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62. (a) The volume rate of ﬂow is related to speed by
R
=
vA
.Thu
s
,
v
1
=
R
1
πr
2
stream
=
7
.
9cm
3
/
s
π
(0
.
13cm)
2
=148
.
8cm
/
s=1
.
5m
/
s
.
(b) The depth
d
of spreading water becomes smaller as
r
(the distance from the impact point) increases
due to the equation of continuity (and the assumption that the water speed remains equal to
v
1
in
this region). The water that has reached radius
r
(with perimeter 2
πr
) is crossing an area of 2
πrd
.
Thus, the equation of continuity gives
R
1
=
v
1
2
πrd
=
⇒
d
=
R
2
πrv
1
.
(c) As noted above,
d
is a decreasing function of
r
.
(d) At
r
=
r
J
we apply the formula from part (b):
d
J
=
R
1
2
πr
J
v
1
=
7
.
9cm
3
/
s
2
π
(2
.
0 cm)(148
.
8cm
/
s)
=0
.
0042 cm
.
(e) We are told “the depth just after the jump is 2
.
0 mm” which means
d
2
=0
.
20 cm, and we are asked
to Fnd
v
2
. We use the equation of continuity:
R
1
=
R
2
=
⇒
2
πr
J
v
1
d
J
=2
πr
0
J
v
2
d
2
where
r
0
J
is some very small amount greater than
r
J
(and for calculation purposes is taken to be
thesamenumer
ica
lva
lue
,2
.
0 cm). This yields
v
2
=
v
1
µ
d
1
d
2
¶
= (148
.
8cm
/
s)
µ
0
.
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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