Physics 231 - Michigan State University€¦ · MSU Physics 231 Fall 2012 6 Rising Submarine A...

MSU Physics 231 Fall 2012 1

Physics 231

Review Session: Chapters 10-13

Wade Fisher

Nov 23 2012


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


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%?


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)


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?


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


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


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.


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


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?


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


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


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?


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


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


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


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


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)


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


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


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)


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...


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?

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