1. science
2. physics

Examination Cover Sheet
Princeton University Undergraduate Honor Committee
January 20, 2008
Course Name & Number: PHY 101
Professors: Galbiati, McDonald, Nappi, Pretorius
Date: January 23, 2007
Time: 9 AM
This examination is administered under the Princeton University Honor Code.
Students should sit one seat apart from each other, if possible, and refrain from
talking to other students during the exam. All suspected violations of the Honor
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•
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•
•
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Write your name in capital letters in the appropriate space in the next page. Also,
dont forget to write and sign the Honor Code pledge in the appropriate line on the
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1
2
Your Name:
PHYSICS 101 FINAL EXAM
January 23, 2006
1
3
5
3 hours
Please Circle your section
9 am
Nappi
2 10 am McDonald
10 am
Galbiati
4 11 am McDonald
12:30 pm Pretorius
Problem
1
2
3
4
5
6
7
8
9
Total
Score
/13
/10
/14
/18
/20
/18
/10
/8
/14
/125
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 23. The exam contains 9 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you
may want to tear off and keep handy.
Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the
Honor Code during this examination.
Signature
3
Problem 1
Circle the true statements. There may be none or more than one true statement
per group.
1. (3 points)
(a) In any perfectly inelastic collision, all the kinetic energy of the bodies
involved is lost.
(b) Mechanical energy is conserved in an elastic collision.
(c) After a perfectly inelastic collision, the bodies involved move with a
common velocity.
Statement a is false. The bodies, after a perfectly inelastic collision, move with the
velocity of the center of mass prior to the collision. Not all the kinetic energy is
necessarily lost.
Statement b is true.
Statement c is true.
2. (3 points)
(a) If the angular velocity of an object is zero at some instant, the net torque
on the object must be zero at that instant.
(b) If the net torque on an object is zero at some instant, the angular velocity
must be zero at that instant.
(c) The moment of inertia of an extended solid object depends on the location
of the axis of rotation.
Statement a is false. There is no such direct relationship between torque and angular
velocity, rather between torque and angular acceleration.
Statement b is false. There is no such direct relationship between torque and angular
velocity, rather between torque and angular acceleration.
Statement b is true.
4
3. (3 points) A horizontal pipe narrows from a diameter of 10 cm to a diameter
of 5 cm. For a non-viscous liquid flowing without turbulence from the larger
diameter to the smaller,
(a) the velocity and the pressure both increase
(b) the velocity increases and the pressure decreases
(c) the velocity decreases and the pressure increases
(d) the velocity and the pressure both decrease
Only statement b is true. A reduction in the cross section implies an increase in the
velocity (due to the conservation of mass) and a decrease in pressure (due to the law
of Bernoulli).
As a consequence, statements a, c, and d are false.
4. (4 points)
(a) work can never be completely turned into heat
(b) heat can never be completely turned into work
(c) all heat engines have the same efficiency
(d) all reversible engines operating between two thermal baths at the same
temperatures have the same efficiency
(e) it is impossible to transfer heat from a cold to a hot reservoir
(f) all Carnot engines are reversible
Statement a is false. Work can be turned completely into heat (by friction, for example).
Statement a is true. Statement c is false. Efficiency of engines depends on many factors,
including reversibility (or lack of thereof) and temperatures of the baths with which heat
is exchanged. Statement d is true. Statement e is false. Heat can be transferred from
cold to hot reservoirs, at the expense of providing work: that’s what a refrigerator does,
for example. Statement f is true.
5
Problem 2
A pendulum bob consists of a string of length L = 1.0 m with a small bob on the
end. The mass of the bob is M = 10 kg. The bob is released from rest at the position
sketched in the drawing, i.e. with the string horizontal. At the lowest point of its
trajectory, the string catches on a thin peg a distance R above the lowest point.
1. (2 points) Calculate the velocity v of the bob at the lowest point of its trajectory.
v=
√
p
2gL =
2 · 9.8 m s−2 · 1.0 m = 4.4 m/s .
6
2. (2 points) Calculate the centripetal acceleration ac of the bob at the lowest point
of its trajectory, just before it hits the peg.
ac =
v2
(4.4 m/s)2
2
=
= 19.6 m/s .
L
1.0 m
3. (6 points) Calculate the largest value of R such that the bob swings around the
peg for one full circle keeping the string taut at all times.
Let’s call u the velocity of the bob at the top of the loop.
conservation:
1
1
mu2 + 2mgR = mv 2 ,
2
2
which can be rearranged into:
Because of energy
mu2 = mv 2 − 4mgR.
If the bob just makes it through a full swing, the centripetal acceleration at the top of
the loop, mu2 /R, is provided solely by the weight force, mg:
mg =
mu2
.
R
Substituting for the previous value of mu2 , we get:
mg =
mv 2 − 4mgR
mv 2
=
− 4mg,
R
R
mv 2
2gL
=
,
R
R
where we used the expression for v as a function of g and L as found in part 1 of this
problem. Therefore:
2
2
R = L = 1.0 m = 0.4 m .
5
5
5mg =
7
Problem 3
A guitar string made of steel (density ρ = 7.85 × 103 kg/m3 ) has a diameter
d = 0.9652 mm.
1. (2 points) What is the linear density µ = m/L of the string?
µ=
m
ρAL
πd2 ρ
π
3
=
=
= (0.9652×10−3 m)2 ·7.85×103 kg/m = 5.74 × 10−3 kg m−3 .
L
L
4
4
2. (4 points) What is the tension FE in a string of length 66.24 cm that has a
fundamental mode of 82.41 Hz (an “E” note)?
The speed of a wave on a string of linear mass density m/L and tension F is:
s
F
,
v=
m/L
and the frequency of its fundamental mode is
f=
v
.
2L
Solving for the tension, and expressing the mass in terms of the density and volume
m = ρπ(d/2)2 L gives:
FE = πρ(df L)2 = π 5.74×10−3
kg
(0.9652×10−3 m 82.41 Hz 0.6624 m)2 = FE = 68.5 N .
m3
8
3. (4 points) Different notes can be played on the same string by pressing down at
different positions on the fretboard while plucking the string, as this shortens
the part of the string that vibrates. Suppose you want to play an “A” note
with a fundamental frequency of 110.0 Hz on the same string of the previous
question. How much shorter does the vibrating part of the string need to be?
Pressing a string does not change its tension, hence the speed of waves on the string
does not change either. Thus:
v
fA =
,
2LA
v
,
fE =
2LE
so the string has to be shortened by
fE
82.41 Hz
= 0.6624 m 1 −
LE − LA = L 1 −
0.166 m .
fA
110.0 Hz
(which is the distance from the head of the fretboard where the string must be pressed).
4. (4 points) Suppose you plucked the string without pressing down on the
fretboard (so the length of the vibrating part is still 66.24 cm) such that only
the third harmonic of the fundamental frequency is heard. Where along the
string will the nodes of vibration be?
The third harmonic has two interior nodes in addition to the nodes at the end points
of the string. All nodes are equidistant to neighbouring nodes, thus there will be an
interior node a distance:
L
0.6624 m
=
= 0.2208 m .
3
3
from either end of the string.
9
Problem 4
The tidal forces between the earth and the moon cause the moon to exert a torque
on the earth. Data: Mearth =5.98 × 1024 kg, and assume the earth to be a uniform
density sphere of radius Rearth =6.36 × 106 m.
1. (3 points) What is the angular frequency ωf of the earth’s rotation today?
Angular frequency is:
2π
.
T
The period of rotation of the earth today is:
ω=
Tf = 1 day = 86400 s.
Thus:
ωf =
2π
2π
=
= 7.28 × 10−5 rad/s .
Tf
86400 s
2. (3 points) The effect of the torque on the earth causes the earth’s rotation to slow
down. If the angular frequency of rotation one century ago was 1.67×10−12 rad/s
less that it is today, what was the average angular acceleration of the earth over
the last century?
Angular acceleration is:
α=
∆ω
.
∆t
The text of the problem specifies that:
∆ω = ωf − ωi = −1.67 × 10−12 rad/s.
Taking into account that:
∆t = 1 century = 3.15 × 109 s.
We get:
α=
∆ω
−1.67 × 10−12 rad/s
=
= −5.4 × 10−22 rad/s2 .
∆t
3.15 × 109 s
10
3. (4 points) What was the magnitude of the average torque exerted on the earth
over this period?
Torque is given by:
τ = Iα.
The momentum of a sphere is given by:
I=
2
2
Mearth Rearth
.
5
Thus:
τ=
2
2
2
Mearth Rearth
α = 5.98×1024 kg (6.36×106 m)2 (−5.4×10−22 rad/s2 ) = −5.2 × 1016 Nm.
5
5
4. (4 points) If this torque had been applied to the earth since its formation
4.5 billion years ago, how long was an earth “day” (the time of one revolution)
then?
The equation for constant angular acceleration is:
ωf = ωi + α∆t,
where ωf is the angular velocity today and ωi is the angular velocity at a time ∆t of
4.5 billion years ago. This gives:
∆t = 4.5 × 109 yr × 365
day
s
86400
= 1.42 × 1017 s,
yr
day
ωi = ωf −α∆t = 7.28×10−5 rad/s−−5.4×10−22 rad/s2 1.42×1017 s = 1.48×10−4 rad/s,
which translates to a period:
Ti =
.
2π
2π
=
= 11.8 hours
ωi
1.48 × 10−4 rad/s
11
5. (4 points) What was the change in the earth’s angular momentum over the 4.5
billion years?
Angular momentum is:
L = Iω,
therefore:
∆L = I∆ω = Iα∆T = τ ∆T.
This gives:
∆L = −5.2 × 1016 Nm · 1.42 × 1017 s = −7.28 × 1033 kg m2 /s .
12
Problem 5
A cycle of a gasoline engine can be approximated by an Otto cycle, which consists
of two reversible adiabatic and two isochoric processes as shown in the figure below.
In this problem, an ideal Otto engine works with 500 moles of a monoatomic gas.
The following data are provided: VA = 3.53 m3 , VB = 1.55 m3 , TA = 301 K and
TC = 615 K.
P
c
B
VB
D
A
VA
V
1. (10 points) Compute V, P and T at the other vertices, and fill the entries in the
table below. For each numerical answer, clearly show your calculations.
13
A
B
C
V [m3 ]
P [Pa]
T [K]
We know that:
VA = VD = 3.53 m3 ,
and:
VB = VC = 1.55 m3 .
By repeatedly using the formulas:
P V = nRT,
Pi Viγ = Pf Vfγ ,
Ti Viγ−1 = Tf Vfγ−1 ,
one gets:
PA = 3.54 × 105 Pa ,
PB = 1.40 × 106 Pa ,
PC = 1.65 × 106 Pa ,
PD = 4.189 × 105 Pa ,
TB = 521 K ,
and
TD = 355 K .
D
14
2. (10 points) Fill now the entries in the table below by computing ∆U , Q, W and
∆S along each part of the cycle. The change of entropy from D to A is given
to you and is -1030 J/K. Show your formulas and calculations clearly.
A→B
B→C
C→D
D→A
∆U [J]
Q [J]
W [J]
∆S [J/K]
To compute ∆U we use the formula:
∆U = 3/2nR(Tf − Ti ).
Using the values of the temperature computed previously, one gets:
∆U A→B = 1.37 × 106 J ,
∆U B→C = 5.86 × 105 J ,
∆U C→D = −1.62 × 106 J ,
∆U D→A = −3.37 × 105 J .
The work is zero along the isochoric lines, and the heat is zero along the adiabatic
curves. Hence, using ∆U = Q − W, one derives:
WA→B = −1.37 × 106 J ,
WC→D = 1.62 × 106 J ,
QB→C = 5.86 × 105 J ,
QD→A = −3.37 × 105 J ,
while the other entries for Q and W are zero.
The entropy change is zero from A to B and from C to D, since these are reversible
transformation with zero heat transfer. Hence:
∆SB→C = 1030 J/K ,
so that the total change of entropy along the closed cycle is zero.
15
Problem 6
A rectangular slab of ice of base area A = 5.30 m2 and height H = 3.20 m floats on
water. Keep in mind that the densities of water and ice are: ρwater = 1.00×103 kg/m3
and ρice = 0.917 × 103 kg/m3 .
1. (2 points) What is the depth of immersion h of the slab?
The weight of the whole ice slab must equal the weight of the water displaced:
Ahgρwater = AHgρice .
Therefore:
3
h=H
ρice
0.917 × 103 kg/m
= 3.2 m
3 = 2.93 m .
ρwater
1.00 × 103 kg/m
2. (2 points) A polar bear climbs on the ice. Now the depth of immersion of the
ice slab is h0 =3.10 m. What is the mass M of the bear?
The weight of the whole ice slab plus the weight of the bear must equal the weight of
the water displaced:
Ah0 gρwater = AHgρice + M g,
from which one derives:
3
3
M = A (h0 ρwater − Hρice ) = 5.3 m2 3.1 m 1.00 × 103 kg/m − 3.2 m 0.917 × 103 kg/m = 878 kg .
16
3. (4 points) Still holding on the ice, the bear (ρbear = 0.872 × 103 kg/m3 ) partially
slides into the water, so that 30 percent of the volume of her body is submerged.
What is the depth of immersion h00 of the block now?
ρwater Ah00 g +
3
M
ρwater
g = ρice AHg + M g,
10
ρbear
from which one computes:
h00 =
or:
h00 = H
ρice AH + M −
3
M
10 ρwater ρbear
ρwater A
,
ρice
M
3M
+
−
= 3.04 m .
ρwater
ρwater A 10ρbear A
4. (4 points) The bear dives off the ice. What is the pressure exerted by the water
on the bear at a depth d = 10.0 m?
P = PAtm + ρwater gd = 1.99 × 105 Pa .
17
5. (4 points) What is the upward push on the bear at that same depth?
ρwater
ρwater
W = Mg −
Mg = Mg 1 −
= -1263 N .
ρbear
ρbear
6. (2 points) The bear emerges to get a deep breath of air and promptly submerges
again. The upward push (apparent weight) is now 1333 N, directed upward. By
how much has the volume of her lungs expanded? (You may neglect the weight
of air.)
The change in apparent weight is due to the expansion of the lungs
∆V gρwater = 70 N,
from which we derive:
∆V = 7.14 × 10−3 m3 .
18
Problem 7
A rectangular block of steel has dimensions 1.0 × 2.0 × 3.0 cm3 , and Young’s modulus
Y = 2.0 × 1011 N/m2 . The mass of this block can be neglected in this problem.
1. (3 points) Suppose the block is welded to a metal ceiling such that the 2.0-cm
length of the block is vertical, and a mass M = 100 kg is welded to the bottom
face of the block. How much does the block stretch vertically?
The block is a spring whose spring constant can be deduced from the definition of the
Young’s modulus:
F ∆L
Y =
,
A L
so that:
YA
F =
∆L = k∆L,
L
which implies that the spring constant is:
2
k=
YA
2 × 1011 N/m · 0.01 m · 0.03 m
=
= 3 × 109 N/m.
L
0.02 m
The vertical force due to M = 100 kg is F = M g = 980 N. Hence, the vertical stretch
of the block is:
∆L = F/k =
980 N
= 0.000327 m = 0.33 mm .
3 × 106 Nm
19
2. (4 points) Suppose the mass M is struck from below with a hammer so that it
oscillates vertically. According to Young’s law, the block behaves like a spring,
where the displacement from the equilibrium position is now ∆L. What is the
frequency of oscillation?
p
The angular frequency of oscillation is ω = 2πf = k/M , so:
s
r
1
k
1
3 × 109 Nm
f=
=
= 872 Hz .
2π M
2π
100 kg
3. (3 points) The block could also be welded so that either 1.0-cm length or its
3.0-cm length were vertical, leading to three different frequencies of vertical
oscillation. What is the ratio of the largest to the smallest of these frequencies?
We could repeat parts a and b√for thepother 2 orientations of the steel block. However,
it is quicker to note that f ∝ k ∝ A/L.
p
√
When the 1.0-cm length is vertical, f1 ∝ p2 · 3/1 = p6.
When the 2.0-cm length is vertical, f1 ∝ p1 · 3/2 = p3/2.
When the 3.0-cm length is vertical, f1 ∝ 1 · 2/3 = 2/3.
Hence, the ratio of the largest to the smallest frequency is:
√ p
√
f1 /f3 = 6/ 2/3 = 9 = 3 .
20
Problem 8
A glider of mass M = 250 g is connected by a string that runs over a pulley to
mass m = 75 g. The glider is connected to the wall by two identical springs of
constant k = 15 N/m.
M
m
1. (4 points) What is the period of oscillation of the glider, assuming that it slides
without friction on an air track? The mass of the springs can be neglected in
this problem.
It takes force F = k∆x to stretch each spring by distance ∆x. Since the two springs
are in paralle, the total total required to stretched them both by ∆x is 2k∆x, which
implies that the total spring constant is ktotal = 2k = 30 N/m. The total oscillating
mass is M + m = 325 g = 0.325 kg. The period of oscillation is therefore:
s
r
Mtotal
0.325 kg
= 2π
T = 2π
= 0.65 s .
ktotal
30 N/m
21
2. (4 points) Two gliders of mass M = 250 g are connected via three springs of
constant k = 15 N/m on an air track, as shown in the figure in the bottom.
What is the period of oscillation of the gliders when they move with the same
velocities, and when the move with opposite velocities?
When the two gliders move with the same velocity, the middle spring is not stretched,
so each glider feels the force of only one spring. In this case the period of oscillation is
simply therefore:
r
p
M
Tsame = 2π
= 2π 0.25 kg15 N/m = 0.81 s .
k
If the two gliders move with opposite velocities, then when the outer springs are stretched
by ∆x, the middle spring is compressed by 2∆x. The total spring force on, say, the left
glider is k∆x + 2k∆x = 3k∆x. The period of oscillation in this case is therefore:
s
r
M
0.25 kg
Topp = 2π
= 2π
= 0.47 s .
3k
45 N/m
M
M
22
Problem 9
A thermodynamic model of a bit of computer memory is a pair of boxes each of volume
V = 10−12 m3 that are in contact with a thermal bath at temperature T = 25 ◦ C.
There is exactly one monoatomic gas molecule present. If it is in the left box, we say
the bit is a 0, while if the molecule is in the right box we say the bit is a 1.
The single-molecule “gas” obeys the ideal gas law if the pressure is taken to be the
average effect of many collisions of the molecule with the wall of the box. You can
treat it as a regular gas where the number of moles is n = 1/NA .
Thermodynamics of Memory Loss. If the partition is removed (and not
reinserted), the usefulness of the memory is irreversibly lost. This operation results
in the free expansion of the gas from volume V to 2V . Although no heat flows
during a free expansion, we cannot say that there is zero entropy change, because free
expansion is not a quasi-equilibrium process during which the laws of thermodynamics
apply.
A reversible process during which these laws apply and which produces the same final
state,and same entropy change, of the gas as the free expansion is to suppose the
partition were a piston that is retracted (sideways) slowly and isothermally until the
volume doubles.
The only formulas needed for this problem are ∆S = Q/T , P V = nRT , ∆U =
nC∆T = Q − W , where Cmono = 3R/2, R = 8.31 J/◦ K, W = nRT ln(Vf /Vi ) during
an isothermal process. Factoid: k = R/NA = 1.38 × 10−23 J/◦ K = Boltzmann’s
constant.
23
1. (4 points) How much work is done by the single molecule during this isothermal
expansion?
W = nRT ln(Vf /Vi ) = (RT ln 2)/NA = kT ln 2,
where NA = 6.023 × 1023 is the Avagadro’s number (number of molecules in a mole).
Noting that T = 273 + 25 = 297K, we have:
W = 1.38 × 10−23 · 295 · 0.69 = 2.82 × 10−21 J .
2. (2 points) How much heat flows into the gas during the isothermal expansion?
The internal energy remains constant during an isothermal process, so ∆U = 0 =
Q − W . Hence:
Q = W = kT ln 2 = 2.82 × 10−21 J
24
3. (2 points) What is the entropy change ∆Sgas of the gas/memory, and ∆Sbath of
the outside thermal bath during the isothermal expansion?
∆Sgas = Q/T = k ln 2 = 9.6 × 10−24 J/K .
Heat Q flows out of the thermal bath into the memory box during this process, so:
∆Sbath = −Q/T = −∆Sgas = −9.6 × 10−24 J/K .
4. (2 points) Now consider the case of memory loss due to removal of the partition.
What is the entropy change ∆Sgas of the gas/memory, and ∆Sbath of the outside
thermal bath during the subsequent (irreversible) free expansion?
No heat flows from the bath to the gas during the free expansion, so:
∆Sbath = 0 .
However, the entropy change of the gas is the same for the free expansion and for the
isothermal compressions since their initial and final states are the same. Hence,
∆Sgas = Q/T = k ln 2 = 9.6 × 10−24 J/K .
The total entropy change, ∆Sgas + ∆Sbath = k ln 2, during the irreversible loss of
memory is positive, which is consistent with the interpretation of entropy as a measure
of disorder.
25
5. (2 points) Thermodynamics of Memory Erasure. We can partially recover
from the “loss of memory” considered in part 4 if, after the free expansion,
the righthand wall/piston compresses the volume 2V back down to V , the
partition is reinserted, and the righthand wall/piston is retracted creating an
empty volume V on the right. The memory is now in the well-defined state 0,
so we have erased the memory (which is more useful than merely “losing” the
memory).
0
0
What is the entropy change ∆Sgas
of the gas/memory, and ∆Sbath
of the outside
thermal bath during the isothermal compression that resets the memory to 0?
The isothermal compression of this part is just the reverse of the isothermal expansion
considered in parts 1-3, so the entropy changes are just the reverse:
0
∆Sgas
= −Q/T = −k ln 2 = −9.6 × 10−24 J/K ,
0
∆Sbath
= Q/T = k ln 2 = 9.6 × 10−24 J/K .
26
6. (2 points) What is the total entropy change of the system of gas + bath during
the erasure of the memory consisting of the free expansion (part 4) followed by
the isothermal compression (part 5)?
The total entropy change is the sum of that in parts 4 and 5. Hence,
0
0
∆Stotal = ∆Sgas +∆Sbath +∆Sgas
+∆Sbath
= k ln 2+0−k ln 2+k ln 2 = k ln 2 = 9.6 × 10−24 J/K .
This method of erasing the memory presumes no knowledge/information as to the
state of the memory prior to its erasure, and the entropy of the system increased
during the erasure. But if we knew the state of the memory, we could reset it to 0
with no increase of entropy. Consistency with the 2nd las of thermodynamics implies
that knowledge/information has a value that can be measured in terms of entropy. In
particular, a system that contains knowledge/information is in a higher state of entropy
than one which does not.
In mechanistic terms, the outside world has knowledge of the computer memory only if
it has a memory of its own that contains a copy of the state of the computer memory.
This knowledge could be used to erase the computer memory at no entropy cost, but
the outside system still retains in its memory the knowledge/copy of the prior state of
the computer memory. If the outside system is to free up its memory by erasing this
now-useless knowledge, it must then pay the entropy cost of doing so.
This problem was first posed in 1867 by Maxwell, who remarked that an intelligent
demon could use its knowledge to reduce the entropy of a system by timely control of
the partition. However, the knowledge of the “demon” implies that the total entropy
of the system + demon is higher than if no knowledge/information were available. The
action of the demon does not result in a lower total entropy of the universe than for
the demon-free case.
27
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
x = x0 + v0 t + 12 at2
F = µFN
s = rθ
p = mv
ω = ω0 + αt
KE = 21 Iω 2
W = F s cos θ
2
ac = vr
τ = F ` sin θ
Ifull cylinder = 21 mr2
Ifull sphere = 52 mr2
F = Y A ∆L
L0
P V = nRT
Q = mL
V
Wisotherm = nRT ln Vfi
Q
= kA
∆T
t
L
Q = σT 4 At
v = λf
v = v0 + at
F = −kx
v = rω
F ∆t = ∆p
P E = 12 kx2
KE = 12 mv 2
L = Iω
ω 2 =p
ω02 + 2α∆θ
ω = k/m√
2πr3/2 = T GM
Ihollow cylinder = mr2
∆L = αL0 ∆T
P1 V1γ = P2 V2γ
P V = nkT
Q = cm∆T
γ =q
CP /CV
v=
v=
√
γkT
m
v2 = v02 + 2ax
F = −G Mr2m
a = rα
P
xcm = M1tot i xi mi
P E = mgh
Wnc = ∆KE + ∆P E
P
τ = Iα
∆θ =pω0 t + 12 αt2
ω = g/l
P
I = mi ri2
Q = πR4 ∆P/8ηL
Q = Av
P + ρgh + 21 ρv 2 = const
∆S = ∆Q
T
∆U = Q − W
U = 32 nRT
q
F
v = m/L
√
v = Yρ
β = (10 dB) log II0
f0 = fs / 1 ∓ vvs
sin θ = Dλ
Bad ρ v
1± 0
f0 = fs 1∓ vvs
v
v
fn = n 2L
sin θ = 1.22 λd
Monatomic:
Diatomic:
CV = 3R/2
CV = 5R/2
CP = 5R/2
CP = 7R/2
R = 8.315 J/K/mol
u = 1.66 × 10−27 kg
σ = 5.67 × 10−8 W/m2 /K4
NA = 6.022 × 1023 mol−1
k = 1.38 × 10−23 J/K
Mearth = 5.98 × 1024 kg
Mmoon = 7.35 × 1022 kg
0◦ C = 273.15 K
Rearth = 6.36 × 106 m
Rmoon = 1.74 × 106 m
1 kcal = 4186 J
GNewton = 6.67 × 10−11 Nm2 /kg2
f0 = fs (1 ± vv0 )
v
fn = n 4L
vsound = 343 m/s