Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Chapter 9 P53
Serway, Faughn and Vuille: College Physics 8th Edition , Thomson Brooks/Cole, Vol I(ISBN #) 978-049511374-3
THE PROBLEM STATEMENT
Ch 9 P53. A jet of water squirts out horizontally from a hole near
the bottom of the tank shown in Figure P9.53. If the hole
has a diameter of 3.50 mm, what is the height h of the
water level in the tank?
.
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Physics 210 Problems - My Solutions
Dr. Hulan E. Jack Jr.
Ch 9 P53. A jet of water squirts out horizontally from a hole near
the bottom of the tank shown in Figure P9.53. If the hole has a
diameter of 3.50 mm, what is the height h of the water level in the
tank?
Basic Solution (Including 1BRAINSTORMING-Definitions, concepts , rinciples and Discussion)
Given: H= 1.00 m, x = 0.600m .
Find: h and vh
Note the pressure top , Ptop ,at the hole Phole, and ouside, Poutside, are all the atmospheric pressure,
Patm. same value Patm
Ptop = Phole = Poutside = Patm .
Physical Principles:
1. The stream from hole to ground is in free fall.
Horizontal
vh = const. , hence x = vht .
so
t = x / vh
.
(1)
(2)
hole
Vertical
av = -g, so y = v0yt – ½ gt2 , with v0y =0.
(3)
Hence Eq. (3) gives
y = ½ gt2 = H
(4)
vh
H
So
t= sqrt(2H/g)
(5)
= sqrt(2*1.00m/9.8m/s2) = 0.452 s
x
Using Eq. (5) in Eq. (1) gives
(6)
vh = x/t = x/sqrt(2H/g)
= 0.600m/ sqrt(2H/g) = 0.600/0.452s = 1.33 m/s .
2. Bernoulli’s Equation in fluid.
Ptop /+ ½ vtop2 + gh = Phole/ + ½ vh2 + 0 and vtop = 0
(7)
gives ,
gh = ½ vh2 =
h =
vh2 /2g =
=
=
=
=
=
(8)
[x/{sqrt(2H/g}]2/(2g)
[x2/{(2H/g}]/(2g) =
[x2g/{(2H}]/(2g)
[x2g/{(2H*2g }]
x2/4H
0.6002 m2 /(4*1) = 0.3600/4 = 0.0900m = 9.00 cm
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