The barometric pressure p at an altitude of h miles above sea level satisfies the differential equation (dp/dh) = −0.2p . If the pressure at sea level is 29.92 inches of mercury, find the barometric preassure at 17,000 ft (A) 56.97 in (B) 15.71 in (C) 7.86 in (D) 1 in


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Question Number 127119 by liberty last updated on 27/Dec/20

$T h e b a r o m e t r i c p r e s s u r e p a t a n a l t i t u d e$ $o f h m i l e s a b o v e s e a l e v e l s a t i s f i e s t h e$ $d i f f e r e n t i a l e q u a t i o n \frac{d p}{d h} = - 0.2 p .$ $I f t h e p r e s s u r e a t s e a l e v e l i s 29.92 i n c h e s$ $o f m e r c u r y, f i n d t h e b a r o m e t r i c$ $p r e a s s u r e a t 17, 000 f t$ $(A) 56.97 i n (B) 15.71 i n$ $(C) 7.86 i n (D) 1 i n$
	Answered by benjo_mathlover last updated on 27/Dec/20

	$\Leftrightarrow \frac{d p}{p} = - 0.2 d h$ $\Rightarrow \ln p = - 0.2 h + c$ $\Rightarrow p = λ e^{- 0.2 h}$ $\Rightarrow p = 29.92 e^{- 0.2 h}$ $h = 17, 000 f t = 3.22 m i l e s$ $\Rightarrow p = 29.92 \times e^{- 0.2 \times 3.22} = 15.71 i n$
	Commented by liberty last updated on 27/Dec/20

	$n o t h i n g c o r r e c t a n s w e r ?$
	Commented by mr W last updated on 27/Dec/20

	$p = 29.92 e^{- 0.2 h}$ $p_{h = 17000 f t} = 29.22 e^{- 0.2 \times \frac{17000}{5280}} = 15.71 i n$ $\Rightarrow a n s w e r B$
	Commented by mr W last updated on 27/Dec/20

	$a t s e a l e v e l : h = 0, n o t h = 1 m i l e!$
	Commented by benjo_mathlover last updated on 27/Dec/20

	$o o h . . t h a n k y o u s i r$
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