Problemas Resueltos Cap 14 Fisica Sears Zemansky

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PROBLEMAS RESUELTOS

FISICA UNIVERSITARIA

CAPITULO 14

VOLUMEN 1

EDICION 11 SEARS ZEMANSKY

Sección 14.1 Densidad

Sección 14.2 Presión de un fluido Sección 14.3 Flotación

Sección 14.4 Fuerzas de flotación y principio de Arquímedes Sección 14.5 Ecuación de BERNOULLI

Sección 14.6 Viscosidad y turbulencia

Erving Quintero Gil Ing. Electromecánico Bucaramanga – Colombia

2010

Para cualquier inquietud o consulta escribir a:

[email protected] [email protected] [email protected]

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Problema 14.1 Sears Zemansky

En un trabajo de medio tiempo, un supervisor le pide traer del almacén una varilla cilíndrica de acero de 85,8 cm de longitud y 2,85 cm de diámetro. ¿Necesitara usted un carrito? (Para contestar calcule el peso de la varilla) g * V * g * m w= = ρ

ρ = 7,8 * 103_kg/m3_{(densidad del acero)}

V = volumen de la varilla de acero g = gravedad = 9,8 m/seg2

V = área de la varilla * longitud de la varilla

d = diámetro de la varilla = 2,85 cm = 0,0285 metros l = longitud de la varilla = 85,8 cm = 0,858 m 2 4 -2 2

m

10 *

6,3793

4 0,002551

4 0,00081225

*

3,14

4 (0,0285)

*

3.14

4 d

varilla

la

de

area

=

π

=

V = 6,3793 * 10-4_{* 0,858 = 5,4735 * 10}-4_m3 g * V * w=ρ Newton 41,83 seg m 9,8 * m 10 * 5,4735 * m kg 10 * 7,8 w -4 3 ₂ 3 3 ₌ = No es necesario el carrito.

El radio de la luna es de 1740 km. Su masa es de 7,35 * 1022_{kg. Calcule su densidad media?}

V = volumen de la luna

(3)

3 r * * 3 4 V=

π

(

₄

)

3 10 * 174 * 3,14 * 3 4 V= 3 12 12 _22066647,3₂_*₁₀ _m 10 * 5268024 * 3,14 * 3 4 V= = m = masa de la luna = 7,35 * 1022_kg.

(

)

3 .

33

10 kg

m

.

m

10 *

32 ,

22066647

kg

10

35 ,

7

3 3 3 12 22

×

=

×

=

V

m

ρ

Imagine que compra una pieza rectangular de metal de 5 * 15 * 30 mm y masa de 0,0158 kg. El vendedor le dice que es de oro. Para verificarlo, usted calcula la densidad media de la pieza. Que valor obtiene? ¿Fue una estafa?

V = volumen de pieza rectangular V = 5 * 15 * 30 mm = 2250 mm3

( )

(

)

₁₀9 0,00000225m3 3 m 2250 3 mm 1000 3 m 1 * 3 mm 2250 V= = =

m = masa de la pieza rectangular de metal = 0,0158 kg.

(

)

₇_,₀₂₂ ₁₀3 _kg _m3_. 3 m 0,00000225 kg 0158 . 0 ₌ _× = = Vm ρ

La densidad del oro = 19,3 * 103_kg/m3

La densidad de la pieza rectangular de metal 7,022 * 103_kg/m3

Por lo anterior fue engañado.

Un secuestrador exige un cubo de platino de 40 kg como rescate . ¿Cuánto mide por lado? L = longitud del cubo

V = volumen del cubo. V = L * L * L = L3

3 V L=

m = masa del platino = 40 kg.

ρ = la densidad del platino = 21,4 * 103_kg/m3

m V ρ = 3 m 3 -10 * 1,869158 3 m kg 3 10 * 21,4 kg 40 m V= = = ρ

(4)

3 V L= m 0,12318 3_1,869158_*₁₀-3_m3 L= =

L = longitud del cubo = 12,31 cm

Una esfera uniforme de plomo y una de aluminio tienen la misma masa. ¿Qué relación hay entre el radio de la esfera de aluminio y el de la esfera de plomo?

mplomo = masa esfera de plomo

maluminio = masa esfera de aluminio

mplomo = Volumen de la esfera de plomo * densidad del plomo

3 plomo r * * 3 4 plomo V = π plomo * 3 plomo r * * 3 4 plomo m = π ρ 3 aluminio r * * 3 4 aluminio V = π aluminio * 3 aluminio r * * 3 4 aluminio m = π ρ mplomo = maluminio aluminio * 3 aluminio r * * 3 4 plomo * 3 plomo r * * 3 4 _π _ρ ₌ _π _ρ

Cancelando términos semejantes aluminio * 3 aluminio r plomo * 3 plomo r ρ = ρ Despejando

(

)

(

)

3 plomo r aluminio r 3 plomo r 3 aluminio r aluminio plomo ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = = ρ ρ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = plomo r aluminio r 3 1 ) aluminio plomo ( ρ ρ

ρ aluminio = 2,7 * 103 kg/m3 (densidad del aluminio)

(5)

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = plomo r aluminio r 3 1 ) 3 10 * 7 , 2 3 10 * 3 , 11 (

(

)

_⎟⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = plomo r aluminio r 3 1 1851 , 4 1,61 plomo r aluminio r = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛

a) Calcule la densidad media del sol. B) Calcule la densidad media de una estrella de neutrones que tiene la misma masa que el sol pero un radio de solo 20 km.

msol = masa del sol = 1,99 * 1030 kg.

Vsol = volumen del sol

rsol = radio del sol = 6,96 * 108 m

( )

3 sol r * * 3 4 sol V = π 3 8 10 * 6,96 * 3,14 * 3 4 V= ⎜_⎝⎛ ⎟_⎠⎞ 3 m 24 10 * 1412,2654 24 10 * 337,15 * 3,14 * 3 4 V= = a) 3 m kg 3 10 * 1,409 3 m 24 10 * 2654 , 1412 kg 30 10 99 . 1 sol V sol ₌ × ₌ = m ρ

m estrella = masa de la estrella de neutrones = 1,99 * 1030 kg.

V estrella = volumen de la estrella de neutrones

rsol = radio de la estrella de neutrones = 20000 m

(

)

3 estrella r * * 3 4 estrella V = π

(

₂₀₀₀₀

)

3 * 3,14 * 3 4 V= 3 m 12 10 * 33,510321 12 10 * 8 * 3,14 * 3 4 V= = b) 0.05938 1018 kg m3 3 m 12 10 * 510321 , 33 kg 30 10 99 . 1 × = × = D =5.93*1016 kg m3

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¿A que profundidad del mar hay una presión manométrica de 1 * 105_Pa?

h * g * 0 p p− =ρ

ρ = 1,03 * 103 kg/m3 (densidad del agua de mar) g = gravedad = 9,8 m/seg2 9,906m 3 m Newton 10,094 2 m Newton 100000 ) 2 s m 80 , 9 ( * ) 3 m kg 3 10 * 03 , 1 ( Pa 5 10 * 1 g * 0 p p h= − = = = ρ

En la alimentación intravenosa, se inserta una aguja en una vena del brazo del paciente y se conecta un tubo entre la aguja y un depósito de fluido (densidad 1050 kg/m3_{) que esta a una altura h sobre el brazo.}

El deposito esta abierto a la atmosfera por arriba. Si la presión manométrica dentro de la vena es de 5980 Pa, ¿Qué valor minino de h permite que entre fluido en la vena?. Suponga que el diámetro de la aguja es lo bastante grande para despreciar la viscosidad (sección 14.6) del fluido.

La diferencia de presión entre la parte superior e inferior del tubo debe ser de al menos 5980 Pa para forzar el líquido en la vena

p – p0 = Pa 5980 h * g * = ρ

ρ = 1050 kg/m3_{(densidad del fluido)}

g = gravedad = 9,8 m/seg2 m 581 , 0 3 m Newton 10290 2 m Newton 5980 ) 2 s m 80 , 9 ( ) 3 m kg 1050 ( 2 m N 5980 g * Pa 5980 h= = = = ρ h = 58,1 cm

Un barril contiene una capa de aceite (densidad de 600 kg/m3_{) de 0,12 m sobre 0,25 m de agua.}

a) Que presión manométrica hay en la interfaz aceite-agua? b) ¿Qué presión manométrica hay en el fondo del barril? ρ aceite = 600 kg/m3 (densidad del aceite)

g = gravedad = 9,8 m/seg2 a) *g*h 600kg m3 9,80m s2⎟

(

0,12m

)

=706Pa. ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = ρ

ρ agua = 1000 kg/m3 (densidad del agua)

g = gravedad = 9,8 m/seg2

haceite = 0,12 m

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b) 706Pa+⎛_⎝⎜1000kg m3⎞_⎠⎟⎛⎜_⎝9,8m s2⎞_⎠⎟

(

0,25m

)

=706Pa+2450Pa=3156Pa. Problema 14.10 Sears Zemansky

Una vagoneta vacía pesa 16,5 KN. Cada neumático tiene una presión manométrica de 205 KPa (29,7 lb/pulg2).

Calcule el área de contacto total de los neumáticos con el suelo. (Suponga que las paredes del neumático son flexibles de modo que la presión ejercida por el neumático sobre el suelo es igual a la presión de aire en su interior.)

Con la misma presión en los neumáticos, calcule el área después de que el auto se carga con 9,1 KN de pasajeros y carga.

The pressure used to find the area is the gauge pressure, and so the total area is W = Peso de la vagoneta vacia = 16,5 KN. = 16500 Newton

P= presión manométrica de cada neumático = 205 KPa = 205000 Pa.

A W P= ⋅ = = = 0,0804m2 ) Pa 205000 ( ) N 16500 ( P W A A = 804 cm2 b) A W P=

(

)

( )

= ⋅ = = + = = 1248cm2 2 m 1 2 cm 100 * 2 m 0,1248 2 m N 205000 N 25600 ) Pa 205000 ( ) N 9100 N 16500 ( P W A A = 1248 cm2

Se esta diseñando una campana de buceo que resista la presión del mar a 250 m de profundidad. a) cuanto vale la presión manométrica a esa profundidad.

b) A esta profundidad, ¿Qué fuerza neta ejercen el agua exterior y el aire interior sobre una ventanilla circular de 30 cm de diámetro si la presión dentro de la campana es la que hay en la superficie del agua? (desprecie la pequeña variación de presión sobre la superficie de la ventanilla)

a)_ρgh₌(1.03_×103kg m3)(9.80m s2)(250m)₌2.52_×106Pa.

b) The pressure difference is the gauge pressure, and the net force due to the water and the air is

N.

10

78 .

1 )

)

m

15 .

0 (

)(

Pa

10

52 .

2 (

_×

6

_π

2

₌

_×

5

¿Qué presión barométrica (en Pa y atm. ) debe producir una bomba para subir agua del fondo del Gran cañón (elevación 730 m) a Indian Gardens (elevación 1370 m)?

atm.

61.9 Pa

10

27 .

6 )

m

640 )(

s

m

80 .

9 )(

m

kg

10

00 .

1 (

_×

3 3 2

₌

_×

6

₌

=

= ρgh

p

(8)

El liquido del manómetro de tubo abierto de la fig 14.8 a es mercurio, y1 = 3 cm y y2 7 cm. La presion

atmosferica es de 980 milibares.

a) ¿Qué presion absoluta hay en la base del tubo en U? b) Y en el tubo abierto 4 cm debajo de la superficie libre? c) Que presion absoluta tiene el aire del tanque?

d) ¿Qué presion manométrica tiene el gas en pascales? a) + =980×102Pa+(13.6×103kg m3)(9.80m s2)(7.00×10−2 m)= 2 a ρgy p Pa. 10 07 .

1 _× 5 _{b) Repeating the calcultion with}

₄

_.

₀₀

1

2

−

=

y

cm instead of

c) The absolute pressure is that found in part (b), 1.03

Pa.

10

1.03 gives

5 2

×

y

Pa. 105 ×

d) (this is not the same as the difference between the results of parts (a) and (b) due to roundoff error).

Pa

10

33 .

5 )

(

3 1 2

−

y

ρg

=

×

y

Hay una profundidad máxima la que un buzo puede respirar por un snorkel (Fig. 14.31) pues, al aumentar la profundidad, aumenta la diferencia de presión que tiende a colapsar los pulmones del buzo. Dado que el snorkel conecta los pulmones con la atmosfera, la presión en ellos es la atmosférica. Calcule la diferencia de presión interna-externa cuando los pulmones del buzo están a 6,1 m de profundidad. Suponga que el buzo esta en agua dulce. (un buzo que respira el aire comprimido de un tanque puede operar a mayores profundidades que uno que usa snorkel, por que la presión del aire dentro de los pulmones aumenta hasta equilibrar la presión externa del agua)

Pa.

10

0 .

6 )

m

1 .

6 )(

s

m

80 .

9 )(

m

kg

10

00 .

1 (

_×

3 3 2

₌

_×

4

=

ρgh

Un cilindro alto con área transversal de 12 cm2_{se lleno parcialmente con mercurio hasta una altura de 5}

cm. Se vierte lentamente agua sobre el mercurio (los dos líquidos no se mezclan). ¿Qué volumen de agua deberá añadirse para aumentar al doble la presión manométrica en la base del cilindro.

With just the mercury, the gauge pressure at the bottom of the cylinder is

p

=

p

₀

+

p

_m

gh

_m_⋅ With the water to a depth

h

_w, the gauge pressure at the bottom of the cylinder is

p

=

p

₀

+

ρ

_m

gh

_m

+

p

_w

gh

_w

.

If this is to be double the first value, then

ρ

_w

gh

_w

=

ρ

_m

gh

_m.

( ) (0.0500m)(13.6 103 1.00 103) 0.680m w m m w =h ρ ρ = × × = h

The volume of water is

_V

₌

_h

_A

₌

_(0.680

_m)(12.0

_×

₁₀

−4

_m

2

₎

₌

_8.16

_×

₁₀

−4

_m

3

₌

₈₁₆

_cm

3

(9)

Un recipiente cerrado se llena parcialmente con agua. en un principio, el aire arriba del agua esta a presion atmosferica (1,01 * 105 Pa) y la presion manometrica en la base del recipiente es de 2500 Pa. Despues, se bombea aire adicional al interior aumentando la presion del aire sobre el agua en 1500 Pa)

(10)

a) Gauge pressure is the excess pressure above atmospheric pressure. The pressure difference between the surface of the water and the bottom is due to the weight of the water and is still 2500 Pa after the pressure increase above the surface. But the surface pressure increase is also transmitted to the fluid, making the total difference from atmospheric 2500 Pa+1500 Pa = 4000 Pa.

b) The pressure due to the water alone is 2500 Pa=ρgh. Thus

m

255 .

0 )

s

m

80 .

9 (

)

m

kg

1000

(

m

N

2500

2 3 2

=

h

To keep the bottom gauge pressure at 2500 Pa after the 1500 Pa increase at the surface, the pressure due to the water’s weight must be reduced to 1000 Pa:

m

102 .

0 )

s

m

80 .

9 )(

m

kg

1000

(

m

N

1000

2 3 2

=

h

Thus the water must be lowered by 0.255m−0.102m=0.153m

The force is the difference between the upward force of the water and the downward forces of the air and the weight. The difference between the pressure inside and out is the gauge pressure, so

N.

10

27 .

2 N

300 )

m

75 .

0 (

m)

30 (

)

s

m

80 .

9 (

)

10

03 .

1 (

)

(

₋

₌

_×

3 2 2

₋

₌

_×

5

=

ρgh

A

w

F

[

130 _×

10

3

Pa

₊

(

1 .

00 _×

10

3

kg

m

3

)(

3 .

71 m

s

2

)(

14 .

2 m)

₋

93 _×

10

3

Pa

]

(2.00

m

2

)

N.

10

79 .

1 _×

5

=

The depth of the kerosene is the difference in pressure, divided by the product

_ρg

₌

mg_V

,

m.

14 .

4 )

m

250 .

0 (

)

s

m

kg)(9.80

205 (

Pa

10

01 .

2 )

m

(0.0700

N)

10

4 .

16 (

3 2 5 2 3

=

×

−

×

=

h

atm.

64 .

1 Pa

10

66 .

1 m)

15 .

0 (

)

s

m

kg)(9.80

1200

(

)

2 (

5 2 2 2

=

×

=

π

d

π

mg

A

F

p

(11)

The buoyant force must be equal to the total weight;

ρ

_water

V

g

=

ρ

_ice

V

g

+

mg

,

so

, m 563 . 0 m kg 920 m kg 1000 kg 0 . 45 3 3 3 water = − = − = ice ρ ρ m V or 3 to two figures. m 56 . 0

14.22: The buoyant force is B=17.50N−11.20N=6.30N, and

. m 10 43 . 6 ) s m 80 . 9 )( m kg 10 00 . 1 ( N) 30 . 6 ( 4 3 2 3 3 water − × = × = = g ρ B V The density is

.

m

kg

10

78 .

2

30 .

6

50 .

17 )

m

kg

10

00 .

1 (

3 3 3 3 water water

×

=

⎟

⎠

⎞

⎜

⎝

⎛

×

=

B

w

ρ

g

ρ

B

g

w

V

m

ρ

Problema 14.23 Sears Zemansky

Un objeto con densidad media ρ flota sobre un fluido de densidad ρfluido

a)?Que relacion debe haber entre las dos densidades?

b) A la luz de su respuesta a la parte(a), ¿Cómo pueden flotar barcos de acero en el agua?

c) En terminos de ρ y ρfluido ¿Qué fraccion del objeto esta sumergida y que fraccion esta sobre el fluido?

Verifique que sus respuestas den el comportamiento correcto en el limite donde ρ → ρfluido y donde ρ →

0.

d) durante un paseo en yate, su primo Tito recorta una pieza rectangular (dimensiones: 5 * 4 *3 cm) de un salvavidas y la tira al mar, donde flota. La masa de la pieza es de 42 gr. ¿Qué porcentaje de su volumen esta sobre la superficie ?

a) The displaced fluid must weigh more than the object, so

ρ

<

ρ

_fluid

.

b) If the ship does not leak, much of the water will be displaced by air or cargo, and the average density of the floating ship is less than that of water. c) Let the portion submerged have volume V, and the total volume be

⋅ = = fluid 0 so , Then, . _o _fluid 0 ρ ρ VV V ρ V

V

ρ

The fraction above the fluid is then

1 .

If

0,

fluid

→

−

_PP

p

_{the entire}

object floats, and if

ρ

→

ρ

_fluid, none of the object is above the surface. d) Using the result of part (c),

%.

32

32 .

0 m

kg

1030

)

m

10

3.0

4.0 (5.0

kg)

042 .

0 (

1

₃ 3 -6 fluid

=

×

−

=

−

ρ

(12)

Un cable anclado al fondo de un lago de agua dulce sostiene una esfera hueca de plástico bajo la superficie. El volumen de la esfera es de 0,650 m3_{y la tensión en el cable es de 900 N.}

a) Calcule la fuerza de flotación ejercida por el agua sobre la esfera. b) ¿Qué masa tiene la esfera

b) El cable se rompe y la esfera sube a la superficie. En equilibrio, ¿Qué fracción de volumen de la esfera estará sumergida?

a)

(

1.00 103kg m3

)(

9.80m s2

)(

0.650m3

)

6370N.

water = × =

= ρ gV B

b)

m

=

_gw

=

B_g−T

=

6370_9.80N_m-900_s2N

=

558 kg.

c) (See Exercise 14.23.) If the submerged volume is V ′,

%. 9 . 85 859 0. N 6370 N 5470 and water water = = = = ′ = ′ gV ρ w V V g ρ w V

Un bloque cubico de Madera de 10 cm. Por lado flota en la interfaz entre aceite y agua con su superficie inferior 1,5 cm. Bajo la interfaz (fig 14.32). La densidad del aceite es de 790 kg/m3

a) ¿Qué presion manometrica hay en la superficie de arriba del bloque? b) ¿y en la cara inferior

c) ¿Qué masa y densidad tiene el bloque? a)

ρ

_oil

gh

_oil

=

116 Pa.

b)

(

790 kg

m

3

)

(

0 .

100 m

)

₊

(

1000

kg

m

3

)

(

0 .

0150

m

)

)(

9 .

80 m

s

2

)

₌

921 Pa.

c)

(

)

(

₍

)(

₎

)

0.822kg. s m 80 . 9 m 100 . 0 Pa 805 2 2 top bottom− ₌ ₌ = = g A p p g w m

The density of the block is ₍ ₎3

822

_m3

.

kg m 10 . 0 kg 822 . 0

₌

=

p

Note that is the same as the average density of the fluid displaced,

(

0 .

85 )

(

790 kg

m

3

)

₊

(

0 .

15 )

(

1000

kg

m

3

)

.

(13)

Un lingote de aluminio sólido pesa 89 N en el aire. a) ¿Qué volumen tiene?

b) el lingote se cuelga de una cuerda y se sumerge por completo en agua. ¿Qué tensión hay en la cuerda (el peso aparente del lingote en agua)?

a) Neglecting the density of the air,

(

)

(

)(

)

3.36 10 m , m kg 10 7 . 2 s m 80 . 9 N 89 3 3 3 3 2 − × = × = = = = gρ w ρ g w ρ m V 3 3_m 10 4 . 3 or _× − _{to two figures.} b)

(

)

56 .

0

N.

7 .

2

00 .

1

1 N

89

1

aluminum water water

⎟

=

⎠

⎞

⎜

⎝

⎛ −

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

=

−

=

−

=

ρ

ω

V

gρ

w

B

w

T

a) The pressure at the top of the block is

p

=

p

₀

+

ρ

gh

,

where is the depth of the top of the block below the surface. is greater for block

Β

, so the pressure is greater at the top of block

Β

.

h

b)B=

ρ

_flV_objg. The blocks have the same volume V_objso experience the same buoyant force. c) T −w+B=0 so T =w−B.

The object have the same

V

but is larger for brass than for aluminum so is larger for the brass block.

. ρVg

w= ρ

w

Bis the same for both, so T is larger for the brass block, block

B.

The rock displaces a volume of water whose weight is 39.2N-28.4N=10.8N. The mass of this much water is thus

10 .

8 N

9.80 m

s

2

₌

1 .

102 kg

_{and its volume, equal to the rock’s volume, is}

3 3 3 3_kg _m 1.102 10 m 10 1.00 kg 102 . 1 ₌ _× − ×

The weight of unknown liquid displaced is 39.2N−18.6N=20.6N,and its mass is

kg.

102 .

2 s

m

9.80 N

6 .

20

2

₌

_{The liquid’s density is thus}

₂

_.

₁₀₂

_kg

_1.102

_×

₁₀

−3

_m

3

₌

₁

_.

₉₁

_×

₁₀

3

_kg

_m

3

_,

_or

roughly twice the density of water.

(14)

)

(

,

₂ ₁ ₁ ₂ 2 2 1 1

Α

v

Α

v

Α

v

=

Α

₁

=

π

(

0 .

80 cm)

2

,

Α

₂

=

20 π

(

0 .

10 cm)

2

9 .

6 m

s

)

10 .

0 (

20 (0.80)

s)

m

0 .

3 (

₂ 2 2

=

π

v

⋅

=

2 3 2 2 2 1 1 2

s

m

245 .

0 )

m

0700

.

0 s)(

m

50 .

3 (

Α

v

a) (i)

0 .

1050

m

,

2 .

33 m

s.

(ii)

0 .

047 m

2

,

₂

5 .

21 m

s.

2 2 2 2

=

v

=

Α

=

v

=

Α

b)

(0.245

m

3

s

)

(

3600

s)

882 m

3

.

2 2 1 1

Α

t

=

υ

Α

t

=

v

a)

16 .

98 .

)

m

150 .

0 (

)

s

m

20 .

1 (

2 3

=

π

A

dt

dV

v

b)

r

₂

=

r

₁

v

₁

v

₂

=

(

dV

dt

)

πv

₂

=

0 .

317 m

.

a) From the equation preceding Eq. (14.10), dividing by the time interval dt gives Eq. (14.12). b) The volume flow rate decreases by 1.50% (to two figures).

Un tanque sellado que contiene agua de mar hasta una altura de 11 metros contiene también aire sobre el agua a una presión manométrica de 3 atmósferas. Sale agua del tanque a través de un agujero pequeño en el fondo. Calcule la rapidez de salida del agua

The hole is given as being “small,”and this may be taken to mean that the velocity of the seawater at the top of the tank is zero, and Eq. (14.18) gives

El agujero se da como "pequeño", y esto puede interpretarse en el sentido que la velocidad del agua de mar en la parte superior del tanque es cero, y la ecuación. (14.18) es

))

(

2 gy

p

ρ

(15)

=

2 ((

9 .

80 m

s

2

)(

11 .

0 m)

₊

(3.00)(1.0

13 _×

10

5

Pa)

(1.03

_×

10

3

kg

m

3

))

=

28 .

4 m

s.

Note that y = 0 and were used at the bottom of the tank, so that p was the given gauge pressure at the top of the tank.

a

p

=

Tenga en cuenta que y = 0 y p = se utilizaron en la parte inferior del tanque, de modo que p es la p0

presión manométrica dada en la parte superior del tanque Problema 14.34 Sears Zemansky

Se corta un agujero circular de 6 mm de diámetro en el costado de un tanque de agua grande, 14 m debajo del nivel del agua en el tanque. El tanque esta abierto al aire por arriba. Calcule

a) la rapidez de salida

b) el volumen descargado por unidad de tiempo. v = velocidad en m/seg. g = gravedad = 9,8 m/seg2 h = altura en metros. h * g * 2 2 v = seg m 16,56 2 seg 2 m 274,4 m 14 * 2 seg m 9,8 * 2 h * g * 2 v= = = =

b) el volumen descargado por unidad de tiempo. V = volumen en m3_{/ seg} V = v * A A = area = π * r2 r = 3 mm = 0,003 m A = π * r2_{= 3,14 * (0,003)}2_{= 2,8274 * 10}-5_m2 V = v * A = 16,56 m/seg * 2,8274 * 10-5_m2 V = 4,68 * 10-4_m3_/seg

¿Que presión manométrica se requiere en una toma municipal de agua para que el chorro de una manguera de bomberos conectada a ella alcance una altura vertical de 15 m?. (Suponga que la toma tiene un diámetro mucho mayor que la manguera).

The assumption may be taken to mean that

v

₁

=

0

in Eq. (14.17). At the maximum height,

v

₂

=

0 ,

and using gauge pressure for

p

₁

and

p

₂

,

p

₂

=

0

(the water is open to the atmosphere),

Pa 5 10 * 1,47 2 m Newton 147000 m 15 * 2 seg m 9,8 * 3 m kg 1000 2 y * g * 1 p = ρ = = = p1 = 1,47 * 105 Pa

Problema 14.36 Sears Zemansky Using ₄1 ₁ 2

v

=

in Eq. (14.17),

_⎥

⎦

⎤

⎢

⎣

⎡

₊

₋

⎟

⎠

⎞

⎜

⎝

⎛

+

=

−

+

−

+

=

(

)

32

15 )

(

)

(

2

1

2 1 2 1 1 2 1 2 2 2 1 1 2

p

ρ

v

ρg

y

p

ρ

υ

g

y

p

(16)

_⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ × + × = (3.00m s) (9.80m s )(11.0m) 32 15 ) m kg 10 00 . 1 ( Pa 10 00 . 5 4 3 3 2 2

₌

1 .

62 _×

10

5

Pa.

Neglecting the thickness of the wing (so that

y

₁

=

y

₂in Eq. (14.17)), the pressure difference is

Pa.

0

78 )

(

2)

1 (

2 1 2 2

−

=

Δ

p

ρ

v

The net upward force is then

N.

496 )

s

m

kg)(9.80

1340

(

)

m

(16.2

Pa)

780 (

_×

2

₋

2

₌

₋

a)(220)(₆₀0_..₀355_s kg)

₌

1 .

30 kg

s.

_{b) The density of the liquid is}

, m kg 1000 3 m 10 0.355 kg 355 . 0 3 3 = −

× and so the volume flow

rate is

1 .

30

10

3

m

3

s

1.30 L

s.

m kg 1000 s kg 30 . 1 3

=

×

=

− _{This result may also be obtained}

from (220)(₆₀0_..₀355_s L)

₌

1 .

30 L

s.

_c) 2 4 3 3 m 10 00 . 2 s m 10 30 . 1 1 − − × ×

=

v

s.

m

63 .

1

4 s,

m

50 .

6

₂

=

₁

=

v

d)

(

2

)

(

₂ ₁

)

1 2 2 2 1 2 1 y y ρg v v ρ p p = + − + −

( )

(

)

(

) (

)

(

1000

kg

m

)(

9 .

80 m

s

)

(

1 .

35 m

)

s

m

50 .

6 s

m

63 .

1 m

kg

1000

2

1 kPa

152

2 3 2 2 3

−

+

−

+

=

=119kPa

The water is discharged at a rate of 3 2

0 .

352 m

s.

3 4 m 10 32 . 1 s m 10 65 . 4 1

=

−

=

− × ×

v

The pipe is given as horizonatal, so the speed at the constriction is 2 2 8.95m s,

1

2 = v + Δp ρ=

v keeping an extra figure, so the cross-section are at the constriction is

5 .

19

10

5

m

2

,

s m 95 . 8 s m 10 65 . 4 _× 4 3 ₋

×

=

−

and the radius is

r

=

A

π

=

0 .

41 cm.

Problema 14.40 Sears Zemansky From Eq. (14.17), with

y

₁

=

y

₂

,

(

)

₁ ₁2 2 1 2 1 1 2 2 2 1 1 2

₈

3

4

2

1

2

1 ρv

p

v

ρ

p

v

ρ

p

_⎟⎟

=

+

⎠

⎞

⎜⎜

⎝

⎛

−

+

=

−

+

=

(

1.00 10 kg m

)

(

2.50m s

)

2.03 10 Pa, 8 3 Pa 10 80 . 1 _× 4 ₊ _× 3 3 2 ₌ _× 4 =

where the continutity relation

2

1 2

v

=

has been used.

(17)

Let point 1 be where

r

₁

=

4 .

00 cm

and point 2 be where

r

₂

=

2 .

00 cm.

The volume flow rate has the value

s

cm

7200

3 _{at all points in the pipe.}

s

m

43 .

1 so

,

cm

7200

3 ₁ 2 1 1 1 1

A

=

v

πr

=

v

=

v

s

m

73 .

5 so

,

cm

7200

3 ₂ 2 2 2 2 2

A

=

v

πr

=

v

=

v

2 2 2 2 2 1 1 1 2 1 2 1 ρv ρgy p ρv ρgy p + + = + +

(

)

2.25 10 Pa 2 1 so Pa, 10 40 . 2 and 2 5 2 2 1 1 2 5 2 2 1= y p = × p = p + ρv −v = × y

a) The cross-sectional area presented by a sphere is _D₄2

,

π

therefore

(

₀

)

_D₄2

.

π

p

F

=

−

b) The force on each hemisphere due to the atmosphere is

_π

(

₅

_.

₀₀

_×

₁₀

−2

_m

)

2

(

₁

_.

₀₁₃

_×

₁₀

5

_Pa

)

(

₀

_.

₉₇₅

)

₌

₇₇₆

_Ν

_.

a)

_ρgh

₌

(

1 .

03 _×

10

3

kg

m

3

)(

9 .

80 _×

m

s

2

)(

10 .

92 _×

10

3

m

)

₌

1 .

10 _×

10

8

Pa.

b) The fractional change in volume is the negative of the fractional change in density. The density at that depth is then

(

)

(

3 3

) (

(

8

)(

11 1

)

01 1.03 10 kg m 1 1.16 10 Pa 45.8 10 Pa − − × × + × = Δ + =ρ k p ρ

₌

1 .

08 _×

10

3

kg

m

3

,

A fractional increase of Note that to three figures, the gauge pressure and absolute pressure are the same.

%.

0 .

5

Problema 14.44 Sears Zemansky a) The weight of the water is

(

1 .

00 _×

10

3

kg

m

3

)(

9 .

80 m

s

2

)

(

5 .

00 m

)(

4 .

0 m

)(

3 .

0 m

)

₌

5 .

88 _×

10

5

N,

=

ρgV

or to two figures. b) Integration gives the expected result the force is what it would be if the pressure were uniform and equal to the pressure at the midpoint;

N

10

9 .

5 _×

5 2 d gA F =

ρ

₌

(

1 .

00 _×

10

3

kg

m

3

)(

9 .

80 m

s

2

)

(

4 .

0 m

)(

3 .

0 m

)

)(

1 .

50 m

)

₌

1 .

76 _×

10

5

N,

or

1 .

8 _×

10

5

N

to two figures.

(18)

)

Let the width be w and the depth at the bottom of the gate be The force on a strip of vertical thickness at a depth is then and the torque about the hinge is

.

H

dh

h

dF

=

ρgh

(

wdh

dτ

=

ρgwh

(

h

−

H

2 dh

)

;

integrating from

h

=

0 to

h

=

H

gives

_τ

₌

_ρ

_g

_ω

_H

3

12 ₌

2 .

61 _×

10

4

N

_⋅

m.

a) See problem 14.45; the net force is

∫

dF

from

_h

₌

0 to

_h

₌

_H

,

_F

₌

_ρ

_g

_ω

_H

2

2 ₌

_ρ

_gAH

2 ,

_where

.

H

A

=

ω

b) The torque on a strip of vertical thickness about the bottom is and integrating from

dh

(

H

h

)

gwh

(

H

h

)

dh

,

dF

dτ

=

−

=

ρ

−

h

=

0 to

h

=

H

gives

_τ

₌

_ρgwH

3

6 ₌

_ρgAH

2

6 .

c) The force depends on the width and the square of the depth, and the torque about the bottom depends on the width and the cube of the depth; the surface area of the lake does not affect either result (for a given width).

The acceleration due to gravity on the planet is d p ρd p g V m Δ = Δ =

and so the planet’s mass is

mGd

pVR

G

gR

M

2 2

_Δ

=

The cylindrical rod has mass M, radius R, and length L with a density that is proportional to the square of the distance from one end,

ρ

=

Cx

2.

a) The volume element Then the integral becomes

Integrating gives

.

2

_dV

Cx

dV

M

=

∫

ρ

=

∫

_dV ₌_πR2_dx. . 2 2 0 Cx R dx M ₌_∫L

_π

_.

3

3 2 2 0 2

_x

_dx

_C

_R

L

R

C

M

₌

_π

_∫

L

₌

_π

_{Solving for}

_C

_,

_C

₌

₃

_M

_π

_R

2

_L

3

_.

b) The density at the

x

=

L

end is

( )

2 3

( )

2

.

3 2 3 2 L πR M L πR M

_L

Cx

ρ

=

The denominator is just the total volume V, so

ρ

=

3 M

V

,

or three times the average density,

M

V

.

So the average density is one-third the density at the

x

=

L

end of the rod.

(19)

.

m

kg

10

3.15 m)

10

37 .

6 )(

m

kg

10

50 .

1 (

m

kg

700 ,

12

3

₋

_×

3 4

_×

6

₌

_×

3 3

=

−

=

_A

_BR

−

ρ

b), c)

∫

⎟⎟

_⎢⎣

⎡ −

_⎥⎦

⎤

⎠

⎞

⎜⎜

⎝

⎛

=

⎥

⎦

⎤

⎢

⎣

⎡

−

=

−

=

dm

π

R

A

Br

r

dr

π

AR

BR

πR

A

BR

M

0 3 4 3 2

4

3

4

3

4 ]

[

4 _⎥

⎦

⎤

⎢

⎣

⎡

₋

×

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

×

=

−

4 m)

10

37 .

6 )(

m

kg

10

50 .

1 (

3 m

kg

700 ,

12

3 m)

10

37 .

6 (

4

3 4 6 3 3 6

π

₌

5 .

99 _×

10

24

kg,

which is within 0.36% of the earth’s mass. d) If is used to denote the mass contained in a sphere of radius

m

(r)

,

r then

_g

₌

_Gm

(

_r

)

_r

2

.

_{Using the same integration as that in part (b), with an upper limit of} rinstead of R gives the result.

e)

_g

₌

0 at

_r

₌

0 ,

and

_g

at

_r

₌

_R

,

_g

₌

_Gm

(

_R

)

_R

2

₌

(

6 .

673 _×

10

−11

N

_⋅

m

2

kg

2

)

.

s

m

85 .

9 m)

10 (6.37

kg)

10

99 .

5 (

_×

24

_×

6 2

₌

2 f)

;

2

3

4

3

4

2

⎥⎦

⎤

⎢⎣

⎡ −

⎟

⎠

⎞

⎜

⎝

⎛

=

⎥

⎦

⎤

⎢

⎣

⎡

−

⎟

⎠

⎞

⎜

⎝

⎛

=

Ar

Br

πG

A

Br

dr

d

πG

dr

dg

setting ths equal to zero gives

_r

₌

2 _A

3 _B

₌

5 .

64 _×

10

6

m

_{, and at this radius}

⎥

⎦

⎤

⎢

⎣

⎡

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

−

⎟

⎠

⎞

⎜

⎝

⎛

⎟

⎠

⎞

⎜

⎝

⎛

=

B

A

B

A

B

A

πG

g

3

2

4

3

2

3

4 B

πGA

9

4

2

=

10 .

02 m

s

.

)

m

kg

10

50 .

1 (

9 )

m

kg

700 ,

12 (

)

kg

m

N

10

673 .

6 (

4

2 4 3 2 3 2 2 11

=

×

⋅

×

=

π

− ₋

a) Equation (14.4), with the radius rinstead of height y, becomes

dp

=

−

ρg

dr

=

−

ρg

_s

(

r

R

)

dr

.

This form shows that the pressure decreases with increasing radius. Integrating, with p=0at r =R,

∫

=

∫

= − − = r R r R R r ρg dr r R ρg dr r R ρg p s s R s₍ 2 2_). 2

b) Using the above expression with

r

=

0 and

ρ

=

_VM

=

₄3_πM_R3

,

Pa.

10

71 .

1 m)

10

38 .

6 (

8 )

s

m

kg)(9.80

10

97 .

5 (

3 )

0 (

11 2 6 2 24

×

=

×

=

π

p

(20)

c) While the same order of magnitude, this is not in very good agreement with the estimated value. In more realistic density models (see Problem 14.49 or Problem 9.99), the concentration of mass at lower radii leads to a higher pressure.

a) (1.00 103kg m3)(9.80m s2)(15.0 10 2 m) 1.47 103 Pa. water water = × × = × − gh ρ

b) The gauge pressure at a depth of 15.0

cm

−

h

below the top of the mercury column must be that found in part (a); ρ_Hgg(15.0cm−h)=ρ_waterg(15.0cm), which is solved for h=13.9cm.

Problema 14.52 Sears Zemansky Following the hint,

∫

=

h o 2

)

)(2

(

ρgy

πR

dy

ρgπRh

F

where R and h are the radius and height of the tank (the fact that

2 R

=

h

is more or less coincidental). Using the given numerical values gives

_F

₌

5 .

07 _×

10

8

N.

14.53: For the barge to be completely submerged, the mass of water displaced would need to be kg. 10 1.056 ) m 12 40 )(22 m kg 10 (1.00 3 3 3 7 waterV = × × × = ×

ρ The mass of the barge itself is

kg,

10

39 .

7 )

m

10

0 .

4 )

40

22

12 )

40

22 (

2 ((

)

m

kg

10

8 .

7 (

_×

3 3

₊

_×

₊

_×

−2 3

₌

_×

5

so the barge can hold of coal. This mass of coal occupies a solid volume of which is less than the volume of the interior of the barge

kg

10

82 .

9 _×

6

,

m

10

55 .

6 _×

3 3

₍

₁

_.

₀₆

_×

₁₀

4

_m

3

_),

but the coal must not be too loosely packed.

The difference between the densities must provide the “lift” of 5800 N (see Problem 14.59). The average density of the gases in the balloon is then

. m kg 96 . 0 ) m 2200 )( s m 80 . 9 ( ) N 5800 ( m kg 23 . 1 3 3 2 3 ave = − = ρ

Problema 14.55 Sears Zemansky a) The submerged volume

is

,

so

waterg ρ w

V ′

(21)

⋅ = = × = = = ′ % 30 30 . 0 ) m (3.0 ) m kg 10 00 . 1 ( ) kg 900 ( 3 3 3 water water V ρ m V g ρ w V V

b) As the car is about to sink, the weight of the water displaced is equal to the weight of the car plus the weight of the water inside the car. If the volume of water inside the car is

V ′′

,

⋅ = = − = − = ′′ ′′ + = ,or 1 1 0.30 0.70 70% g Vp w V V g p V w g Vρ water water water

a) The volume displaced must be that which has the same weight and mass as the

ice, 3 cm gm 00 . 1 gm 70 . 9

₉

_.

₇₀

_cm

3

=

(note that the choice of the form for the density of water avoids conversion of

units). b) No; when melted, it is as if the volume displaced by the of melted ice displaces the same volume, and the water level does not change. c)

gm 70 . 9

⋅

=

3 cm gm 05 . 1 gm 70 . 9

cm

24 .

9

3 d) The melted water takes

up more volume than the salt water displaced, and so flows over. A way of considering this situation (as a thought experiment only) is that the less dense water “floats” on the salt water, and as there is insufficient volume to contain the melted ice, some spills over.

3

cm

46 .

0

The total mass of the lead and wood must be the mass of the water displaced, or

;

)

(

_Pb _wood _water wood wood Pb Pb

ρ

V

ρ

V

ρ

V

+

=

+

solving for the volume

V

_Pb

,

water Pb wood water wood Pb

_ρ

ρ

−

= V

V

₃ ₃ ₃ ₃ 3 3 3 3 2

m

kg

10

00 .

1 m

kg

10

3 .

11 m

kg

600 m

kg

10

00 .

1 )

m

10

2 .

1 (

×

−

×

−

×

=

−

₌

4 .

66 _×

10

−4

m

3

,

which has a mass of 5.27 kg.

The fraction f of the volume that floats above the fluid is 1 ,

fluid

ρ

− =

f where is the average density of

the hydrometer (see Problem 14.23 or Problem 14.55), which can be expressed as ρ

.

1

fluid

f

ρ

−

=

(22)

Thus, if two fluids are observed to have floating fraction . 1 1 , and 2 1 1 2 2 1 f f ρ ρ f f − −

= In this form, it’s clear that a larger

f

₂ corresponds to a larger density; more of the stem is above the fluid. Using

097 .

0 ,

242 .

0

₂ (3.20₍cm)(0.400₁₃_.₂_cm ₎cm ) ) cm 2 . 13 ( ) cm cm)(0.400 00 . 8 ( 1 3 2 3 2

=

f

gives (0.839 ) 839kg m3. water alcohol = ρ = ρ

Problema 14.59 Sears Zemansky a) The “lift” is V(ρair−ρH₂)g, from which

. m 10 0 . 11 ) s m 80 . 9 )( m kg 0899 . 0 m kg (1.20 N 000 , 120 3 3 2 3 3 ₋ = × = V

b) For the same volume, the “lift” would be different by the ratio of the density differences,

N. 10 2 . 11 N) 000 , 120 ( 4 H air He air 2 × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − ρ ρ ρ ρ

This increase in lift is not worth the hazards associated with use of hydrogen.

a) Archimedes’ principle states

,

so

.

A

M

L

Mg

gLA

ρ

=

b) The buoyant force is

ρ

gA(L+x)=Mg+F, and using the result of part (a) and solving for x gives

_x

₌

_ρgAF

.

c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F, is k = ρgA, and the period of oscillation is

. 2 2 ρgA M π k M π T = =

1: a)

₍

(

₎

)

(

0.450m

)

0.107m. m kg 10 03 . 1 kg 0 . 70 2 3 3 = × = = = = π ρA m ρgA mg ρgA w x

b) Note that in part (c) of Problem 14.60 ,M is the mass of the buoy, not the mass of the man, and Ais the cross-section area of the buoy, not the amplitude. The period is then

(23)

(

)

(

1.03

10 kg

m

)(

9 .

80 m

s

)

(

0 .

450 m

)

2 .

42 s.

kg

950

2

₃ ₃ ₂ ₂

=

×

=

π

T

To save some intermediate calculation, let the density, mass and volume of the life preserver

be

ρ

₀

,

m

and

v

,

and the same quantities for the person be Then, equating the buoyant force and the weight, and dividing out the common factor of

.

and

,

1

M

V

ρ

, g

(

)

(

0 .

80 )

₀ ₁

,

water

V

v

ρ

v

ρ

V

ρ

+

=

+

Eliminating

V

in favor of

ρ

₁ and M, and eliminating m in favor of

ρ

₀ and v,

(

0 .

80 )

.

1 water 0

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

+

v

ρ

M

ρ

M

v

ρ

Solving for

ρ

₀

,

(

)

_⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + = v M ρ M ρ v ρ 1 water 0 0.80 1

(

)

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−

=

1 water water

1

0 .

80 ρ

ρ

v

M

ρ

_⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

₋

×

−

×

=

3 3 ₃ 3 ₃ 3

m

kg

980 m

kg

10

1.03 (0.80)

1 m

0.400 kg

0 .

75 m

kg

10

03 .

1 =

732 kg

m

3

⋅

To the given precision, the density of air is negligible compared to that of brass, but not compared to that of the wood. The fact that the density of brass may not be known the three-figure precision does not matter; the mass of the brass is given to three figures. The weight of the brass is the difference between the weight of the wood and the buoyant force of the air on the wood, and canceling a common factor of

and )

(

,V_wood ρ_wood ρ_air M_brass,

g − = 1 wood air brass air wood wood brass wood wood wood 1 − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = − = = ρ ρ M ρ ρ ρ M V ρ M kg. 0958 . 0 m kg 150 m kg 20 . 1 1 ) kg 0950 . 0 ( 1 3 3 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = −

(24)

The buoyant force on the mass A, divided byg ,must be7.50kg−1.00kg−1.80kg=4.70kg(see Example 14.6), so the mass block is 4.70kg+3.50kg=8.20kg.a) The mass of the liquid displaced by the block is 4.70kg,so the density of the liquid is

1 .

24

10

3

kg

m

3

.

m 10 3.80 kg 70 . 4 3 3 -

=

×

× b) Scale D will read the

mass of the block, as found above. Scale E will read the sum of the masses of the beaker and liquid, kg, 20 . 8 kg. 80 . 2

Neglecting the buoyancy of the air, the weight in air is

N.

0 .

45 )

(

ρ

_Au

V

_Au

+

ρ

_A1

V

_A1

=

g

and the buoyant force when suspended in water is

N.

6.0 N

39.0 N

45.0 )

(

_Au _A1 water

V

+

V

g

=

−

=

ρ

These are two equations in the two unknowns

V

_Au

and

V

_A1

.

Multiplying the second by

ρ

_A1 and the first by

water

ρ

and subtracting to eliminate the

V

_A1term gives

ρ

_water

V

_Au

g

(

ρ

_Au

−

ρ

_A1

)

=

ρ

_water

(

45 .

0 N)

−

ρ

_A1

(

6 .

0 N)

( (45.0N) (6.0)) ) ( _Au _A1 water Au water Au Au Au Au ρ ρ ρ ρ ρ ρ gV ρ w − − = =

((

1 .

00 )(

45 .

0 N)

(

2 .

7 )(

6 .

0 N))

)

7 .

2

3 .

19 )(

00 .

1 (

)

3 .

19 (

₋

−

=

=33.5N.

Note that in the numerical determination of

w

_Au

,

specific gravities were used instead of densities.

Problema 14.66 Sears Zemansky The ball’s volume is

3 3 3 ₍₁₂_.₀_cm) ₇₂₃₈_cm 3 4 3 4 ₌ ₌ = πr π V

As it floats, it displaces a weight of water equal to its weight. a) By pushing the ball under water, you displace an additional amount of water equal to 84% of the ball’s volume or

This much water has a mass of

.

cm

6080

)

cm

7238

)(

84 .

0 (

3

₌

3 ₆₀₈₀_g ₌₆_.₀₈₀_kg_{and weighs}

N,

6 .

59 )

s

m

kg)(9.80

080 .

6 (

2

₌

_{which is how hard you’ll have to push to submerge the ball.}

b) The upward force on the ball in excess of its own weight was found in part (a): The ball’s mass is equal to the mass of water displaced when the ball is floating: