Unformatted text preview: at the shallow end, P2 is
3 3 the pressure at the deep end, and ρ is the density of water (1.00 × 10 kg/m , see Table
11.1). We seek a value for the pressure at the deep end minus the pressure at the shallow
end.
SOLUTION Using Equation 11.4, we find
PDeep = PShallow + ρgh or PDeep − PShallow = ρgh The drawing at the right shows that a value for h can
be obtained from the 15m length of the pool by using
the tangent of the 11° angle: tan11° = h
15 m or 15 m
11° h = (15 m ) tan11° PDeep − P
Shallow = ρ g (15 m ) tan11° ( )( ) = 1.00 × 103 kg/m3 9.80 m/s 2 (15 m ) tan11° = 2.9 × 104 Pa h 576 FLUIDS 24. REASONING AND SOLUTION The gauge pressure of the solution at the location of the
vein is
3 2 3 P = ρgh = (1030 kg/m )(9.80 m/s )(0.610 m) = 6.16 × 10 Pa
Now
1.013 × 105 Pa = 760 mm Hg so 1 Pa = 7.50 × 10–3 mm Hg Then 7.50 × 10−3 mm Hg P = 6.16 × 103 Pa = 46.2 mm Hg 1 Pa ( ) 25. REASONING Since the diver uses a snorkel, the pressure in her lungs is atmospheric
pressure. If she is swimming at a depth h below the surface, the pressure outside her lungs is
atmospheric pressure plus that due to the water. The water pressure P2 at the depth h is related to the pressure P1 at the surface by Equation 11.4, P2 = P1 + ρ gh, where ρ is the
density of the fluid, g is the magnitude of the acceleration due to gravity, and h is the depth.
This relation can be used directly to find the depth. SOLUTION We are given that the maximum difference in pressure between the outside
and inside of the lungs is onetwentieth of an atmosphere, or P2 − P1 = 1
20 (1.01×105 Pa ) . Solving Equation 11.4 for the depth h gives ( ) 1 1.01× 105 Pa
P2 − P
20
1=
h=
= 0.50 m
ρg
1025 kg/m3 9.80 m/s 2 ( )( ) 26. REASONING The pressure difference across the diver’s eardrum is Pext − Pint , where Pext
is the external pressure and Pint is the internal pressure. The diver’s eardrum ruptures when
Pext − Pint = 35 kPa . The exterior pressure Pext on the diver’s eardrum increases with the
depth h of the dive according to Pext = Patm + ρ gh (Equation 11.4), where Patm is the
atmospheric pressure at the surface of the water, ρ is the density of seawater, and g is the
magnitude of the acceleration due to gravity. Solving Equation 11.4 for the depth h yields
h= Pext − Patm ρg (1) SOLUTION When the diver is at the surface, the internal pressure is equalized with the
external pressure: Pint = Patm . If the diver descends slowly from the surface, and takes steps
to equalize the exterior and interior pressures, there is no pressure difference across the
eardrum, regardless of depth: Pext = Pint . But if the diver sinks rapidly to a depth h without Chapter 11 Problems 577 equalizing the pressures, then the internal pressure is still atmospheric pressure. In this case,
Patm in Equation (1) equals Pint, so the eardrum is at risk for rupture at the depth h given by
h= 35 kPa = 35 × 103 Pa (1025 kg/m3 )( 9.80 m/s2 ) (1025 kg/m3 )( 9.80 m/s2 ) = 3.5 m 27. SSM REASONING AND SOLUTION
a. The pressure at the level of house A is given by Equation 11.4 as P = Patm + ρ gh . Now
the height h consists of the 15.0 m and the diameter d of the tank. We first calculate the
radius of the tank, from which we can infer d. Since the tank is spherical, its full mass is
given by M = ρV = ρ[(4 / 3)π r 3 ] . Therefore, 3M
r=
4π ρ
3 1/ 3 or 3M r = 4π ρ 1/3 3(5.25 × 105 kg) =
3
3 4π (1.000 ×10 kg/m ) = 5.00 m Therefore, the diameter of the tank is 10.0 m, and the height h is given by
h = 10.0 m + 15.0 m = 25.0 m According to Equation 11.4, the gauge pressure in house A is, therefore, P − Patm = ρ gh = (1.000 × 103 kg/m3 )(9.80 m/s 2 )(25.0 m) = 2.45 × 105 Pa
b. The pressure at house B is P = Patm + ρ gh , where
h = 15.0 m + 10.0 m − 7.30 m = 17.7 m According to Equation 11.4, the gauge pressure in house B is P − Patm = ρ gh = (1.000 × 103 kg/m3 )(9.80 m/s 2 )(17.7 m) = 1.73 × 105 Pa 28. REASONING In each case the pressure at point A in Figure 11.11 is atmospheric pressure
and the pressure in the tube above the top of the liquid column is P. In Equation 11.4 (P2 =
P1 + ρgh), this means that P2 = Patmosphere and P1 = P. With this identification of the
pressures and a value of 13 600 kg/m3 for the density of mercury (see Table 11.1),
Equation 11.4 provides a solution to the problem. 578 FLUIDS SOLUTION Rearranging Equation 11.4, we have P2 – P1 = ρgh. Applying this expression
to each setup gives
P2 − P1 Mercury = PAtmosphere − P = ρgh Mercury ch
P
c −P h
2 1 Unknown = PUnknown bg
− P = bgh g
ρ Unknown Since the left side of each of these equations is the same, we have ρ Mercury gh Mercury = ρ Unknown gh Unknown
ρ Unknown = ρ Mercury h
F
G
h
H Mercury Unknown F
I = c 600 kg / m h h
13
Gh
J
16
K
H
3 Mercury
Mercury I=
J
K 850 kg / m 3 29. SSM REASONING Let the length of the tube be denoted by L, and let the length of the
liquid be denoted by l . When the tube is whirled in a circle at a constant angular speed
about an axis through one...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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