Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.
<u>the liquid limit</u>
LL=-0.8078x*49.503
LL=-0.8078(25)+49.503
LL=29.308
<u>plasticity index of the soil</u>
PL=217.2%+17.8%/2
PL=17.5
PI=LL-PL
PI=29.308-17.5
PL=11.808
Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great rest of Black History Month! :-)
- Cutiepatutie ☺❀❤
Answer:
modulus of elasticity for the nonporous material is 340.74 GPa
Explanation:
given data
porosity = 303 GPa
modulus of elasticity = 6.0
solution
we get here modulus of elasticity for the nonporous material Eo that is
E = Eo (1 - 1.9P + 0.9P²) ...............1
put here value and we get Eo
303 = Eo ( 1 - 1.9(0.06) + 0.9(0.06)² )
solve it we get
Eo = 340.74 GPa
Answer:
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