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Physics Solution Manual for 1100 and 2101

# 1 solution substituting m v into the relation w mg

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Unformatted text preview: 10.2). Solving Equation 10.2 for x and using Fx = max , we obtain x=− Fx k =− ma x k =− ( 92 kg ) ( 0.30 m/s2 ) 2300 N/m = −0.012 m The amount that the spring stretches is 0.012 m . ______________________________________________________________________________ 6. REASONING The restoring force of the spring and the static frictional force point in opposite directions. Since the box is in equilibrium just before it begins to move, the net force in the horizontal direction is zero at this instant. This condition, together with the expression for the restoring force (Equation 10.2) and the expression for the maximum static frictional force (Equation 4.7), will allow us to determine how far the spring can be stretched without the box moving upon release. SOLUTION The drawing at the right shows the four forces that act on the box: its weight mg, the normal force FN, the restoring force Fx exerted by the spring, and the maximum +y FN MAX Fx fs MAX static frictional force fs . Since the box is not moving, it is in equilibrium. Let the x axis be parallel to the table top. According to Equation 4.9a, the net force ΣFx in the x direction must be zero, ΣFx = 0. mg +x CHAPTER 11 FLUIDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS ______________________________________________________________________________ 1. (b) According to the relation P2 = P + ρ gh (Equation 11.4), the pressure P2 at the bottom 1 of the container depends on the height h of the fluid above it. Since this height is the same for all three containers, the pressure at the bottom is the same for each container. 2. (d) According to the relation P2 = P + ρ gh (Equation 11.4), the pressure at any point 1 depends on the height h of the fluid above it. Since this height is the same for the ceiling of chamber 1 and the floor of chamber 2, the pressure at these two locations is the same. 3. F = 3.0 × 105 N 4. (c) The pressure at the top of each liquid is the same, since the U-tube is open at both ends. Also, the pressure at the location of the dashed line is the same in both the left and right sides of the U tube, since these two locations are at the same level. Thus, the pressure increment ρ1gh1 for liquid 1 must be equal to the pressure increment ρ2gh2 for liquid 2, where h1 and h2 are the heights of the liquids above the dashed line. Since h1 is greater than h2, ρ1 must be less than ρ2. 5. (b) The pressure at the top of each liquid is the same, since the U-tube is open at both ends. Also, the pressure at the location of the dashed line is the same in both the left and right sides of the U tube, since these two locations are at the same level. Thus, the pressure increment ρ1gh1 for liquid 1 must be equal to the pressure increment ρ2gh2 for liquid 2, where h1 and h2 are the heights of the liquids above the dashed line. Since h1 is 3 times as great as h2, ρ1 must be one-third that of ρ2. 6. (b) According to the relation, P2 = P + ρ gh (Equation 11.4), a drop in the pressure P1 at 1 the top of the pool produces an identical drop in the pressure P2 at the bottom of the pool. 7. W = 14 000 N 8. (a) According to Archimedes’ principle, the buoyant force equals the weight of the fluid that the object displaces. Both objects displace the same weight of fluid, since they have the same volume. The buoyant force does not depend on the depth of an object. 9. (d) The buoyant force (19.6 N) is less than the weight (29.4 N) of the object. Therefore, the object sinks. 562 FLUIDS ( ) 10. (e) When an object floats, its weight ρobjectVobject g equals the buoyant force ( ρfluidVdisplaced g ) , where Vdisplaced is the volume of fluid displaced by the object. Thus, ρobjectVobject g = ρfluidVdisplaced g , so the density of the object is ( ) ρobject = ρfluid Vdisplaced / Vobject . Thus, the density of the object is proportional to the ratio Vdisplaced / Vobject of the volumes. This ratio is greatest for object C and least for B. 11. (a) The beaker with the ball contains less water, because part of the ball is below the water line. According to Archimedes’ principle, the weight of this “missing” (or displaced) water is equal to the magnitude of the buoyant force that acts on the ball. Since the ball is floating, the magnitude of the buoyant force equals the weight of the ball. Thus, the weight of the missing water is exactly equal to the weight of the ball, so the two beakers weight the same. 12. (e) The volume flow rate is equal to the speed of the water times the cross-sectional area through which the water flows (see Equation 11.10). If the speed doubles and the cross-sectional area triples, the volume flow rate increases by a factor of six (2 × 3). 13. (c) Because water is incompressible and no water accumulates within the pipe, the volume of water per second flowing through the wide section is equal to that flowing through the narrow section. Thus, the volume flow rate is the same in both sections. 14. (b) Water is incompressible, so it cannot accumulate anywhere within the pipe. Thus, the volume flow rate is t...
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