Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch14

Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch14

(Parte 2 de 3)

41. For our estimate of Vsubmerged we interpret “almost completely submerged” to mean

3 submerged 4 where 60 cm .3 o oV r rpi≈ =

Thus, equilibrium of forces (on the iron sphere) leads to

3 3 iron water submerged iron where ri is the inner radius (half the inner diameter). Plugging in our estimate for Vsubmerged as well as the densities of water (1.0 g/cm3) and iron (7.87 g/cm3), we obtain the inner

7.87 g/cmir r

42. From the “kink” in the graph it is clear that d = 1.5 cm. Also, the h = 0 point makes it clear that the (true) weight is 0.25 N. We now use Eq. 14-19 at h = d = 1.5 cm to obtain

Fb = (0.25 N – 0.10 N ) = 0.15 N.

Thus, ρliquid g V = 0.15, where V = (1.5 cm)(5.67 cm2) = 8.5 × 10−6 m3. Thus, ρliquid = 1800 kg/m3 = 1.8 g/cm3.

43. The volume Vcav of the cavities is the difference between the volume Vcast of the casting as a whole and the volume Viron contained: Vcav = Vcast – Viron. The volume of the iron is given by Viron = W/gρiron, where W is the weight of the casting and ρiron is the density of iron. The effective weight in water (of density ρw) is Weff = W – gρw Vcast. Thus, Vcast = (W – Weff)/gρw and

W W WV g gρ ρ

4. Due to the buoyant force, the ball accelerates upward (while in the water) at rate a given by Newton’s second law:

ρwaterVg – ρballVg = ρballVa⇒ ρball = ρwater (1 + “a”)

where – for simplicity – we are using in that last expression an acceleration “a” measured in “gees” (so that “a” = 2, for example, means that a = 2(9.80 m/s2) = 19.6 m/s2). In this

CHAPTER 14 problem, with ρball = 0.300 ρwater, we find therefore that “a” = 7/3. Using Eq. 2-16, then the speed of the ball as it emerges from the water is v = , were a = (7/3)g and ∆y = 0.600 m. This causes the ball to reach a maximum height hmax

(measured above the water surface) given by hmax = v2/2g (see Eq. 2-16 again). Thus, hmax = (7/3)∆y = 1.40 m.

45. (a) If the volume of the car below water is V1 then Fb = ρwV1g = Wcar, which leads to

1000kg m 9.8m sw

(b) We denote the total volume of the car as V and that of the water in it as V2. Then car 2b w wF Vg W V gρ ρ= = + which gives

1000kg mw

46. (a) Since the lead is not displacing any water (of density ρw), the lead’s volume is not contributing to the buoyant force Fb. If the immersed volume of wood is Vi, then

wood wood wood 0.900 0.900 ,b w i w w mF V g V g gρ ρ ρ ρ which, when floating, equals the weights of the wood and lead:

wood wood lead wood 0.900 ( ) .b w mF g m m gρ ρ

lead wood 3 wood

(b) In this case, the volume Vlead = mlead/ρlead also contributes to Fb. Consequently,

wood lead wood lead wood lead

0.900 ( ) ,w b w mF g m g m m gρρ which leads to wood wood wood

w w m m ρ ρ

47. (a) When the model is suspended (in air) the reading is Fg (its true weight, neglecting any buoyant effects caused by the air). When the model is submerged in water, the reading is lessened because of the buoyant force: Fg – Fb. We denote the difference in readings as ∆m. Thus, which leads to Fb = ∆mg. Since Fb = ρwgVm (the weight of water displaced by the model) we obtain

1000 kg/m w mV ρ

(b) The 1 20 scaling factor is discussed in the problem (and for purposes of significant figures is treated as exact). The actual volume of the dinosaur is

(c) Using ρ ≈ ρw = 1000 kg/m3, we find

3 3dino dino

which yields 5.102 × 103 kg for the T. rex mass.

48. Let ρ be the density of the cylinder (0.30 g/cm3 or 300 kg/m3) and ρFe be the density of the iron (7.9 g/cm3 or 7900 kg/m3). The volume of the cylinder is

Vc = (6×12) cm3 = 72 cm3 = 0.000072 m3, and that of the ball is denoted Vb . The part of the cylinder that is submerged has volume

Vs = (4 × 12) cm3 = 48 cm3 = 0.000048 m3. Using the ideas of section 14-7, we write the equilibrium of forces as

ρgVc + ρFe gVb = ρw gVs + ρw gVb⇒ Vb = 3.8 cm3

CHAPTER 14 where we have used ρw = 998 kg/m3 (for water, see Table 14-1). Using Vb = pir3 we find r = 9.7 m.

49. We use the equation of continuity. Let v1 be the speed of the water in the hose and v2 be its speed as it leaves one of the holes. A1 = piR2 is the cross-sectional area of the hose.

If there are N holes and A2 is the area of a single hole, then the equation of continuity becomes

A Rv A v NA v v v

where R is the radius of the hose and r is the radius of a hole. Noting that R/r = D/d (the ratio of diameters) we find

Dv v

which leads to h = 4.0 m.

51. This problem involves use of continuity equation (Eq. 14-23): 1122AvAv=.

(a) Initially the flow speed is 1.5m/siv= and the cross-sectional area is iAHD=. At point a, as can be seen from Fig. 14-47, the cross-sectional area is

( ) ( )aA H h D b h d= − − − . Thus, by continuity equation, the speed at point a is i i i a

Av HDvv

(b) Similarly, at point b, the cross-sectional area isbAHDbd=−, and therefore, by continuity equation, the speed at point b is i i i b

Av HDvv

52. The left and right sections have a total length of 60.0 m, so (with a speed of 2.50 m/s) it takes 60.0/2.50 = 24.0 seconds to travel through those sections. Thus it takes (8.8 – 24.0) s = 64.8 s to travel through the middle section. This implies that the speed in the middle section is vmid = (110 m)/(64.8 s) = 0.772 m/s. Now Eq. 14-23 (plus that fact that A = pir2) implies rmid = rA where rA = 2.0 cm. Therefore, rmid = 3.60 cm.

53. Suppose that a mass ∆m of water is pumped in time ∆t. The pump increases the potential energy of the water by ∆mgh, where h is the vertical distance through which it is lifted, and increases its kinetic energy by 212mv∆, where v is its final speed. The work it does is 212Wmghmv∆=∆+∆ and its power is

W mP gh v t t

Now the rate of mass flow is ∆m/ ∆t = ρwAv, where ρw is the density of water and A is the area of the hose. The area of the hose is A = pir2 = pi(0.010 m)2 = 3.14 × 10–4 m2 and

2 2 P Av gh vρ

54. (a) The equation of continuity provides (26 + 19 + 1) L/min = 56 L/min for the flow rate in the main (1.9 cm diameter) pipe.

(b) Using v = R/A and A = pid 2/4, we set up ratios:

5. (a) We use the equation of continuity: A1v1 = A2v2. Here A1 is the area of the pipe at the top and v1 is the speed of the water there; A2 is the area of the pipe at the bottom and v2 is the speed of the water there. Thus

(b) We use the Bernoulli equation:

where ρ is the density of water, h1 is its initial altitude, and h2 is its final altitude. Thus


= × 56. We use Bernoulli’s equation:

where ρ = 1000 kg/m3, D = 180 m, v1 = 0.40 m/s and v2 = 9.5 m/s. Therefore, we find ∆p = 1.7 × 106 Pa, or 1.7 MPa. The SI unit for pressure is the Pascal (Pa) and is equivalent to N/m2.

57. (a) The equation of continuity leads to rv A v A v v r

(b) With h = 7.6 m and p1 = 1.7 × 105 Pa, Bernoulli’s equation reduces to

58. (a) We use Av = const. The speed of water is

(b) Since 212const.,pvρ+= the pressure difference is

59. (a) We use the Bernoulli equation:

where h1 is the height of the water in the tank, p1 is the pressure there, and v1 is the speed of the water there; h2 is the altitude of the hole, p2 is the pressure there, and v2 is the speed of the water there. ρ is the density of water. The pressure at the top of the tank and at the hole is atmospheric, so p1 = p2. Since the tank is large we may neglect the water speed at the top; it is much smaller than the speed at the hole. The Bernoulli equation then

(b) We use the equation of continuity: A2v2 = A3v3, where 1 322AA= and v3 is the water speed where the area of the stream is half its area at the hole. Thus

The water is in free fall and we wish to know how far it has fallen when its speed is doubled to 4.84 m/s. Since the pressure is the same throughout the fall,

v vh h g

60. (a) The speed v of the fluid flowing out of the hole satisfies 212 or 2vghvghρρ== . Thus, ρ1v1A1 = ρ2v2A2, which leads to

2 2 2.AghA ghA A ρρ ρ ρ

R v A A

61. We rewrite the formula for work W (when the force is constant in a direction parallel to the displacement d) in terms of pressure:

CHAPTER 14 where V is the volume of the water being forced through, and p is to be interpreted as the pressure difference between the two ends of the pipe. Thus,

62. (a) The volume of water (during 10 minutes) is

V v t A pi

(b) The speed in the left section of pipe is

A dv v v A d atmospheric pressure,

Thus, the gauge pressure is (1.97 atm – 1.0 atm) = 0.97 atm = 9.8 × 104 Pa. 63. (a) The friction force is

(b) The speed of water flowing out of the hole is v = Thus, the volume of water flowing out of the pipe in t = 3.0 h is

64. (a) We note (from the graph) that the pressures are equal when the value of inverse-area-squared is 16 (in SI units). This is the point at which the areas of the two pipe sections are equal. Thus, if A1 = 1/ when the pressure difference is zero, then A2 is 0.25 m2.

(b) Using Bernoulli’s equation (in the form Eq. 14-30) we find the pressure difference may be written in the form a straight line: mx + b where x is inverse-area-squared (the horizontal axis in the graph), m is the slope, and b is the intercept (seen to be –300 kN/m2). Specifically, Eq. 14-30 predicts that b should be – ρ v22. Thus, with ρ = 1000 kg/m3 we obtain v2 = m/s. Then the volume flow rate (see Eq. 14-24) is

If the more accurate value (see Table 14-1) ρ = 998 kg/m3 is used, then the answer is 6.13 m3/s.

65. (a) Since Sample Problem 14-8 deals with a similar situation, we use the final equation (labeled “Answer”) from it:

02 for the projectile motion.vghvv=⇒=

The stream of water emerges horizontally (θ0 = 0° in the notation of Chapter 4), and setting y – y0 = –(H – h) in Eq. 4-2, we obtain the “time-of-flight”

Using this in Eq. 4-21, where x0 = 0 by choice of coordinate origin, we find

(b) The result of part (a) (which, when squared, reads x2 = 4h(H – h)) is a quadratic equation for h once x and H are specified. Two solutions for h are therefore mathematically possible, but are they both physically possible? For instance, are both solutions positive and less than H? We employ the quadratic formula:

which permits us to see that both roots are physically possible, so long as x < H. Labeling the larger root h1 (where the plus sign is chosen) and the smaller root as h2 (where the minus sign is chosen), then we note that their sum is simply


Thus, one root is related to the other (generically labeled h' and h) by h' = H – h. Its numerical value is '40cm 10 cm 30 cm.h=−=

(c) We wish to maximize the function f = x2 = 4h(H – h). We differentiate with respect to h and set equal to zero to obtain

or h = (40 cm)/2 = 20 cm, as the depth from which an emerging stream of water will travel the maximum horizontal distance.

6. By Eq. 14-23, we note that the speeds in the left and right sections are vmid and vmid, respectively, where vmid = 0.500 m/s. We also note that 0.400 m3 of water has a mass of 399 kg (see Table 14-1). Then Eq. 14-31 (and the equation below it) gives

W = m vmid2 = –2.50 J .

67. (a) The continuity equation yields Av = aV, and Bernoulli’s equation yields

2 21 1 22pvVρρ∆+=, where ∆p = p1 – p2. The first equation gives V = (A/a)v. We use this to substitute for V in the second equation, and obtain ()21 22pvAavρρ∆+=. We solve for v. The result is

( / ) 1 p a pv A a A aρ ρ

(b) We substitute values to obtain

68. We use the result of part (a) in the previous problem.

(a) In this case, we have ∆p = p1 = 2.0 atm. Consequently,

(b) And the equation of continuity yields V = (A/a)v = (5a/a)v = 5v = 21 m/s. (c) The flow rate is given by

69. (a) This is similar to the situation treated in Sample Problem 14-7, and we refer to some of its steps (and notation). Combining Eq. 14-35 and Eq. 14-36 in a manner very similar to that shown in the textbook, we find

2 pR A A A Aρ

for the flow rate expressed in terms of the pressure difference and the cross-sectional areas. Note that this reduces to Eq. 14-38 for the case A2 = A1/2 treated in the Sample square root is well defined. Therefore, we obtain R = 0.0776 m3/s.

(b) The mass rate of flow is ρR = 69.8 kg/s.

70. (a) Bernoulli’s equation gives 21 air2But ABABppvpppghρρ=+⋅∆=−=in order to balance the pressure in the two arms of the U-tube. Thus 21 air2ghvρρ=, or


(b) The plane’s speed relative to the air is

3 air

71. We use the formula for v obtained in the previous problem:

3 air

0.031kg/m pv ρ

When the water level rises to height h2, just on the verge of flooding, 2v, the speed of water in pipe M , is given by


By continuity equation, the corresponding rainfall rate is

73. The normal force NFr exerted (upward) on the glass ball of mass m has magnitude

0.0948 N. The buoyant force exerted by the milk (upward) on the ball has magnitude

Fb = ρmilk g V where V = pi r3 is the volume of the ball. Its radius is r = 0.0200 m. The milk density is ρmilk = 1030 kg/m3. The (actual) weight of the ball is, of course, downward, and has magnitude Fg = mglass g. Application of Newton's second law (in the case of zero acceleration) yields

FN + ρmilk g V − mglass g = 0 which leads to mglass = 0.0442 kg. We note the above equation is equivalent to Eq.14-19 in the textbook.

74. The volume rate of flow is R = vA where A = pir2 and r = d/2. Solving for speed, we obtain

R R Rv

(a) With R = 7.0 × 10–3 m3/s and d = 14 × 10–3 m, our formula yields v = 45 m/s, which is about 13% of the speed of sound (which we establish by setting up a ratio: v/vs where vs = 343 m/s).

(b) With the contracted trachea (d = 5.2 × 10–3 m) we obtain v = 330 m/s, or 96% of the speed of sound.

75. If we examine both sides of the U-tube at the level where the low-density liquid (with ρ = 0.800 g/cm3 = 800 kg/m3) meets the water (with ρw = 0.998 g/cm3 = 998 kg/m3), then the pressures there on either side of the tube must agree:

ρgh = ρwghw where h = 8.0 cm = 0.080 m, and Eq. 14-9 has been used. Thus, the height of the water column (as measured from that level) is hw = (80/998)(8.0 cm) = 6.41 cm. The volume of water in that column is therefore

V = pir2hw = pi(1.50 cm)2(6.41 cm) = 45.3 cm3.

76. Since (using Eq. 5-8) Fg = mg = ρskier g V and (Eq. 14-16) the buoyant force is Fb = ρsnow g V, then their ratio is

= = = = 0.094 (or 9.4%).

7. (a) We consider a point D on the surface of the liquid in the container, in the same tube of flow with points A, B and C. Applying Bernoulli’s equation to points D and C, we obtain

2 2D D D C C Cp v gh p v ghρ ρ ρ ρ+ + = + + which leads to

p pv g h h v g d hρ where in the last step we set pD = pC = pair and vD/vC ≈ 0. Plugging in the values, we obtain

Since vB = vC by equation of continuity, and pC = pair, Bernoulli’s equation becomes

(c) Since pB ≥ 0, we must let pair – ρg(h1 + d + h2) ≥ 0, which yields air air


78. To be as general as possible, we denote the ratio of body density to water density as f

(so that f = ρ/ρw = 0.95 in this problem). Floating involves equilibrium of vertical forces acting on the body (Earth’s gravity pulls down and the buoyant force pushes up). Thus, b g w wF F gV gVρ ρ= ⇒ = where V is the total volume of the body and Vw is the portion of it which is submerged. (a) We rearrange the above equation to yield w w

V fV ρ ρ which means that 95% of the body is submerged and therefore 5.0% is above the water surface.

(b) We replace ρw with 1.6ρw in the above equilibrium of forces relationship, and find

V fV ρ ρ

which means that 59% of the body is submerged and thus 41% is above the quicksand surface.

(Parte 2 de 3)