Physics Solution Manual for 1100 and 2101

11 point 1 is where the water leaves 1 1 2 1 2 the

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Unformatted text preview: If we take up as the positive direction, and we apply Newton's second law, we find that T − mg = 0 , or m = T / g . Therefore, the density can be written as 3m 3T ρ= = 3 4π r 4 πr 3 g This expression can now be solved for r. SOLUTION We find that r3 = 3T 4 πρg Taking the cube root of both sides of this result and using a value of ρ = 8470 kg/m3 for the density of brass (see Table 11.1), we find that r= 3 3T 4 πρg = 3 3(120 N) = 4 π ( 8470 kg / m 3 )(9.80 m / s 2 ) 7.0 × 10 –2 m 92. REASONING In this problem, we are treating air as a viscous fluid. According to Poiseuille's law, a fluid with viscosity η flowing through a pipe of radius R and length L has a volume flow rate Q given by Equation 11.14: Q = πR 4 ( P2 – P1 ) / ( 8ηL ) . This expression can be solved for the quantity P2 – P1 , the difference in pressure between the ends of the air duct. First, however, we must determine the volume flow rate Q of the air. 618 FLUIDS SOLUTION Since the fan forces air through the duct such that 280 m 3 of air is replenished every ten minutes, the volume flow rate is Q= F280 m IF min I = 0.467 m G G min J 1.0 s J H H K 60 K 10.0 3 3 /s The difference in pressure between the ends of the air duct is, according to Poiseuille's law, P2 – P1 = 8η LQ 8(1.8 × 10 –5 Pa ⋅ s)(5.5 m)(0.467 m 3 / s) = = 4.4 Pa π R4 π (7.2 × 10 –2 m) 4 The drawing at the 93. REASONING Evacuated so that right shows the situation. As discussed pressure is zero, in Conceptual Example 6, the job of P1 = 0 the pump is to draw air out of the pipe that dips down into the water. The atmospheric pressure in the well then pushes the water upward into the pipe. In the drawing, the best the pump can do is to remove all of the air, in which h case, the pressure P1 at the top of the water in the pipe is zero. The pressure P2 at the bottom of the pipe at point A B 5 is the same as that at the point B, P =1.013×10 Pa 2 namely, it is equal to atmospheric A pressure ( 1.013 × 10 5 Pa ), because the two points are at the same elevation, and point B is open to the atmosphere. Equation 11.4, P2 = P1 + ρgh can be applied to obtain the maximum depth h of the well. } SOLUTION Setting P1 = 0 Pa, and solving Equation 11.4 for h, we have h= P1 ρg = 1.013 × 10 5 Pa = 10.3 m (1.000 × 10 3 kg / m 3 ) (9.80 m / s 2 ) 94. REASONING The paperweight weighs less in water than in air, because of the buoyant force FB of the water. The buoyant force points upward, while the weight points downward, leading to an effective weight in water of WIn water = W – FB. There is also a buoyant force when the paperweight is weighed in air, but it is negligibly small. Thus, from the given weights, we can obtain the buoyant force, which is the weight of the displaced water, according to Archimedes’ principle. From the weight of the displaced water and the density Chapter 11 Problems 619 of water, we can obtain the volume of the water, which is also the volume of the completely immersed paperweight. SOLUTION We have WIn water = W − FB FB = W − WIn water or According to Archimedes’ principle, the buoyant force is the weight of the displaced water, which is mg, where m is the mass of the displaced water. Using Equation 11.1, we can write the mass as the density times the volume or m = ρV. Thus, for the buoyant force, we have FB = W − WIn water = ρVg Solving for the volume and using ρ = 1.00 × 103 kg/m3 for the density of water (see Table 11.1), we find V= W − WIn water ρg = 1.00 c × 10 6. 9 N − 4 . 3 N 3 kg / m 3 = 2 .7 × 10 −4 m 3 h9.80 m / s h c 2 95. SSM REASONING AND SOLUTION a. The volume flow rate is given by Equation 11.10. Assuming that the line has a circular cross section, A = π r 2 , we have Q = Av = (π r 2 )v = π (0.0065 m) 2 (1.2 m/s) = 1.6 × 10 –4 m3 / s b. The volume flow rate calculated in part (a) above is the flow rate for all twelve holes. Therefore, the volume flow rate through one of the twelve holes is Qhole = Q 2 = (π rhole )vhole 12 Solving for vhole we have vhole = Q 12π 2 rhole = 1.6 × 10 –4 m3 /s 12π (4.6 × 10 −4 m) 2 = 2.0 × 101 m/s 620 FLUIDS 96. REASONING Pressure is defined in Equation 11.3 as the magnitude of the force acting perpendicular to a surface divided by the area over which the force acts. The force acting perpendicular to the slope is due to the component of the skier’s weight W that is directed perpendicular to the slope. In the drawing at the right, this component is labeled Wperpendicular. Note that the fact that the skier is moving is of no importance. SOLUTION 1 W 2 perpendicular Wperpendicular W 35° 35° We assume that each ski bears the same amount of force, namely (see the drawing). According to Equation 11.3, the pressure that each ski applies to the snow is P= 1 W 2 perpendicular A where A is the area of each ski in contact with the snow. From the drawing, we see that Wperpendicular = W cos35° , so that the pressure exerted by each ski on the snow is P= 1 W 2 perpendicular A ( ) ) 2 W cos 35° ( 58 kg ) 9.80 m/s cos 35° = = = 1.8 ×103 Pa 2 2A 2 0.13 m ( where we have used the fact that...
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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.

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