Physics Solution Manual for 1100 and 2101

80 ms2 1025 kgm3 980 ms2 35 m 27 ssm reasoning

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Unformatted text preview: ity. Taking upwards as the positive direction, Newton’s 2nd law yields ΣF = FFonS − mg = ma (1) Solving Equation (1) for FFonS, and noting that by Newton’s 3rd law, FFonS = FSonF, we obtain 572 FLUIDS FFonS = mg + ma = m( g + a ) = FSonF Substituting Equation (2) into P = P= FSonF A = FSonF A (2) (Equation 11.3), we find that m ( g + a ) (16 kg ) ( 9.80 m/s 2 + 1.5 m/s2 ) = = 2400 Pa ( 0.15 m )( 0.50 m ) A 18. The tension forces T acting on the edges of the square section of the bladder wall add up to give a total inward force Fin. Because the section is in equilibrium, the net force acting on it must be zero. Therefore, the inward force Fin balances the outward force Fout due to the internal pressure P of the bladder: Fin = Fout. The magnitude Fout of the outward force is found from Fout = P A (Equation 11.3), where A is the area of the outer surface of the bladder wall. Thus, we have that Fin = Fout = PA θ Fout (1) T Thoriz θ θ T θ T θ T Tin T Fin SOLUTION We note that the horizontal components Thoriz = T cos θ of the tension forces acting on opposite edges of the section balance each other out. Thus, there is no net horizontal force on the section. The inward components Tin of the four tension forces are equal to one another, and each is given by Tin = T sin θ The magnitude Fin of the net inward force is, therefore, Fin = 4Tin = 4T sin θ or T= Substituting Fin = PA [Equation (1)] into Equation (2) yields Fin 4 sin θ (2) Chapter 11 Problems T= PA 4 sin θ 573 (3) The area A of the square is the product of the lengths (l = 0.010 m) of two of its sides: A = l 2 . Making this substitution into Equation (3), we obtain ( 3300 Pa )( 0.010 m ) = 0.95 N PA Pl 2 T= = = 4sin θ 4sin θ 4sin 5.0o 2 19. SSM REASONING AND SOLUTION Both the cylinder and the hemisphere have circular cross sections. According to Equation 11.3, the pressure exerted on the ground by the hemisphere is W W P = h = h2 Ah π rh where Wh and rh are the weight and radius of the hemisphere. Similarly, the pressure exerted on the ground by the cylinder is P= Wc = Ac Wc π rc2 where Wc and rc are the weight and radius of the cylinder. Since each object exerts the same pressure on the ground, we can equate the right-hand sides of the expressions above to obtain Wh W = c2 2 π rh π rc Solving for rh 2 , we obtain Wh rh2 = rc2 (1) Wc The weight of the hemisphere is Wh = ρgV h = ρg 1 2 cπ r h= 4 3 3 h 2 3 ρgπ rh3 where ρ and V h are the density and volume of the hemisphere, respectively. The weight of the cylinder is Wc = ρgV c = ρgπ rc2 h 574 FLUIDS where ρ and Vc are the density and volume of the cylinder, respectively, and h is the height of the cylinder. Substituting these expressions for the weights into Equation (1) gives rh2 = rc2 Wh Wc = rc2 ρg π rh3 ρg π rc2 h 2 3 Solving for rh gives 3 3 rh = 2 h = 2 ( 0.500 m ) = 0.750 m 20. REASONING The magnitude of the force that would be exerted on the window is given by Equation 11.3, F = PA , where the pressure can be found from Equation 11.4: P2 = P1 + ρgh . Since P1 represents the pressure at the surface of the water, it is equal to atmospheric pressure, Patm . Therefore, the magnitude of the force is given by F = ( Patm + ρgh ) A where, if we assume that the window is circular with radius r, its area A is given by A = π r . 2 SOLUTION a. Thus, the magnitude of the force is F = 1.013 × 10 5 Pa + (1025 kg / m 3 )( 9.80 m / s 2 )( 11 000 m) π (0.10 m) 2 = 3.5 × 10 6 N b. The weight of a jetliner whose mass is 1.2 × 10 5 kg is W = mg = (1.2 × 10 5 kg)(9.80 m / s 2 ) = 1.2 × 10 6 N Therefore, the force exerted on the window at a depth of 11 000 m is about three times greater than the weight of a jetliner! 21. REASONING AND SOLUTION The pump must generate an upward force to counteract the weight of the column of water above it. Therefore, F = mg = (ρhA)g. The required pressure is then P = F/A = ρgh = (1.00 × 103 kg/m3)(9.80 m/s2)(71 m) = 7.0 × 105 Pa 22. REASONING The atmospheric pressure outside the tube pushes the sauce up the tube, to the extent that the smaller pressure in the bulb allows it. The smaller the pressure in the bulb, the higher the sauce will rise. The height h to which the sauce rises is related to the Chapter 11 Problems 575 atmospheric pressure PAtmospheric outside the tube, the pressure PBulb in the bulb, and the density ρ of the sauce by PAtmospheric = PBulb + ρ gh (Equation 11.4). SOLUTION a. Solving Equation (11.4) for the absolute pressure in the bulb when the height of the sauce is 0.15 m, we find that PBulb = PAtmospheric − ρ gh ( )( ) = 1.013 ×105 Pa − 1200 kg/m3 9.80 m/s 2 ( 0.15 m ) = 9.95 ×104 Pa b. When the height of the sauce is 0.10 m, the absolute pressure in the bulb is PBulb = PAtmospheric − ρ gh ( )( ) = 1.013 ×105 Pa − 1200 kg/m3 9.80 m/s 2 ( 0.10 m ) = 1.001×105 Pa 23. SSM REASONING As the depth h increases, the pressure increases according to Equation 11.4 (P2 = P1 + ρgh). In this equation, P1 is the pressure...
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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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