MSU Physics 231 Fall 2012 1
Physics 231
Review Session: Chapters 10-13
Wade Fisher
Nov 23 2012
MSU Physics 231 Fall 2012 2
A helium balloon in an air-
filled glass jar floats to the top.
If the air is replaced with
helium, what will happen to
the helium balloon?
a) It still floats at the top because it has positive buoyancy
b) It stays in the middle because it has neutral buoyancy
c) It sinks to the bottom because it has negative buoyancy
d) The balloon shrinks in size due to the surrounding helium
d) The balloon grows in size due to the lack of surrounding air
The balloon floats initially because the displaced air weighs more than the
balloon, so the buoyant force provides a net upward force. When the
balloon is in the lighter helium gas (instead of air), the displaced helium
gas does not provide enough of an upward buoyant force to support the
weight of the balloon.
A Helium Balloon
MSU Physics 231 Fall 2012 3
Arterial Blood Flow By what fraction would the inside diameter of an arterial wall have to decrease for blood flow to be reduced by 10%?
MSU Physics 231 Fall 2012 4
Arterial Blood Flow By what fraction would the inside diameter of an arterial wall have to decrease for blood flow to be reduced by 10%?
The radius (and diameter too!) would have to decrease by just 2.6%.
𝑄 = 𝜋𝑅4(𝑃1 − 𝑃2)
8𝐿
𝑄𝑓𝑖𝑛𝑎𝑙
𝑄𝑖𝑛𝑖𝑡𝑖𝑎𝑙= (
𝑅𝑓𝑖𝑛𝑎𝑙
𝑅𝑖𝑛𝑖𝑡𝑖𝑎𝑙)4= 0.9
𝑅𝑓𝑖𝑛𝑎𝑙 = 𝑅𝑖𝑛𝑖𝑡𝑖𝑎𝑙 (0.974)
MSU Physics 231 Fall 2012 5
Rising Submarine A submarine remains submerged by holding excess seawater in its bilge tank. Suppose a submarine with volume 135 m3 is submerged at rest when it expels 1.5 m3 of seawater from its tank. What is its subsequent upward acceleration?
MSU Physics 231 Fall 2012 6
Rising Submarine A submarine remains submerged by holding excess seawater in its bilge tank. Suppose a submarine with volume 135 m3 is submerged at rest when it expels 1.5 m3 of seawater from its tank. What is its subsequent upward acceleration?
Initial volume: 135-1.5 m3
Final volume: 135 m3 Initially, the sub is at rest: Buoyant force: (9.81 m/s2)(1.5 m3) = (133.5 m3) a a = 0.11 m/s2
𝜌𝑜 = 𝜌𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 = 𝑀𝑠𝑢𝑏𝑚𝑎𝑟𝑖𝑛𝑒
𝑉𝑜
𝐹𝐵 = 𝜌𝑓𝑙𝑢𝑖𝑑 𝑔 𝑉𝑓𝑙𝑢𝑖𝑑
𝐹𝐵 = 𝑀𝑎 = 𝜌𝑜 𝑉𝑜 𝑎
𝜌𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 𝑔 𝑉𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 = 𝜌𝑜 𝑉𝑜 𝑎 = 𝜌𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 𝑉𝑜 𝑎
𝑔 𝑉𝑠𝑒𝑎𝑤𝑎𝑡𝑒𝑟 = 𝑉𝑜 𝑎
MSU Physics 231 Fall 2012 7
Guitar Strings
a) They are the same.
b) String G has a higher
fundamental frequency.
c) String D has a higher
fundamental frequency.
Two guitar strings (G and D) have length 1m,
mass density =1 kg/m3, and are each under
tension T=150 N. String D has a radius that is
20% larger than string G. Which string has a
higher fundamental frequency?
𝒗𝒔𝒕𝒓𝒊𝒏𝒈 =𝑻
𝝁 𝛍 =
𝑴
𝑳
1) Each string has the same fundamental frequency =2L
fundamental = 2L
Cross sectional area
D G
2) The linear density of string D is larger than that of
string G: μG = π(rG)2
μD = π(rD)2 = π(1.2rG)
2 = 1.44 μG
3) v=f, thus the string with the larger wave velocity
has the larger fundamental frequency: String G
MSU Physics 231 Fall 2012 8
Fetal Heartbeat Obstetricians use ultrasound to monitor fetal heartbeat. If 5.0-MHz ultrasound reflects off the moving heart wall with a total 100-Hz frequency shift between heart beats, what’s the speed of the heart wall? Assume the velocity of sound in water is 1480 m/s.
MSU Physics 231 Fall 2012 9
Fetal Heartbeat Obstetricians use ultrasound to monitor fetal heartbeat. If 5.0-MHz ultrasound reflects off the moving heart wall with a total 100-Hz frequency shift between heart beats, what’s the speed of the heart wall? Assume the velocity of sound in water is 1480 m/s.
Heart moving toward the observer: Heart moving away from the observer: Difference in in vs out frequencies: If heart wall moves at the same speed in and out: Rearrange: (1480 m/s)(1-5000000/5000050) = 0.0148 m/s
𝑓𝑜𝑢𝑡 = 𝑓𝑜(𝑣
𝑣 − 𝑣𝑠)
𝑓𝑖𝑛 = 𝑓𝑜(𝑣
𝑣 + 𝑣𝑠)
𝑓𝑜𝑢𝑡 − 𝑓𝑖𝑛 = 100 𝐻𝑧
𝑣 1 −𝑓𝑜
𝑓𝑜𝑢𝑡= 𝑣𝑠
𝑓𝑜𝑢𝑡 = 5000050 𝐻𝑧
MSU Physics 231 Fall 2012 10
Hearing Loss Dave can barely hear a sound at a particular frequency with an intensity level of 2.4 dB. Steve, who has hearing loss, can barely hear the same frequency at 9.4 dB. How far away from Steve must Dave stand to simulate his hearing loss if they are listening to the same source of sound?
MSU Physics 231 Fall 2012 11
Hearing Loss Dave can barely hear a sound at a particular frequency with an intensity level of 2.4 dB. Steve, who has hearing loss, can barely hear the same frequency at 9.4 dB. How far away from Steve must Dave stand to simulate his hearing loss if they are listening to the same source of sound?
β = 10 log(I/Io) βSteve – βDave = 10log(IS/Io) – 10log(ID/Io) = 10 log(IS/ID) 9.4-2.4 = 10log(IS/ID) 7/10 = log(IS/ID) IS/ID = 10^(0.7) = 5.0 I = P/A = P/(4πR2) IS/ID = RD
2/RS2 = 5
RD = sqrt(5) RS = 2.24 RS
MSU Physics 231 Fall 2012 12
Ideal gas law: PV = nRT
Solve for temperature:
For constant V and P, the one with less gas (the smaller
value of n) has the higher temperature T.
Ideal Gas Law
a) cylinder A
b) cylinder B
c) both the same
d) it depends on the
pressure P
Two identical cylinders at the same
pressure contain the same gas. If A
contains three times as much gas as
B, which cylinder has the higher
temperature?
PVT =
nR
MSU Physics 231 Fall 2012 13
Diving Bell A cylindrical diving bell (diameter=3m & height=4m), with an open bottom is submerged to a depth of 220m in the sea. The surface temperature is 250C and at 220m, T=50C. The density of sea water is 1025 kg/m3. How high does the sea water rise in the bell when it is submerged?
MSU Physics 231 Fall 2012 14
A cylindrical diving bell (diameter=3m & height=4m), with an open bottom is submerged to a depth of 220m in the sea. The surface temperature is 250C and at 220m, T=50C. The density of sea water is 1025 kg/m3. How high does the sea water rise in the bell when it is submerged?
Consider the air in the bell. Psurf = 1.0x105Pa Vsurf = r2h = 28.3m3 Tsurf = 25+273 = 298K Psub = P0+wgdepth = 2.3E6 Pa Vsub=? Tsub= 5+273 = 278K Next, use PV/T=constant PsurfVsurf/Tsurf=PsubVsub/Tsub plug in the numbers and find: Vsub=1.15m3 (this is the amount of volume taken by the air left) Vtaken by water = 28.3-1.15 = 27.15m3 = r2h h = 27.15/r2 = 3.8m rise of water level in bell.
Diving Bell
MSU Physics 231 Fall 2012 15
Global Warming One danger of global warming is a rise in sea level that could inundate coastal areas. A primary cause of this rise is thermal expansion of water. Estimate the sea-level rise resulting from each 1°C rise in average ocean temperature. Assume a uniform ocean depth of 3.8 km and a coefficient of volume expansion of salt water @20°C β = 2.5E-4/°C.
Earth
Water
MSU Physics 231 Fall 2012 16
Global Warming One danger of global warming is a rise in sea level that could inundate coastal areas. A primary cause of this rise is thermal expansion of water. Estimate the sea-level rise resulting from each 1°C rise in average ocean temperature. Assume a uniform ocean depth of 3.8 km and a coefficient of volume expansion of salt water @20°C β = 2.5E-4/°C.
For volume expansion: Vwater = Ad A=area, d=depth V = VinitialβT = AdinitialβT V = Vinitia – Vinitia = A(dinitia-dinitia) = AD AD = AdinitialβT D = dinitialβT = (3800km)(2.5E-4/°C) (1°C) = 0.95 m
∆𝑽
𝑽= 𝜷∆𝑻
Earth
Water
MSU Physics 231 Fall 2012 17
Two objects are made of the
same material, but have different
masses and temperatures. If the
objects are brought into thermal
contact, which one will have the
greater temperature change?
a) the one with the higher initial temperature
b) the one with the lower initial temperature
c) the one with the greater mass
d) the one with the smaller mass
e) the one with the higher specific heat
Because the objects are made of the same material, the only difference
between them is their mass. Clearly, the object with less mass will change
temperature more easily because not much material is there (compared to
the more massive object).
Thermal Contact
𝐐 = 𝐜 ∙ 𝐦 ∙ ∆𝐓
MSU Physics 231 Fall 2012 18
Heating your Home A house has a floor area of 190 m2 and 2.3-m ceilings. (A)What energy (in kWh) is required to raise the air
temperature by from 20°C to 21°C at constant pressure? (specific heat of air cP=30 J/mol°C and 1 mol of air = 24 L)
(B)At $0.16/kWh, what’s the cost to heat this air by 1°C? (assume a 80% heating efficiency)
MSU Physics 231 Fall 2012 19
Heating your Home A house has a floor area of 190 m2 and 2.3-m ceilings. (A)What energy (in kWh) is required to raise the air
temperature by from 20°C to 21°C at constant pressure? (specific heat of air cP=30 J/mol°C and 1 mol of air = 24 L)
(B)At $0.16/kWh, what’s the cost to heat this air by 1°C? (assume a 80% heating efficiency)
A) V = (190 m2)(2.3m) = 437 m3 = 4.37E5 Liters 1 mol = 30 L, n = (4.37E5)/(3E1) = 1.82E4 mol Q = n cPT = (1.82E4 mol)(30 J/mol °C)(1 °C) = 5.46E5 Joule 1 kWh = 1000 (J/s)x(3600s) = 3.6E6 J Thus, the heat required is (5.46E5)/(3.6E6) = 0.152 kWh B) The cost is $0.16/kWh at 80% efficiency. Cost = (0.152 kWh)($0.16/kWh)/(0.80) = $0.03
MSU Physics 231 Fall 2012 20
Testing an Unknown Substance A block of unknown substance with a mass of 8 kg, initially at T=280K is thermally connect to a block of copper (5 kg) that is at T=320 K (ccopper=0.093 cal/g
0C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown material in cal/goC?
????
copper
MSU Physics 231 Fall 2012 21
Testing an Unknown Substance A block of unknown substance with a mass of 8 kg, initially at T=280K is thermally connect to a block of copper (5 kg) that is at T=320 K (ccopper=0.093 cal/g
0C). After the system has reached thermal equilibrium the temperature T equals 290K. What is the specific heat of the unknown material in cal/goC?
????
copper
Qcold+Qhot=0 Q=cmT Qcold=-Qhot
munknowncunknown(Tfinal-Tunknown)=-mcopperccopper (Tfinal-Tcopper)
cunknown=-mcopperccopper(Tfinal-Tcopper) munknown (Tfinal-Tunknown)
cunkown=-50000.093(290-320) = 0.174 cal/goC 8000(290-280)
MSU Physics 231 Fall 2012 22
Although the water is indeed hot, it releases only 1 cal/g of heat as it
cools. The steam, however, first has to undergo a phase change into
water and that process releases 540 cal/g, which is a very large amount of
heat. That immense release of heat is what makes steam burns so
dangerous.
Hot Water
Which will cause more severe burns
to your skin: 100°C water or 100°C
steam?
a) water
b) steam
c) both the same
d) it depends...
MSU Physics 231 Fall 2012 23
Aluminum is the only material that has a larger b value than the
steel ring, so that means that the aluminum rod will expand more
than the steel ring. Thus, only in that case does the rod have a
chance of reaching the top of the steel ring.
Thermal Expansion
Coefficient of volume expansion b (1/°C )
Glass Hg Quartz Air
Al Steel
a) aluminum
b) steel
c) glass
d) aluminum and steel
e) all three
A steel ring stands on edge with a rod of
some material inside. As this system is
heated, for which of the following rod
materials will the rod eventually touch the
top of the ring?