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

solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.

54.864 meters

step-by-step explanation: