The position of a particle is given by s = 0.27t
1 answer:
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Answer: hello the complete question is attached below
answer:
A) Group symbol = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Explanation:
<u>A) Classifying the soil according to USCS system</u>
( using 2nd image attached below )
<em>description of sand</em> :
The soil is a coarse sand since ≤ 50% particles are retained on No 200 sieve, also
The soil is a sand given that more than 50% particles passed from No 4 sieve
The soil can be a clean sand given that fines ≤ 12%
The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time
Group symbol as per the 2nd image attached below = SW
B) Group name = well graded sand , fine to coarse sand
C) It is not a clean sand given that ≤ 50% particles are retained on No 200
Answer:
135 hour
Explanation:
It is given that a carburizing heat treatment of 15 hour will raise the carbon concentration by 0.35 wt% at a point of 2 mm from the surface.
We have to find the time necessary to achieve the same concentration at a 6 mm position.
we know that where x is distance and t is time .As the temperature is constant so D will be also constant
So
then we have given
and we have to find
putting all these value in equation
so