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### ch1121

Course: CHEM 101, Spring 2009
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Principles Giancoli Physics: with Applications, 6th Edition 80. For the pebble to lose contact with the board means that there is no normal force of the board on the pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This is the maximum downward acceleration that the...

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Principles Giancoli Physics: with Applications, 6th Edition 80. For the pebble to lose contact with the board means that there is no normal force of the board on the pebble. If there is no normal force on the pebble, then the only force on the pebble is the force of gravity, and the acceleration of the pebble will be g downward, the acceleration due to gravity. This is the maximum downward acceleration that the pebble can have. Thus if the boards downward acceleration exceeds g, then the pebble will lose contact. The maximum acceleration and the amplitude are related by amax 4 2 f 2 A . amax 4 2 f 2A g A g 4 2 9.8 m s2 f 2 4 2 1.5 Hz 2 1.1 10 1 m 81. For a resonant condition, the free end of the string will be an antinode, and the fixed end of the string will be a node. The minimum distance from a node to an antinode is 4 . Other wave patterns that fit the boundary conditions of a node at one end and an antinode at the other end include 3 4,5 4, . See the diagrams. The general relationship is L 2n 1 4 , n 1, 2, 3, . Solving for the wavelength gives 4L 2n 1 , n 1, 2, 3, . 1 n=1 0 0 n=3 n=5 1 -1 82. period The of a pendulum is given by T (a) LAustin LParis LParis 2 L g , and so the length is L T 2g 4 2 . T 2 gAustin 4 4 LAustin 2 2 2.000 s 2.000 s 2 2 9.793 m s2 2 2 4 4 2 2 0.9922 m 0.9939 m 1.6 mm T 2 g Paris 9.809 m s 2 (b ) 0.9939 m 0.9922 m 0.0016 m (c) LMoon T 2 gMoon 4 2.00 s 1.62 m s2 2 4 0.164 m 83. The spring, originally of length l0 , will be stretched downward to a new equilibrium length L when the mass is hung on it. The amount of downward stretch L l0 is found from setting the spring force upward on the mass equal to the weight of the mass: k L l0 of the pendulum is then L Tver 2 mg L l0 mg k . The length l0 mg k . The period of the vertical oscillations is given by 2 L g . Now m k , while the period of the pendulum oscillations is given by Tpen compare the periods of the two motions. 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 289
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Arcadia University - CHEM - 101
Chapter 11Vibrations and WavesTpen Tver Tpen2l0 2mg k mkgl0mg k mg k1l0 k mg1Tver , by a factor of 1l0 k mg84. Block m stays on top of block M (executing SHM relative to the ground) without slipping due to static friction. The maximum static
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition87. The car on the end of the cable produces tension in the cable, and stretches the cable according to 1F Lo , where E is Youngs modulus. Rearrange this equation to see that Equation (9-4), L EA
Arcadia University - CHEM - 101
CHAPTER 12: Sound Answers to Questions1. Sound exhibits several phenomena that give evidence that it is a wave. The phenomenon of interference is a wave phenomenon, and sound produces interference (such as beats). The phenomenon of diffraction is a wave
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionparticular pipe is fixed, the frequency is also a function of temperature. Thus when the temperature changes, the resonant frequencies of the organ pipes change as well. Since the speed of sound
Arcadia University - CHEM - 101
Chapter 12Soundamplitude (anti-node). Thus the interference can be described as interference in time. To experience the full range from constructive interference to destructive interference, the time of observation must change, but all observations coul
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition2.The round trip time for sound is 2.5 seconds, so the time for sound to travel the length of the lake is 1.25 seconds. Use the time and the speed of sound in water to determine the depth of the
Arcadia University - CHEM - 101
Chapter 12SoundThe speed of sound in concrete is obtained from Equation (11-14a), Table (9-1), and Table (10-1).vconcreteE20 109 N m2 2.3 103 kg m32949 m sdvair tair343 m s2949 m s 2949 m s 343 m s1.1 s427m4.3 102 m7.The 5 second rule says
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition13. (a) According to Table 12-2, the intensity in normal conversation, when about 50 cm from the speaker, is about 3 10 6 W m2 . The intensity is the power output per unit area, and so the power
Arcadia University - CHEM - 101
Chapter 12Sound17. The intensity is proportional to the square of the amplitude. I A2. A A2.0 2.0 dB 10 log 2.0 10 log 220 20 log 2.0 100.1 1.259 I0 A0 A0 A01.318. (a) The intensity is proportional to the square of the amplitude, so if the amplitude i
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition24. For a vibrating string, the frequency of the fundamental mode is given by fv 2L1 2LFT mL.f1 2LFT mLFT =4Lf 2 m4 0.32 m 440 Hz23.5 10 4 kg87 N25. (a) If the pipe is closed at one
Arcadia University - CHEM - 101
Chapter 12Sound29. (a) We assume that the speed of waves on the guitar string does not change when the string is v fretted. The fundamental frequency is given by f , and so the frequency is inversely 2L proportional to the length. 1 f fL constant L f 33
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionfv 2LLv 2f343 m s 2 294 Hz0.583 m33. (a) At T20o C , the speed of sound is 343 m s . For an open pipe, the fundamental frequency is v given by f . 2L v v 343 m s f L 0.583 m 2L 2 f 2 294 H
Arcadia University - CHEM - 101
Chapter 12Sound(b) The harmonics for the closed pipe are f n, n odd. Again, they must be below 20 kHz. 4L 4 2.14 m 2 10 4 Hz nv 4 2 10 Hz n 499.1 4L 343 m s The values of n must be odd, so n = 1, 3, 5, , 499. There are 250 harmonics, and so there are2
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionf f1 2L f f0.98 FT mL f10.98 0.981 2LFT mL0.98 f 0.98 3.0 Hz294 Hz 144. Beats will be heard because the difference in the speed of sound for the two flutes will result in two different fr
Arcadia University - CHEM - 101
Chapter 12Sound48. To find the beat frequency, calculate the frequency of each sound, and then subtract the two frequencies. vv 1 1 f beat f1 f 2 343 m s 5.649 5.6 Hz 2.64 m 2.76 m 1 2 49. (a) Observer moving towards stationary source. v 30.0 m s f 1 ob
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(c) For the 300 m/s relative velocity: 1 fsource f 2000 Hz vsrc moving 1 vsnd1 1 300 m s 343 m s 300 m s 343 m s16.0 103 Hzf observermovingf1vsrc vsnd2000 Hz13.75 103 HzThe difference i
Arcadia University - CHEM - 101
Chapter 12SoundThen the wall can be treated as a stationary source emitting the frequency f wall , and the bat as a moving observer, flying toward the wall. v v v v 1 f bat f wall 1 bat f bat 1 bat f bat snd bat vsnd vsnd vsnd vbat v 1 bat vsnd 3.00 10
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionvbloodvsndf 2 f original f1.54 103 m s500 Hz 2 2.25 106 Hz 500 Hz0.171 m sIf instead we had assumed that the heart was moving towards the original source of sound, we would f get vblood vsn
Arcadia University - CHEM - 101
Chapter 12Sound61. (a) The Mach number is the ratio of the objects speed to the speed of sound. 1m s 1.5 104 km hr 3.6 km hr vobs M 119.05 120 vsound 35 m s (b) Use Eq. 12.5 to find the angle. v 1 1 sin 1 snd sin 1 sin 1 0.48o vobj M 119.05 62. From Eq.
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition66. Each octave is a doubling of frequency. The number of octaves, n, can be found from the following. 20, 000 Hz 2 n 20 Hz 1000 2 n log1000 nlog2n log 1000 log 2 9.97 10 octaves67. Assume that
Arcadia University - CHEM - 101
Chapter 12Sound71. Relative to the 1000 Hz output, the 15 kHz output is 10 dB. P P P 5 kHz 1 10 dB 10 log 15 kHz 1 log 15 kHz 0.1 150 W 150 W 150 WP kHz 1515 W72. The 140 dB level is used to find the intensity, and the intensity is used to find the p
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionL f1 22L 343 m s 0.540 m2 0.395 m 0.125 m0.540 mv635 Hz77. The frequency of the guitar string is to be the same as the third harmonic (n = 3) of the closed tube. nv The resonance frequenci
Arcadia University - CHEM - 101
Chapter 12Sound82. The sound is Doppler shifted up as the car approaches, and Doppler shifted down as it recedes. The observer is stationary in both cases. The octave shift down means that f approach 2 f recede .f approach f enginef engine 1 vcar vsnd
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionff originalf detectorf originalf originalvsnd vsndvblood vbloodf original2vblood vsnd vblood5.50 106 Hz2 0.32 m s 1.54 10 m s 0.32 m s32.29 103 Hz86. Use Eq. 12-4, which applies when
Arcadia University - CHEM - 101
Chapter 12Sound(b) The wavelength of sound in the rod is twice the length of the rod, 1.80 m . (c) The wavelength of the sound in air is determined by the frequency and the speed of sound in air. v 34 3 m s 0 .1 2 m f 2833 Hz 91. Eq. 11-18 gives the rel
Arcadia University - CHEM - 101
CHAPTER 13: Temperature and Kinetic Theory Answers to Questions1. Because the atomic mass of aluminum is smaller than that of iron, an atom of aluminum has less mass than an atom of iron. Thus 1 kg of aluminum will have more atoms than 1 kg of iron. Prop
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory10. The lead floats in the mercury becauseHgPb. As the substances are heated, the density ofboth substances will decrease due to volume expansion (see problem 17 for the derivation of this result). The density
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition18. Charless law states that the volume of a fixed mass of gas increases proportionately to the absolute temperature, when the pressure is held constant. As the temperature increases, the molecul
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryAlso, the violent bursting forth of steam propels some of the overheated water out of the radiator as well, which can spray onto the person opening the cap and again cause serious burns. 27. Exhaled air contains w
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition6.Assume that the temperature and the length are linearly related. The change in temperature per unit length change is as follows. T 100.0o C 0.0o C 9.066 Co cm L 22.85 cm 11.82 cm Then the temp
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory13. The amount of water that can be added to the container is the final volume of the container minus the final volume of the water. Also note that the original volumes of the water and the container are the same.
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) The fractional change in density is T 0 T 87 100 06oC40o C 25o C5.7 103This is a 0.57% increase . 18. Assume that each dimension of the plate changes according to Equation 13-1a. A A
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryOriginal volume for glass bulb and Hg in bulb: V0bulb Change in glass bulb volume: Change in Hg volume in glass bulb:VglassV0bulbglassTVHgVbulb 0HgTNow find the additional volume of Hg, and use that to f
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(c) For concrete, repeat the calculation with the expansion coefficient and elastic modulus for concrete.Stress FA ET 12 106Co20 109 N m260 C o1.4 107 N m26 2 The ultimate tensile strength
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory29. Assume the gas is ideal. Since the amount of gas is constant, the value ofPV Tis constant.PV1 1 T1P2V2 T2V2V1P T2 1 P2 T13.00m31.00 atm 3.20 atm273 38 K 273 K1.07 m330. Assume the air is an ideal g
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition34. (a) Assume that the helium is an ideal gas, and then use the ideal gas law to calculate the volume. Absolute pressure must be used, even though gauge pressure is given. 18.75 mol 8.315 J mol
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic TheoryOne factor limiting the maximum altitude would be that as the balloon rises, the density of the air decreases, and thus the temperature required gets higher. Eventually the air would be too hot and the balloon fab
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition(b) 6.55 1022 moles6.02 1023 molecules 1 mol4 1046 molecules44. The net force on each side of the box will be the pressure difference between the inside and outside of the box, times the area
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory48. The rms speed is given by Equation 13-9, vrms3kT m .vrms vrms2 13kT2 m 3kT1 mT2 T1373 K 273 K1.1749. The rms speed is given by Equation 13-9, vrms 3kT m . Since the rms speed is proportional to the squ
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition55. The temperature of the nitrogen gas is found from the ideal gas law, and then the rms speed is found from the temperature. 2.1 atm 1.013 105 Pa atm 8.5 m3 PV PV nRT T 167.3 K nR 1300 mol 8.31
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory59. The pressure can be stated in terms of the ideal gas law, PNkT V . Substitute for the temperaturefrom the expression for the rms speed, vrms 3kT m T mvr2ms 3k . The mass of the gas is the mass of a molecule
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition66. The relative humidity of 40% with a partial pressure of 530 Pa of water gives a saturated vapor pressure of 530 Pa 0.40 Psaturated 530 Pa Psaturated 1325 Pa 0.40 From Table 13-3, the temperat
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory71. From Example 13-19, we have an expression for the time to diffuse a given distance. Divide the distance by the time to get the average speed.t C C x Dx2 1 21.00 0.40 mol m3 1.00 0.40 mol m315 10 6 m 95 10
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionThus 6 290.20721% of the original gas remains in the cylinder.76. Assume the air is an ideal gas, and that the pressure is 1.0 atm. PV NkTN PV kT 1.013 105 Pa 6.5 3.1 2.5 m3 1.38 1023J K 27
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory80. The temperature can be found from the rms speed by Equation 13-9, vrmsvrms 3kT m23kT m .T2 mvrms28 1.66 10 kg274 10 km h2341m s 3.6 km h3k3 1.38 10JK1.4 105 K81. From the ideal gas law, if the
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionM N2 4 REarth P46.38 106 m21.01 105 Pag 5.27 1018 kg 1 mole9.80 m s 2 29 10 3 kg 1 mole5.27 1018 kg 1.1 1044 molecules6.02 1023 molecules84. The temperature of the nitrogen gas is found
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theory87. (a) Assume that a mass M of gasoline with volume V0 at 0oC is under consideration, and so its density is volume V0M V0 . At a temperature of 32oC, the same mass has aT.M V0 1 T 10V0 1M V0.68 103 kg m3
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Editionbelow the surface of the water is discussed in chapter 10, and is given by PP0 is atmospheric pressure andP0gh , whereis the density of the sea water.P2V2 T2 V2PV1 1 T1 11.3 LV2V1P T2 1
Arcadia University - CHEM - 101
Chapter 13Temperature and Kinetic Theorydsurfaced depthPdepth Psurface1/ 33.00 cm1.01 105 Pa1.0 103 kg m3 1.01 105 Pa9.8 m s 2 14.0 m1/ 33.99 cm95. (a) At a temperature of 30oC, the saturated vapor pressure, from Table 13-6, is 4.24 103 Pa . I
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th EditionButVV0 T , and so V293 K , 1 T0 1 293 KV0 T0TV0 T1 T0For T03.4 10 3 K , which agrees well with Table 13-1.(b) Assume the temperature and amount of gas are held constant, and so P0V0 nRT
Arcadia University - CHEM - 101
CHAPTER 14: Heat Answers to Questions1. The work goes primarily into increasing the temperature of the orange juice, by increasing the average kinetic energy of the molecules comprising the orange juice. When a hot object warms a cooler object, energy is
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition9.The potatoes will not cook faster if the water is boiling faster. The boiling water is the same temperature whether it is boiling fast or slow.10. An ordinary fan does not cool the air direct
Arcadia University - CHEM - 101
Chapter 14Heat17. When the garment is fluffed up, it will have the most air trapped in its structure. The air has a low thermal conductivity, and the more the garment can be fluffed, the more air it will trap, making it a better insulator. The loft valu
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition24. Shiny surfaces absorb very little of the radiation that is incident on them they reflect it back towards the source. Thus the liner is silvered to reduce radiation energy transfer (both into
Arcadia University - CHEM - 101
Chapter 14Heat2.The kcal is the heat needed to raise 1 kg of water by 1 Co. Use that definition to find the temperature change. Then the final temperature can be found. o 7700 J 1 kcal 1kg 1C 0.61Co Final Temperature 10.6o C 3.0 kg 4186 J 1 kcal3.(a)
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition10. The heat absorbed by all three substances is given by Eq. 14-2, Q mc T . Thus the amount of Q mass can be found as m . The heat and temperature change are the same for all three cT substances
Arcadia University - CHEM - 101
Chapter 14Heat15. The heat must warm both the water and the pot to 100oC. The heat is also the power times the time. Q Pt mAl cAl mH O cH O TH O2 2 2tmAl cAl 425 smH O cH O2 2TH O20.36 kg 900 J kg Co0.75 kg 4186 J kg Co 750 W92CoP 7 min16. T
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition22. Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of vaporization at room temperature (585 kcal/kg), since the persons temperature is closer to r
Arcadia University - CHEM - 101
Chapter 14Heat28. The heat lost by the steam condensing and then cooling to 20oC must be equal to the heat gained by the ice melting and then warming to 20oC. msteam Lvap cH O Ti steam Teq mice Lfus cH O Teq Ti ice2 2msteammiceLfus LvapcH O Teq Ti
Arcadia University - CHEM - 101
GiancoliPhysics: Principles with Applications, 6th Edition34. The heat conduction rate is given by Eq. 14-4.Q t kA T1 T2 l 0.84 J s m Co3.0 m215.0o C35o C3.2 10 m1.6 104 W35. (a) The power radiated is given by Eq. 14-5. The temperature of the
Arcadia University - CHEM - 101
Chapter 14Heat40. (a) The cross-sectional area of the Earth, perpendicular to the Sun, is a circle of radius REarth , and2 so has an area of REarth . Multiply this area times the solar constant to get the rate at which the Earth is receiving solar ener