A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and loosest state of this soil are 0.51 and 0.87. Find the relative density and degree of saturation.

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, = 0.51

Void ratio in the loosest state, = 0.87

Now,

Dry density,

= 17.14 kN/m³

Also,

here, G = Specific gravity = 2.7 for sand

or

e = 0.545

Relative density =

=

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

answer:

okay

explanation:

answer:

the average distance would be 20ly

Explanation:

100,000ly times 5,000ly=20ly