Download F - SPS186.org

5. (II) A bottle has a mass of 35.00 g when empty
and 98.44 g when filled with water. When filled
with another fluid, the mass is 88.78 g. What is
the specific gravity of this other fluid?
5.  Take the ratio of the density of the fluid to that
of water, noting that the same volume is used
for both liquids (so V cancels).
SG
1
ΔP = ρgh
= (1.05 x 103 kg/m3)(9.8m/s2)(1.60m)
= 1.646 x
104
N/m2
1mmHg
×
133N /m 2
= 124 mmHg
€
2
9.  (I) (a) Calculate the total force of the
atmosphere acting on the top of a table that
measures
(b) What is the total force acting upward on the
underside of the table?
9. (a)
The total force of the atmosphere on
the table will be the air pressure times the area
of the table.
(b) Since the atmospheric pressure is the same on the underside of the
table (the height difference is minimal), the upward force of air pressure is
the same as the downward force of air on the top of the table,
3
10.(II) In a movie, Tarzan evades his captors by
hiding underwater for many minutes while
breathing through a long, thin reed. Assuming
the maximum pressure difference his lungs can
manage and still breathe is 85 mmHg, calculate
the deepest he could have been.
10.The pressure difference on the lungs is the
pressure change from the depth of water
4
Atmospheric Pressure
•  Earth’s atmosphere: A fluid.
•  ⇒ Change in pressure: ΔP = ρgΔh
•  At sea level: P0 ≡ 1.013 × 105 N/m2
= 101.3 kPa ≡ 1 atm
The cause of (air) pressure at any height
in atmosphere is:
weight of air above that height!
Measurement of Pressure; Gauges and the
Barometer
There are a number of different types of
pressure gauges. This one is an opentube manometer. The pressure in the
open end is atmospheric pressure; the
pressure being measured will cause
the fluid to rise until
the pressures on both
sides at the same
height are equal.
6
Pressure beneath surface of liquid open to the atmosphere
Gauge Pressure
Initial Pressure (atmospheric pressure)
ABSOLUTE PRESSURE
Gauge Pressure = The amount by which P
exceeds atmospheric pressure Po
7
Gauge Pressure
•  Pressure gauges (like tire gauges, etc.)
measure difference between
atmospheric pressure P0 & internal
pressure (of tire, for example).
•  Gauge pressure: PG = P – P0
Measurement of Pressure; Gauges and the
Barometer
Here are two more devices for
measuring pressure: the tire
pressure gauge and the
aneroid gauge.
9
Measurement of Pressure; Gauges and the Barometer
This is a mercury barometer,
developed by Torricelli to measure
atmospheric pressure. The height of
the column of mercury is such that
the pressure in the tube at the
surface level is 1 atm.
Therefore, pressure is often quoted
in millimeters (or inches) of
mercury.
10
Measurement of Pressure; Gauges and the
Barometer
Any liquid can serve in a
Torricelli-style barometer, but
the most dense ones are the
most convenient.
(Density of Hg is 13.5 g/cm3)
This barometer uses water.
(Density of H2O is 1 g/cm3)
11
Mercury Barometer
•  Weather reports:
•  Barometric pressure (atmospheric
pressure): 28-32 inches Hg
76 cm = 760 mm = 29.29 inches
When h = 760 mm,
P = ρHg gh = 1.013 × 105 N/m2 =1atm
•  If we use water instead of mercury,
P = 1atm = ρH2O gh
⇒ h ≈ 10 m ≈ 30 feet!
Example
How high would the level
be in an alcohol barometer
at normal atmospheric
pressure? (The density of
the alcohol is 0.79 g/cm3)
o
A closer look at Pressure vs. Depth
Depth below surface
Initial Pressure – May or MAY NOT be atmospheric pressure
ABSOLUTE PRESSURE
14
Example
(a) What are the total force and the absolute
pressure on the bottom of a swimming pool
22.0 m by 8.5 m whose uniform depth is 2.0 m?
The total force is the absolute pressure times the area of the
bottom of the pool.
(b) What will be the pressure against the side of
the pool at the bottom?
The pressure against the side of the pool, near the
bottom, will be the same as the pressure at the bottom,
Pascal’s Law
•  Experimental fact:
An external pressure P applied to a
confined fluid increases the pressure
throughout a fluid by P
≡ Pascal’s Law
•  Simple example: Water in a lake (at rest).
At depth h below surface, pressure is
P = P0 + ρgh
(P0 = atmospheric pressure)
If P0 changes, P changes evenly at a given height!
Pascal’s Law
•  The volume of liquid pushed
down on left must equal volume
pushed up on right. So,
Pascal’s Law
•  The volume of liquid pushed
down on left must equal volume
pushed up on right. So,
•  Combining this with
Pascal’s Law
•  The volume of liquid pushed down
on left must equal volume pushed up
on right. So,
•  Combining this with
•  And knowing V = AΔx  A = V/Δx
F1Δx1 F2 Δx 2
=
V
V
Pascal’s Law
•  The volume of liquid pushed
down on left must equal volume
pushed up on right. So,
•  Combining this with
•  Gives
Pascal’s Law
•  The volume of liquid pushed
down on left must equal volume
pushed up on right. So,
•  Combining this with
•  Gives
⇒ This means that
Work1 = Work2
Pascal’s Law
•  The volume of liquid pushed
down on left must equal volume
pushed up on right. So,
•  Combining this with
•  Gives
⇒ This means that
Work1 = Work2
So, Pascal’s Law is consequence of Conservation of Mechanical Energy
A closed system
If you take a liquid and place it in a system
that is CLOSED like plumbing, for
example, or a car’s brake line, the
PRESSURE is the same everywhere.
Since this is true, if you apply a force at
one part of the system, the pressure is
the same at the other end of the
system. The force, on the other hand
MAY or MAY NOT equal the initial force
applied. It depends on the AREA.
You can take advantage of the fact that
the pressure is the same in a closed
system as it has MANY applications.
The idea behind this is called PASCAL’S
PRINCIPLE
23
Pascal’s Principle
If an external pressure is applied to a confined
fluid, the pressure at every point within the fluid
increases by that amount.
Pin = Pout
Hydraulic press
24
Hydraulic press
An applied force F1 can
be “amplified”:
Examples: hydraulic brakes,
forklifts, car lifts, etc.
Pascal’s Law, Example #3
•  Car lift in a service station. A large output force can be
applied by means of a small input force. Volume of
liquid pushed down on left must equal volume pushed
up on right.
•  Circular cross section system.
On left, r1 = 5 cm = 0.05 m. On right r2 = 15
cm = 0.15 m. Car’s weight mg = 13,300 N.
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5 cm = 0.05 m. On
right r2 = 15 cm = 0.15 m. Car’s
weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5.00 cm = 0.0500 m.
On right r2 = 15.0 cm = 0.150 m.
Car’s weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Pascal’s Law ⇒
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5.00 cm = 0.0500 m.
On right r2 = 15.0 cm = 0.150 m.
Car’s weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Gives
Pascal’s Law ⇒
F1 = (A1/A2)F2
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5.00 cm = 0.0500 m.
On right r2 = 15.0 cm = 0.150 m.
Car’s weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Gives
Pascal’s Law ⇒
F1 = (A1/A2)F2
2
2
A
π
r
r
where A=πr2 ... so 1 = 1 2 = 1 2
A2
πr2
r2
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5.00 cm = 0.0500 m.
On right r2 = 15.0 cm = 0.150 m.
Car’s weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Gives
Pascal’s Law ⇒
F1 = (A1/A2)F2 = 1,476.7 N =1480N
Pascal’s Law, Example
•  Car lift in a service station. See
figure. A large output force can be
applied by means of a small input force.
Volume of liquid pushed down on left
must equal volume pushed up on right.
•  Circular cross section system.
On left r1 = 5.00 cm = 0.0500 m.
On right r2 = 15.0 cm = 0.150 m.
Car’s weight mg = 13,300 N.
N’s 2nd Law on right: ∑Fy = 0 = F2 – mg. Or F2 = 13,300 N
Calculate minimum F1 to lift the car & pressure P in the system.
Gives
Pascal’s Law ⇒
F1 = (A1/A2)F2 = 1,476.7 N =1480N
P = (F1/A1) = 1.88 × 105 Pa
Example: What is the resulting force on the
bottom of the jug?
F1
F2 = A2
A1
€
10N
2
=
(500cm
)
2
5cm
1m 2
2
×
=
0.0005m
100 2 cm 2
€
(since units on A
cancel, no need to
convert cm2 to m2)
= 1000 N
Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is
found to be the upward
force on the same volume
of water:
Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is found to
be the upward force on the
same volume of water:
= P2 A - P1 A = ρf gh2 A-ρf gh1 A
Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is
found to be the upward
force on the same volume
of water:
Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is
found to be the upward
force on the same volume
of water:
Buoyancy and Archimedes’ Principle
This is an object submerged in a fluid. There is a
net force on the object because the pressures at
the top and bottom of it are different.
The buoyant force is
found to be the upward
force on the same volume
of water:
Archimedes Principle
Any object completely or partially
submerged in a fluid is buoyed up by a
force whose magnitude is equal to the
weight of the fluid displaced by the
object.
39
Buoyancy and Archimedes’ Principle
The new (“apparent”) weight of the object is then
the difference between the buoyant force and the
gravitational force.
Example in extra
chapter hand-out,
p.284
Buoyancy and Archimedes’ Principle
Applied to Floating Objects
If the object’s density is less than that of water,
there will be an upward buoyant force on it that
is greater than the downward force of its weight,
and it will rise until it is partially out of the water.
41
Floating objects
If the log is fully submerged, it will displace a mass
of water mF = ρFV = (1000 kg/m3)(2.0 m3) = 2000kg
So the buoyant force on the log will be greater than its
weight, and it will float to the surface.
It will come to equilibrium when it displaces 1200 kg
of water. (At this time, FB has gone down).
In general, for a floating object, the buoyant force will
equal the weight of the object.
FB = w
42
Buoyancy and Archimedes’ Principle
This principle also works in
the air; this is why hot-air and
helium balloons rise.
43
Checking Understanding
Two blocks of identical size are submerged
in water. One is made of lead (heavy), the
other of aluminum (light). Upon which is the
buoyant force greater?
A.  On the lead block.
B.  On the aluminum block.
C.  They both experience the same buoyant force.
Answer
Two blocks of identical size are submerged
in water. One is made of lead (heavy), the
other of aluminum (light). Upon which is the
buoyant force greater?
A.  On the lead block.
B.  On the aluminum block.
C.  They both experience the same buoyant force.
Checking Understanding
Two blocks are of identical size. One is made
of lead, and sits on the bottom of a pond; the
other is of wood and floats on top. Upon
which is the buoyant force greater?
A.  On the lead block.
B.  On the wood block.
C.  They both experience the same buoyant force.
Answer
Two blocks are of identical size. One is made
of lead, and sits on the bottom of a pond; the
other is of wood and floats on top. Upon
which is the buoyant force greater?
A.  On the lead block.
B.  On the wood block.
C.  They both experience the same buoyant force.
Checking Understanding
A barge filled with ore floats in a canal lock. If
the ore is tossed overboard into the lock, the
water level in the lock will
A.  rise.
B.  fall.
C.  remain the same.
Answer
A barge filled with ore floats in a canal lock. If
the ore is tossed overboard into the lock, the
water level in the lock will
A.  rise.
B.  fall.
C.  remain the same.
4913-30
Slide
A helicopter lowers a probe into Lake Michigan which
is suspended on a cable. The probe has a mass of 500 kg
and its average density is 1400 kg/m3. What is the
tension in the cable?
1400 N
a) mg upward
b) ρH20Vg upward
c) (ρH20-ρwood) Vg upward
a) mg upward
b) ρH20Vg upward
c) (ρsteel-ρH20) Vg upward
40 cm
p. 323/ #54, 57, 69, 72, 74, 75
due Tues
Fluids in Motion; Flow Rate and the
Equation of Continuity
If the flow of a fluid is smooth, it is called streamline or
laminar flow (a).
Above a certain speed, the flow becomes turbulent (b).
Turbulent flow has eddies; the viscosity of the fluid is much
greater when eddies are present.
55
Bernoulli’s Principle & Fluid
Continuity
56
Principles of Fluid Flow
•  The speed of a fluid depends on the cross
sectional area
–  This is because the same volume of fluid must pass
through the given object per unit of time.
•  Fluid Continuity
A1v1 = A2v2
•  When you put your finger over the end of the
hose, what happens?
–  Animation
57
The Equation of Continuity
Have you ever used your thumb to control the water flowing
from the end of a hose?
58
The Equation of Continuity
Have you ever used your thumb to control the water flowing
from the end of a hose?
When the end of a hose is partially closed off, thus reducing
its cross-sectional area, the fluid velocity increases.
This kind of fluid behavior is described by the equation of
continuity.
59
Equation of Continuity
60
If there are no losses of
fluid within a uniform
tube, the mass of fluid
flowing into the tube in
a given time must be
equal to the mass
flowing out of the tube
in the same time. So
the mass Δm1 entering
the tube during time Δt
is:
Δm1 = ρ1ΔV1 = ρ1(A1Δx1 ) = ρ1(A1v1Δt)
61
Equation of Continuity
62
Equation of Continuity
So the mass entering is
Δm1 = ρ1ΔV1 = ρ1(A1Δx1 ) = ρ1(A1v1Δt)
And the mass leaving is
Δm2 = ρ2ΔV2 = ρ2(A2Δx2 ) = ρ2(A2v2Δt)
And since mass is conserved, Δm1 = Δm2 and
ρ1A1v1 = ρ2A2v2
And for an incompressible fluid, the density is constant, so
A1v1 = A2v2
63
Equation of Continuity
64
Equation of Continuity
65
Fluids in Motion; Flow Rate and the
Equation of Continuity
The mass flow rate is the mass that passes a
given point per unit time. The flow rates at any
two points must be equal, as long as no fluid is
being added or taken away.
This gives us the equation of continuity:
66
Fluids in Motion; Flow Rate and the
Equation of Continuity
If the density doesn’t change – typical for
liquids – this simplifies to
.
Where the pipe is wider, the flow is slower.
67
Flow Rate Equation
•  The continuity equation is sometimes called
the flow rate equation. Av is called the
volume rate of flow and is the volume of fluid
that passes by a point in the tube per unit
time.
•  Av has units: m2 m/s = m3/s, volume per time.
Flow rate = Q = Av = Vol/time
Example:
High cholesterol in the blood can cause fatty deposits
called plaques to form on the walls of blood vessels.
Suppose a plaque reduces the effective radius of an
artery by 25%. How does this partial blockage affect
the speed of blood through the artery?
A1v1 = A2v2
π r12v1 = π r22v2
v2 = (r1 / r2)2v1
Since r2 = .75 r1 , then r1 / r2 = 1 / 0.75
So v2 = (1 / 0.75)2 v1 = 1.8v1
The speed through the clogged artery increases by 80%
69
Example 2:
Blood flows at a rate of 5.00 L/min through
an aorta with a radius of 1.00 cm. What is
the speed of blood flow in the aorta?
Flow rate = Av
70
Example 2:
Blood flows at a rate of 5.00 L/min through an
aorta with a radius of 1.00 cm. What is the
speed of blood flow in the aorta?
5L
min 1000cm 3
13 m 3
×
×
×
Flow rate = Av =
min 60sec
L
100 3 cm 3
€
71
Example 2:
Blood flows at a rate of 5.00 L/min through an
aorta with a radius of 1.00 cm. What is the
speed of blood flow in the aorta?
5L
min 1000cm 3
13 m 3
×
×
×
Flow rate = Av =
min 60sec
L
100 3 cm 3
= 8.3 x 10-5 m3/s
€
72
Example 2:
Blood flows at a rate of 5.00 L/min through an
aorta with a radius of 1.00 cm. What is the
speed of blood flow in the aorta?
5L
min 1000cm 3
13 m 3
×
×
×
Flow rate = Av =
min 60sec
L
100 3 cm 3
= 8.3 x 10-5 m3/s
€
v = 8.3 x 10-5 m3/s / A
73
Example 2:
Blood flows at a rate of 5.00 L/min through an
aorta with a radius of 1.00 cm. What is the
speed of blood flow in the aorta?
5L
min 1000cm 3
13 m 3
×
×
×
Flow rate = Av =
min 60sec
L
100 3 cm 3
= 8.3 x 10-5 m3/s
€
v = 8.3 x 10-5 m3/s / A = 8.3 x 10-5 m3/s / π r 2
74
Example 2:
Blood flows at a rate of 5.00 L/min through an
aorta with a radius of 1.00 cm. What is the
speed of blood flow in the aorta?
5L
min 1000cm 3
13 m 3
×
×
×
Flow rate = Av =
min 60sec
L
100 3 cm 3
= 8.3 x 10-5 m3/s
€
v = 8.3 x 10-5 m3/s / A = 8.3 x 10-5 m3/s / π r 2
v = 0.265 m/s
75
Bernoulli’s Equation
For steady flow, the speed, pressure, and elevation of an
incompressible and nonviscous fluid are related by an equation
discovered by Daniel Bernoulli (1700–1782).
76
Bernoulli’s Principle
•  Pressure also changes with velocity.
•  Bernoulli’s Principle
–  When the speed of a fluid increases, the pressure within
the fluid decreases.
•  This has to do with conservation of energy.
–  If the velocity increases, the KE increases.
–  In order to keep the total energy constant, a quantity
must decrease.
–  Since PE is constant, work must be done.
–  W = Fd and P = F/A  W = PAd = PV
•  The volume stays the same, therefore the pressure decreases.
77
Bernoulli’s Equation
In the steady flow of a nonviscous, incompressible fluid of
density ρ, the pressure P, the fluid speed v, and the elevation
y at any two points (1 and 2) are related by
78
Bernoulli’s Equation
A fluid can also change its
height. By looking at the
work done as it moves, we
find:
This is Bernoulli’s
equation. One thing it tells
us is that as the speed
goes up, the pressure
79
goes down.
It’s a conservation of energy equation!
Applications of Bernoulli
•  Why during storm might a
roof blow off?
•  There is fast moving air above
(bunched up streamlines) the
roof. Faster moving air means
less pressure above than
inside.
81
Applications of Bernoulli
•  A small jeep has a soft, ragtop roof. When the jeep is at rest
the roof is flat. When the jeep is cruising at highway speeds
with its windows rolled up, does the roof
a. Bow upward
b. Remain flat
c.  Bow downward
•  A. The roof bows upward
–  When the jeep is in motion air flows over the top of the
roof, while the air inside the jeep is at rest—since the
windows are closed. Thus, there is less pressure over the
roof than under it. As a result the roof bows upward.
82
Bernoulli’s Equation example
Water circulates throughout a house in a
hot-water heating system. If the water is
pumped at a speed of 0.50 m/s through a
4.0-cm-diameter pipe in the basement
under a pressure of 3.0 atm, what will be
the flow speed and pressure in a 2.6-cmdiameter pipe on the second floor 5.0 m
above? Assume these pipes do not divide
into branches.
83
Solution
First, calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
84
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
We can call the basement point 1.
85
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
We can call the basement point 1.
v2A2 = v1A1
86
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
We can call the basement point 1.
v2 = (v1A1) / A2
= (v1π(r1)2) / (π(r2)2)
= (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s
87
Bernoulli’s Equation,
Is often more conveniently rearranged to solve for P2.
And re-labeling the
ys as hs, we write:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
We can call the basement point 1.
v2 = (v1A1) / (A2)
= (v1π(r1)2) / (π(r2)2)
= (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s
To find pressure, we use Bernoulli’s equation:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)
89
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
v2 = (v1A1) / (A2)
= (v1π(r1)2) / (π(r2)2)
= (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s
To find pressure, we use Bernoulli’s equation:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)
P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m)
+ ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2]
90
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
v2 = (v1A1) / (A2)
= (v1π(r1)2) / (π(r2)2)
= (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s
To find pressure, we use Bernoulli’s equation:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)
P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m)
+ ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2]
P2 = (3.0 x 105 N/m2) – (4.9 x 104 N/m2) – (6.0 x 102 N/m2)
91
Solution
First calculate the flow speed on the second floor, calling it v2,
using the equation of continuity.
v2 = (v1A1) / (A2)
= (v1π(r1)2) / (π(r2)2)
= (0.50 m/s)(0.020 m)2 /(0.013 m)2 = 1.2 m/s
To find pressure, we use Bernoulli’s equation:
P2 = P1 + ρg(h1– h2) + ½ρ((v1)2 – (v2)2)
P2 = (3.0 x 105 N/m2) + (1.0 x 103 kg/m3)(9.8 m/s2)(-5.0m)
+ ½ (1.0 x 103 kg/m3)[ (0.50 m/s)2 – (1.2 m/s)2]
P2 = (3.0 x 105 N/m2) – (4.9 x 104 N/m2) – (6.0 x 102 N/m2)
P2 = 2.5 x 105 N/m2
92
Applications of Bernoulli’s Principle:
Using Bernoulli’s principle, we find that the speed
of fluid coming from a spigot on an open tank is:
This is called
Torricelli’s theorem.
(Since P1=P2, Bernoulli’s
equation becomes
½ ρv12 + ρgy = ρgy2) …
solve for v1
93