solucionario capitulo 9 física serway and faughn

7/26/2019 Solucionario Capitulo 9 Física Serway and Faughn

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CHAPTER NINE SOLUTIONS

1


Chapter Nine Readings

Denton, E., "The Buoyancy of Marine Animals," Scientific American, July 1960,

p. 118.

Gilman, J.J., "Fracture in Solids," Scientific American, February 1960, p. 94.

Smith, N., "Bernoulli and Newton in Fluid Mechanics," The Physics Teacher ,

10, 1972, p. 451.

Morrison, P., "Under Pressure", Scientific American, v. 274 (June '96) p. 113.

9.1 Stress = F

A , where F = 0.3(weight) = 0.3(480 N) = 144 N,

and A = r 2 = (0.50 x 10-2 m)2 = 7.85 x 10-5 m2. Thus,

Stress =

144 N

7.85 x 10-5 m2 = 1.8 x 10

6

Pa

9.2 Using Y = FLo

A( L) , we get A =

FLo

Y ( L) = (d /2)2.

So, d =4mgLo

Y ( L) =

4(380 kg)(9.80 m/s2)(18.0 m)

(2 x 1011 Pa)(0.009 m) = 6.89 mm.

9.3 Using Y = FLo

A( L) with A = (d /2)2 and F = mg , we get

Y =mgLo

(d /2)2 L =

4(90 kg)(9.80 m/s2)(50 m)

(0.01 m)2(1.6 m) = 3.5 x 108 Pa.

9.4 The maximum stress = F / Amin = 5.0 x 108 Pa.

Therefore, Amin = (70 kg)(9.80 m/s2)/5.0 x 108 Pa = 1.37 x 10-6 m2.

But, Amin =d 2

4 , from which, d min = 1.3 mm.

9.5 We assume a length for the femur of 0.50 m. The amount of compression L is given by:

L = L(stress)

Y =

(5.0 x 10-1 m)(160 x 106 Pa)

14.5 x 109 Pa = 5.52 x 10-3 m = 5.5 mm.

9.6 We know that the shear modulus is given by

S =shear stress

shear strain =

stress

( x/h)

or, Stress = S xh = (1.5 x 1010 Pa)

5.0 m104 m

) = 7.5 x 106 Pa.

9.7 From the defining equation for the shear modulus, we find x as

x =h( F / A)

S =

(5.0 x 10-3 m)(20 N)/(14 x 10-4 m2)

3.0 x 106 Pa = 2.4 x 10-5 m.

or x = 2.4 x 10-2 mm.




2

9.8 (a) When at rest, the tension in the cable is equal to the weight of the 800 kg mass, 7840 N.

Thus, from the definition of Young's modulus, we find the amount the cable is stretched:

L =( F / A) L

Y =

( )(7840 N)/(4.00 x 10-4 m2) (25.0 m)

(20 x 1010 Pa) = 2.45 x 10-3 m,

or L = 2.45 mm.(b) We write down Newton's second law for the block when it is accelerating upward.

T - mg = ma, or T = m( g + a). (1)

When a = 3 m/s2, we find: T = 1.02 x 104 N, so

Lnew = F L

Y A =

(1.02 x 104 N)(25.0 m)

(20 x 1010 Pa)(4.00 x 10-4 m2) = 3.20 x 10-3 m = 3.20 mm

Therefore, the increase in elongation is:

Lnew - L = 3.20 mm - 2.45 mm = 0.75 mm.

(c) If the stress ( F / A) is not to exceed 2.2 x 108 Pa, the maximum force allowed is: F = T =

(2.2 x 108 Pa)(4.00 x 10-4 m2) = 8.8 x 104 N.

From (1) we find the largest mass to be:

m =T

a + g =

8.8 x 104 N

(3.00 + 9.80)m/s2 = 6.9 x 103 kg.

9.9 Applying Newton's second law to the dancer gives: N - mg = ma

where N is the normal force the floor exerts on the dancer and a is theupward acceleration (if any) the dancer is given. Thus,

N = m( g + a)

is the force exerted on the dancer by the floor.

(a) In this case, a = 0 and N = mg = 490 N.

Therefore, P = F

A =

490 N

26.0 x 10-4 m2 = 1.88 x 105 Pa.

(b) Here, a = + 4.00 m/s2. Thus, N = m( g + a) = 690 N, or P = 2.65 x 105 Pa.

9.10 Let W be its weight. Then each tire supports W /4, so that P = F

A =

W

4 A ,

yielding:W = 4 AP = 4(0.024 m2)(2.0 x 105 N/m2) = 1.9 x104 N.

9.11 The area of one of the legs is A = r 2 = (10-2 m)2 = x 10-4 m2.

The force exerted by one leg on the floor is

F =1

2 (weight of man + weight of chair) =

1

2 (75 kg)(9.80 m/s2) = 368 N.

Thus, P = F / A = 368 N/( x 10-4)m2 = 1.2 x 106 Pa.

9.12 Let H be the height of the pillar, and let A be its cross-sectional area.

Then, the pressure at the base is:

P =mg

A =

( AH ) g

A = ( g ) H = (weight per unit volume) H = (5.0 x 104 N)H

With a maximum pressure of P max = 1.7 x 107 Pa, the maximum height is

H max = P max

5.0 x 104 N/m3 =1.7 x 107 Pa

5.0 x 104 N/m3 = 3.4 x 102 m.

9.13 P g = P - P atm = gh = (103 kg/m3)(9.80 m/s2)(1200 ft)(1m/3.281 ft), or

P g = 3.58 x 106 Pa.

9.14 The gauge pressure of the fluid at the level of the needle must equal the gauge pressure in the artery.

m

N

mg




3

P gauge = gh = 1.33 x 104 Pa, so

h =1.33 x 104 Pa

(1.02 x 103 kg/m3)(9.80 m/s2) = 1.33 m

9.15 We use P = P o + gh, where = 806 km/m3 for ethyl alcohol, and P o = 1.10 atm =

1.114 x 105 Pa.

Thus,

P = 1.114 x 105 Pa + (806 kg/m3)(9.80 m/s2)(4.0 m) = 1.43 x 105 Pa, or

P = 1.4 atm.

9.16 The pressure at the upper surface of each liquid is given by

P = P atm - w ghw = P atm - gh. Therefore, =

hw

h w.

9.17 We first find the absolute pressure at the interface between oil and water: P t = P atm + gh, or

P t = 1.013 x 105 Pa + (7.00 x 102 kg/m3)(9.80 m/s2)(0.300 m) = 1.03 x 105 Pa.

This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use:

P = P t + gh, or

P = 1.03 x 105 Pa + (1.00 x 103 kg/m3)(9.80 m/s2)(0.200 m) = 1.05 x 105 Pa.

9.18 First, use Pascal's principle, F 2

A2 =

F 1

A1 ,

to find the piston 1 will exert on the handle:

F 1 = A1

A2 F 2 =

d 12

4

d 22

4

F 2 =(0.25 in)2

(1.5 in)2 (500 lb)

= 13.9lb

Now, consider torques on the jack handle with the pivot point at the left end.

= 0 = (13.9 lb)(2 in) - F (12 in) = 0,or F = 2.31 lb.

9.19 F 2

A2 =

F 1

A1 Pascal's principle becomes:

F brake = A brake cylinder

Amaster cylinder ( F pedal) =

1.75 cm2

6.4 cm2 (44 N) = 12.0 N

This is the normal force exerted on the brake shoe. The frictional force is: f = N =0.50(12.0 N) = 6.0 N, and the torque is

= f · r tire = (6.0 N)(0.34 m) = 2.0 N m.

9.20 Since the frog floats, the buoyant force = the weight of the frog. Also, the weight of the displaced

fluid = weight of the frog, so

fluidVg = mfrog g , or,

mfrog = fluidV = (1.35 x 103 kg/m3) 1

2

4(6.00 x 10-2 m)3

3.00 .

Hence, mfrog = 0.611 kg.

9.21 The weight of the truck,W , is equal to the weight of the additional water displaced when the truck

drives onto the boat. This is:

W = w(V ) g = (103 kg/m3)(4.00m)(6.00 m)(0.0400 m)(9.80 m/s2) = 9.41 x 103 N.




4

9.22 The buoyant force, B, on the iceberg must be equal to its weight, w, in order for it to float. Thus:

B = w.

But, the buoyant force is equal to the weight of the water displaced. Therefore, wV uw g =

iceV total g where V uw is the volume of the iceberg under water and V total is the total volume of

the berg.

We have,

V uw

V total =

ice

w =

920

1030 = 0.89.

Therefore, 89% of the volume is submerged and 11% is exposed.

9.23 The balloon is in equilibrium under the action of three forces, F b, the buoyant force on the balloon,

w, its weight, and T , the tension in the string. We have: F y = 0 T = F b - w.

(1)

F b = weight of displaced air = air Vg , or

F b = (1.29 kg/m3)(9.80 m/s2)4

3 (0.500 m)3 = 6.62 N.

w = weight of empty balloon + weight of enclosed helium, or

w = (0.012 kg)(9.80 m/s2) + (0.181 kg/m3)4

3 (0.500 m)3 (9.80 m/s2)

= 1.05 NThen from equation (1), T = 6.62 N - 1.05 N = 5.57 N.

9.24 At equilibrium, we must have F y = B - F spring - W = 0, or F spring = k ( ) L = B - W ,

where B is the buoyant force, k ( ) L is the downward spring force, and W is the weight of the

block of wood.

But, W = mg = (5.00 kg)(9.80 m/s2) = 49.0 N and

B = ( wV ) g = ( w(mwood

wood )) g = (1000

5.00 kg

650(9.80 m/s2) = 75.4 N

Therefore, F spring = k ( ) L = B - W = 75.4 N - 49.0 N = (160 N/m)(L),

which yields L =26.4 N

160 N/m = 0.165 m = 16.5 cm

9.25 (a) The forces on the object are the tension, T , in the string connecting it to a balance, The weight

of the object, w, and the buoyant force on it, B. Because the object is in equilibrium when

immersed in the alcohol, we have F y = 0, which gives,

T + B = w , or 200 N + B = 300 N. Thus, B = 100 N.

We also know that the buoyant force is equal to the weight of the displaced alcohol. So, B =

alcoholV alcohol g .

But, V alcohol is equal to the volume of the object because the object is completely submerged.

Thus,

V object = B

alcohol g =

100 N

(700 kg/m3)(9.80 m/s2) = 1.46 x 10-2 m3.

(b) The mass of the 300 N object is 30.6 kg, and now that we know its volume, its density can be

found as =

m

V = 2.10 x 103 kg/m3.




5

9.26 (a) The forces acting on the object when suspended in water

are the tension in the string, T 1, the weight of the object,

and the buoyant force, Bw. At equilibrium, we have

Bw = w - T 1 = 300 N - 265 N = 35 N.

Also, Bw = wVg . Therefore,

V =

35 N

(103 kg/m3)(9.80 m/s2)

= 3.57 x 10-3 m3.

The mass of the 300 N object is 30.6 kg, and the volume V found above is the volume of water

displaced which is also the volume of the object. Thus, the density of the object is:

object = mobject/V = 30.6 kg/3.57 x 10-3 m3 = 8.57 x 103 kg/m3.

(b) When submerged in the oil, the forces on the object are T 2, the tension in the string, the

weight of the object, w, and the buoyant force of the oil, Bo. For equilibrium, we have

T 2 + Bo = w, or

Bo = w - T 2 = 300 N - 275 N = 25 N. However, the buoyant force exerted by the oil is also

equal to the weight of the oil displaced. The volume of the oil displaced is equal to the

volume of the object. Thus, the density of the oil is

oil =m

oilV =w

oil gV = 25 N(9.80 m/s2)(3.57 x 10-3 m3)

= 714 kg/m3.

9.27 Applying Newton's second law:

F y = B - W total = B - (mshell + m) g = (mshell + m)a

Therefore,

a= B

(mshell + m) - g . (1)

The volume is V =4

3 r 3 =

4

3 (0.10 m)3 = 4.19 x 10-3 m3. Thus,

B = (weight of displaced water) = water Vg

= (1000 kg/m3)(4.19 x 10-3 m3)(9.80 m/s2) = 41.1 N,

and m = V = (806 kg/m3)(4.19 x 10-3 m3) = 3.38 kg.

Equation (1) for the acceleration then gives:

a =41.1 N

(0.400 kg + 3.38 kg) - 9.80 m/s2 = 1.07 m/s2

9.28 When the system floats, F B = w, or the weight of the displaced water equals the weight of the

object. Therefore, w(r 2 z ) g = 1.96 N ,

where ( )r 2 z = volume of displaced water. The depth of the bottom end is thus,

z =1.96 N

(103 kg/m3)(2.00 x 10-2 m)2(9.80 m/s2) = 0.159 m = 15.9 cm.

9.29 When the mattress is totally submerged, the buoyant force exerted by the water (and hence the total

weight that can be supported) is:

F m = wVg = (103 kg/m3)(2.0 m)(0.50 m)(0.08 m)(9.80 m/s2) = 784 NThe total mass supported is the sum of the mass of the mattress and the mass of the load, or

M = mmattress + m =784 N

9.80 m/s2 = 80 kg.

The load mass is therefore: m = M - mmattress = 80 kg - 2.0 kg = 78 kg.

9.30 Looking first at the top scale and the iron block, we have: T 1 + B = W iron,

where T 1 is the tension in the spring scale, B is the buoyant force and W iron is the weight of the iron

block. The volume, V , of the iron block, and hence of the displaced oil, is:

T

Bw

300 N(water)

1T

Bo

2

300 N(Oil)




6

V =miron

iron =

2.00 kg

7860 kg/m3 = 2.54 x 10-4 m3.

Then B = oilVg = (916 kg/m3)(2.54 x 10-4 m3)()9.80 m/s2) = 2.28 N. Therefore,

T 1 = W iron - B = 19.6 N - 2.28N = 17.3 N.

Next we look at the bottom scale which reads T 2. (i.e., exerts an upward force T 2 on the system)

Consider the external vertical forces acting on the beaker-oil-iron combination. F y = 0 gives T 1 + T 2 - w beaker - woil - wiron = 0, or

T 2 = (m beaker + moil + miron) g - T 1 = (5.00 kg)(9.80 m/s2) - 17.3 N.

Thus, T 2 = 31.7 N is the lower scale reading.

9.31 The volume rate of flow =Volume

time = vA, where A is the cross-sectional

area of the pipe and v is the average velocity of the water in the pipe.

Therefore, the average velocity is

v =Volume

At =

20.0 gal

(1.00 in2)(30.0 s) (

231 in3

1 gal ) = 154 in/s

9.32 The cross-sectional area of the 2.0 cm diameter hose is A = x 10-4 m2, and the flow rate is: Av =

( x 10-4 m2)(1.5 m/s) = 1.5 x 10-4 m3/s.

The volume to be filled is (1.5 m)(0.6 m)(0.4 m) = 3.6 x 10 -1 m3.The time required to fill the trough is:

time =volume

flow rate =

3.6 x 10-1 m3

1.5 x 10-4 m3/s = 7.6 x 102 s = 13 min

9.33 (a) Flow Rate = Av = (2.0 cm2)(40 cm/s) = 80 cm3/s.

However, since blood has a mass of 1 g/cm3, this is equivalent to a mass flow rate of 80 g/s.(b) From the equation of continuity, we have:

v2 = A1v1

A2 =

2

3000 (40 cm/s) = 2.7 x 10-2 cm/s.

9.34 We select point 1 just above the wing and point 2 just below it. As a result, the difference invertical heights between these two points is negligible, and Bernoulli's equation reduces to

P 2 - P 1 =1

2 v1

2 - v22) =

1

2 (1.29 kg/m3) [(300 m/s)2 - (280 m/s)2] , or

P 2 - P 1 = 7480 Pa. The net upward force is therefore

F = ( P 2 - P 1) A = (7480 Pa)(20.0 m2) = 1.50 x 105 N upward.

9.35 (a) We find the flow velocity in the second section from the continuity equation: v2 = A1v1

A2 =

10

2.5 v1 = 4(2.75 m/s) = 11.0 m/s.

(b) Choosing the zero level for y along the common center line of the pipes, we have:

P 1 +1

2 v12 = P 2 +1

2 v22, or P 2 = P 1 +1

2 v12 - v22), giving

P 2 = (1.20 x 105 Pa) +1

2 (1650 kg/m3) [(2.75 m/s)2 - (11.0 m/s)2], and

P 2 = 2.64 x 104 Pa.

9.36 P 1gauge = P 1 - P atm = F

A1 =

2.00 N

2.50 x 10-5 m2 = 8.00 x 104 Pa.

We write Bernoulli's equation as:




7

1

2 v2

2 = ( P 1 - P 2) +1

2 v1

2 + g ( y1 - y2).

The last term goes to zero because the syringe is in a horizontal position. Also, we realize that P 1

- P 2 = P 1 - P atm = P 1gauge = 8.00 x 104 Pa.

Finally, we assume v1 = 0 in comparison to the speed inside the needle. Thus, with these

substitutions, we find v2 = 12.6 m/s.

9.37 P = P 2 - P 1 =1

2 v1

2 - v22) (ignoring differences in height), so

P =1

2 (1.29 kg/m3)( )((0.15 m/s)2 - (0.30 m/s)2) = -4.4 x 10-2 Pa

9.38 First, consider the path from the standpoint of projectile motion to find the speed at which the water

emerges from the tank. The time to drop one meter with an intial vertical velocity of zero is:

y = voyt +1

2 at 2 1.00 m = 0 +

1

2 (9.80 m/s2) t 2, or t = 0.452 s,

and from the horizontal motion: vx = vo = xt

=0.600 m

0.452 s = 1.33 m/s.

We now use Bernoulli's equation, with point 1 at the top of the tank and point 2 at the level of the

hole. With P 1 = P 2 = P atm and v1 approximately equal to zero, we have: 12 v2

2 = g ( y1 - y2) =

gh, giving

h =vo

2

2 g =

(1.33 m/s)2

2(9.80 m/s2) = 9.00 x 10-2 m = 9.00 cm.

9.39 (a) We choose point 1 at the surface of the tank and point 2 at the hole. Both of these points are

at atmospheric pressure, so the pressure cancels from Bernoulli's equation. We also assume

that v1 is negligibly small. Finally, we choose the zero level for y at the level of the hole.

Under these conditions, we have:

v2 = 2 gy = 2 g (16.0 m) = 17.7 m/s.

(b) The area of the hole is found from the flow rate as:

A = flow ratev2

= 4.17 x 10-5 m3/s17.7 m/s

= 2.35 x 10-6 m2.

From which the diameter is easily found to be 1.73 mm.

9.40 First, find the velocity inside the larger portions:

v1 =

flow rate

A1 =

1.80 x 10-4 m3/s

4.91 x 10-4 m2 = 0.367 m/s.

The absolute pressure in the large section (to left) is

P 1 = P atm + gh1 = P atm + (1000 kg/m3)(9.80 m/s2)(0.10 m)

= P atm + 980 Pa.

In the absolute pressure in the constriction is:

P 2 = P atm + (1000 kg/m3)(9.80 m/s2)(0.050 m) = P atm + 490 Pa.

Now use Bernoulli's equation:

v22 = v1

2 +2

( P 1 - P 2) = (0.367 m/s)2 +

2(490 Pa)

1000 kg/m2

where we have chosen y = 0 at the level of the centerline of the pipe.

This yields v2 = 1.06 m/s, and from the flow rate, we find:

A2 =

flow rate

v2 =

1.80 x 10-4 m3/s

1.06 m/s = 1.71 x 10-4 m2 =

d 22

4 ,

from which d 2 = 1.5 x 10-2 m = 1.5 cm.




8

9.41 (a) The flow rate, Av, as given may be expressed as follows:

25 liters/30 s = 0.833 liters/s = 833 cm3/s.

The area of the faucet tap is cm2, so we can find the velocity as

v2 =flow rate

A2 =

833 cm3/s

cm2 = 265 cm/s = 2.65 m/s.

(b) Similarly, the velocity in the main pipe is:

v1 =flow rate

A1 =

833 cm3/s

(3.0 cm)2 = 29.5 cm/s = 0.295 m/s.

Now, we use Bernoulli's equation with point 1 to be in the main pipe and point 2 to be at the

faucet tap:

P 1 - P 2 =1

2 v2

2 - v12) + g ( y2 - y1), which gives

P 1 - P 2 =1

2 (103 kg/m3) [(2.65 m/s)2 - (0.295 m/s)2] +

(103 kg/m3)(9.80 m/s2)(2.00 m), or

P gauge = P 1 - P 2 = 2.3 x 104 Pa.

9.42 (a) Using Bernoulli's equation with point 1 at the top of the tank and point 2 at the exit from thetube, we have:

P 1 +1

2 v1

2 + gy1 = P 2 +1

2 v2

2 + gy2, where P 1 = P 2 = P atm.

For a large tank v1 is approximately equal to zero, and y1 - y2 = h. This gives v2 =

2 gh .

(b) Use Bernoulli's equation with point 1 at the top of the tank and point 2 at the highest point in

the tube:

P 1 +1

2 v1

2 + gy1 = P 2 +1

2 v2

2 + gy2, where P 1 = P atm.

When the siphon ceases to work, the fluid will be at rest at point 2,

or v2 = v1 = 0, and y2 - y1 = ymax, so P atm = P + gymax.

The minimum value of P is 0. Therefore, ymax = P atm g

.

9.43 Because there are two edges (the inside and outside of the ring) we have,

= F

Ltotal =

F

2(circumference) =

F

4r =

1.61 x 10-2 N

2.20 x 10-1 N = 7.32 x 10-2 N/m

9.44 The tension in the string attaching the sheet to the balance is equal to the sum of the vertical

component of the surface force plus the weight of the sheet. This is a two sided surface, so the

surface force is F = 2 L. Since L = 3.0 x 10-2 m, the vertical component of this force is:

F v = 2 Lcos = (6 x 10-2 m) cos When = 0°, the tension measures 0.4 N and we have:

T = F v + w becoming 0.40 N = +(6 x 10-2

m) + w. (1)When = 180°, T = 0.39 N, and we have

T = F v + w becoming 0.39 N = -(6 x 10-2 m) + w. (2)

Solving equations (1) and (2) simultaneously gives = 8.3 x 10-2 N/m.

9.45 The height the blood can rise is given by

h =2 cos

gr =

2(5.8 x 10-2 N/m)

(1050 kg/m3)(9.80 m/s2)(2.0 x 10-6 m) = 5.6 m




9

9.46 The vertical component of the surface force is equal to the weight of water inside the capillary tube:

F v = w, or

Lcos = (2r )cos = w = (r 2)hg , where h is the height of the water in the tube. We solve

the above for the surface tension.

= rhg

2cos =

(1080 kg/m3)(5.0 x 10-4 m)(2.1 x 10-2 m)(9.80 m/s2)

2

which yields = 5.6 x 10-2 N/m

9.47 We have

h = 5 x 10-2 m =2 cos

gr =

2(8.80 x 10-2 N/m)

(1035 kg/m3)(9.80 m/s2)r

From which, we find r = 3. 47 x 10-4 m, or a diameter of 0.694 mm.

9.48 From the definition of the coefficient of viscosity, we have

F = Av

L =

(1.79 x 10-3 Ns/m2)(0.80 m)(1.2 m)(0.5 m/s)

10-4 m = 8.6 N

9.49 From the definition of the coefficient of viscosity, we have

F = Av

L =

(1500 x 10-3 Ns/m2)(4.00 x 10-4 m2)(0.30 m/s)

1.50 x 10-3 m = 0.12 N

9.50 From Poiseuille's law: P 1 - P 2 =(flow rate)8 L

R4

P 1 - P 2 =(8.6 x 10-5 m3/s)8(0.12 Ns/m2)(50 m)

(5.0 x 10-3 m)4

= 2.1 x 106 Pa = 20.7 atm.

Also, since P 2 = 1.0 atm, this is also the gauge pressure at the inlet point of the pipe.

9.51 Flow rate =( P ) R4

8 L

=(400 Pa)(2.6 x 10-3 m)4

8(2.7 x 10-3

Pa s)(8.4 x 10-2

m)

= 3.16 x 10-5 m3/s

Then v =flow rate

area =

3.16 x 10-5 m3/s

( 2.6 x 10-3 m)2 = 1.5 m/s

9.52 From Poiseuille's law: P =(flow rate)8L

R 4 .

For water, 1 g/s = 1 cm3/s = 10-6 m3/s. Therefore,

P =(10-6 m3/s)8(0.03 m)(1.0 x 10-3 N s/m2)

(1.5 x 10-4 m)4 = 1.5 x 105 Pa

9.53 The required flow rate =500 cm3

1800 s = 2.78 x 10-1 cm3/s = 2.78 x 10-7 m3/s

If the solution is elevated 1 m, the pressure differential across the needle is P = gy = (1000

kg/m3)(9.80 m/s2)(1.0 m) = 9800 Pa

We find the radius via Poiseuille's law. R4 =8 L (flow rate)

P , or

R4 =8(2.5 x 10-2 m)(1.0 x 10-3 N s/m2)(2.78 x 10-7 m3/s)

(9800 Pa)




10

From which R = 2.06 x 10-4 m = 0.206 mm, or diameter = 0.41 mm.

9.54 The Reynold's number is

RN = vd

(1050 kg/m3)(0.55 m/s)(2.0 x 10-2 m)

2.7 x 10-3 N s/m2 = 4.3 x 103.

In this region ( RN > 3000), the flow is turbulent.

9.55 From the definition of the Reynolds's number,

vmax =( RN max)

d =

(2000)(10-3 Ns/m2)

(103 kg/m3)(2.5 x 10-2 m) = 8.0 x 10-2

m

s = 8.0

cm

s

9.56 Fick's law enables us to find the difference in concentration as

C =(diffusion rate) L

DA =

(5.33 x 10-15 kg/s)(0.10 m)

(5.0 x 10-10 m2/s)(6.0 x 10-4 m2) = 1.8 x

10-3 kg/m3

9.57 We use Fick's law to find the diffusion coefficient.

D =

diffusion rate

A

C

L =

5.7 x 10-15 kg/s

(2.0 x 10-4 m2)(3.0 x 10-2 kg/m4)

= 9.5 x 10-10 m2/s

9.58 From Stoke's law, F = 6 rv. Therefore,

= F

6rv =

3.0 x 10-13 N

6(2.5 x 10-6 m)(4.5 x 10-4 m/s) = 1.4 x 10-5 N s/m2

9.59 We use vt =2r 2 g

9 ( - f )

or ( - f ) =9 vt

2r 2

g

=9(1.0 x 10-3 N s/m2)(1.10 x 10-2 m/s)

2(5.0 x 10-4

)2

(9.80 m/s2

)

= 20.2 kg/m3.

Thus, = f + 20.2 kg/m3 = 1000 kg/m3 + 20.2 kg/m3 = 1.02 x 103 kg/m3

9.60 If at the end of one hour a particle is still in suspension, then its terminal velocity must be less than

5.0 cm/h = 1.39 x 10-5 m/s. Thus, we use vt =2r 2 g

9 ( - f ) to find

r 2 =9 vt

2 g ( - f ) =

9(1.00 x 10-3 N s/m2)(1.39 x 10-5 m/s)

2(9.80 m/s2)(800 kg/m3)

and r = 2.82 x 10-6 m = 2.82 microns is the size of the largest particles that can stillremain in suspension.

9.61 A1 =

A2v2

v1 , where A2 = 1.96 x 10

-5

m

2

= area of aorta.Then A1 = total capillary cross-section needed.

A1 =(1.96 x 10-5 m2)(1.0 m/s)

10-2 m/s = 1.96 x 10-3 m2

But, A1 = (number of capillaries) Asingle capillary, and

Asingle capillary = 7.85 x 10-11 m2.

Thus, number of capillaries =1.96 x 10-3 m2

7.85 x 10-11 m2 = 2.5 x 107 = 25 million.




11

9.62 (a) The speed at the narrow section is found from the equation of continuity:

v2 = A1v1

A2 =

d 12

d 22 v1 = 16 v1 = 16.00 m/s.

(b) We choose the zero level for y at the common center line of the horizontal pipes, and solve the

resulting form of Bernoulli's equation for the pressure at the narrow section, point 2.

P 2 = P 1 + 12 (v1

2 - v22)

= (3.00 x 105 Pa) +1

2 (1000 kg/m3)(1.00 m/s) 2 - (16.00 m/s)2)

= 1.73 x 105 Pa

9.63 When the sinker alone is submerged, the forces on the system are T 1, the tension in the string

attached to the block, wB, the weight of the block, ws, the weight of the sinker, and Bs, the buoyant

force on the sinker. Since equilibrium exists, we have

T 1 = wB + ws - Bs. (1)

When both are submerged, the forces on the system are T 2, the tension in the string attached to the

block, the weight of the block, the buoyant force on the block, BB, and the buoyant force on the

sinker. From the first condition for equilibrium, we haveT 2 = wB + ws - BB - Bs. (2)

Subtract (2) from (1) to give: T 1 - T 2 = BB.

Thus, the buoyant force on the block is: BB = 200 N - 140 N = 60.0 N.

However, the buoyant force on the block is equal to the weight of water displaced by the block:

BB = 60 N = wV block g.

Thus, V block =60.0 N

(1000 kg/m3)(9.80 m/s2) = 6.12 x 10-3 m3.

The mass of the 50 N block is 5.10 kg. So, its density is

=m

V =

5.10 kg

6.12 x 10-3 m3 = 833 kg/m3.

9.64 The forces on the balloon while in flight are Ba, the buoyant force of the air, wHe, the weight of the

helium, wB, the weight of the balloon, and wL, the weight of the load. These quantities are found as

follows,

Ba = aV balloon g , wHe = HeV balloon g , wB = (600 kg) g , and wL = (4000 kg) g .

When floating in equilibrium, we have: Ba = wHe + wB + wL, or aV balloon g =

HeV balloon g + (600 kg) g + (4000 kg) g .

The density of Helium is 0.179 kg/m3 and the density of air is

1.29 kg/m3. Thus, we can solve for the volume of the balloon to find

V balloon = 4.14 x 103 m3.

9.65 When the balloon comes into equilibrium, we must have

F y = B - w balloon - wHe - wstring = 0

where wstring is the weight of the string above the ground, and B is the buoyant force. Thus, wstring = B - w balloon - wHe. (1)

The mass per unit length of the string is:

=m

L =

0.05 kg

2.0 m = 2.5 x 10-2 kg/m.

Thus, wstring = hg = (0.025 kg/m)(9.80 m/s2)h; B = air Vg = 3.39 N;

w balloon = m balloon g = 2.45 N; and wHe = HeVg = 0.470 N.

Equation (1) above then becomes:




12

h =3.39 N - 2.45 N - 0.470 N

(2.5 x 10-2 kg/m)(9.80 m/s2) =

0.470 N

0.245 kg/s2 = 1.9 m

9.66 Let us find the tension to stretch the wire by 0.1 mm.

The area = A =d 2

4 = 3.80 x 10-8 m2. Thus, the force is

F = YA( L) Lo

= (18 x 1010 Pa)(3.80 x 10-8 m2)(10-4 m)3.1 x 10-2 m

= 22 N.

We have an equilibrium situation, so

F x = 0 becomes F cos30° - F cos 30° = 0,

F y = 0 becomes 2 F sin30° = 2(22.1 N)sin30° = 22 N directed down the page in the textbook

figure.

9.67 Solve by work-energy methods taking where the object is released from rest 10 m above the surface as

the initial state and where the object comes to rest distance d below the surface of the water as the

final state. The work-energy equation becomes:

W nc = ( KE f - KE i) + ( PE f - PE i) = (0 - 0) + (mgyf - mgyi).

If we neglect friction effects and energy loss upon impact, the only non-conservative force doing

work is the buoyant force, which acts as a retarding force after the object is in the water. Thus,

W nc = - F bd = -( water V o g )d .

Here V o is the volume of the object. Also, m = oV o where o is the density of the object.

Choosing the zero of gravitional potential energy at the water surface, the work-energy equation

becomes:

-( water V o g )d = ( oV o ) g [(-d ) - 10 m]. This reduces to

d = ( o / water )(d + 10 m). With o/ water = 0.60, this last equation yields d = 15 m.

9.68 When the bar of soap, of cross-sectional area A, is in water only, the forces on it are the buoyant

force of the water and its weight, w. Because it is floating, we know Bw = w. The buoyant

force, Bw , is

Bw = w( A)(1.5 x 10-2 m) g = w

From this equation, we find

w

Ag = w(1.5 x 10-2 m) (1)

When the bar is floating in both water and oil, the

forces on it are the buoyant force of the oil, the buoyant

force of the water, and its weight. We have

Bo + Bw = w. (2)

Let us call x the height of the bar that is in oil. Thus,

the portion of the height of the bar which is in water is

(2 x 10-2 m - x). Equation (2) becomes

o( A)( x) g + w A(2.0 x 10-2 m - x) g = w

or, o x + w(2.0 x 10-2 m - x) =w

Ag .

From (1) above, this is o x + w(2.0 x 10-2 m - x) =

w(1.5 x 10-2 m).

When we substitute 1000 kg/m3 for the density of water and 600 kg/m3 for the density of oil, we

find x = 1.25 x 10-2 m = 1.3 cm.

9.69 Let s stand for the edge of the cube, h for the depth of immersion, ice stand for the density of the

ice, w stand for the density of water, and a stand for density of the alcohol.

(a) According to Archimedes' principle, at equilibrium we have

W

Bw

(Water)

W 2 - X(Water)

X

BBwo

Oil

(in water alone)

(in oil and water )

1.5 cm2 cm




13

ice gs3 = w ghs2

giving h = s ice

w

With ice = 0.917 x 103 kg/m3, w = 1.00 x 103 kg/m3, and s = 20 mm

we get h = 20(0.917) = 18.34 mm

(b) We assume that the top of the cube is still above the alcohol surface. Letting ha stand for thethickness of the alcohol layer, we have

a gs2ha + w gs2hw = ice gs3

giving hw = ice

w s -

a

w ha

With a = 0.806 x 103 kg/m3, and ha = 5 mm we obtain

hw = 18.34 - (0.806)(5) = 14.31 mm

(c) Here hw' = s - ha' , so Archimedes' principle gives

a gs2ha' + w gs2( s - ha') = ice s3

Leading to aha' + w( s - ha') = ice s

From which, ha' =

s( w - ice)

( w - a) = 20

1.000 - 0.917

1.000 - 0.806 = 8.557 mm

9.70 (a) P = E

t =

mgh

t =

m

t gh = Rgh

(b) P EL = 0.85(8.50 x 105)(9.80)(87) = 616 MW

9.71 (a) Consider the pressure at points A and B (at the same level inthe two tubes).

Using the left tube:

P A = P atm + a gh + w g ( L - h),

where the second term is due to the variation of air pressure

with altitude.

Using the right tube: P B = P atm + 0 gL.

But Pascal's principle says that P A = P B. Therefore,

P atm + o gL = P atm + a gh + w g ( L - h), or

( w - a)h = ( w - o) L, giving

h =

w - o

w - a L =

1000 - 750

1000- 1.29(5.00 cm) = 1.25 cm.

(b) Consider the diagram at the right showing the

situation when the air flow over the left tube

equalizes the fluid levels in the two tubes. First,

apply Bernoulli's equation to points A and B ( yA

= yB, vA = v, and vB = 0). This gives:

P A

+1

2

av2 +

a gy

A = P

B +

1

2

a(0)2 +

a gy

B

and since yA = yB, this reduces to:

P B - P A =1

2 av2. (1)

Now consider points C and D, both at the level of

the oil-water interface in the right tube. Using the variation of pressure with depth in static

fluids, we have:

P C = P A + a gH + w gL, and P D = P B + a gH + o gL.

But Pascal's principle says that P C = P D. Equating these two gives:

air

water

h

L = 5 cm

Point A

Point B

PatmPatm

oil




14

P B + a gH + o gL = P A + a gH + w gL, or

P B - P A = ( w - o) gL. (2)

Substitute equation (1) for P B - P A into (2) to obtain

1

2 av2 = ( w - o) gL, or

v =2 gL(

w -

o)

a = 2(9.80 m/s2)(0.05 m) 1000 - 7501.29 = 13.8 m/s.

9.72 Consider the diagram and apply Bernoulli's equation to

points A and B, taking y = 0 at the level of point B, and

recognizing that vA is approximately zero. This gives:

P A +1

2 w0)2 + w g (h - Lsin ) =

P B +1

2 wvB)2 + w g (0).

Now, recognize that P A = P B = P atmosphere since both

points are open to the atmosphere (neglecting variation of atmospheric pressure with altitude).

Thus, we obtain

vB = 2 g (h - Lsin ) = 2(9.80 m/s2)(10.0m - (2.00 m)sin30°) = 13.3 m/s.

Now the problem reduces to one of projectile motion.

voy = vBsin30° = 6.64 m/s. Then v2y = v2

oy + 2a( y) gives at the top of the arc (where y =

ymax and vy = 0)

0 = (6.64 m/s)2 + 2(-9.80 m/s2)(ymax - 0), or

ymax = 2.3 m (above the level where the water emerges)

9.73 The pressure on the ball is given by: P = P atm + w gh

so the change in pressure on the ball when it is on the surface of the ocean to when it is at the

bottom of the ocean is P = w gh.

In addition, from the definition of bulk modulus, this increase in pressure will change the volume ofthe ball as follows:

V =-V P

B , where B is the bulk Modulus. This change in volume can be written as V = Vf -

Vi =4

3 r f

3 -4

3 r i

3 =4

3 r f 3 - r i

3). Equating the expressions for V gives4

3 r f 3 - r i

3) =4

3 r i

3

P

B ,

or r f 3 = r i

3(1 -P

B ). Thus, the final radius of the ball is

r f = r i

1 -

P

B 1/3

The decrease in diameter of the ball is D = 2(r i - r f )

or D = 2r i 1 - 1 -P

B1/3

= Di 1 - 1 -P

B1/3

= (3.00 m)

1 -

1 -1.01 X 108 PA

14 X 1010 PA

1/3 = 7.2 X 10-4 m = 0.72 mm

9.74 Let V =4

3 r 3 be the volume of the ball, D the depth of the water, m the mass of the ball. Then

while the ball is under the water

(m + mair )a = B - (m + mair ) g , or

ymax

h L

L sin

vB

Point BPoint A




15

(m + air V )a = water Vg - (m + air V ) g , so a =( water - air )V - m

m + air V g .

a =

(1000 - 1.29)(kg/m3)

4

3 (0.10 m)3 - 1.0 kg

1.0 kg + 1.29 (kg/m3) 4

3 (0.10)3

(9.80 m/s2) = 31.0 m/s2.

Then we use v2 = v2o + 2a( y) = 0 + 2( )31.0 m/s2(2.0 m) = 124.1 m2/s2,

to get the velocity of the ball as it emerges from the pool.

This gives: v = 11.1 m/s.

Finally, we use1

2 (m + mair ) v2 = (m + mair ) gh to get h = 6.3 m.

9.75 (a) The pressure on the surface of the two hemispheres is constant at all points and the force on

each element of surface area is directed along the radius of the hemispheres. The applied forcealong the axis must balance the force on the "effective" area which is the projection of the

actual surface onto a plane perpendicular to the x axis, A = R2. Therefore,

F = ( P atm - P ) R2.

(b) For the values given

F = ( P atm - 0.10 P atm)(0.30 m)2 = 0.254 P atm = 2.6 x 104 N

9.76 We call the position of the lower hole point 1 and the position of the higher hole point 2. Both of

these points are at atmospheric pressure.

From our earlier study of projectile motion, the range is given by

R = v1t 1 = v2t 2. (1)

An expression for the time for the water to reach the floor can also be found through the projectile

motion equations. We have

y = voyt +1

2 at 2 = 0 +

1

2 gt 2, or t =

2

g h where h is the height through which the

projectile falls before it reaches ground level. Thus, we have the time for the water from the upper

hole to reach the floor to be:

t 2 =

2

g(0.120 m) , and the time for the water from the lower hole as

t 1 =2

g (0.0500 m) .

Substitute these two values for the times into (1) above to give

v12 = 2.4v2

2. (2)

Since the pressure cancels from Bernoulli's equation, it reduces to

v12 - v2

2) = 2 g ( y2 - y1). (3)

Given: y2 = 0.120 m, y1= 0.0500 m. Then, with the use of (2) and (3) we find

v2 = 0.99 m/s.

Let us now consider Bernoulli's Equation once again, with a point 3 at the top of the tank and point

2 still at the position of the top hole. Both these points are at atmospheric pressure, and the velocity

of fall of water at the top of the tank is negligibly small. We find:

gy3 =1

2 v2

2 + gy2, or g ( y3 - y2) = gh =1

2 v2

2.

From which, h =v2

2

2 g =

(0.99 m/s)2

2(9.80 m/s2) = 5 x 10-2 m = 5.0 cm.

Thus, the water surface is 5.0 cm above the top hole or 17.0 cm above the bottom of the tank.




16

9.77 [Refer to the above diagram for this problem.]

Since the Block floats, Total buoyant force = weight of block, or

F B(oil) + F B(water) = weight of block. This becomes:

oil( ) A(4.00 x 10-2 m - x) g + water ( Ax) g = block ( ) A(4.00 x 10-2 m) g

where A(4.00 x 10-2 m - x) = volume of oil displaced, and Ax = volume of water displaced.

Canceling A, and substituting into the equation above, we have:

(930 kg/m3)(4.00 x 10-2 m - x) + (1000 kg/m3) x

= (960 kg/m3)(4.00 x 10-2 m)

Solving for x: x = 1.71 x 10-2 m = 1.71 cm.

9.78 To have the object float fully submerged in the fluid, its average density must be the same as that of

the fluid. Therefore, we must add ethanol to water until the density of the mixture is 900 kg/m3. If

m = mass of mixture, me = mass of ethanol, and mw = mass of water, we obtain:

m = me + mw. (1)

For the volumes: V = V e + V w . (2)

Equation (1) becomes: V = eV e + wV w and with the use of equation ( 2), this reduces to

w(V e + V w) =

e

w(V e) + V w, or

0.9(V e + V w) = (0.806)(V e) + V w

which yields: Ve = 1.064Vw = (1.064)(500 cc) = 532 cm3.

ANSWERS TO CONCEPTUAL QUESTIONS

2. Both must have the same strength. The force on the back of each dam is the average pressure of the water times the area of the dam. If both reservoirs are equally deep, the

force is the same.

4. The external pressure on the chest cavity makes it difficult to take a breath while under

water. Thus, a snorkel will not work in deep water.

6. The buoyant force depends on the amount of air displaced by the objects. Since they

have the same dimensions, the buoyant force will be the same on each.

8. The larger the density of a fluid, the higher does an object float in it. Thus, an objectwill float lower in low density alcohol.

10. The water level on the side of the glass stays the same. The floating ice cube displacesits own weight of liquid water, and so does the liquid water into which it melts.

Area of top = A

4 cm = 0.04 m

X

.04 m - X Oil

Water




17

12. A breeze from any direction speeds up to go over the mound, and the air pressuredrops at this opening. Air then flows through the burrow from the lower to the upper

entrance.

14. No. The somewhat lighter barge will float higher in the water.

16. The rapidly moving air over the roof of the house reduces the downward pressure on it.However, the stagnant air inside the house keeps the pressure the same as it was in theabsence of the storm. This pressure differential can be large enough to push off the roof.

Opening windows helps to equalize the pressure inside and out, providing minimal protection in a storm.

solucionario capitulo 9 física serway and faughn

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