Answer:
<h2>698.3Kpa</h2>
Explanation:
Step one:
given data
V1=0.25m^3
T1=290k
P1=100kPa
V2=0.5m^2
T2=405k
P2=? final pressure
Step two:
The combined gas equation is given as
P1V1/T1=P2V2/T2
Substituting we have
(100*0.25)/290=P2*0.05/405
25/290=0.5P2/405
0.086=0.05P2/405
cross multiply
0.086*405=0.05P2
34.9=0.05P2
divide both sides by 0.05
P2=34.9/0.05
P2=698.3Kpa
<u>Therefore the new pressure is 698.3Kpa when the gas is compressed</u>
Answer:
Explanation:
From the information given:

The total load is distributed across both the rod and tube:

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.






Replace
into equation (1)

Finally, to determine the normal stress in aluminum rod:


Thus, the normal stress = 23.523 MPa in compression.
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