Exam (elaborations)
Solution Manual For Introduction to Flight, 8th Edition by Anderson
- Course
- Aviation
- Institution
- University Of North Dakota
Chapter 2 2.1 ρ= p RT/ = (1.2)(1.01 10 )/(287)(300)× 5 ρ=1.41kg/m2 v=1/ρ=1/1.41= 0.71m /kg3 3 kT= 3 (1.38× 10−23)(500)=1.035× 10−20J 2.2 Mean kinetic energy of each atom = 2 2 One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has (6.02 ×1026)=1.505 �...
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