REVIEW AND SYNTHESIS: CHAPTERS 9 12

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1 REVIEW AND SYNTHESIS: CHAPTERS 9 Reiew Exercises. Strategy The magnitude of the buoyant force on an object in water is equal to the weight of the water displaced by the object. (a) Lead is much denser than aluminum, so for the same mass, its olume is much less. Therefore, aluminum has the larger buoyant force acting on it; since it is less dense it occupies more olume. (b) Steel is denser than wood. Een though the wood is floating, it displaces more water than does the steel. Therefore, wood has the larger buoyant force acting on it; since it displaces more water than the steel. mpb (c) Lead:.0 kg wgvpb wg (.00 0 kg m )(9.80 m s ) 0.87 N Pb,00 kg m m Aluminum: Al.0 kg wgval wg (.00 0 kg m )(9.80 m s ).6 N 70 kg m Steel: gv Al m.0 kg (.00 0 kg m )(9.80 m s ). N Steel w Steel w g Steel 7860 kg m Wood: mg (.0 kg)(9.80 m s ) 9.8 N (Since the wood is floating, the buoyant force is equal to its weight.). (a) Strategy The relationship between the fraction of a floating object s olume that is submerged to the ratio of the object s density to the fluid in which it floats is V f V o o f. Find the percentage of the plastic that is submerged in the water. V submerged plastic 890 kg m 0.89, or 89% V plastic water.00 0 kg m (b) Strategy and In Figure C, the plastic will actually rise. Think about it this way. In Figure B, the part of the plastic aboe the water has a buoyant force from the air lifting it up. Since air is not ery dense, this force is not ery strong. Diagram C is correct. The oil is ery dense so the buoyant force from the oil is strong and the plastic rises. (c) Strategy The buoyant forces of the water and oil on the plastic are equal to the total buoyant force on the plastic. Let Vw and V o be the olume of the plastic within the water and oil, respectiely. Then, the total olume is V p V o + V w. Find the percentage of the plastic submerged in the water. wgvw + ogvo pgvp wvw + o( Vp Vw) pvp wvw ovw pvp ovp V w p o , or 5% Vp w o

2 Physics Reiew and Synthesis: Chapters 9. Strategy Use the continuity equation for incompressible fluids and Bernoulli s equation. Let the entrance point be and the faucet be. Find, the speed of the water in the main pipe., so A r A A A. r P + gy+ P + gy + ( P P) r + g( y y) + r r ( P P) g( y y) r 4 ( P P) r g( y y) r ( Pa) ( m s )(4.0 m m) 0.6 m s.00 0 kg m.0 4. (a) Strategy As Arnold falls, graity does mgh (8 kg)(9.80 m s )(0 m) 8.0 kj of work on him, so each of his legs must absorb about 4.0 kj of energy when he lands; that is, each leg must do work W Fa y to bring him to rest. Compute the compressie stress on Arnold s legs. W 4000 J 5 The aerage force on each femur is Fa N. So, the compressie stress on each y m F 5 a N 9 femur is 0 N m. Since the compressie stress on each femur exceeds the A m maximum ultimate strength for compression, Arnold s femur will break. (b) Strategy and This time, Arnold has 0 cm instead of 5 mm to come to a stop, and it is the snow 4000 J that does the work. So, the aerage force is. 04 N and the compressie stress is 0.0 m. 04 N 7 0 N m. This time, the compressie stress is less than the maximum ultimate strength m for compression, therefore, Arnold s femur will not break. 557

3 Reiew and Synthesis: Chapters 9 Physics 5. Strategy Use Eqs. (0-0a), (0-), and (0-), and Newton s second law. The normal force on the.0-kg block m is N m g. So, the force of friction on m is f µ N µ m g. Find the maximum acceleration that the top block can experience before it starts to slip. f µ Σ F f ma, so a mg µ g. m m For SHM, the maximum acceleration is k am ω A A, which in this case is equal to m ka. Equate the accelerations and sole for A. m + m ka µ ( m+ m) g µ g, so A. m + m k The maximum speed is m ω A. Compute the maximum speed that this set of blocks N.0 kg m g can hae without the top block slipping. k µ ( m+ m) g m+ m.0 kg kg m ω A µ g 0.45(9.80 m s ) 0.88 m s m + m k k 50 N m m m 6. Strategy Use Eq. (-) and conseration of energy. Find the maximum speed of the child, which occurs at the lowest point of her swing. As the child swings toward the whistle (), o < 0. As she swings away (), o > 0. o o + o fs fs o f.040 +, so f (4 m s).040( o ) + o, or o 6.7 m s At the lowest point of the swing, the kinetic energy is at its maximum and the potential energy is at its minimum. Find how high the child is swinging; that is, her maximum height. max (6.7 m s) max max max mgh m, so h. m. g (9.80 m s ) 7. (a) Strategy Use Eq. (-7). Compute the wae speed for each traeling wae. ω I 6.00 s ω I.50 cm s and II.00 s k I 4.00 cm II.00 cm s. k II.00 cm Eq. I has the fastest wae speed of.50 cm s. (b) Strategy Use Eq. (-7). Compute the waelengths for each traeling wae. π π π π λi.57 cm and λ k 4.00 cm II.09 cm. k.00 cm I Eq. II has the longest waelength of.09 cm. II 558

4 Physics Reiew and Synthesis: Chapters 9 (c) Strategy Use Eq. (0-). Compute the maximum speed of a point in the medium for each traeling wae. Im ωi AI (6.00 s )(.50 cm) 9.00 cm s and IIm ωii AII (.00 s )(4.50 cm).5 cm s. Eq. II has the fastest maximum speed of a point in the medium of.5 cm s. (d) Strategy and kx ωt indicates a wae is moing in the positie x-direction and kx + ωt indicates a wae is moing in the negatie x-direction. So, Eq. II is moing in the positie x-direction. 8. (a) Strategy Use Eq. (0-6a) and ω π f. Compute the frequency of the Foucault pendulum. ω g 9.80 m s f 0. Hz π π L π 4.0 m (b) Strategy Find the amplitude of the oscillation and use it and the frequency to find the maximum speed. Use Eq. (0-). θ The amplitude of the oscillation is A Lsin θ. The maximum speed of the pendulum is m ωa g Lsinθ glsin θ (9.80 m s )(4.0 m) sin m s. L A L (c) Strategy The maximum speed and tension occur at the equilibrium position. Use Newton s second law. Find the maximum tension. Σ F m y F mg mar m m, so r L m (.4 m s) F m g+ (5.0 kg) 9.80 m s + 49 N. L 4.0 m F mg (d) Strategy Use Eqs. (-) and (-). Find the fundamental frequency of the wire. FL F 49 N f L L m 4Lm 4(4.0 m)(0.000 kg) 6. Hz 9. (a) Strategy Use Eqs. (-) and (-). Find the tension in the guitar string. FL F f, so F 4Lmf 4(0.655 m)(0.00 kg)(8 Hz) 58 N. L L m 4Lm (b) Strategy Use Eqs. (-4) and (-). Find the length of the lowest frequency string when it is fingered at the fifth fret. 58 N f F, so L F 49 cm. L L µ f µ (0 Hz) 0.00 kg (0.655 m) 559

5 Reiew and Synthesis: Chapters 9 Physics 0. Strategy Use Eq. (-4) and x t. Find the time it takes for the wae pulse to trael from one child to the other. x µ 0.00 kg m t x ( m) 0.5 s F 8.0 N. Strategy For two adjacent steps, the extra distance traeled by the wae reflected from the upper step is twice the tread depth. For the waes to cancel, they must be one-half waelength out of phase or an odd multiple of onehalf waelength. Therefore, the minimum tread depth is half of one-half waelength or one-quarter waelength. Find the minimum tread depth d min. λ 4 m s λ, so dmin.4 cm. f 4 4 f 4(400.0 Hz). (a) Strategy First, treat the brick wall as the obserer. Next, treat the wall as the source. Use Eq. (-). The motion of Akiko is in the direction of sound propagation, so Akiko > 0. f f f f wall o s s Akiko Akiko Now, since the wall and Haruki are stationary, there is no further Doppler effect. So, fharuki fwall fakiko (5.0 Hz) 5.7 Hz. Akiko 7.00 m s 4 m s (b) Strategy The situation is similar to that of part (a), but instead of a stationary Haruki, we hae Junichi as an obserer moing in the direction of propagation of the sound reflected from the wall ( Junichi > 0). Use Eq. (-). o Junichi.00 m s fjunichi fo fs fwall (5.67 Hz) 59.6 Hz 4 m s. Strategy The frequency of the sound is increased by a factor equal to the number of holes in the disk. Use Eq. (-6). The frequency of the sound is f 5(60.0 Hz) 500 Hz. Compute the waelength that corresponds to this frequency. 4 m s λ.9 cm f.50 0 Hz 560

6 Physics Reiew and Synthesis: Chapters 9 4. Strategy Use Eqs. (-4) and (-). Find the frequency in terms of the tension. F f L L µ Form a proportion to find the new fundamental frequency. f F µ L, so f f (847 Hz) 470 Hz. f L µ F 5. Strategy and There are 000 cm in L. The heart pumps blood at a rate of 80 cm 60 s 4800 cm min. Since s min to trael throughout the body. 6. Strategy Use Eq. (-6). 5 L 5000 cm 4800 cm, it takes about minute for the medicine 4 m s (a) The initial waelength is λ f Hz 7.7 cm. (b) The frequency of the sound in the wall is the same as it was in the air. Therefore, the waelength of the sound 60 m s in the wall is λ 50 cm. f Hz (c) The frequency and the speed of sound are the same as before they entered the wall, so the waelength is 7.7 cm. 7. Strategy The source is moing in the direction of propagation of the sound, so s > 0. The obserer is moing in the direction opposite the propagation of the sound, so o < 0. Use Eq. (-4). Find the frequency heard by the passenger in the oncoming boat. o 4 m s ( 5.6 m s) fo fs ( Hz) 46 Hz 4 m s 0. m s s 8. (a) Strategy The waelength of the diaphragm is 0.0 m. The frequency of the gas in the tube is the same as that of the diaphragm. Use Eq. (-6). Find the speed of sound in the gas. fλ (457 Hz)(0.0 m) 90 m s (b) Strategy and The piles of sawdust represent displacement nodes: regions where the air remains at rest. They also represent pressure antinodes. 56

7 Reiew and Synthesis: Chapters 9 Physics 9. (a) Strategy Use the continuity equation for incompressible fluids and Bernoulli s equation. Let the lower end be and the upper end be. Find the speed of the water as it exits the pipe. A r A r 0.0 A A, so (5.0 cm s) 4.7 cm s Find the pressure at the lower end. P+ gy+ P + gy + P P + g( y y) + ( ) P 0. kpa + (.00 0 kg m ) (9.80 m s )(.70 m) + ( 6.00 ) (0.50 m s) P 8 kpa (b) Strategy Use the equations of motion for a changing elocity. Find the time it takes for the water to fall to the ground. y i sin θ t g( t) 0 g( t) i sinθ t+ y isinθ ± 4i sin θ 8g y t g (0.467 m s)sin 60.0 ± 4(0.467 m s) sin60.0 8(9.80 m s )( 0.00 m) t (9.80 m s ) t 0.87 s ( 0. s is rejected, since time is positie.) The horizontal distance from the pipe outlet where the water lands is θ x i cos t (4.7 cm s)cos 60.0 (0.87 s) 5.98 cm. 0. Strategy The standing wae is the first harmonic, f. The wae is in position B one-quarter period after it is in position A. It is in position C one-half period after it is in position A. Use Eqs. (-4) and (-). F F The speed of the wae on the string is, so f. µ L L µ µ kg m The period is T L (0.70 m) 7.0 ms. f F.00 N Since the string is at A when t 0, it will be at least one period before a photo of the string can be taken at position A; then it will be another period before the second photo can be taken. Therefore, the two earliest times after t 0 that the string can be photographed in position A are 7.0 ms and 4.4 ms. Unlike positions A and C, which only appear once per period, position B occurs twice per period. The string is at position B one-quarter period ( T 4) after it is at position A, and it will be at position B again an additional one-half period later ( T 4+ T T 4); therefore, the two earliest times after t 0 that the string can be photographed in position B are.80 ms and 5.40 ms. The string is at position C one-half period ( T ) after it is at position A, and it will be at position C again an additional period later ( T + T T ); therefore, the two earliest times after t 0 that the string can be photographed in position C are.60 ms and 0.8 ms. 56

8 Physics Reiew and Synthesis: Chapters 9. Strategy Use the equations describing waes on a string fixed at both ends. (a) The waelength of the fundamental mode is λ L (0.640 m).8 m. (b) The wae speed on the string is λ f (.8 m)(0.0 Hz) 4 m s. (c) Find the linear mass density of the string. F F N, so µ 6.7 g m. µ (40.8 m s) (d) Find the maximum speed of a point on the string. ωa π fa π(0.0 Hz)(0.000 m) 0.5 m s (e) The frequency in air is that same as that in the bridge and body of the guitar, which is the same as the frequency of the string: 0.0 Hz. (f) Find the speed of sound in air. ms + (0.60 ms)(0) 4 ms Find the waelength of the sound wae in air. λ 4 m s. m f 0.0 Hz. Strategy Use conseration of energy, conseration of momentum, and Newton s second law. (a) Find the speed of the oranges just before they land on the pan. m mgh, so gh (9.80 m s )(0.00 m).45 m s. Find the speed of the oranges and the pan immediately after the oranges land on the pan. m.0 kg moi ( m + M ), so oi (.45 ms).8 ms. m+ M.0 kg kg (b) Find the new equilibrium point. mg (.0 kg)(9.80 m s ) Σ F k y mg, so y 4.8 cm. k 450 N m (c) Find the amplitude of the oscillations. mtot g y+ k( y) + m ka, so A m tot g y+ k( y) + mtot k (.45 kg)(9.80 m s )( m) + (450 N m)( m) + (.45 kg)(.8 m s) 450 N m 0.8 m (d) Find the frequency of the oscillations. k 450 N m f. Hz. π m π.45 kg 56

9 Reiew and Synthesis: Chapters 9 Physics. (a) Strategy Find the mass of the balloon and helium. Then use Newton s second law. Find the mass. 4 m m He + mb HeV + mb (0.79 kg m ) π(0.0 m) kg kg Find the tension in the ribbon. Σ F FB mg T 0, so 4 T FB mg airvg mg π (.9 kg m ) (0.0 m) kg (9.80 m s ) 5. 0 N. (b) Strategy Use the small angle approximation for sin θ. Consider Hooke s law. Find the period of oscillation. x FT F FTsin θ FTθ FT, which implies that k. Thus, we can find the period using L L m ml (4.0 0 kg)(.0 m) T π.69 s. k π F π T 5. 0 N 4. Strategy Use Bernoulli s equation and the pressure difference of a static liquid. We hae the two relations: P P liquid gh and P + air P. Eliminate the pressure difference and find the minimum speed of the air. liquid gh (800 kg m )(9.8 m s )(0.00 m) air liquid gh, so 9 m s. air.9 kg m 5. Strategy Use Newton s second law and Hooke s law. (a) Find the tension in the string when hanging straight down. Σ F T mg 0, so T mg. Find the tension in the string while it is swinging. mg Σ Fy Tcosθ mg 0, so T. cosθ Find the stretch of the string. T A Y, so LL L T Lmg (.00 m)(0.4 kg)(9.8 m s ) L m. AY AY cosθ π (0.005 m) ( Pa) cos 65.0 (b) Find the kinetic energy. m K Σ Fx T sin θ max, so r r mg rt sin sin sinθ L θ θ cosθ mgl tanθsin θ (0.4 kg)(9.80 m s )(.00 m) tan 65.0 sin 65.0 K 8.6 J 564

10 Physics Reiew and Synthesis: Chapters 9 (c) Find the time it takes a transerse wae pulse to trael the length of the string. Neglect the small change due to the stretch when finding the linear mass density. T T and L t, so µ A L+ L A Acos θ (50 kg m ) π(0.005 m) cos 65.0 t ( L+ L) ( L+ L) (.006 m) T mg (0.4 kg)(9.80 m s ) s 6. Strategy The tension is 0.9 of the tensile strength. Use Eqs. (0-), (-), and (-). Set the stress equal to the strength. T F strength 0.9 n TL, so T 0.9 A(strength). fn and. Find A A L m f. TL TL 0.9 A(strength) L 0.9(strength) V 0.9(strength) f L L m ml ml ml L 8 0.9(6. 0 Pa).4 khz (8500 kg m )(0.094 m) MCAT Reiew. Strategy The buoyant force on the brick is equal in magnitude to the weight of the olume of water it displaces. The brick is completely submerged, so its olume is equal to that of the displaced water. The weight of the displaced water is 0 N 0 N 0 N. Find the olume of the brick. 0 N 0 N wgvw wgvbrick 0 N, so Vbrick 0 m w g (000 kg m )(0 m s ) The correct answer is A.. Strategy The expansion of the cable obeys F k L. Compute the expansion of the cable. F 5000 N L 0 m k N m The correct answer is A.. Strategy Sole for the intensity in the definition of sound leel. Find the intensity of the fire siren. SL 0 log I SL , so 0 0 (.0 0 W m )0.0 0 I I I W m. 0 The correct answer is D. 4. Strategy and The two centimeters of liquid with a specific graity of 0.5 are equialent to one centimeter of water; that is, the 6-cm column of liquid is equialent to a 5-cm column of water. Therefore, the new gauge pressure at the base of the column is fie-fourths the original. The correct answer is C. 565

11 Reiew and Synthesis: Chapters 9 Physics 5. Strategy The waelength is inersely proportional to the index n. Let λ n 8 m and λ n 4.8 m. + Find n. nλn ( n+ ) λn+ λn n + λn L, so +, or n. 4 4 λn+ n n λ n+ Compute L. nλn λn λn 8 m L (8 m) 6 m 4 λ n m The correct answer is C. 6. Strategy The wae may interfere within the range of possibility of totally constructie or totally destructie interference. For totally constructie interference, the amplitude of the combined waes is units. For totally destructie interference, the amplitude of the combined waes is 5 units. The correct answer is B. 7. Strategy and As the bob repeatedly swings to and fro, it speeds up and slows down, as well as changes direction. Therefore, its linear acceleration must change in both magnitude and direction. The correct answer is D. 8. Strategy Since K and r 4, K r 8. Compute the ratio of kinetic energies. 8 K or 4 K 4 : K : 4. K 56 4 The correct answer is B. 9. Strategy The buoyant force on a ball is equal to the weight of the olume of water displaced by that ball. Since B is not fully submerged and B and B are, the buoyant force on B is less than the buoyant forces on the other two. Since B and B are fully submerged, the buoyant forces on each are the same. The correct answer is B. 0. Strategy and Ball is floating, so its density is less than that of the water. Since Ball is submerged within the water, its density is the same as that of the water. Ball is sitting on the bottom of the tank, so its density is greater than that of the water. The correct answer is A.. Strategy The supporting force of the bottom of the tank is equal to the weight of Ball less the buoyant force of the water. Compute the supporting force on Ball. Ball gvball water gvball gvball ( Ball water ) (9.80 m s )( m )(7.8 0 kg m.0 0 kg m ) N The correct answer is B. 566

12 Physics Reiew and Synthesis: Chapters 9. Strategy The relationship between the fraction of a floating object s olume that is submerged to the ratio of the object s density to the fluid in which it floats is V f V o o f. So, the fraction of the object that is not submerged is Vns Vf Vo o f. Find the fraction of the olume of Ball that is aboe the surface of the water. Ball kg m water.0 0 kg m 5 The correct answer is D.. Strategy The pressure difference at a depth d in water is gien by water gd. Compute the approximate difference in pressure between the two balls. water gd (.0 0 kg m )(9.80 m s )(0.0 m).0 0 N m The correct answer is C. 4. Strategy For the force exerted on Ball by the bottom of the tank to be zero, Ball must hae the same density as the water. Find the olume of the hollow portion of Ball, V H. mball FeVFe Ball water and VH VBall VFe, so VBall VBall water water kg m V 6 H VBall VBall VBall (.0 0 m ) m. Fe Fe kg m The correct answer is C. 567