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:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
T₂ =93.77 °C
Explanation:
Initial temperature ,T₁ =27°C= 273 +27 = 300 K
We know that
Absolute pressure = Gauge pressure + Atmospheric pressure
Initial pressure ,P₁ = 300+1=301 kPa
Final pressure ,P₂= 367+1 = 368 kPa
Lets take temperature=T₂
We know that ,If the volume of the gas is constant ,then we can say that


Now by putting the values in the above equation we get

The temperature in °C
T₂ = 366.77 - 273 °C
T₂ =93.77 °C
Answer:
In general a cache memory is useful because the speed of the processor is higher than the speed of the ram . so reducing the number of memory is desirable to increase performance .
Explanation:
.
.
#hope it helps you ..
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