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Unformatted text preview: 50. (a) The speed v of the ﬂuid ﬂowing out of the hole satisﬁes 1 ρv 2 = ρgh or v =
2
ρ2 v2 A2 , which leads to
ρ1 2ghA1 = ρ2 2ghA2 =⇒ √
2gh. Thus, ρ1 v1 A1 = ρ1
A2
=
=2.
ρ2
A1 (b) The ratio of volume ﬂow is
R1
v1 A1
A1
1
=
=
=.
R2
v2 A2
A2
2
(c) Letting R1 /R2 = 1, we obtain v1 /v2 = A2 /A1 = 2 = h1 /h2 . Thus h2 = h1 /4. ...
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 Fall '08
 SPRUNGER
 Physics

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