ft

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft

answer: 10 turtles

step-by-step explanation: as turtles have four legs so 4.10=40 therefore birds are 4 which sue counted