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- New-jersey
- Rutgers University - New Brunswick/Piscataway
- Physics
- Physics 124
- Gilman
- phys124s10-hw08.pdf

Danny Y.

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HW 8 Solutions A Submerged Ball A ball of mass m b and volume V is lowered on a string into a fluid of density ρ f . Assume that the object would sink to the bottom if it were not supported by the string. What is the tension T in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Express your answer in terms of any or all of the given quantities and g, the magnitude of the acceleration due to gravity. ( ) bf T weight buoyancy m V gρ=− =− Pressure in the Ocean The pressure at 10 m below the surface of the ocean is about 2.00×10 5 Pa. a) Which of the following statements is true? The weight of a column of seawater 1 m 2 in cross section and 10 m high is about 2.00×10 5 N. The weight of a column of seawater 1 m 2 in cross section and 10 m high plus the weight of a column of air with the same cross section extending up to the top of the atmosphere is about 2.00×10 5 N. The weight of 1 m 3 of seawater at 10 m below the surface of the ocean is about 2.00×10 5 N. The density of seawater is about 2.00×10 times the density of air at sea level. 5 The pressure at a given level in a fluid is caused by the weight of the overlying fluid plus any external pressure applied to the fluid. In this case, the external pressure is the atmospheric pressure. b) Now consider the pressure at 20 m below the surface of the ocean. This pressure is twice that at a depth of 10 m. the same as that at a depth of 10 m. equal to that at a depth of 10 m plus the weight of a column of seawater 1 m 2 in cross section and 10 m high. equal to the weight of a column of seawater 1 m 2 in cross section and 20 m high. 1 Block Suspended in Water Conceptual Question A flask of water rests on a scale that reads 100 N. Then, a small block of unknown material is held completely submerged in the water. The block does not touch any part of the flask, and the person holding the block will not tell you whether the block is being pulled up (keeping it from falling further) or pushed down (keeping it from bobbing back up). a) What is the new reading on the scale? Regardless of whether the block is held up or pushed down, the water level in the flask goes up. This means that the pressure on the bottom of the flask increases. greater than 100 N 100 N less than 100 N It is impossible to determine. b) The experiment is repeated with the six different blocks listed below. In each case, the blocks are held completely submerged in the water. Mass (g) Volume (cm 3 ) A 100 50 B 100 200 C 200 50 D 50 100 E 200 100 F 400 50 Rank these blocks on the basis of the scale reading when the blocks are completely submerged. Rank from largest to smallest. To rank items as equivalent, overlap them. B, D and E, A and C and F c) If the blocks were released while submerged, which, if any, would sink to the bottom of the flask? A,C,E, and F - for these blocks, bwater mg Vgρ> Exercise 14.11 In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 kg/m 3 ) located a height h above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 Pa, what is the minimum value of h that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity of the fluid. 2 IDENTIFY: Apply 0 p pghρ=+ . SET UP: Gauge pressure is air p p− . EXECUTE: The pressure difference between the top and bottom of the tube must be at least 5980 Pa in order to force fluid into the vein: and 5980 Paρgh= 2 32 5980 Pa 5980 N/m 0.581 m (1050 kg/m )(9.80 m/s ) h gρ == =. EVALUATE: The bag of fluid is typically hung from a vertical pole to achieve this height above the patient’s arm. Exercise 14.17 There is a maximum depth at which a diver can breathe through a snorkel tube because as the depth increases, so does the pressure difference, tending to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1 m? Assume that the diver is in fresh water. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.) IDENTIFY: Apply 0 p pghρ=+ 1.00=× air pp−= . SET UP: For water, ρ . 33 10 kg/m EXECUTE: 33 2 4 (1.00 10 kg/m )(9.80 m/s )(6.1 m) 6.0 10 Pa.ρgh = × =× EVALUATE: The pressure difference increases linearly with depth. Exercise 14.36 Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.07 m 2 , and the magnitude of the fluid velocity is 3.5 m/s. a) What is the fluid speed at point in the pipe where the cross-sectional area is 0.105 m 2 ? IDENTIFY: . The volume flow rate is vA. 11 2 2 vA vA= EXECUTE: (a) 2 1 21 2 2 0.070 m (3.50 m/s) 2.33 m/s 0.105 m A vv A ⎛⎞ ⎛⎞ == = ⎜⎟ ⎜⎟ ⎝⎠⎝⎠ b) What is the fluid speed at point in the pipe where the cross-sectional area is 0.047 m 2 ? 3 (b) 2 1 21 2 2 0.070 m (3.50 m/s) 5.21 m/s 0.047 m A vv A ⎛⎞ ⎛⎞ == = ⎜⎟ ⎜⎟ ⎝⎠⎝⎠ c) Calculate the volume of water discharged from the open end of the pipe in 1 hour. SET UP: . 1.00 h 3600 s= (c) . 23 11 (3.50 m/s)(0.070 m )(3600 s) 882 mVvAt== = EVALUATE: The equation of continuity says the volume flow rate is the same at all points in the pipe. Exercise 14.40 A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. a) What is the speed of efflux? IDENTIFY: Toricelli’s theorem says the speed of efflux is 2vg= h, where h is the distance of the small hole below the surface of the water in the tank. The volume flow rate is vA. SET UP: , with . 2 /4ADπ= 3 6.00 10 mD − =× EXECUTE: (a) 2 2(9.80 m/s )(14.0 m) 16.6 m/sv == b) What is the volume discharged per unit time? (b) . A volume of is discharged each second. 32 43 (16.6 m/s) (6.00 10 m) / 4 4.69 10 m /svA π − =×=× − 43 4.69 10 m 0.469 L − ×= EVALUATE: We have assumed that the diameter of the hole is much less than the diameter of the tank. Exercise 14.42 At one point in a pipeline the water's speed is 3 m/s and the gauge pressure is 5×10 4 Pa. Find the gauge pressure at a second point in the line, 11 m lower than the first, if the pipe diameter at the second point is twice that at the first. IDENTIFY: Apply Bernoulli’s equation to the two points. SET UP: The continuity equation says . In Eq.(14.17) either absolute or gauge pressures can be used at both points. 11 2 2 v Α v Α= EXECUTE: Using 1 214 vv= , 22 2 2 1 12 12 1 1 12 115 ()() ( 23 pp )ρ vv ρgy y p ρυgy y ⎡ ⎤⎛⎞ =+ −+ − =+ + − ⎜⎟⎢ ⎥ ⎝⎠⎣ ⎦ 433 22 2 15 5.00 10 Pa (1.00 10 kg/m ) (3.00 m/s) (9.80 m/s )(11.0 m) 1.62 10 Pa 32 p =× +× + =× 5 . EVALUATE: The decrease in speed and the decrease in height at point 2 both cause the pressure at point 2 to be greater than the pressure at point 1. 4 Problem 14.57 A U-shaped tube open to the air at both ends contains some mercury. A quantity of water is carefully poured into the left arm of the U- shaped tube until the vertical height of the water column is 15.0 cm (the figure). a) What is the gauge pressure at the water- mercury interface? (a) IDENTIFY and SET UP: Apply 0 p pghρ=+ to the water in the left-hand arm of the tube. See Figure 14.57. Figure 14.57 EXECUTE: 0a ,p p= so the gauge pressure at the interface (point 1) is 32 a (1000 kg/m )(9.80 m/s )(0.150 m) 1470 Papp ghρ−= = = b) Calculate the vertical distance h from the top of the mercury in the right-hand arm of the tube to the top of the water in the left-hand arm. (b) IDENTIFY and SET UP: The pressure at point 1 equals the pressure at point 2. Apply Eq.(14.6) to the right- hand arm of the tube and solve for h. EXECUTE: and 1aw (0.150 m)pppg=+ 2aHg (0.150 m )p pgρ h= +− 12 p p= implies wHg (0.150 m) (0.150 m )g ghρρ=− 3 w 33 Hg (0.150 m) (1000 kg/m )(0.150 m) 0.150 m 0.011 m 13.6 10 kg/m h ρ ρ −= = = × 0.150 m 0.011 m 0.139 m 13.9 cmh =−== EVALUATE: The height of mercury above the bottom level of the water is 1.1 cm. This height of mercury produces the same gauge pressure as a height of 15.0 cm of water. 5 Mike Gershenson HW 1 Solutions

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