SOLUTION According to Equation 11.3, pressure is defined as P= F/ A; therefore, the magnitude of the force on the lid due to the air pressure is

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1 PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, 69.. ssm An airtight box has a remoable lid of area.3 0 m and negligible weight. The box is taken up a mountain where the air pressure outside the box is Pa. The inside box is completely eacuated. What is the magnitude force required to pull the lid off the box? REASONING Since the inside box is completely eacuated; there is no air to exert an upward force on the lid from the inside. Furthermore, since the weight of the lid is negligible, there is only one force that acts on the lid; the downward force caused by the air pressure on the outside lid. In order to pull the lid off the box, one must supply a force that is at least equal in magnitude and opposite in direction to the force exerted on the lid by the outside air. SOLUTION According to Equation.3, pressure is defined as P F/ A; therefore, the magnitude force on the lid due to the air pressure is 5 3 F ( N/m )(.3 0 m ). 0 N 7. ssm A water tower is a familiar sight in many towns. The purpose of such a tower is to proide storage capacity and to proide sufficient pressure in the pipes that delier the water to customers. The drawing shows a spherical reseroir that contains kg of water when full. The reseroir is ented to the atmosphere at the top. For a full reseroir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter deliery pipes. REASONING The pressure P at a distance h beneath the water surface at the ented top water tower is P Patm gh (Equation.4). We note that the alue for h in this expression is different for the two houses and must take into account the diameter spherical reseroir in each case. SOLUTION a. The pressure at the leel of house A is gien by Equation.4 as PA Patm gha. Now the height h A consists 5.0 m plus the diameter d of the tank. We first calculate the radius tank, from which we can infer d. Since 3 the tank is spherical, its full mass is gien by M ρv ρ[(4 / 3) π r ]. Therefore, r 3 /3 5 3M 3M 3(5.5 0 kg) or r 5.00 m 4πρ 4πρ π ( kg/m ) /3

2 PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, 69. Therefore, the diameter tank is 0.0 m, and the height h A is gien by h 0.0 m m 5.0 m A According to Equation.4, the gauge pressure in house A is, therefore, PA Patm ρ gha ( kg/m )(9.80 m/s )(5.0 m).45 0 Pa b. The pressure at house B is PB Patm ghb, where h 5.0 m m 7.30 m 7.7 m B According to Equation.4, the gauge pressure in house B is PB Patm ρ ghb ( kg/m )(9.80 m/s )(7.7 m).73 0 Pa 40. The density of ice is 97 kg/m 3, and the density of seawater is 05 kg/m 3. A swimming polar bear climbs onto a piece of floating ice that has a olume of 5. m 3. What is the weight heaiest bear that the ice can support without sinking completely beneath the water? REASONING The ice with the bear on it is floating, so that the upward-acting buoyant force balances the downward-acting weight W ice ice and weight W bear bear. The magnitude F B buoyant force is the weight W HO displaced water, according to Archimedes principle. Thus, we hae F W W + W, B H O ice bear the expression with which we will obtain W bear. We can express each weights W HO and W ice as mass times the magnitude acceleration due to graity (Equation 4.5) and then relate the mass to the density and the displaced olume by using Equation.. SOLUTION Since the ice with the bear on it is floating, the upward-acting buoyant force F B balances the downward-acting weight W ice ice and the weight W bear bear. The buoyant force has a magnitude that equals the weight W HO displaced water, as stated by Archimedes principle. Thus, we hae F W W + W or W W W () B H O ice bear bear H O ice In Equation (), we can use Equation 4.5 to express the weights W HOand W ice as mass m times the magnitude g acceleration due to graity. Then, the each mass

3 PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, 69. can be expressed as m ρv (Equation.). With these substitutions, Equation () becomes ( ρ ) W m g m g ( ρ V ) g V g () bear H O ice H O H O ice ice When the heaiest possible bear is on the ice, the ice is just below the water surface and displaces a olume of water that is V V. Substituting this result into H O Equation (), we find that W ( ρ V ) g ρ V g ( ρ ρ ) V g ( ) bear H O ice ice ice H O ice ice ( )( )( ) 05 kg/m 97 kg/m 5. m 9.80 m/s 5500 N 44. A paperweight, when weighed in air, has a weight of W 6.9 N. When completely immersed in water, howeer, it has a weight of W in water 4.3 N. Find the olume paperweight. REASONING The paperweight weighs less in water than in air, because buoyant force F B water. The buoyant force points upward, while the weight points downward, leading to an effectie weight in water of W In water W F B. There is also a buoyant force when the paperweight is weighed in air, but it is negligibly small. Thus, from the gien weights, we can obtain the buoyant force, which is the weight displaced water, according to Archimedes principle. From the weight displaced water and the density of water, we can obtain the olume water, which is also the olume completely immersed paperweight. SOLUTION We hae ice W W F or F W W In water B B In water According to Archimedes principle, the buoyant force is the weight displaced water, which is mg, where m is the mass displaced water. Using Equation., we can write the mass as the density times the olume or m ρv. Thus, for the buoyant force, we hae F W W ρ Vg B In water Soling for the olume and using ρ kg/m 3 for the density of water (see Table.), we find

4 PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, Water flowing out of a horizontal pipe emerges through a nozzle. The radius pipe is.9 cm, and the radius nozzle is 0.48 cm. The speed water in the pipe is 0.6 m/s. Treat the water as an ideal fluid, and determine the absolute pressure water in the pipe. REASONING The absolute pressure in the pipe must be greater than atmospheric pressure. Our solution proceeds in two steps. We will begin with Bernoulli s equation. Then we will incorporate the equation of continuity. SOLUTION According to Bernoulli s equation, as gien by Equation., we hae The pipe and nozzle are horizontal, so that y y and Bernoulli s equation simplifies to ( ) or P P P P where P is the absolute pressure water in the pipe. We hae alues for the pressure P (atmospheric pressure) at the nozzle opening and the speed in the pipe. Howeer, to sole this expression we also need a alue for the speed at the nozzle opening. We obtain this alue by using the equation of continuity, as gien by Equation.9: r or π π or r A A r r Here, we hae used that fact that the area of a circle is A π r. Substituting this result for into Bernoulli s equation, we find that P P ρ 4 r + 4 r Taking the density of water to be ρ kg/m 3 (see Table.), we find that the absolute pressure water in the pipe is P P 4 r 4 r ( ) ( ) ( ).9 0 m 3 ( m).0 0 Pa.00 0 kg/m 0.6 m/s.48 0 Pa

5 PHYS 3 Fall 07 Week Recitation: Chapter :, 7, 40, 44, 64, 69. *69. ssm A Venturi meter is a deice that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed υ through a horizontal section of pipe whose cross-sectional area is A m. The gas has a density of ρ.30 kg/m 3. The Venturi meter has a cross-sectional area of A m and has been substituted for a section larger pipe. The pressure difference between the two sections is P P 0 Pa. Find (a) the speed υ gas in the larger, original pipe and (b) the olume flow rate Q gas. REASONING Since the pressure difference is known, Bernoulli s equation can be used to find the speed gas in the pipe. Bernoulli s equation also contains the unknown speed gas in the Venturi meter; therefore, we must first express in terms of. This can be done by using Equation.9, the equation of continuity. SOLUTION a. From the equation of continuity (Equation.9) it follows that. Therefore, m (. 40) m Substituting this expression into Bernoulli s equation (Equation.), we hae P (. 40 ) P Soling for, we obtain ( P P ) 40 ρ (. ) ( 0 Pa) 3 (.30 kg / m ) (.40) 4 m / s b. According to Equation.0, the olume flow rate is Q 3 A ( m )( 4 m / s) 0.98 m / s