is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:

First has a natural water content of 25% =
= 0.25
Shrinkage limit, 

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
![V \propto[1+e]](https://tex.z-dn.net/?f=V%20%5Cpropto%5B1%2Be%5D)
------> eq 1

The above equation is at
,

Applying the given values, we get

Shrinkage limit is lowest water content

Applying the given values, we get

Applying the found values in eq 1, we get


Answer: 150m
Explanation:
The following can be depicted from the question:
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Therefore, the plinth area of the building will be:
= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
= 10 × 15
= 150m
You need to explain it more simple as everyone is clueless
Answer:
Explanation:
Run the code given in text file following instructions.
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by


78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa