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Unformatted text preview: quation (2) yields V oil = 5.38 × 10 –3 m 3 . SOLUTION The height of the cylinder that is in the oil is, therefore, hoil = Voil πr2 = 5.38 × 10 –3 m 3
= 7.6 × 10 –2 m
π ( 0.150 m) 2 52. REASONING AND SOLUTION Only the weight of the block compresses the spring.
Applying Hooke's law gives W = kx. The spring is stretched by the buoyant force acting on
the block minus the weight of the block. Hooke's law again gives FB – W = 2kx.
Eliminating kx gives FB = 3W. Now FB = ρwgV, so that the volume of the block is 594 FLUIDS 3 3 V = 3M/ρw = 3(8.00 kg)/(1.00 × 10 kg/m ) = 2.40 × 10 –2 3 m The volume of wood in the block is
3 Vw = M/ρb = (8.00 kg)/(840 kg/m ) = 9.52 × 10 –3 3 m The volume of the block that is hollow is V – Vw = 1.45 × 10−2 m3. The percentage of the
block that is hollow is then
–2 –2 100(1.45 × 10 )/(2.40 × 10 ) = 60.3 % . 53. REASONING AND SOLUTION The figure at the right shows the
forces that act on the balloon as it holds the passengers and the ballast
stationary above the earth. W0 is the combined weight of the balloon,
the load of passengers, and the ballast. The quantity FB is the buoyant
force provided by the air outside the balloon and is given by F B W FB = ρairgVballoon (1) 0 Since the balloon is stationary, it follows that FB − W0 = 0, or
FB = W0 (2) Then the ballast is dropped overboard, the balloon accelerates upward through a distance y
2y
in a time t with acceleration ay, where (from kinematics) ay = 2 .
t
The figure at the right shows the forces that act on the balloon while it
accelerates upward. F B Applying Newton's second law, we have FB − W = may, where W is the
weight of the balloon and the load of passengers. Replacing m by (W/g)
we have FB − W = W
a
gy Thus, FB = W +
Solving for W gives W
g 2y 2y 2 = W 1 + 2 t gt W Chapter 11 Problems W= 595 gt 2 = FB 2 gt + 2 y 2y 1+ 2 gt FB The amount of ballast that must be thrown overboard is therefore [using Equations (1) and
(2)] gt 2 gt 2 ∆W = W0 − W = FB − FB 2
ρ gV = 1 − 2 gt + 2 y gt + 2 y air balloon ∆W = rair g ( 4
3 ) 3
ρ rballoon 1 − gt + 2 y gt 2 2 (9.80 m/s 2 )(15.0 s) 2
∆W = (1.29 kg/m3 )(9.80 m/s 2 ) 4 π (6.25 m)3 1 − 3
2
2 (9.80 m/s )(15.0 s) + 2(105 m) = 1120 N 54. REASONING The speed v of the gasoline in the fuel line is related to its mass flow rate,
the density ρ of the gasoline, and the crosssectional area A of the fuel line by
Mass flow rate = ρ Av (Equation 11.7). Solving Equation 11.7 for the speed v, we obtain v= Mass flow rate
ρA (1) SOLUTION The fuel line has a circular crosssection with a radius r. Therefore, its crosssectional area is A = π r 2 . Equation (1) then yields the speed of the gasoline in the fuel line: v= Mass flow rate
5.88 ×10−2 kg/s
=
ρπ r 2
735 kg/m3 π 3.18 ×10−3 m ( )( ) 2 = 2.52 m/s 55. SSM REASONING AND SOLUTION The mass flow rate Qmass is the amount of fluid
mass that flows per unit time. Therefore, Qmass = m ρV (1030 kg/m3 )(9.5 × 10 –4 m3 ) 1.0 h −5
=
= = 4.5 × 10 kg/s
t
t
6.0 h
3600 s 596 FLUIDS 56. REASONING The volume flow rate (in cubic meters per second) of the falling water is
the same as it was when it left the faucet. This is because no water is added to or taken out
of the stream after the water leaves the faucet. With the volume flow rate unchanging, the
equation of continuity applies in the form A1v1 = A2v2 (Equation 11.9). We will assign A1
and v1 to be the crosssectional area and speed of the water at any point below the faucet,
and A2 and v2 to be the crosssectional area and speed of the water at the faucet.
SOLUTION Using the equation of continuity as given in Equation 11.9, we have
= A2v2
{ A1v1
{
Below faucet A1 = or At faucet A2v2
v1 (11.9) Since the effects of air resistance are being ignored, the water can be treated as a
freelyfalling object, as Chapter 2 discusses. Thus, the acceleration of the water is that due
to gravity. To find the speed v1 of the water, given its initial speed v2 as it leaves the faucet, ( ) 2
2
2
we use the relation v1 = v2 + 2ay or v1 = v2 + 2ay from Equation 2.9 of the equations of kinematics. Substituting Equation 2.9 into Equation 11.9 gives
A1 = A2v2
=
v1 A2v2
2
v2 + 2ay Choosing downward as the positive direction, so y = +0.10 m and a = +9.80 m/s2, the
crosssectional area of the stream at a distance of 0.10 m below the faucet is A1 = A2v2
2
v2 + 2ay = (1.8 × 10−4 m2 ) ( 0.85 m/s)
=
2
0.85 m/s ) + 2 ( 9.80 m/s2 ) ( 0.10 m )
( 9.3 × 10−5 m 2 57. REASONING The length L of the side of the square can be obtained, if we can find a value
for the crosssectional area A of the ducts. The area is related to the volume flow rate Q and
the air speed v by Equation 11.10 (Q = Av). The volume flow rate can be obtained from the
volume V of the room and the replacement time t as Q = V/t.
SOLUTION For a square cross section with sides of length L, we have A = L2. And we
know that the volume flow rate is Q = V/t. Therefore, using Equation 11.10 gives Q = Av
Solving for L shows that or V
= L2 v
t Chapter 11 Problems (a) Air speed = 3.0 m / s L= V
=
tv (b) Air speed = 5.0 m / s L= V
=
tv 597 120 m 3
= 0.18 m
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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