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Solution Manual For Introduction to Flight, 8th Edition by Anderson
- Exam (elaborations) • 9 pages • 2021
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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 ×1026 atoms. 
Totalinternalenergy = (energyperatom)(numberof atoms) 
 	= (1.035´10-20)(1.505´1026) = 1.558 ´106J 
	RTp	(1716)( +	0.00237 slug3 
2.3 ρ=	=	= 
	59)	ft 
Volumeof theroom = (20)(15)(8...
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