Ch 12 Hw 2u2w39
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3/8/2017
Ch 12 HW
Ch 12 HW Due: 11:59pm on Tuesday, March 7, 2017 You will receive no credit for items you complete after the assignment is due. Grading Policy
Exercise 12.6
Part A What is the average density of the sun? ANSWER: ρav
= 1410 kg/m3
Correct
Part B What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km? ANSWER: ρ
av
= 5.94×1016 kg/m3
Correct
Floating Cream In the 1950s, fresh unhomogenized milk in glass bottles was delivered to suburbanites' back doorsteps well before dawn. When delivered, the milk was thoroughly mixed, so that it appeared homogenized, but anyone rising much after sunrise would find that the milk had separated, the cream having risen to the top. Cream and milk are immiscible (like oil and water), and the total volume of liquid does not change when they separate. The top part of the bottle was intentionally given a much smaller diameter than the bottom, so that the cream, typically 3 percent of the total volume, occupied much more than 3% of the total vertical height of the milk bottle. For this problem, assume that the total height of the milk bottle is h and the depth of the cream layer is d.
Typesetting math: 70%
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Part A Assume that before separation, the weight of the milk bottle's contents (mixed milk and cream) is wmix . How does the combined weight of the milk and cream after separation, wsep , compare to wmix ? ANSWER:
wsep > wmix
wsep = wmix wsep < wmix
Correct
Part B Assume that before separation, the pressure at the bottom of the milk bottle is pmix . How does the pressure at the bottom of the milk bottle after separation, psep , compare to pmix ? For simplicity, you may assume that the weight and density of the cream is negligible compared to that of the milk.
Hint 1. Pressure when the bottle contents are mixed For simplicity, assume that the density of the cream is zero, and the density of the rest of the milk is ρ (in general, cream is less dense than the rest of the milk). If the cream represents 3 percent of the total volume of the milk bottle, what is the gauge pressure pmix at the bottom of the milk bottle when the milk and cream are homogeneously mixed? Express your answer in of ρ, h, and any physical constants needed.
Hint 1. Formula for gauge pressure (at depth h) Gauge pressure in a fluid of uniform density ρ is p crosssection of the container.
. This pressure is independent of the shape of the
= ρgh
Hint 2. Density of the milk and cream mixture. If the density of the cream (3% by volume) is zero, and the density of the rest of the milk is ρ, what is the density of the milk and cream mixture, ρmix ? ANSWER: ρ
mix
=
0.97ρ
ANSWER: p
mix
=
0.97ρgh
Typesetting math: 70%
Hint 2. What is the pressure after the bottle contents have separated
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Now compute psep , the gauge pressure on the bottom of the milk bottle after the milk and cream have separated. Assume that the depth of the layer of cream is 20% of the total height of the milk bottle (d
).
= 0.20h
Express your answer in of the density of the milk (not including cream) ρ, h, and any physical constants needed.
Hint 1. Formula for gauge pressure (at depth h) Gauge pressure in a fluid of uniform density ρ is p container.
. This pressure is independent of the shape of the
= ρgh
Hint 2. The role of the cream Because we have assumed that the cream has zero density (which implies zero weight), it contributes nothing to the pressure at the bottom of the milk bottle.
Hint 3. Find the depth of the milk What is hmilk the depth of milk (not including the cream) in the bottle after the milk and cream have separated? Express your answer in of h, the height of the milk bottle. ANSWER: hmilk
=
0.8h
ANSWER: p
sep
=
0.8ρgh
ANSWER:
p
p
p
sep
sep
sep
> p
= p
< p
mix
mix
mix
Correct If you really understood this and got it right the first time, congratulations! The main idea is that if you have a narrow container and a wide container with the same amount of milk in them, then the narrow container has a greater pressure exerted on it's base, because pressure is the force per unit area. So think of the situation in reverse, say you had the container partially filled with milk since the cream's weight is ignored, the cream floating on top can be ignored. Now if you mix the cream in, it forces the milk to rise into the narrow portion of the container , increasing the pressure. However, you need to do a little calculation to check that the effect of the height increase is greater than the effect of the reduction in density of the milk. Another important thing to realize is that there is another force on the fluid that balances any changes in force from the base and this is the reaction from the sloping walls of the container. Typesetting math: 70%
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Video Tutor: Water Level in Pascal's Vases First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question on the right. You can watch the video again at any point.
Part A Consider the set of tubes shown in the figure. Each tube contains an unknown fluid that is less dense than the water it floats on top of. Rank the four unknown fluids from least dense to most dense.
To rank items as equivalent, overlap them.
Hint 1. How to approach the problem Recall that the pressure at a point in a fluid is determined by the weight per unit area of the fluid above that point. In this problem, pressure is constant along the bottom of the manifold. What does that imply about the weight per unit area of the fluid in each tube? ANSWER:
Typesetting math: 70%
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Reset
Help
Least dense
Most dense A B
D
C
The correct ranking cannot be determined.
Correct
Archimedes' Principle Learning Goal: To understand the applications of Archimedes' principle. Archimedes' principle is a powerful tool for solving many problems involving equilibrium in fluids. It states the following: When a body is partially or completely submerged in a fluid (either a liquid or a gas), the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. As a result of the upward Archimedes force (often called the buoyant force), some objects may float in a fluid, and all of them appear to weigh less. This is the familiar phenomenon of buoyancy. Quantitatively, the buoyant force can be found as F buoyant = ρ gV , f luid where Fbuoyant is the force, ρf luid is the density of the fluid, g is the magnitude of the acceleration due to gravity, and V is the volume of the displaced fluid.
In this problem, you will be asked several qualitative questions that should help you develop a feel for Archimedes' principle. An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. (Note that for Parts A through D, you should assume that the object has settled in equilibrium.)
Part A Consider the following statement: The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true?
Hint 1. Archimedes' principle The statement of Part A is one way of expressing Archimedes' principle. Typesetting math: 70%
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ANSWER: for every object submerged partially or completely in a fluid only for an object that floats only for an object that sinks for no object submerged in a fluid
Correct Use Archimedes' principle to answer the rest of the questions in this problem.
Part B Consider the following statement: The magnitude of the buoyant force is equal to the weight of the amount of fluid that has the same total volume as the object. Under what circumstances is this statement true?
Hint 1. Consider Archimedes' principle Archimedes' principle deals with the displaced volume. When is the displaced volume equal to the total volume of the object? ANSWER: for an object that is partially submerged in a fluid
only for an object that floats
for an object completely submerged in a fluid for no object partially or completely submerged in a fluid
Correct
Part C Consider the following statement: The magnitude of the buoyant force equals the weight of the object. Under what circumstances is this statement true?
Hint 1. Forces and equilibrium If the buoyant force and the weight of the object are equal, the object can be in equilibrium without any other forces acting on it. Typesetting math: 70%
ANSWER:
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for every object submerged partially or completely in a fluid for an object that floats only for an object that sinks for no object submerged in a fluid
Correct
Part D Consider the following statement: The magnitude of the buoyant force is less than the weight of the object. Under what circumstances is this statement true? ANSWER: for every object submerged partially or completely in a fluid for an object that floats for an object that sinks for no object submerged in a fluid
Correct Now apply what you know to some more complicated situations.
Part E An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a denser liquid. What would you observe?
Hint 1. Density and equilibrium If the second liquid is denser than the first one, the fluid would exert a greater upward force on the object if it were held submerged as deeply as in the first case. ANSWER:
Typesetting math: 70%
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The object would sink all the way to the bottom.
The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container.
More than one of these outcomes is possible.
Correct
Part F An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe?
Hint 1. Density and equilibrium If the second liquid is less dense than the first one, the liquid would exert a smaller upward force on the object if it were held submerged as deeply as in the first case. ANSWER: The object would sink all the way to the bottom.
The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container.
More than one of these outcomes is possible.
Correct If the fluid in the second container is less dense than the object, then the object will sink all the way to the bottom. If the fluid in the second container is denser than the object (though less dense than the fluid in the original container), the object will still float, but its depth will be greater than it was in the original container.
Part G Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true? ANSWER: Object T has a greater density than object B. Object B has a greater density than object T.
Both objects have the same density. Typesetting math: 70%
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Correct Since both objects float, the buoyant force in each case is equal to the object's weight. Block B displaces more fluid, so it must be heavier than block T. Given that the two objects have the same volume, block B must also be denser. In fact, since the weight equals the buoyant force, and B is fully submerged, ρB V g = ρliquid V g, where all the symbols have their usual meaning. From this equation, one can see that the density of B must equal the density of the fluid.
Video Tutor: Weighing Weights in Water First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question on the right. You can watch the video again at any point.
Part A Suppose that we repeat the experiment shown in the video, but we replace one of the cylinders with a cylinder that has twice the radius (and use larger containers of water). If the height of the original cylinder is h, how deeply must we submerge the new cylinder to get the same weight reduction as in the video?
Hint 1. How to approach the problem How does the reduction in the cylinder’s weight relate to the amount of water the cylinder displaces? How does the volume of a cylinder depend on the cylinder’s radius and height? ANSWER: 4h 1 2 1 4
h
h
h
2h
Typesetting math: 70%
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Correct To get the same reduction in weight, the same volume of cylinder must be submerged. The volume of a cylinder is 2 V = πr h , so if we double the radius, we must reduce the depth of immersion by a factor of four.
A Submerged Ball A ball of mass mb and volume V is lowered on a string into a fluid of density ρf . Assume that the object would sink to the bottom if it were not ed by the string.
Part A What is the tension T in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Express your answer in of any or all of the given quantities and g , the magnitude of the acceleration due to gravity.
Hint 1. Equilibrium condition Although the fact may be obscured by the presence of a liquid, the basic condition for equilibrium still holds: The net force on the ball must be zero. Draw a freebody diagram and proceed from there.
Hint 2. Find the magnitude of the buoyant force Find Fbuoyant, the magnitude of the buoyant force. Express your answer in of any or all of the variables ρf , V , mb , and g .
Hint 1. Archimedes' principle Quantitatively, the buoyant force can be found as F buoyant = ρ gV , f luid where Fbuoyant is the force, ρf luid is the density of the fluid, g is the magnitude of the acceleration due to gravity, and V is the volume of the displaced fluid.
Hint 2. Find the mass of the displaced fluid
Typesetting math: 70%
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Express your answer in of any or all of the variables ρf , V , mb , and g . ANSWER: mf
=
ρ V f
ANSWER: F buoyant
=
ρ Vg f
ANSWER: T
=
mb g − ρ gV f
Correct
Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, W actual , and the apparent weight of the crown when it is submerged in water, W apparent . See whether you can follow in Archimedes' footsteps. The figure shows what is meant by weighing the crown while it is submerged in water.
Typesetting math: 70%
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Part A Take the density of the crown to be ρc. What is the ratio of the crown's apparent weight (in water) W apparent to its actual weight W actual ? Express your answer in of the density of the crown ρc and the density of water ρw .
Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume V . Find the actual weight W actual of the crown. Express W actual in of V , g (the magnitude of the acceleration due to gravity), and ρc. ANSWER: W actual
=
ρ Vg c
Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume V , and take the density of water to be ρw . Find the apparent weight W apparent of the crown submerged in water. Express your answer in of V , g (the magnitude of the acceleration due to gravity), ρw , and ρc.
Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity.
Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude Fbuoyant of the buoyant force on the crown when it is completely submerged in water. Express your answer in of ρw , V , and the gravitational acceleration g . ANSWER: F buoyant
=
ρ
w
Vg
ANSWER: W apparent
=
(ρ
c
−ρ
w
)gV
ANSWER: Wapparent Wactual
=
1−
ρw ρc
Correct Typesetting math: 70%
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Part B Imagine that the apparent weight of the crown in water is W apparent = 4.50 N, and the actual weight is W actual = 5.00 N. Is the crown made of pure (100%) gold? The density of water is ρ = 1.00 grams per cubic w centimeter. The density of gold is ρg = 19.32 grams per cubic centimeter.
Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for
Wapparent Wactual
, what should the ratio of weights be if the crown is made of pure
gold? Express your answer numerically, to two decimal places. ANSWER: Wapparent Wactual
= 0.95
ANSWER: Yes No
Correct For the values given,
Wapparent Wactual
= 4.50/5.00 = 0.90 , whereas for pure gold,
Wapparent Wactual
= 1−
ρw ρg
= 0.95 .
Thus, you can conclude that the the crown is not pure gold but contains some lessdense metal. The goldsmith made sure that the crown's (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been.
Exercise 12.33 A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.3 N . When the rock is totally immersed in water, the tension is 27.6 N . When the rock is totally immersed in an unknown liquid, the tension is 20.1 N .
Part A What is the density of the unknown liquid? ANSWER: = 1480 kg/m3
ρ
Correct Typesetting math: 70%
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± Playing with a Water Hose Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away.
Part A Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed v 0 of 3.5 meters per second? Assume that the hose is pointed parallel to the ground and take the magnitude of the acceleration g due to gravity to be 9.81 meters per second, per second.
Hint 1. General approach: considerations on particle motion Consider the water flowing out of the hose to be composed of independent particles in projectile motion. Recall that projectile motion is described as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Note that the hose is pointed parallel to the ground, so the initial velocity of the water is purely horizontal.
Hint 2. Projectile motion A projectile is a body with an initial velocity that follows a path determined entirely by the effects of gravitational acceleration and air resistance. If the effects due to air resistance are ignored, the motion of a projectile can be analyzed as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Consider a particle undergoing projectile motion in the xy coordinate plane, with the x axis horizontal and the y axis pointing vertically upward. Let v x and v y represent the components of the initial velocity of the particle. Let x0 and y represent the initial horizontal and vertical positions of the particle. The equations describing the position of the 0 particle as a function of time t are then x = x0 + v x t
y= y
0
+ vy t −
1 2
2
gt ,
where g is the magnitude of the acceleration due to gravity. ANSWER: Yes No
Correct
Part B To increase the range of the water, Isabella places her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the crosssectional area of the hose hole does she have to cover to be able to spray her friend? Assume that the cross section of the hose opening is circular with a radius of 1.5 centimeters. Express your answer as a percentage to the nearest integer. Typesetting math: 70% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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Hint 1. General approach: considerations on fluid mechanics By restricting the area across which the water is flowing, Isabella forces the water to flow out at a higher speed. After estimating the initial velocity that the water must have in order to reach a distance of 10 meters, you can determine the crosssectional area of the hose hole that corresponds to such an initial velocity of the water by applying the continuity equation.
Hint 2. Find the outflow speed needed Find the initial speed v that the water should have in order to reach a distance of 10 meters. Consider the water flowing out of the hose as composed of independent particles in projectile motion. Recall that projectile motion is described as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Note that the hose is pointed parallel to the ground, so the initial velocity of water is purely horizontal. Thus v = v x. Express your answer numerically in meters per second to three significant figures.
Hint 1. Projectile motion If the effects due to air resistance are ignored, as they should be in this problem, the motion of a projectile can be analyzed as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Consider a particle undergoing projectile motion in the xy coordinate plane, with the x axis horizontal and the y axis pointing vertically upward. Let v x and v y represent the components of the initial velocity of the particle. Let x0 and y0 represent the initial horizontal and vertical positions of the particle. The equations describing the position of the particle as a function of time t are then x = x0 + v x t
y = y0 + vy t −
1 2
gt
2
,
where g is the magnitude of the acceleration due to gravity. In this part, you are asked to find the outflow speed, which is horizontal, and thus corresponds to v x is the above equation. Consider carefully what the values of the other constants x0 , y0 , and v y are. ANSWER: vx
= 22.1 m/s
Hint 3. Find the crosssectional area needed Once you know the horizontal velocity of the water that would be needed for the emerging stream to reach a distance of 10 meters, find the crosssectional area A that corresponds to that outflow speed. Assume that the flow rate remains steady after the partial blockage of the hose hole, so that the continuity equation applies. Express your answer numerically in centimeters squared to three significant figures.
Hint 1. Apply the continuity equation In an incompressible fluid moving steadily, the mass of fluid flowing along a flow tube is constant. In particular, consider a flow tube between two stationary cross sections with areas A1 and A2 . Let the fluid speeds at these sections be v 1 and v 2 , respectively. Then conservation of mass takes the form Typesetting math: 70% A1 v 1 = A2 v 2
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which is known as the continuity equation. If initially the water is flowing through the hose of crosssectional area A0 at a speed v 0 , and you want it to flow through a smaller area at a higher speed v , what should the new crosssectional area A of the hose be? Express your answer in of A0 , v 0 , and v . ANSWER: A
=
A 0v 0 v
ANSWER: A
= 1.12 cm2
ANSWER: f
= 84 %
Correct
Exercise 12.37 Water is flowing in a circular pipe varying crosssectional area, and at all points the water completely fills the pipe.
Part A At one point in the pipe the radius is 0.245 m . What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.85 m3 /s ? ANSWER: v1
= 9.81 m/s
Correct
Part B At a second point in the pipe the water speed is 2.30 m/s . What is the radius of the pipe at this point? ANSWER: r2
= 0.506 m
Correct
Typesetting math: 70%
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Streamlines and Fluid Flow Learning Goal: To understand the continuity equation. Streamlines represent the path of the flow of a fluid. You can imagine that they represent a timeexposure photograph that shows the paths of small particles carried by the flowing fluid. The figure shows streamlines for the flow of an incompressible fluid in a tapered pipe of circular cross section. The speed of the fluid as it enters the pipe on the left is v 1 . Assume that the cross sectional areas of the pipe are A1 at its entrance on the left and A2 at its exit on the right.
Part A Find Q1 , the volume of fluid flowing into the pipe per unit of time. This quantity is also known as the volumetric flow rate. Express the volumetric flow rate in of any of the quantities given in the problem introduction.
Hint 1. Find the volume of fluid entering the pipe The volumetric flow rate has units of volume per unit time (cubic meters per second). What is the volume ΔV of fluid entering the pipe in time Δt? Express your answer in of any or all of the following quantities: v 1 , A1 , and Δt.
Hint 1. How far does the fluid move in time Δt? The figure shows a snapshot of the pipe at two instants in time: t0 and t0 + Δt . In the time interval Δt, some additional fluid (shown in blue) has entered the pipe from the left. The speed of the lefthand boundary of the bluecolored fluid is v 1 . What is the length L 1 of the region of additional fluid that has entered the pipe in the time interval Δt? Express your answer in of v 1 and Δt.
Typesetting math: 70%
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ANSWER: L1
=
v 1 Δt
ANSWER: ΔV
=
A 1 v 1 Δt
ANSWER: Q
1
=
v 1 A1
Correct
Part B Because the fluid is assumed to be incompressible and mass is conserved, at a particular moment in time, the amount of fluid that flows into the pipe must equal the amount of fluid that flows out. This fact is embodied in the continuity equation. Using the continuity equation, find the velocity v 2 of the fluid flowing out of the right end of the pipe. Express your answer in of any of the quantities given in the problem introduction.
Hint 1. Find the volumetric flow rate out of the pipe Find Q2 , the volume of fluid flowing out of the pipe per unit of time. Express your answer in of v 2 and A2 . ANSWER: Q2
=
v 2 A2
Hint 2. Apply the continuity equation The continuity equation states that the volumetric flows into and out of the pipe must be the same. Fill in the right hand side of the continuity equation for this problem. Express your answer in of v 2 and any of the quantities given in the problem introduction. ANSWER: v 1 A1
=
v 2 A2
ANSWER:
v2
=
v1
A1 A2
Typesetting math: 70%
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Correct
Part C If you are shown a picture of streamlines in a flowing fluid, you can conclude that the __________ of the fluid is greater where the streamlines are closer together. Enter a oneword answer. ANSWER: speed
Correct Thus the velocity of the flow increases with increasing density (number per unit area) of streamlines.
Exercise 12.46 A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the crosssectional area is 8.00 cm2 . At point 1, 1.35 m above point 2, the crosssectional area is 2.00 cm2 .
Part A Find the mass flow rate. ANSWER: M
= 1.30 kg/s
Correct
Part B Find the volume flow rate. ANSWER: = 1.30 L/s
Q
Correct
Part C Find the flow speed at point 1. Typesetting math: 70% ANSWER:
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= 6.50 m/s
Correct
Part D Find the flow speed at point 2. ANSWER: v2
= 1.63 m/s
Correct
Part E Find the gauge pressure at point 1. ANSWER: P
= 119 kPa
Correct
Venturi Meter with Two Tubes A pair of vertical, openended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density ρ. The fluid rises to heights h1 and h2 in the two openended tubes (see figure). The crosssectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Typesetting math: 70%
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Ch 12 HW
Part A Find p1 , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in of quantities given in the problem introduction and \texttip{g}{g}, the magnitude of the acceleration due to gravity.
Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density.
Hint 2. Simplified Bernoulli's equation For a fluid of uniform density \texttip{\rho }{rho} that is not flowing, the pressure \texttip{p}{p} at a depth \texttip{h}{h} below the surface is given by p = p_0 + \rho g h, where \texttip{p_{\rm 0}}{p_0} is the pressure at the surface and \texttip{g}{g} is the magnitude of the acceleration due to gravity. ANSWER: \texttip{p_{\rm 1}}{p_1} = {\rho}gh_{1}
Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube.
Part B Find \texttip{v_{\rm 1}}{v_1}, the speed of the fluid in the left end of the main pipe. Express your answer in of \texttip{h_{\rm 1}}{h_1}, \texttip{h_{\rm 2}}{h_2}, \texttip{g}{g}, and either \texttip{A_{\rm 1}}{A_1} and \texttip{A_{\rm 2}}{A_2} or \texttip{\gamma }{gamma}, which is equal to \large{\frac{A_1}{A_2}}.
Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of tube 1, and the fluid pressure and velocity at the bottom of tube 2.
Hint 2. Find \texttip{p_{\rm 2}}{p_2} in of \texttip{h_{\rm 2}}{h_2} What is \texttip{p_{\rm 2}}{p_2}, the pressure at the bottom of tube 2? Express your answer in of \texttip{h_{\rm 2}}{h_2}, \texttip{g}{g}, and any other given quantities.
Hint 1. Recall Part A Typesetting math: 70% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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Obtain the solution for this part in the same way that you found an expression for \texttip{p_{\rm 1}}{p_1} in of \texttip{h_{\rm 1}}{h_1} in Part A of this problem. ANSWER: \texttip{p_{\rm 2}}{p_2} = {\rho} g h_{2}
Hint 3. Find \texttip{p_{\rm 2}}{p_2} in of given quantities Find \texttip{p_{\rm 2}}{p_2}, the fluid pressure at the bottom of tube 2. Express your answer in of \texttip{p_{\rm 1}}{p_1}, \texttip{v_{\rm 1}}{v_1}, \texttip{\rho }{rho}, \texttip{A_{\rm 1}}{A_1}, and \texttip{A_{\rm 2}}{A_2}.
Hint 1. Find the pressure at the bottom of tube 2 Find \texttip{p_{\rm 2}}{p_2}, the fluid pressure at the bottom of tube 2. Express your answer in of \texttip{p_{\rm 1}}{p_1}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2}. ANSWER: \texttip{p_{\rm 2}}{p_2} = \large{p_{1} + \frac{1}{2} {\rho} \left({v_{1}}^{2} {v_{2}}^{2}\right)}
Hint 2. Find \texttip{v_{\rm 2}}{v_2} in of \texttip{v_{\rm 1}}{v_1} The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2} in of the geometry of the pipe. Find \texttip{v_{\rm 2}}{v_2}, the fluid velocity at the bottom of tube 2, in of \texttip{v_{\rm 1}}{v_1}. Your answer may include \texttip{A_{\rm 1}}{A_1} and \texttip{A_{\rm 2}}{A_2}, the crosssectional areas of the pipe. ANSWER: \texttip{v_{\rm 2}}{v_2} = \large{\frac{A_{1}}{A_{2}} v_{1}}
ANSWER: \texttip{p_{\rm 2}}{p_2} \large{p_{1} + \frac{1}{2} {\rho} \left({v_{1}}^{2}\right)\left(1\left(\frac{A_{1}} = {A_{2}}\right)^{2}\right)}
ANSWER: \texttip{v_{\rm 1}}{v_1} = \large{\sqrt{2g{\frac{h_{1}h_{2}}{{\gamma}^{2}1}}}}
Typesetting math: 70%
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Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves.
Problem 12.92 In 1993 the radius of Hurricane Emily was about 350 {\rm km}. The wind speed near the center ("eye") of the hurricane, whose radius was about 30 {\rm km}, reached about 200 {\rm km}/{\rm h}. As air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant.
Part A Estimate the wind speed at the rim of the hurricane. Express your answer using two significant figures. ANSWER: v = 17 {\rm km/h}
Correct
Part B Estimate the pressure difference at the earth's surface between the eye and the rim. (Hint: The density of air (1 {\rm atm}, 20 ^\circ {\rm C}) is 1.20 {\rm kg}/{\rm m}^{3}.) Express your answer using two significant figures. ANSWER: \Delta P = 1800 {\rm Pa}
Correct
Part C Where is the pressure greater? ANSWER: at the eye at the rim
Correct Typesetting math: 70%
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Part D If the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go? Express your answer using two significant figures. ANSWER: h = 160 {\rm m}
Correct
Problem 12.95 A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v_0 and the radius of the stream of liquid is r_0.
Part A Find an equation for the speed of the liquid as a function of the distance y it has fallen. Express your answer in of the given quantities and appropriate constants. ANSWER: v = \sqrt{v_{0}{^2}+2gy}
Correct
Part B Combining this with the equation of continuity, find an expression for the radius of the stream as a function of y. Express your answer in of the given quantities and appropriate constants. ANSWER: r = \large{\sqrt{r_{0}{^2}{\frac{v_{0}}{\sqrt{v_{0}{^2}+2gy}}}}}
Correct
Part C If water flows out of a vertical pipe at a speed of 1.20 \rm m/s, how far below the outlet will the radius be onehalf the original radius of the stream? Express your answer using three significant figures. ANSWER:
Typesetting math: 70%
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y = 1.10 {\rm m}
Correct
Problem 12.57 A 0.170{\rm kg} cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.80 {\rm cm} . The level of the glycerine is well below the top of the cylinder.
Part A If the ice completely melts, by what distance does the height of liquid in the cylinder change? Express your answer with the appropriate units. ANSWER: d = 0.773 {\rm cm}
Correct
Part B Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted? ANSWER: The level of liquid in the cylinder rises. The level of liquid in the cylinder falls.
Correct
Exercise 12.3 You purchase a rectangular piece of metal that has dimensions 5.2 {\rm {\rm mm}} \times 15.2 {\rm {\rm mm}} \times 33.7 {\rm {\rm mm}} and mass 1.86×10−2 {\rm {\rm kg}} .
Part A The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? ANSWER: Typesetting math: 70%
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\rho = 6980 {\rm kg/m^3}
Correct
Part B Were you cheated? ANSWER: yes no
Correct
Exercise 12.6
Part A What is the average density of the sun? ANSWER: \rho_{\rm av} = 1410 {\rm kg}/{\rm m}^{3}
Correct
Part B What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km? ANSWER: \rho_{\rm av} = 5.94×1016 {\rm kg}/{\rm m}^{3}
Correct
Exercise 12.12 A barrel contains a 0.125m layer of oil floating on water that is 0.280 {\rm {\rm m}} deep. The density of the oil is 650 {\rm {\rm kg/m^3}} Typesetting math: 70%
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Part A What is the gauge pressure at the oilwater interface? ANSWER: P = 796 {\rm Pa}
Correct
Part B What is the gauge pressure at the bottom of the barrel? ANSWER: P = 3540 {\rm Pa}
Correct
Exercise 12.18 The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 \rm m above the surface of the water in the cup.
Part A What is the lowest gauge pressure that the average person can achieve inside his lungs? Express your answer with the appropriate units. ANSWER: p_{\rm g} = −1.1×104 {\rm Pa}
Correct
Part B Why is your answer in the previous part negative? ANSWER:
Typesetting math: 70%
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In order for water to go up the straw, the pressure at the top of the straw must be greater than the atmospheric pressure. In order for water to go up the straw, the gauge pressure of the person must be greater than the pressure due to the column of water in the straw. In order for water to go up the straw, the pressure at the top of the straw must be less than the atmospheric pressure. In order for water to go up the straw, the gauge pressure of the person must be less than the pressure due to the column of water in the straw.
Correct
Exercise 12.32 A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.660 {\rm \; m^3} and the tension in the cord is 2120 {\rm \; N} .
Part A Calculate the buoyant force exerted by the water on the sphere. Express your answer with the appropriate units. ANSWER: F = 6470 {\rm N}
Correct
Part B What is the mass of the sphere? Express your answer with the appropriate units. ANSWER: m = 444 {\rm kg}
Correct
Part C The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? ANSWER:
Typesetting math: 70%
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\large{\frac{V_{\rm sub}}{V_{\rm obj}}} = 67.2 \%
Correct
Exercise 12.37 A shower head has 20 circular openings, each with radius 0.90 {\rm mm} . The shower head is connected to a pipe with radius 0.98 {\rm cm} .
Part A If the speed of water in the pipe is 3.4 {\rm m/s} , what is its speed as it exits the showerhead openings? Express your answer using two significant figures. ANSWER: v = 20 \rm m/s
Correct
Exercise 12.43
Part A What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 12.0 {\rm m} ? (Assume that the mains have a much larger diameter than the fire hose.) ANSWER: P = 1.18×105 \rm Pa
Correct
Exercise 12.48 A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 L cans per minute. At point 2 in the pipe, the gauge pressure is 152\;{\rm kPa} and the crosssectional area is 8.00 \;{\rm cm}^{2}. At point 1, 1.35 \;{\rm m} above point 2, the crosssectional area is 2.00 \;{\rm cm}^{2}.
Part A Find the mass flow rate. ANSWER:
Typesetting math: 70%
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M = 1.30 {\rm kg/s}
Correct
Part B Find the volume flow rate. ANSWER: Q = 1.30 {\rm L/s}
Correct
Part C Find the flow speed at point 1. ANSWER: v_1 = 6.50 {\rm m/s}
Correct
Part D Find the flow speed at point 2. ANSWER: v_2 = 1.63 {\rm m/s}
Correct
Part E Find the gauge pressure at point 1. ANSWER: P = 119 {\rm kPa}
Correct Typesetting math: 70%
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Problem 12.82 In 1993 the radius of Hurricane Emily was about 350 {\rm km}. The wind speed near the center ("eye") of the hurricane, whose radius was about 30 {\rm km}, reached about 200 {\rm km}/{\rm h}. As air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant.
Part A Estimate the wind speed at the rim of the hurricane. Express your answer using two significant figures. ANSWER: v = 17 {\rm km/h}
Correct
Part B Estimate the pressure difference at the earth's surface between the eye and the rim. (Hint: The density of air (1 {\rm atm}, 20 ^\circ {\rm C}) is 1.20 {\rm kg}/{\rm m}^{3}.) Express your answer using two significant figures. ANSWER: \Delta P = 1800 {\rm Pa}
Correct
Part C Where is the pressure greater? ANSWER: at the eye at the rim
Correct
Part D If the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go? Express your answer using two significant figures. Typesetting math: 70%
ANSWER:
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h = 160 {\rm m}
Correct
Problem 12.88 A siphon, as shown in the figure , is a convenient device for removing liquids from containers. To establish the flow, the tube must be initially filled with fluid. Let the fluid have density \rho, and let the atmospheric pressure be p_{\rm{a}} {\kern 1pt}. Assume that the crosssectional area of the tube is the same at all points along it.
Part A If the lower end of the siphon is at a distance h below the surface of the liquid in the container, what is the speed of the fluid as it flows out the lower end of the siphon? (Assume that the container has a very large diameter, and ignore any effects of viscosity.) Express your answer in of the given quantities and appropriate constants. ANSWER: v = \sqrt{2gh}
Correct
Part B A curious feature of a siphon is that the fluid initially flows "uphill." What is the greatest height H that the high point of the tube can have if flow is still to occur? Express your answer in of the given quantities and appropriate constants. ANSWER: H = \large{\frac{p_{a}}{{\rho}g}}
Correct
Typesetting math: 70%
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Score Summary: Your score on this assignment is 99.9%. You received 21.97 out of a possible total of 22 points.
Typesetting math: 70%
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Ch 12 HW Due: 11:59pm on Tuesday, March 7, 2017 You will receive no credit for items you complete after the assignment is due. Grading Policy
Exercise 12.6
Part A What is the average density of the sun? ANSWER: ρav
= 1410 kg/m3
Correct
Part B What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km? ANSWER: ρ
av
= 5.94×1016 kg/m3
Correct
Floating Cream In the 1950s, fresh unhomogenized milk in glass bottles was delivered to suburbanites' back doorsteps well before dawn. When delivered, the milk was thoroughly mixed, so that it appeared homogenized, but anyone rising much after sunrise would find that the milk had separated, the cream having risen to the top. Cream and milk are immiscible (like oil and water), and the total volume of liquid does not change when they separate. The top part of the bottle was intentionally given a much smaller diameter than the bottom, so that the cream, typically 3 percent of the total volume, occupied much more than 3% of the total vertical height of the milk bottle. For this problem, assume that the total height of the milk bottle is h and the depth of the cream layer is d.
Typesetting math: 70%
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Part A Assume that before separation, the weight of the milk bottle's contents (mixed milk and cream) is wmix . How does the combined weight of the milk and cream after separation, wsep , compare to wmix ? ANSWER:
wsep > wmix
wsep = wmix wsep < wmix
Correct
Part B Assume that before separation, the pressure at the bottom of the milk bottle is pmix . How does the pressure at the bottom of the milk bottle after separation, psep , compare to pmix ? For simplicity, you may assume that the weight and density of the cream is negligible compared to that of the milk.
Hint 1. Pressure when the bottle contents are mixed For simplicity, assume that the density of the cream is zero, and the density of the rest of the milk is ρ (in general, cream is less dense than the rest of the milk). If the cream represents 3 percent of the total volume of the milk bottle, what is the gauge pressure pmix at the bottom of the milk bottle when the milk and cream are homogeneously mixed? Express your answer in of ρ, h, and any physical constants needed.
Hint 1. Formula for gauge pressure (at depth h) Gauge pressure in a fluid of uniform density ρ is p crosssection of the container.
. This pressure is independent of the shape of the
= ρgh
Hint 2. Density of the milk and cream mixture. If the density of the cream (3% by volume) is zero, and the density of the rest of the milk is ρ, what is the density of the milk and cream mixture, ρmix ? ANSWER: ρ
mix
=
0.97ρ
ANSWER: p
mix
=
0.97ρgh
Typesetting math: 70%
Hint 2. What is the pressure after the bottle contents have separated
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Now compute psep , the gauge pressure on the bottom of the milk bottle after the milk and cream have separated. Assume that the depth of the layer of cream is 20% of the total height of the milk bottle (d
).
= 0.20h
Express your answer in of the density of the milk (not including cream) ρ, h, and any physical constants needed.
Hint 1. Formula for gauge pressure (at depth h) Gauge pressure in a fluid of uniform density ρ is p container.
. This pressure is independent of the shape of the
= ρgh
Hint 2. The role of the cream Because we have assumed that the cream has zero density (which implies zero weight), it contributes nothing to the pressure at the bottom of the milk bottle.
Hint 3. Find the depth of the milk What is hmilk the depth of milk (not including the cream) in the bottle after the milk and cream have separated? Express your answer in of h, the height of the milk bottle. ANSWER: hmilk
=
0.8h
ANSWER: p
sep
=
0.8ρgh
ANSWER:
p
p
p
sep
sep
sep
> p
= p
< p
mix
mix
mix
Correct If you really understood this and got it right the first time, congratulations! The main idea is that if you have a narrow container and a wide container with the same amount of milk in them, then the narrow container has a greater pressure exerted on it's base, because pressure is the force per unit area. So think of the situation in reverse, say you had the container partially filled with milk since the cream's weight is ignored, the cream floating on top can be ignored. Now if you mix the cream in, it forces the milk to rise into the narrow portion of the container , increasing the pressure. However, you need to do a little calculation to check that the effect of the height increase is greater than the effect of the reduction in density of the milk. Another important thing to realize is that there is another force on the fluid that balances any changes in force from the base and this is the reaction from the sloping walls of the container. Typesetting math: 70%
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Video Tutor: Water Level in Pascal's Vases First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question on the right. You can watch the video again at any point.
Part A Consider the set of tubes shown in the figure. Each tube contains an unknown fluid that is less dense than the water it floats on top of. Rank the four unknown fluids from least dense to most dense.
To rank items as equivalent, overlap them.
Hint 1. How to approach the problem Recall that the pressure at a point in a fluid is determined by the weight per unit area of the fluid above that point. In this problem, pressure is constant along the bottom of the manifold. What does that imply about the weight per unit area of the fluid in each tube? ANSWER:
Typesetting math: 70%
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Reset
Help
Least dense
Most dense A B
D
C
The correct ranking cannot be determined.
Correct
Archimedes' Principle Learning Goal: To understand the applications of Archimedes' principle. Archimedes' principle is a powerful tool for solving many problems involving equilibrium in fluids. It states the following: When a body is partially or completely submerged in a fluid (either a liquid or a gas), the fluid exerts an upward force on the body equal to the weight of the fluid displaced by the body. As a result of the upward Archimedes force (often called the buoyant force), some objects may float in a fluid, and all of them appear to weigh less. This is the familiar phenomenon of buoyancy. Quantitatively, the buoyant force can be found as F buoyant = ρ gV , f luid where Fbuoyant is the force, ρf luid is the density of the fluid, g is the magnitude of the acceleration due to gravity, and V is the volume of the displaced fluid.
In this problem, you will be asked several qualitative questions that should help you develop a feel for Archimedes' principle. An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. (Note that for Parts A through D, you should assume that the object has settled in equilibrium.)
Part A Consider the following statement: The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true?
Hint 1. Archimedes' principle The statement of Part A is one way of expressing Archimedes' principle. Typesetting math: 70%
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ANSWER: for every object submerged partially or completely in a fluid only for an object that floats only for an object that sinks for no object submerged in a fluid
Correct Use Archimedes' principle to answer the rest of the questions in this problem.
Part B Consider the following statement: The magnitude of the buoyant force is equal to the weight of the amount of fluid that has the same total volume as the object. Under what circumstances is this statement true?
Hint 1. Consider Archimedes' principle Archimedes' principle deals with the displaced volume. When is the displaced volume equal to the total volume of the object? ANSWER: for an object that is partially submerged in a fluid
only for an object that floats
for an object completely submerged in a fluid for no object partially or completely submerged in a fluid
Correct
Part C Consider the following statement: The magnitude of the buoyant force equals the weight of the object. Under what circumstances is this statement true?
Hint 1. Forces and equilibrium If the buoyant force and the weight of the object are equal, the object can be in equilibrium without any other forces acting on it. Typesetting math: 70%
ANSWER:
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for every object submerged partially or completely in a fluid for an object that floats only for an object that sinks for no object submerged in a fluid
Correct
Part D Consider the following statement: The magnitude of the buoyant force is less than the weight of the object. Under what circumstances is this statement true? ANSWER: for every object submerged partially or completely in a fluid for an object that floats for an object that sinks for no object submerged in a fluid
Correct Now apply what you know to some more complicated situations.
Part E An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a denser liquid. What would you observe?
Hint 1. Density and equilibrium If the second liquid is denser than the first one, the fluid would exert a greater upward force on the object if it were held submerged as deeply as in the first case. ANSWER:
Typesetting math: 70%
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The object would sink all the way to the bottom.
The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container.
More than one of these outcomes is possible.
Correct
Part F An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, filled with a less dense liquid. What would you observe?
Hint 1. Density and equilibrium If the second liquid is less dense than the first one, the liquid would exert a smaller upward force on the object if it were held submerged as deeply as in the first case. ANSWER: The object would sink all the way to the bottom.
The object would float submerged more deeply than in the first container. The object would float submerged less deeply than in the first container.
More than one of these outcomes is possible.
Correct If the fluid in the second container is less dense than the object, then the object will sink all the way to the bottom. If the fluid in the second container is denser than the object (though less dense than the fluid in the original container), the object will still float, but its depth will be greater than it was in the original container.
Part G Two objects, T and B, have identical size and shape and have uniform density. They are carefully placed in a container filled with a liquid. Both objects float in equilibrium. Less of object T is submerged than of object B, which floats, fully submerged, closer to the bottom of the container. Which of the following statements is true? ANSWER: Object T has a greater density than object B. Object B has a greater density than object T.
Both objects have the same density. Typesetting math: 70%
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Correct Since both objects float, the buoyant force in each case is equal to the object's weight. Block B displaces more fluid, so it must be heavier than block T. Given that the two objects have the same volume, block B must also be denser. In fact, since the weight equals the buoyant force, and B is fully submerged, ρB V g = ρliquid V g, where all the symbols have their usual meaning. From this equation, one can see that the density of B must equal the density of the fluid.
Video Tutor: Weighing Weights in Water First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question on the right. You can watch the video again at any point.
Part A Suppose that we repeat the experiment shown in the video, but we replace one of the cylinders with a cylinder that has twice the radius (and use larger containers of water). If the height of the original cylinder is h, how deeply must we submerge the new cylinder to get the same weight reduction as in the video?
Hint 1. How to approach the problem How does the reduction in the cylinder’s weight relate to the amount of water the cylinder displaces? How does the volume of a cylinder depend on the cylinder’s radius and height? ANSWER: 4h 1 2 1 4
h
h
h
2h
Typesetting math: 70%
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Correct To get the same reduction in weight, the same volume of cylinder must be submerged. The volume of a cylinder is 2 V = πr h , so if we double the radius, we must reduce the depth of immersion by a factor of four.
A Submerged Ball A ball of mass mb and volume V is lowered on a string into a fluid of density ρf . Assume that the object would sink to the bottom if it were not ed by the string.
Part A What is the tension T in the string when the ball is fully submerged but not touching the bottom, as shown in the figure? Express your answer in of any or all of the given quantities and g , the magnitude of the acceleration due to gravity.
Hint 1. Equilibrium condition Although the fact may be obscured by the presence of a liquid, the basic condition for equilibrium still holds: The net force on the ball must be zero. Draw a freebody diagram and proceed from there.
Hint 2. Find the magnitude of the buoyant force Find Fbuoyant, the magnitude of the buoyant force. Express your answer in of any or all of the variables ρf , V , mb , and g .
Hint 1. Archimedes' principle Quantitatively, the buoyant force can be found as F buoyant = ρ gV , f luid where Fbuoyant is the force, ρf luid is the density of the fluid, g is the magnitude of the acceleration due to gravity, and V is the volume of the displaced fluid.
Hint 2. Find the mass of the displaced fluid
Typesetting math: 70%
Compute the mass mf of the fluid displaced by the object when it is entirely submerged. https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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Express your answer in of any or all of the variables ρf , V , mb , and g . ANSWER: mf
=
ρ V f
ANSWER: F buoyant
=
ρ Vg f
ANSWER: T
=
mb g − ρ gV f
Correct
Crown of Gold? According to legend, the following challenge led Archimedes to the discovery of his famous principle: Hieron, king of Syracuse, was suspicious that a new crown that he had received from the royal goldsmith was not pure gold, as claimed. Archimedes was ordered to determine whether the crown was in fact made of pure gold, with the condition that only a nondestructive test would be allowed. Rather than answer the problem in the politically most expedient way (or perhaps extract a bribe from the goldsmith), Archimedes thought about the problem scientifically. The legend relates that when Archimedes stepped into his bath and caused it to overflow, he realized that he could answer the challenge by comparing the volume of water displaced by the crown with the volume of water displaced by an amount of pure gold equal in weight to the crown. If the crown was made of pure gold, the two volumes would be equal. If some other (less dense) metal had been substituted for some of the gold, then the crown would displace more water than the pure gold. A similar method of answering the challenge, based on the same physical principle, is to compute the ratio of the actual weight of the crown, W actual , and the apparent weight of the crown when it is submerged in water, W apparent . See whether you can follow in Archimedes' footsteps. The figure shows what is meant by weighing the crown while it is submerged in water.
Typesetting math: 70%
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Part A Take the density of the crown to be ρc. What is the ratio of the crown's apparent weight (in water) W apparent to its actual weight W actual ? Express your answer in of the density of the crown ρc and the density of water ρw .
Hint 1. Find an expression for the actual weight of the crown Assume that the crown has volume V . Find the actual weight W actual of the crown. Express W actual in of V , g (the magnitude of the acceleration due to gravity), and ρc. ANSWER: W actual
=
ρ Vg c
Hint 2. Find an expression for the apparent weight of the crown Assume that the crown has volume V , and take the density of water to be ρw . Find the apparent weight W apparent of the crown submerged in water. Express your answer in of V , g (the magnitude of the acceleration due to gravity), ρw , and ρc.
Hint 1. How to approach the problem The apparent weight of the crown when it is submerged in water will be less than its actual weight (weight in air) due to the buoyant force, which opposes gravity.
Hint 2. Find an algebraic expression for the buoyant force. Find the magnitude Fbuoyant of the buoyant force on the crown when it is completely submerged in water. Express your answer in of ρw , V , and the gravitational acceleration g . ANSWER: F buoyant
=
ρ
w
Vg
ANSWER: W apparent
=
(ρ
c
−ρ
w
)gV
ANSWER: Wapparent Wactual
=
1−
ρw ρc
Correct Typesetting math: 70%
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Part B Imagine that the apparent weight of the crown in water is W apparent = 4.50 N, and the actual weight is W actual = 5.00 N. Is the crown made of pure (100%) gold? The density of water is ρ = 1.00 grams per cubic w centimeter. The density of gold is ρg = 19.32 grams per cubic centimeter.
Hint 1. Find the ratio of weights for a crown of pure gold Given the expression you obtained for
Wapparent Wactual
, what should the ratio of weights be if the crown is made of pure
gold? Express your answer numerically, to two decimal places. ANSWER: Wapparent Wactual
= 0.95
ANSWER: Yes No
Correct For the values given,
Wapparent Wactual
= 4.50/5.00 = 0.90 , whereas for pure gold,
Wapparent Wactual
= 1−
ρw ρg
= 0.95 .
Thus, you can conclude that the the crown is not pure gold but contains some lessdense metal. The goldsmith made sure that the crown's (true) weight was the same as that of the amount of gold he was allocated, but in so doing was forced to make the volume of the crown slightly larger than it would otherwise have been.
Exercise 12.33 A rock is suspended by a light string. When the rock is in air, the tension in the string is 43.3 N . When the rock is totally immersed in water, the tension is 27.6 N . When the rock is totally immersed in an unknown liquid, the tension is 20.1 N .
Part A What is the density of the unknown liquid? ANSWER: = 1480 kg/m3
ρ
Correct Typesetting math: 70%
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± Playing with a Water Hose Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away.
Part A Will Isabella be able to spray Ferdinand if the water is flowing out of the hose at a constant speed v 0 of 3.5 meters per second? Assume that the hose is pointed parallel to the ground and take the magnitude of the acceleration g due to gravity to be 9.81 meters per second, per second.
Hint 1. General approach: considerations on particle motion Consider the water flowing out of the hose to be composed of independent particles in projectile motion. Recall that projectile motion is described as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Note that the hose is pointed parallel to the ground, so the initial velocity of the water is purely horizontal.
Hint 2. Projectile motion A projectile is a body with an initial velocity that follows a path determined entirely by the effects of gravitational acceleration and air resistance. If the effects due to air resistance are ignored, the motion of a projectile can be analyzed as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Consider a particle undergoing projectile motion in the xy coordinate plane, with the x axis horizontal and the y axis pointing vertically upward. Let v x and v y represent the components of the initial velocity of the particle. Let x0 and y represent the initial horizontal and vertical positions of the particle. The equations describing the position of the 0 particle as a function of time t are then x = x0 + v x t
y= y
0
+ vy t −
1 2
2
gt ,
where g is the magnitude of the acceleration due to gravity. ANSWER: Yes No
Correct
Part B To increase the range of the water, Isabella places her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the crosssectional area of the hose hole does she have to cover to be able to spray her friend? Assume that the cross section of the hose opening is circular with a radius of 1.5 centimeters. Express your answer as a percentage to the nearest integer. Typesetting math: 70% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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Hint 1. General approach: considerations on fluid mechanics By restricting the area across which the water is flowing, Isabella forces the water to flow out at a higher speed. After estimating the initial velocity that the water must have in order to reach a distance of 10 meters, you can determine the crosssectional area of the hose hole that corresponds to such an initial velocity of the water by applying the continuity equation.
Hint 2. Find the outflow speed needed Find the initial speed v that the water should have in order to reach a distance of 10 meters. Consider the water flowing out of the hose as composed of independent particles in projectile motion. Recall that projectile motion is described as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Note that the hose is pointed parallel to the ground, so the initial velocity of water is purely horizontal. Thus v = v x. Express your answer numerically in meters per second to three significant figures.
Hint 1. Projectile motion If the effects due to air resistance are ignored, as they should be in this problem, the motion of a projectile can be analyzed as a combination of horizontal motion with constant velocity and vertical motion with constant acceleration. Consider a particle undergoing projectile motion in the xy coordinate plane, with the x axis horizontal and the y axis pointing vertically upward. Let v x and v y represent the components of the initial velocity of the particle. Let x0 and y0 represent the initial horizontal and vertical positions of the particle. The equations describing the position of the particle as a function of time t are then x = x0 + v x t
y = y0 + vy t −
1 2
gt
2
,
where g is the magnitude of the acceleration due to gravity. In this part, you are asked to find the outflow speed, which is horizontal, and thus corresponds to v x is the above equation. Consider carefully what the values of the other constants x0 , y0 , and v y are. ANSWER: vx
= 22.1 m/s
Hint 3. Find the crosssectional area needed Once you know the horizontal velocity of the water that would be needed for the emerging stream to reach a distance of 10 meters, find the crosssectional area A that corresponds to that outflow speed. Assume that the flow rate remains steady after the partial blockage of the hose hole, so that the continuity equation applies. Express your answer numerically in centimeters squared to three significant figures.
Hint 1. Apply the continuity equation In an incompressible fluid moving steadily, the mass of fluid flowing along a flow tube is constant. In particular, consider a flow tube between two stationary cross sections with areas A1 and A2 . Let the fluid speeds at these sections be v 1 and v 2 , respectively. Then conservation of mass takes the form Typesetting math: 70% A1 v 1 = A2 v 2
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which is known as the continuity equation. If initially the water is flowing through the hose of crosssectional area A0 at a speed v 0 , and you want it to flow through a smaller area at a higher speed v , what should the new crosssectional area A of the hose be? Express your answer in of A0 , v 0 , and v . ANSWER: A
=
A 0v 0 v
ANSWER: A
= 1.12 cm2
ANSWER: f
= 84 %
Correct
Exercise 12.37 Water is flowing in a circular pipe varying crosssectional area, and at all points the water completely fills the pipe.
Part A At one point in the pipe the radius is 0.245 m . What is the speed of the water at this point if water is flowing into this pipe at a steady rate of 1.85 m3 /s ? ANSWER: v1
= 9.81 m/s
Correct
Part B At a second point in the pipe the water speed is 2.30 m/s . What is the radius of the pipe at this point? ANSWER: r2
= 0.506 m
Correct
Typesetting math: 70%
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Streamlines and Fluid Flow Learning Goal: To understand the continuity equation. Streamlines represent the path of the flow of a fluid. You can imagine that they represent a timeexposure photograph that shows the paths of small particles carried by the flowing fluid. The figure shows streamlines for the flow of an incompressible fluid in a tapered pipe of circular cross section. The speed of the fluid as it enters the pipe on the left is v 1 . Assume that the cross sectional areas of the pipe are A1 at its entrance on the left and A2 at its exit on the right.
Part A Find Q1 , the volume of fluid flowing into the pipe per unit of time. This quantity is also known as the volumetric flow rate. Express the volumetric flow rate in of any of the quantities given in the problem introduction.
Hint 1. Find the volume of fluid entering the pipe The volumetric flow rate has units of volume per unit time (cubic meters per second). What is the volume ΔV of fluid entering the pipe in time Δt? Express your answer in of any or all of the following quantities: v 1 , A1 , and Δt.
Hint 1. How far does the fluid move in time Δt? The figure shows a snapshot of the pipe at two instants in time: t0 and t0 + Δt . In the time interval Δt, some additional fluid (shown in blue) has entered the pipe from the left. The speed of the lefthand boundary of the bluecolored fluid is v 1 . What is the length L 1 of the region of additional fluid that has entered the pipe in the time interval Δt? Express your answer in of v 1 and Δt.
Typesetting math: 70%
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ANSWER: L1
=
v 1 Δt
ANSWER: ΔV
=
A 1 v 1 Δt
ANSWER: Q
1
=
v 1 A1
Correct
Part B Because the fluid is assumed to be incompressible and mass is conserved, at a particular moment in time, the amount of fluid that flows into the pipe must equal the amount of fluid that flows out. This fact is embodied in the continuity equation. Using the continuity equation, find the velocity v 2 of the fluid flowing out of the right end of the pipe. Express your answer in of any of the quantities given in the problem introduction.
Hint 1. Find the volumetric flow rate out of the pipe Find Q2 , the volume of fluid flowing out of the pipe per unit of time. Express your answer in of v 2 and A2 . ANSWER: Q2
=
v 2 A2
Hint 2. Apply the continuity equation The continuity equation states that the volumetric flows into and out of the pipe must be the same. Fill in the right hand side of the continuity equation for this problem. Express your answer in of v 2 and any of the quantities given in the problem introduction. ANSWER: v 1 A1
=
v 2 A2
ANSWER:
v2
=
v1
A1 A2
Typesetting math: 70%
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Correct
Part C If you are shown a picture of streamlines in a flowing fluid, you can conclude that the __________ of the fluid is greater where the streamlines are closer together. Enter a oneword answer. ANSWER: speed
Correct Thus the velocity of the flow increases with increasing density (number per unit area) of streamlines.
Exercise 12.46 A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 L cans per minute. At point 2 in the pipe, the gauge pressure is 152 kPa and the crosssectional area is 8.00 cm2 . At point 1, 1.35 m above point 2, the crosssectional area is 2.00 cm2 .
Part A Find the mass flow rate. ANSWER: M
= 1.30 kg/s
Correct
Part B Find the volume flow rate. ANSWER: = 1.30 L/s
Q
Correct
Part C Find the flow speed at point 1. Typesetting math: 70% ANSWER:
https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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= 6.50 m/s
Correct
Part D Find the flow speed at point 2. ANSWER: v2
= 1.63 m/s
Correct
Part E Find the gauge pressure at point 1. ANSWER: P
= 119 kPa
Correct
Venturi Meter with Two Tubes A pair of vertical, openended glass tubes inserted into a horizontal pipe are often used together to measure flow velocity in the pipe, a configuration called a Venturi meter. Consider such an arrangement with a horizontal pipe carrying fluid of density ρ. The fluid rises to heights h1 and h2 in the two openended tubes (see figure). The crosssectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Typesetting math: 70%
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Part A Find p1 , the gauge pressure at the bottom of tube 1. (Gauge pressure is the pressure in excess of outside atmospheric pressure.) Express your answer in of quantities given in the problem introduction and \texttip{g}{g}, the magnitude of the acceleration due to gravity.
Hint 1. How to approach the problem Use Bernoulli's law to compute the difference in pressure between the top and bottom of tube 1. The pressure at the top of the tube is defined to be atmospheric pressure. Note: Inside the tube, since the fluid is not flowing, the involving velocity in Bernoulli's equation can be ignored. Thus, Bernoulli's equation reduces to the formula for pressure as a function of depth in a fluid of uniform density.
Hint 2. Simplified Bernoulli's equation For a fluid of uniform density \texttip{\rho }{rho} that is not flowing, the pressure \texttip{p}{p} at a depth \texttip{h}{h} below the surface is given by p = p_0 + \rho g h, where \texttip{p_{\rm 0}}{p_0} is the pressure at the surface and \texttip{g}{g} is the magnitude of the acceleration due to gravity. ANSWER: \texttip{p_{\rm 1}}{p_1} = {\rho}gh_{1}
Correct The fluid is pushed up tube 1 by the pressure of the fluid at the base of the tube, and not by its kinetic energy, since there is no streamline around the sharp edge of the tube. Thus energy is not conserved (there is turbulence at the edge of the tube) at the entrance of the tube. Since Bernoulli's law is essentially a statement of energy conservation, it must be applied separately to the fluid in the tube and the fluid flowing in the main pipe. However, the pressure in the fluid is the same just inside and just outside the tube.
Part B Find \texttip{v_{\rm 1}}{v_1}, the speed of the fluid in the left end of the main pipe. Express your answer in of \texttip{h_{\rm 1}}{h_1}, \texttip{h_{\rm 2}}{h_2}, \texttip{g}{g}, and either \texttip{A_{\rm 1}}{A_1} and \texttip{A_{\rm 2}}{A_2} or \texttip{\gamma }{gamma}, which is equal to \large{\frac{A_1}{A_2}}.
Hint 1. How to approach the problem Energy is conserved along the streamlines in the main flow. This means that Bernoulli's law can be applied to obtain a relationship between the fluid pressure and velocity at the bottom of tube 1, and the fluid pressure and velocity at the bottom of tube 2.
Hint 2. Find \texttip{p_{\rm 2}}{p_2} in of \texttip{h_{\rm 2}}{h_2} What is \texttip{p_{\rm 2}}{p_2}, the pressure at the bottom of tube 2? Express your answer in of \texttip{h_{\rm 2}}{h_2}, \texttip{g}{g}, and any other given quantities.
Hint 1. Recall Part A Typesetting math: 70% https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5168450
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Obtain the solution for this part in the same way that you found an expression for \texttip{p_{\rm 1}}{p_1} in of \texttip{h_{\rm 1}}{h_1} in Part A of this problem. ANSWER: \texttip{p_{\rm 2}}{p_2} = {\rho} g h_{2}
Hint 3. Find \texttip{p_{\rm 2}}{p_2} in of given quantities Find \texttip{p_{\rm 2}}{p_2}, the fluid pressure at the bottom of tube 2. Express your answer in of \texttip{p_{\rm 1}}{p_1}, \texttip{v_{\rm 1}}{v_1}, \texttip{\rho }{rho}, \texttip{A_{\rm 1}}{A_1}, and \texttip{A_{\rm 2}}{A_2}.
Hint 1. Find the pressure at the bottom of tube 2 Find \texttip{p_{\rm 2}}{p_2}, the fluid pressure at the bottom of tube 2. Express your answer in of \texttip{p_{\rm 1}}{p_1}, \texttip{v_{\rm 1}}{v_1}, and \texttip{v_{\rm 2}}{v_2}. ANSWER: \texttip{p_{\rm 2}}{p_2} = \large{p_{1} + \frac{1}{2} {\rho} \left({v_{1}}^{2} {v_{2}}^{2}\right)}
Hint 2. Find \texttip{v_{\rm 2}}{v_2} in of \texttip{v_{\rm 1}}{v_1} The fluid is incompressible, so you can use the continuity equation to relate the fluid velocities \texttip{v_{\rm 1}}{v_1} and \texttip{v_{\rm 2}}{v_2} in of the geometry of the pipe. Find \texttip{v_{\rm 2}}{v_2}, the fluid velocity at the bottom of tube 2, in of \texttip{v_{\rm 1}}{v_1}. Your answer may include \texttip{A_{\rm 1}}{A_1} and \texttip{A_{\rm 2}}{A_2}, the crosssectional areas of the pipe. ANSWER: \texttip{v_{\rm 2}}{v_2} = \large{\frac{A_{1}}{A_{2}} v_{1}}
ANSWER: \texttip{p_{\rm 2}}{p_2} \large{p_{1} + \frac{1}{2} {\rho} \left({v_{1}}^{2}\right)\left(1\left(\frac{A_{1}} = {A_{2}}\right)^{2}\right)}
ANSWER: \texttip{v_{\rm 1}}{v_1} = \large{\sqrt{2g{\frac{h_{1}h_{2}}{{\gamma}^{2}1}}}}
Typesetting math: 70%
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Correct Note that this result depends on the difference between the heights of the fluid in the tubes, a quantity that is more easily measured than the heights themselves.
Problem 12.92 In 1993 the radius of Hurricane Emily was about 350 {\rm km}. The wind speed near the center ("eye") of the hurricane, whose radius was about 30 {\rm km}, reached about 200 {\rm km}/{\rm h}. As air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant.
Part A Estimate the wind speed at the rim of the hurricane. Express your answer using two significant figures. ANSWER: v = 17 {\rm km/h}
Correct
Part B Estimate the pressure difference at the earth's surface between the eye and the rim. (Hint: The density of air (1 {\rm atm}, 20 ^\circ {\rm C}) is 1.20 {\rm kg}/{\rm m}^{3}.) Express your answer using two significant figures. ANSWER: \Delta P = 1800 {\rm Pa}
Correct
Part C Where is the pressure greater? ANSWER: at the eye at the rim
Correct Typesetting math: 70%
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Part D If the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go? Express your answer using two significant figures. ANSWER: h = 160 {\rm m}
Correct
Problem 12.95 A liquid flowing from a vertical pipe has a very definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed v_0 and the radius of the stream of liquid is r_0.
Part A Find an equation for the speed of the liquid as a function of the distance y it has fallen. Express your answer in of the given quantities and appropriate constants. ANSWER: v = \sqrt{v_{0}{^2}+2gy}
Correct
Part B Combining this with the equation of continuity, find an expression for the radius of the stream as a function of y. Express your answer in of the given quantities and appropriate constants. ANSWER: r = \large{\sqrt{r_{0}{^2}{\frac{v_{0}}{\sqrt{v_{0}{^2}+2gy}}}}}
Correct
Part C If water flows out of a vertical pipe at a speed of 1.20 \rm m/s, how far below the outlet will the radius be onehalf the original radius of the stream? Express your answer using three significant figures. ANSWER:
Typesetting math: 70%
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y = 1.10 {\rm m}
Correct
Problem 12.57 A 0.170{\rm kg} cube of ice (frozen water) is floating in glycerine. The gylcerine is in a tall cylinder that has inside radius 3.80 {\rm cm} . The level of the glycerine is well below the top of the cylinder.
Part A If the ice completely melts, by what distance does the height of liquid in the cylinder change? Express your answer with the appropriate units. ANSWER: d = 0.773 {\rm cm}
Correct
Part B Does the level of liquid rise or fall? That is, is the surface of the water above or below the original level of the gylcerine before the ice melted? ANSWER: The level of liquid in the cylinder rises. The level of liquid in the cylinder falls.
Correct
Exercise 12.3 You purchase a rectangular piece of metal that has dimensions 5.2 {\rm {\rm mm}} \times 15.2 {\rm {\rm mm}} \times 33.7 {\rm {\rm mm}} and mass 1.86×10−2 {\rm {\rm kg}} .
Part A The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? ANSWER: Typesetting math: 70%
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\rho = 6980 {\rm kg/m^3}
Correct
Part B Were you cheated? ANSWER: yes no
Correct
Exercise 12.6
Part A What is the average density of the sun? ANSWER: \rho_{\rm av} = 1410 {\rm kg}/{\rm m}^{3}
Correct
Part B What is the average density of a neutron star that has the same mass as the sun but a radius of only 20.0 km? ANSWER: \rho_{\rm av} = 5.94×1016 {\rm kg}/{\rm m}^{3}
Correct
Exercise 12.12 A barrel contains a 0.125m layer of oil floating on water that is 0.280 {\rm {\rm m}} deep. The density of the oil is 650 {\rm {\rm kg/m^3}} Typesetting math: 70%
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Part A What is the gauge pressure at the oilwater interface? ANSWER: P = 796 {\rm Pa}
Correct
Part B What is the gauge pressure at the bottom of the barrel? ANSWER: P = 3540 {\rm Pa}
Correct
Exercise 12.18 The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 \rm m above the surface of the water in the cup.
Part A What is the lowest gauge pressure that the average person can achieve inside his lungs? Express your answer with the appropriate units. ANSWER: p_{\rm g} = −1.1×104 {\rm Pa}
Correct
Part B Why is your answer in the previous part negative? ANSWER:
Typesetting math: 70%
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In order for water to go up the straw, the pressure at the top of the straw must be greater than the atmospheric pressure. In order for water to go up the straw, the gauge pressure of the person must be greater than the pressure due to the column of water in the straw. In order for water to go up the straw, the pressure at the top of the straw must be less than the atmospheric pressure. In order for water to go up the straw, the gauge pressure of the person must be less than the pressure due to the column of water in the straw.
Correct
Exercise 12.32 A hollow, plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.660 {\rm \; m^3} and the tension in the cord is 2120 {\rm \; N} .
Part A Calculate the buoyant force exerted by the water on the sphere. Express your answer with the appropriate units. ANSWER: F = 6470 {\rm N}
Correct
Part B What is the mass of the sphere? Express your answer with the appropriate units. ANSWER: m = 444 {\rm kg}
Correct
Part C The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged? ANSWER:
Typesetting math: 70%
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\large{\frac{V_{\rm sub}}{V_{\rm obj}}} = 67.2 \%
Correct
Exercise 12.37 A shower head has 20 circular openings, each with radius 0.90 {\rm mm} . The shower head is connected to a pipe with radius 0.98 {\rm cm} .
Part A If the speed of water in the pipe is 3.4 {\rm m/s} , what is its speed as it exits the showerhead openings? Express your answer using two significant figures. ANSWER: v = 20 \rm m/s
Correct
Exercise 12.43
Part A What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of 12.0 {\rm m} ? (Assume that the mains have a much larger diameter than the fire hose.) ANSWER: P = 1.18×105 \rm Pa
Correct
Exercise 12.48 A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 L cans per minute. At point 2 in the pipe, the gauge pressure is 152\;{\rm kPa} and the crosssectional area is 8.00 \;{\rm cm}^{2}. At point 1, 1.35 \;{\rm m} above point 2, the crosssectional area is 2.00 \;{\rm cm}^{2}.
Part A Find the mass flow rate. ANSWER:
Typesetting math: 70%
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M = 1.30 {\rm kg/s}
Correct
Part B Find the volume flow rate. ANSWER: Q = 1.30 {\rm L/s}
Correct
Part C Find the flow speed at point 1. ANSWER: v_1 = 6.50 {\rm m/s}
Correct
Part D Find the flow speed at point 2. ANSWER: v_2 = 1.63 {\rm m/s}
Correct
Part E Find the gauge pressure at point 1. ANSWER: P = 119 {\rm kPa}
Correct Typesetting math: 70%
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Problem 12.82 In 1993 the radius of Hurricane Emily was about 350 {\rm km}. The wind speed near the center ("eye") of the hurricane, whose radius was about 30 {\rm km}, reached about 200 {\rm km}/{\rm h}. As air swirled in from the rim of the hurricane toward the eye, its angular momentum remained roughly constant.
Part A Estimate the wind speed at the rim of the hurricane. Express your answer using two significant figures. ANSWER: v = 17 {\rm km/h}
Correct
Part B Estimate the pressure difference at the earth's surface between the eye and the rim. (Hint: The density of air (1 {\rm atm}, 20 ^\circ {\rm C}) is 1.20 {\rm kg}/{\rm m}^{3}.) Express your answer using two significant figures. ANSWER: \Delta P = 1800 {\rm Pa}
Correct
Part C Where is the pressure greater? ANSWER: at the eye at the rim
Correct
Part D If the kinetic energy of the swirling air in the eye could be converted completely to gravitational potential energy, how high would the air go? Express your answer using two significant figures. Typesetting math: 70%
ANSWER:
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Ch 12 HW
h = 160 {\rm m}
Correct
Problem 12.88 A siphon, as shown in the figure , is a convenient device for removing liquids from containers. To establish the flow, the tube must be initially filled with fluid. Let the fluid have density \rho, and let the atmospheric pressure be p_{\rm{a}} {\kern 1pt}. Assume that the crosssectional area of the tube is the same at all points along it.
Part A If the lower end of the siphon is at a distance h below the surface of the liquid in the container, what is the speed of the fluid as it flows out the lower end of the siphon? (Assume that the container has a very large diameter, and ignore any effects of viscosity.) Express your answer in of the given quantities and appropriate constants. ANSWER: v = \sqrt{2gh}
Correct
Part B A curious feature of a siphon is that the fluid initially flows "uphill." What is the greatest height H that the high point of the tube can have if flow is still to occur? Express your answer in of the given quantities and appropriate constants. ANSWER: H = \large{\frac{p_{a}}{{\rho}g}}
Correct
Typesetting math: 70%
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Ch 12 HW
Score Summary: Your score on this assignment is 99.9%. You received 21.97 out of a possible total of 22 points.
Typesetting math: 70%
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