The field identification test results for a soil are given below. Wh.pdf

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The field identification test results for a soil are given below. What is the Unified Soil Classification System (USCS) for this soil? More than 90% by wieght larger than No. 200 sieve More than half of the course fraction is smaller than 1/4 inch There is a wide range of grain sizes, with all intermediate sizes present Solution If more than 50% of soil is retained on the Seive No.200, the soil is designated as coarse grained, otherwise fine grained. In this case 90% soil is retained on No 200 seive. Hence the soil is coarse grained Coarse grained soils can be further calssified in to gravel and sand according to the particle size. If 50% or more of coarse fraction is retained on No.4 (4.75mm) seive,soil is designated as gravel. If more than 50% pass through the seive No.4, the soil is sand. In this case more than 50% coarse fraction pass through 1/4 inch (6.35 mm). As no other datas are given, we may have to assume that more than 50% coarse fraction pass through 4.75 mm also. Hence the soil can be classified as sand. The soil is containing wide range of grain size, with all intermediate sizes. Hence the soil is well graded. (No details about the % of fines is given in the question) The soil according to USCS classification an be designated as WELL GRADED SAND (SW)..

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$The field identification test results for a soil are given below. What is the Unified Soil Classification System (USCS) for this soil? More than 90% by wieght larger than No. 200 sieve More than half of the course fraction is smaller than 1/4 inch There is a wide range of grain sizes, with all intermediate sizes present Solution If more than 50% of soil is retained on the Seive No.200, the soil is designated as coarse grained, otherwise fine grained. In this case 90% soil is retained on No 200 seive. Hence the soil is coarse grained Coarse grained soils can be further calssified in to gravel and sand according to the particle size. If 50% or more of coarse fraction is retained on No.4 (4.75mm) seive,soil is designated as gravel. If more than 50% pass through the seive No.4, the soil is sand. In this case more than 50% coarse fraction pass through 1/4 inch (6.35 mm). As no other datas are given, we may have to assume that more than 50% coarse fraction pass through 4.75 mm also. Hence the soil can be classified as sand. The soil is containing wide range of grain size, with all intermediate sizes. Hence the soil is well graded. (No details about the % of fines is given in the question) The soil according to USCS classification an be designated as WELL GRADED SAND (SW).$

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The field of quotients of any fieldDis isomorphic toD.Why? Solution Let K be the field of quotients of D. Define the following function f: D --> K such that f(x)=x First this is clearly a ring homomorphism: f(x+y)=x+y=f(x)+f(y) f(xy)=xy=f(x)f(y) f(1)=1 It is also easily seen to be one to one: if f(x)=f(y) then x=y So far all this is saying is that D is contained in its field of quotients. But this is true for any integral domain. Next we show f is onto which will say that F and K are actually the same field. In other words the field of quotients of a field is just the field itself. Let x be in K. Now by definition x=a/b for some elements a and b in D such that b is nonzero. But if b is nonzero then since D is a field, b has a multiplicative inverse 1/b in D. Then a*(1/b)=a/b is in D and f(a/b)=a/b=x. So f is onto. In short, the field of quotients of an integral domain is the smallest field in which the integral domain can be embedded. So if D is a field then it\'s field of quotients must be at least as big as D (since D has to be embedded into it). But D is a field so it is thus the smallest field in which D can be embedded. In other words, the field of quotients of a field F is just F. Now of course you could say that they are not exactly the same field since the field of quotients is actually comprised of equivalence classes of elements a/b where b is nonzero. But if D is a field and a,b are elements of F with b nonzero then what is the equivalence class of a/b? Well, suppose c/d is in the same equivalence class, then ad=bc. But b is nonzero so since we are in a field we can multiply by its inverse to obtain. (ad)/b = c But d is also nonzero so multiplying by its inverse gives a/b=c/d (now we are talking about elements of D and not equivalence classes) So the equivalence class of a/b is really just a/b. So D and its field of quotients really are the same field..

The field of quotients of any fieldDis isomorphic toD.WhySoluti.pdf

The field identification test results for a soil are given below. Wh.pdf

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The field identification test results for a soil are given below. Wh.pdf