Physics Solution Manual for 1100 and 2101

Using equation 1 to substitute for p gives 580 fluids

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 15-m 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 one-twentieth 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...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online