Has luster
2 answers:
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Answer:
a)ΔS₁ = - 9.9 J/K
ΔS₂ = 69 J/K
b)The entropy change for the rod = 0 J/K
c)ΔS = 59.1 J/K
Explanation:
Given that
T₁ = 699 K
T₂= 101 K
Q= 6970 J
Change in entropy given as
For 699 K:
ΔS₁ = - 9.9 J/K ( Negative because heat is leaving from the system)
For 101 K;
ΔS₂ = 69 J/K
The entropy change for the rod = 0 J/K
Entropy change for the system
ΔS = ΔS₂ + ΔS₁
ΔS = 69 -9.9 J/K
ΔS = 59.1 J/K
Answer:
1)Wing ratio=2.935
2)Lift force=l=222264 N
3)Drag coefficient=Cd=0.0237
4)Drag force generated by aircraft=Cd=2388.189 N
Explanation:
Solution explanation is given in attachments.
