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Solution Manual For College Physics: A Strategic Approach, 4th Edition

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    College Physics: A Strategic Approach Fourth Edition Randall D. Knight Brian Jones Stuart Field Instructor’s Solutions Manual for C ollege P hysics : A S trategic A pproach Chris Porter

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    I-1 PptI.1. Reason: The orbital speed depends on the mass of the central body. If there were no dark matter then the stars would orbit around the Milky Way slower so the period would be longer. The answer is A. Assess: If the sun’s mass were smaller then the earth would take more than 365 days to orbit. PptI.2 . Reason: Assume circular orbits. The gravitational force is centripetally directed. If Saturn is 10 farther  than the earth then Kepler’s third law says its period T is 1000 that  of the earth. Use the formula for centripetal acceleration. 2 2 2 2 S S S e e S S S e 2 2 2 e e e e e e e e e (2 ) (2 10 1000 ) 10 100 10 1 1000 1000 10 10 100 (2 ) (2 ) v R /T R / T a R R R a v R /T R /T R R R         = = = = = =   The answer is B. Assess: Another way is to say the gravitational field is 100 weaker  at that distance, so the acceleration will be 100 less  (from net ) F ma =  PptI.3 . Reason: If the orbital speeds are the same, then we simply compare the centripetal accelerations. 2 2 S e S S e 2 2 e e e e e 10 1 10 v v a R R a v v R R = = = The answer is A. Assess: The acceleration is greater with the dark matter than without. PptI.4 . Reason: Newton’s law for the gravitational force depends on the mass. More mass means more gravitational force. The answer is A. Assess: We expect more massive things to pull harder on each other. PptI.5 . Reason: Solve a v/ t =   for t   2 20 m/s 3 2 s 6 0 m/s v t a   = = =   The correct answer is C. Assess: It would take a greyhound less time to reach top speed, but 3.2 s seems reasonable for a horse. F ORCE AND M OTION P A R T I

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    I - 2 Part I PptI.6 . Reason: Solve 2 2 f i 2 v v a x = +  for x   The equation is simple because i 0 v =  2 2 f 2 (20 m/s) 33 m 2 2(6 0 m/s ) v x a  = = =  The correct choice is B. Assess: It would take the horse 33 m to reach top speed; this seems reasonable. PptI.7 . Reason: Solve 2 c / a v r = for r  2 2 2 c (15 m/s) 31 7 m 7 1 m/s v r a = = =   The correct answer is B. Assess: From the photograph it appears 32 m is in the ballpark for the radius of the turn. PptI.8 . Reason: Use net by block on foot , F F ma = = where 2 12 m/s a = (twice that of a horse). 2 by block on foot (70 kg)(12 m/s ) 840 N F = = The answer is B. Assess: Since the acceleration is moderately close to , g we expect the answer to be moderately close to the runner’s weight. PptI.9 . Reason: Because net F ma = the net force on the dog must be in the same direction as the dog’s acceleration. In uniform circular motion the acceleration is toward the center of the circle, so the answer is D. Assess: The force is provided by friction of the ground acting on the dog’s paws. PptI.10 . Reason: Let us call the direction of motion . x + Solve net, x x F ma = (where net, x F is just the drag force 6 ) x D rv  = − for x a . 2 2 11 6 6 (0 0010 N s/m )(25 m)(0 25 mm/s) 1 812 m/s 6 5 10 kg x x D rv a m m    − − −    = = = = −    The magnitude of this is 0 2 , g   so the correct choice is A. Assess: This is a reasonable acceleration through a viscous fluid. PptI.11 . Reason: The acceleration is constant so we can use kinematics. Solve 2 2 f i 2 v v a x = +  for x   The equation is simple because f 0 v =  2 2 8 i 2 (0 25 mm/s) 1 72 10 m 0 02 m 2 2( 1 812 m/s ) v x a  − − −   = = =     −  The answer is A. Assess: The paramecium comes to rest in a distance much less than its own length. PptI.12 . Reason: The drag force 6 D rv  = is directly proportional to the speed, so doubling the speed would double the drag force. The answer is C. Assess: This is straight - forward given the model of the drag force as proportional to speed. PptI.13 . Reason: At terminal speed the acceleration is zero, so the net force is also zero. The answer is B. Assess: At terminal speed the magnitude of the drag force is the same as the magnitude of the gravitational force.

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    Force a nd Motion I - 3 PptI.14 . Reason: Because net F ma = the net force on the falcon must be in the same direction as the falcon’s acceleration. In uniform circular motion the acceleration is toward the center of the circle, so the answer is C. Assess: This centripetally - directed acceleration is provided by the lift. PptI.15 . Reason: The mass of the falcon is (8 0 N) 0 816 kg /g  =   2 2 net lift lift ( ) (0 816 kg)(15 m/s 9 8 m/s ) 20 N F F mg ma F m a g = − =  = + =  +  = The correct answer is D. Assess: The gravitational force cannot be neglected in this problem. PptI.16 . Reason: Combine the kinematic equation with the centripetal acceleration equation. 2 2 2 c 2 2(9 8 m/s )(30 m) 11 76 m/s 50 m v a x a r r   = = = =  This value of acceleration is greater than the acceleration during free - fall 2 ( 9 8 m/s ), g =  so the answer is B. Assess: The centripetal acceleration is still less than the stated maximum of 2 15 m/s , so this is reasonable. PptI.17 . Reason: F and d are directly proportional, so if the force (due to the weight of the person) is only half, then the deflection will also be only half. The answer is B. Assess: This makes intuitive sense since the variables are proportional. PptI.18 . Reason: Since deflection is proportional to the force, then the force is three times as great when the deflection is. So when the deflection is 12 cm the force must be 2 3 3(70 kg)(9 8 m/s ) 2058 N 2100 N mg =  =   The correct answer is C. Assess: The springboard must be able to withstand forces much greater than the weight of the person. PptI.19 . Reason: With all other variables held constant, the deflection is proportional to 3 L  3 3 4 d F L Ywt   =     If the length is decreased to half, then the deflection will only be 3 (1 2) 1 8 / / = as much. 1 (4 0 cm) 0 50 cm 8 d  =  =  The correct choice is A. Assess: This makes sense since the variables are proportional. PptI.20 . Reason: Use Newton’s second law in both directions. Position the x - axis down the slide. 40  = J 0 45  = 

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    I - 4 Part I y cos 0 cos F n mg n mg    = − =  = x sin F mg n ma    = − = Substitute in for n , cancel m , and solve for a . sin cos mg mg ma    − = (sin cos ) a g    = − Assume the acceleration is constant so we can use the kinematic equations with i 0 v =  2 1 ( ) 2 x a t  =   2 2 2 2(3 0 m) 1 4 s (sin cos ) (9 8 m/s )(sin 40 0 45cos40 ) x x t a g        = = = =  −   −   Assess: The slide is short, so it doesn’t take very long to slide down. PptI.21 . Reason: (a) For uniform circular motion there must be a net force toward the center of the circle. (b) The new reading is , n and the amount the reading is reduced is mg n −  The radius of the earth is 6 6 37 10 m    The mass of the person is 2 (800 N)/(9 8 m/s ) 81 6 kg  =   1d 86400 s = 2 2 2 2 2 r v t mg n F ma m m m r r r t              − =  = = = = =            2 6 2 (81 6 kg) (6 37 10 m) 2 8 N 86400 s       =      Assess: The reading is only 2.8 N less than 800 N. PptI.22. Reason : We will need the speed of the dolphin as it enters the water. We compute this from the 7.0 m free fall with i 0 v =  2 2 f f 2 2 2(9 8 m/s )(7 0 m) 11 71 m/s v a x v a x =   =  =   =  (a) Using other kinematic equations for the time in the water: 2 2 2(1 5 m) 0 2561s 0 26 s 11 71 m/s ( ) /2 v v v x a t t a v v x      =   = = = = =        

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    Force a nd Motion I - 5 (b) When the dolphin is in the water, there is an upward buoyant force due to the water that mostly cancels the downward force of gravity on the dolphin; that is why the dolphin can generally float while underwater without accelerating toward the bottom. Thus we will ignore both the buoyant and weight forces of the dolphin while it is underwater. Then we can use Newton’s second law to write , a F/m = where F is the force of the water on the dolphin. We find a from the kinematic equation 2 2 f i 2 v v a y = +  Here, i 11 7 m/s v =  is the speed just before the dolphin hits the water, f 0 m/s, v = and y  is the stopping distance of the dolphin. Solving for a we get 2 2 2 i (11 7 m/s) 45 6 m/s 2 2(1 5 m) v a y  = = =    so that 2 (210 kg)(45 6 m/s ) 9.6 kN F ma = =  =  Assess: It takes a lot of force to stop a massive object in a short distance.

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    II-1 PptII.1 . Reason: The water molecules that become the snowflake become more ordered, so the entropy of the water decreases. The answer is A . Assess: But the entropy of the entire universe increases. The entropy of the air increases by more than the entropy of the water decreases. PptII.2. Reason: The overall entropy always increases, so the answer is B. Assess: It is possible for the entropy of an isolated system to decrease, but it is fantastically improbable for a reasonably sized system. PptII.3. Reason: When energy is transferred between hot and cold (as in convection cells) the entropy increases, so the process is not reversible. The answer is B. Assess: The entropy of the universe never decreases. PptII.4 . Reason: In an isolated system the energy is conserved, but the entropy never decreases, so the answer is C. Assess: The other options are easily eliminated. PptII.5. Reason: The rubber bands slow down the rider over a longer period of time so the force is reduced. net F t p  =  shows this. The answer is D. Assess: This is the same reason it doesn’t hurt to land on a trampoline from a large height. PptII.6. Reason: The height the rider will go is proportional to the maximum spring energy in the bands. 2 1 2 . kx mgh = But the energy in the bands is proportional to the square of the stretch. 2 1 s 2 . U kx  = So half the stretch will produce a height 2 1 2 as high. 1 4 (2 0 m) 0 5 m.  =  The answer is D . Assess: Half the stretch gives less than half the energy, so also less than half the height. PptII.7. Reason: Solve 2 1 2 kx mgh = for , h so the height the rider will go is inversely proportional to the mass (given that that spring energy in the numerater is the same) 2 1 2 / , h kx mg = so the rider will go twice as high with half the mass. The answer is C. Assess: Four meters is dangerously high. C ONSERVATION L AWS P A R T I I

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    II - 2 Part II PptII.8. Reason: By Newton’s third law the force the pogo stick exerts on the ground is the same as the force the ground exerts on the pogo stick. The work - kinetic energy theorem says the net force on the stick times the distance the force acts ( 1 4 m) y  =  is 2 1 2 mv mgh = where h is the 1.6 m free fall. The net force on the stick is the normal force up (which we seek) minus the gravitiational force down: net F . n mg = − Put this all together. net ( ) mgh W F y n mg y mgh n mg y =  = −  =  − =  2 2 (80 kg)(9 8 m/s )(1 6 m) (80 kg)(9 8 m/s ) 4000 N 0 4 m mgh n mg y   = + = +  =   The answer is A. Assess: We can’t forget the gravitational force during the slowing phase. PptII.9. Reason: Since F kx = reducing the spring constant will reduce the force on the rider. And since 2 1 2 kx mgh = reducing the spring constant will also reduce the final height (all else remaining equal). So the answer is A. Assess: It makes sense that a smaller force would produce a lower jump height. PptII.10. Reason: The elastic potential energy is what makes the ball spring back up. There is some thermal energy, but there must be more elastic energy than thermal because the ball bounces back up to more than half its original height. The answer is D. Assess: The other choices are eliminated because 0 h = and 0. v = PptII.11. Reason: The force is what makes the ball change momentum and bounce back up, so if it doesn’t bounce as high it must be because the force wasn’t as great. The answer is C. Assess: The force must also be less to slow the ball down over a longer time. PptII.12. Reason: Assume the ball is in free fall. (a) 2 2 f f 2 2 2( 9 8 m/s )( 2 5 m) 7 0 m/s v a y v a y =   =  = −  −  =  (b) 2 2 i i 2 2 2( 9 8 m/s )(1 4 m) 5 2 m/s v a y v a y = −   = −  = − −   =  (c) The “lost” energy is transformed into thermal energy. (d) i f ( ) (0 057 kg)(5 2 m/s 7 0 m/s) 120 N 6 0 ms p m v v F t t  −   +  = = = =    Assess: These numbers all seem reasonable. PptII.13. Reason: Use conservation of momentum, with the initial momentum of the system zero. w s s s w w s s (0 30 kg)(10 m/s) 0 75 m/s 4 0 kg m v m v m v v m − −  = −  = = = −   The squid’s speed is thus 0.75 m/s, s o the answer is D. Assess: Because the water’s mass is less than the squid’s, we expect the squid’s speed to be less than the water’s. PptII.14. Reason: 2 0 75 m/s 7 5 m/s 0 20 s v a t   = = =    2 2 (7 5 m/s ) (9 8 m/s ) 0 75, a g  =      so the answer is D. Assess: The first two choices seem to o big, so we are happy with an answer in the range of 1 . g

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    Conservation Laws II - 3 PptII.15. Reason: Compute the average force from net (0 30 kg)(10 m/s) 30 N 0 10 s p m v F t t    = = = =    So the answer is B. Assess: This force is not very big. PptII.16. Reason: Define “what you get” as the kinetic energy of the squid and “what you had to pay” as the sum of the kinetic energies of the squid and the water. 2 2 1 s s 2 2 2 2 2 1 1 s s w w 2 2 (4 0 kg)(0 75 m/s) 7 0% (4 0 kg)(0 75 m/s) (0 30 kg)(10 m/s) m v e m v m v   = = =  +   +  Assess: In this case speed must be more important evolutionarily than efficiency. PptII.17. Reason: (0 046 kg)(63 m/s) 2.9 kg m/s p m v  =  =  =  The answer is C. Assess: Momentum is conserved for the club - ball system. PptII.18. Reason: Use conservation of momentum of the club - ball system. c c i c c f b b f ( ) ( ) ( ) m v m v m v = + c c i b b f c f c ( ) ( ) (0 30 kg)(40 m/s) (0 046 kg)(63 m/s) ( ) 30.3 m/s 30 m/s 0 30 kg m v m v v m −  −  = = =   The answer is A. Assess: We expect the club to be slower after than right before. PptII.19. Reason: See if the kinetic energy is the same before and after. 2 2 i c c i 1 1 ( ) (0 30 kg)(40 m/s) 240 J 2 2 K m v = =  = 2 2 2 2 f c c f b b f 1 1 1 1 ( ) ( ) (0 30 kg)(30.3 m/s) (0 046 kg)(63 m/s) 229 J 2 2 2 2 K m v m v = + =  +  = Kinetic energy is not conserved, so the answer is B . Assess: We would be worried if f i . K K  PptII.20. Reason: 2 2 1 b b f 2 2 2 1 c c i 2 ( ) (0 046 kg)(63 m/s) 0 38 ( ) (0 30 kg)(40 m/s) m v e m v  = = =   The anwer is C . Assess: This process is fairly efficient. PptII.21. Reason: 2 2 1 1 2 2 (80 kg)(11 m/s) 1180 W 1200 W 4 1 s mv E P t t  = = = =     The answer is D. Assess: It is hard to select between the choices based on ballpark expectations because they are all in a reasonable range.

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    II - 4 Part II PptII.22. Reason: 2 1 kg 1609 m 1 h (0 10)(1200 lb) (9 8 m/s )(3 0 mi/h) 717 W 720 W 2 2 lb 1 mi 3600 s P Fv      = =    =             1 hp 720 W 720 W 0 96 hp 746 W   = =      Assess: We expect a horse to be able to exert about a horsepower of power. PptII.23. Reason: Momentum is conserved in this inelastic collision. 1 1 i 2 2 i 1 2 f ( ) ( ) ( ) m v m v m m v + = + 1 1 i 2 2 i f 1 2 ( ) ( ) (100 kg)(6 0 m/s E) (130 kg)(5 0 m/s W) (0 22 m/s W) 230 kg m v m v v m m +   +   = = =   + Assess: Since they are about equal in momentum the final speed is small. PptII.24. Reason: (a) The kinetic energy of the hand needs to be sufficient to bend the board 1.2 cm, s o we set the kinetic energy of the hand equal to the s pring potential energy of the b o a rd at the breaking point. The spring constant k (800 N) (1 2 cm) 6667 N/m. =   = 2 2 1 1 66667 N/m (0 012 m) 4.38 m/s 4 4 m/s 2 2 0 50 kg k mv kx v x m −  = =  =    (b) ( ) 2 2 4.38 m/s (0.50 kg) 400 N 2 2(0.012 m) v F ma m x       = = = =          Another way to look at this is when the had just touches the board its force on the board is 0 N; just before the board breaks the force is 800 N. Thus the average force is 400 N. Assess: That’s a fast speed to swing a hand, but possible in karate. PptII.25. Reason: (a) Use conservation of momentum. L L i L L i L s f f L s ( ) (30 kg)(4 0 m/s) ( ) ( ) 3 0 m/s 40 kg m v m v m m v v m m  = +  = = =  + (b) (10 kg)(3 0 m/s) 120 N 0 25 s p F t   = = =   (c) Use 2 1 2 . K mv = 2 2 i f 1 1 (30 kg)(4 0 m/s) 240 J (40 kg)(3 0 m/s) 180 J 2 2 K K =  = =  = f i 180 K 240 J 60 J K K K  = − = − = − This energy was dissipated as thermal energy. Assess: In part (c) it seems reasonable to “lose” a quarter of the energy.

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    III-1 PptIII.1. Reason: The scaling law for specific metabolic rate vs. body mass is 0 25 . M −  Because the wolf has a mass 16 times as great as the jackrabbit’s we expect the specific metabolic rate of the jackrabbit to be 0 25 (16) 2  = times the wolf’s. The answer is A. Assess: Remember , the specific metabolic rate is the power used per kilogram of tissue. PptIII.2. Reason: The wolf weighs 16 times more than the jackrabbit, but the jackrabbit’s specific metabolic rate is twice as high as the wolf’s, so the total energy used by the wolf per day would be 16/ 2 8 = times as much as the jackrabbit. The answer is C. Assess: The energy used per day is the specific metabolic rate times the mass. PptIII.3. Reason: The specific metabolic rate for a rat is 5 W/kg so the metabolic rate is (5 W/kg)(0 20 kg) 1W 1J/s.  = = In a day the rat would use 1 cal 1 Cal 24 h 3600 s (1 J/s)(1 d) 20 Cal 4 19 J 1000 cal 1 d 1 h                  The answer is B. Assess: This is about 1/100 what a human would use; this makes sense. PptIII.4. Reason : The passage talks about heat transfer and that larger animals have a smaller surface - to - volume ratio, so a large animal loses less heat to the cold environment. The answer is D. Assess: Choice D is also the one that best addresses the issue. PptIII.5. Reason: Jump height is the answer that can be explained by scaling laws. We know that all animals can’t jump exactly the same height, but the variation in jump height is tiny compared to the ratio of the masses of the animals. The scaling laws must introduce offsetting factors that leave the jump height approximately the same. The answer is D. Assess: The other choices are true but don’t reflect scaling laws. PptIII.6. Reason: Assume no friction or air resistance, and use conservation of energy. 15 km/h 4 17 m/s =  2 2 2 1 2 2 2 1 (4 17 m/s) 1 m 2 2 2(9 8 m/s ) mv v mgh mv h mg g  =  = = =   The answer is D. Assess: The other options seem too big. P ROPERTIES OF M ATTER P A R T III

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    III - 2 Part III PptIII.7. Reason: If 25% of the 480 W is converted to mechanical energy of motion, then the other 75% is converted to thermal energy in his body. So (0.75)(480 W) = 360 W. The answer is B. Assess: It takes some effort to stay cool on a strenuous bike ride. PptIII.8. Reason: The heat of vaporization v L of water is 5 22 6 10 J/kg.   Use v Q ML = to calculate the amount of water one would need to evaporate to get rid of 360 J of thermal energy every second for two hours. 5 v (360 J/s)(2 h) 3600 s 1 1 kg 22.6 10 J/kg 1 h Q M L   = = =       Because one liter of water has a mass of one kilogram, the answer is C. Assess: The seems to be in the reasonable range for the amount of water a cyclist would drink in two hours. PptIII.9. Reason: If the cyclist is unable to get rid of this thermal energy then Q Mc T =  will tell us how much the temperature will go up. For mammalian bodies 3400 J (kg K). c =  (360 J/s)(10 min) 60 s 0 9 K 0 9 C (68 kg)(3400 J/(kg K)) 1 min Q T Mc    = = =  =       The answer is C. Assess: Less than one degree doesn’t seem like much, but that is a dangerous temperature rise for humans. PptIII.10. Reason: The buoyant force needs to be about the same magnitude as the weight force, so we compute the weight force on the balloon by assuming its mass is about the mass of the displaced air. 3 3 2 b (1 2 kg/m )(4 m )(9.8 m/s ) = 47 N 50 N F mg Vg  = = =   The answer is A. Assess: If the buoyant force is a little bigger than the weight force, then the balloon accelerates upward. PptIII.11. Reason: The buoyant force needs to be about the same magnitude as the weight force, so we compute the weight force on the balloon by assuming its mass is about the same as the mass of the displaced air. Read from the figure that the density of air at 10 km altitude is 0.4 3 kg/m . 3 3 2 b (0 4 kg/m )(12 m )(9.8 m/s ) 47 N 50 N F mg Vg  = = =  =  The answer is A. Assess: The weight of the balloon has hardly changed at all (being a little bit farther from the earth), so the buoyant force doesn’t need to either. PptIII.1 2 . Reason: If the temperature were unchanged, then as the pressure halved the volume would double to 3 8.0 m . But because the temperature in the balloon drops, this will somewhat shrink the gas volume, leading to a volume less than 3 8.0 m . Or we can use an equation. Since there is no heat transfer ( = 0), Q then the process is adiabatic. The air is diatomic, so we use the adiabatic equation with 1.4.  = 1 1.4 3 3 3 1.4 i f f i i f i f 2(4.0 m ) 2 (4.0 m ) 8.0 m p p V pV V V p    =  = = =  Either way, the answer is C. Assess: We do not need to actually compute 1 1.4 2 to know it is less than 2.

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    Properties of Matter III - 3 PptIII.13. Reason: If the temperature were unchanged, then as the pressure halved the volume would double to 3 8.0 m . But because the temperature in the balloon drops, this will somewhat shrink the gas volume, leading to a volume less than 3 8.0 m . T he answer is C. Assess: We do not need to actually compute 1 1.4 2 to know it is less than 2. PptIII.14. Reason: Examine 4 term mg A v  = to see that term v decreases as  increases if , m , g and A remain the same. The answer is C. Assess: This is in contradiction to the statement in the book “once an object has reached terminal speed, it will continue falling at that speed until it hits the ground” because that statement assumes  isn’t changing. PptIII.15. Reason: Assume the helium is an ideal gas. If f i T T = then f f i i , p V pV = so reducing the pressure by a factor of three must triple the volume. The answer is D. Assess: The ratios confirm our intuition. PptIII.16. Reason: Assume the air is an ideal gas. We are told the volume and the temperature of the air are constant, but n can change. 1 2 2 2 1 1 2 1 p RT p p n n n V n p = =  = Since / m V  = with V constant, the density is proportional to m which is proportional to the number of molecules. Since 2 1 1 3 p p  =  then n (and therefore the density) also decreases by a factor of 3. So the answer is D. Assess: It is intuitive that as the balloon rises the density of the air decreases. PptIII.17. Reason: The buoyancy force is in the opposite direction from the weight, so it is up. The drag force is in the opposite direction from the motion, so it is up too. The answer is A. Assess: Since it is descending at a constant rate the sum of the three forces is zero. PptIII.18. Reason: To get the air through a smaller cross section requires it to move faster. So the air moves slower in the trachea than in the nostrils. Assess: If this weren’t the case then air would pile up at the nostrils. PptIII.19. Reason: Estimate the area of the diaphragm to be 2 15 cm 30 cm 0 045 m .  =  Since pressure is force/area then 2 (7 0 kPa)(0 045 m ) 315 N 300 N F PA = =   =  Assess: The estimate is probably good to only one significant figure. PptIII.20. Reason: The force the block exerts on the bottom is the same size as the normal force the tank exerts on the block. In a free body diagram of the block we have the weight force down, and the buoyant force and the normal force up. The buoyant force is the weight of the water displaced. b w , F m g = but because the density of aluminum is 3 2 7 kg/m  the buoyant force is 1 b Al 2 7 . F m g  = 2 Al b Al Al Al 1 1 1 (20 kg)(9 8 m/s )(0 6296) 120 N 2 7 2 7 n m g F m g m g m g   = − = − = − =   =       Assess: This is less than the weight of the aluminum, which we expected.

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    III - 4 Part III PptIII.21. Reason: (a) Assume the air in the bladder is an ideal gas. We solve the ideal gas equation for . n The pressure at a depth d is 0 . p p gd  = + 80 ft 24 384 m. =  15 C 288 K. = 3 (0 070)(7 0 L) 0 00049 m . V =   =  3 2 3 0 ( )( ) [101 3 kPa (1030 kg/m )(9 8 m/s )(24 384 m)](0 00049 m ) 0.07113 mol (8 31 J/(mol K))(288 K) pV p gd V n RT RT  +  +    = = = =   Which we report as 0.071 mol to two significant figures. (b) 50 ft 15 24 m. =  3 4 3 3 2 (0.07113 mol)(8 310 J/(mol K))(288 K) 0 000667 m 6 7 10 m 101 3 kPa (1030 kg/m )(9 8 m/s )(15 24 m) nRT V p −   = = =      +   (c) Find the new number o f moles and then subtract. 3 2 3 [101 3 kPa (1030 kg/m )(9 8 m/s )(15 24 m)](0 00049 m ) 0 05226 mol (8 31 J/(mol K))(288 K) pV n RT  +    = = =    So 0 07113 mol 0 05226 mol 0 01887 mol 0 019 mol  −  =    need to be removed. Assess: We expect the needed moles at 50 ft to be less than at 80 ft.

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    IV-1 PptIV.1 . Reason: Nothing goes faster than light in a vacuum, so it is first. Waves through the solid earth are next fastest, followed by sound in air. According to the passage, tsunamis travel at hundreds of kilometers per hour, (about 200 m/s), but not quite as fast as the speed of sound in air. light earthquake sound tsunami v v v v    Assess: This ranking matches expectations. PptIV.2. Reason: As the front of the wave nears shore, its speed decreases, and the back of the wave moves faster than the front, so the water piles up. The answer is C. Assess: This is as the passage explained it. PptIV.3. Reason: Use 1/ T f = in . v f  = 150 000 m 750 s 12 5 min 15 min 200 m s T v  = = = =    The answer is D. Assess: This is a long period, but it makes sense given the long wavelength. PptIV.4. Reason: The frequency is unaffected. Since the speed decreases, so does the wavelength. The answer is C. Assess: The peak - to - peak distance gets smaller. PptIV.5. Reason: A tsunami is not a standing wave, but it does reflect as shown in the simulation. The reflected wave can interfere with the primary wave to make extra large crests. The answer is B. Assess: A standing wave would need a continuous source of new pulses. PptIV.6. Reason: Use 1/ T f = in v f  = and the given equation for v in deep water. 2 75 m 6 9 s 7 s /2 (9.8 m/s )(75 m)/((2 ) T v g      = = = =   The answer is C. Assess: Imagining oneself on a boat, this sounds reasonable. O SCILLATIONS AND W AVES P A R T IV

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    I V - 2 Part IV PptIV.7 . Reason: Use 1/ T f = in v f  = and the given equation for v in deep water. The effective speed with which the waves pass the ship is the speed of the waves plus the speed of the ship toward the waves. ( ) 2 ship 75 m 4.9 s 5 s (9.8 m/s ) 75 m 4.5 m/s 2 2 T v g v      = = = =  + + The answer is B . Assess: We expect the waves to reach the ship more frequently if the ship is sailing toward them. PptIV.8. Reason: Frequency and wavelength are inversely related, so higher frequency means shorter wavelength. Therefore, the answer is C. Assess: The concept here is similar to the dispersion of light in glass where the speed of blue (short wavelength) light is slower than the speed of red (long wavelength) light. PptIV.9. Reason: We need to know how far 40 wavelengths is, so we find .  In Table 15.1 find the speed of ultrasound in human tissue: 1540 m/s. v = 1540 m/s 0 00154 m 1 0 MHz v f  = = =   Now find the number of wavelengths in 12 cm. ( ) 1 wavelength 0.00154 m 12 cm 12 cm 78 wavelengths. = = This is about twice the 40 - wavelength half - distance, so the intensity is halved twice. In other words, the new intensity is 250 2 W/m . The answer is C. Assess: The penetration depth depends on the frequency, but this is reasonable for 1 MHz ultrasound. PptIV.10. Reason: Doubling the frequency halves the wavelength; so twice as many wavelengths fit in the same distance. So instead of 40 wavelengths, there would be 80; this means the intensity is halved twice, so the final intensity is 2 250 W/m . The answer is C. Assess: Putting more wavelengths in the same distance means the intensity would have to be less, so this eliminates choices A and B. PptIV.11. Reason: Higher frequency provides less penetration, as we saw in the previous problem, but with smaller wavelength it provides better resolution. See Example 15.7. The correct answer is B. Assess: There are tradeoffs in deciding which frequency to use. PptIV.12. Reason: The beat frequency is the difference between the original frequency and the reflected frequency, but this is just o 0 2 . v v f f  =  So increasing 0 f will also increase the beat frequency. The answer is A. Assess: Increasing 0 f would allow the physician to detect slower movements of the heart. PptIV.13. Reason: There are three full oscillations in 0.004 s so the period is 0 00133 s.  The frequency is the inverse of this. 1 3 750 Hz 0.004 s f T = = = The answer is C. Assess: All of the choices are plausible answers, so we must have faith in the math.

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    Oscillations a nd Waves I V - 3 PptIV.14. Reason: The second batch of wiggles (after reflection from the closed end) is not inverted; but the third batch is, after reflecting from the open end, and then the closed end again. The answer is C. Assess: This is similar to a wave pulse on a string inverting upon reflection if the end is tied down, but not if the end is free. PptIV.15. Reason: The wave pulse travels down the tube and back in 0.008 s dist speed time 2 (343 m/s)(0 008s) 1 4 m L L =   =   =  The answer is C. Assess: This is a reasonable length for a tube for this measurement. PptIV.16. Reason: In the fundamental or lowest resonance mode only half a wavelength can fit in the tube. In other words, 1 4 , L  = so 4 . L  = See Figure 16.19. The answer is D. Assess: An open - closed organ pipe sounds an octave lower than one of the same length open at both ends. PptIV.17. Reason: It will take half a period to return to the starting spot. 2 5.0 m 2.2 s 2 9.8 m/s T L g   = = = The answer is C. Assess: This seems about right for a 5.0 m long swing. PptIV.18. Reason: The period is independent of the mass of a simple pendulum, which is our model for this problem. So the time is still 2.2 s. The answer is C. Assess: Try this on a long swing by timing the period for different swingers. PptIV.19. Reason: At the farthest point you are 1.0 m higher than at the bottom. See figure. We now use conservation of energy to find the speed at the bottom. 2 2 1 2 2(9 8 m/s )(1 0 m) 4 4 m/s 2 mv mgh v gh =  = =   =  The correct answer is C. Assess: This seems like a reasonable speed for a big swing.

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    I V - 4 Part IV PptIV.20. Reason: (a) The instant the kangaroo touches ground the spring is still unstretched. When the kangaroo leaves the ground the spring is unstretched again. This is the bottom half of a typical mass - on - spring motion (we ignore the kangaroo’s trajectory through the air), so the full period is twice the time the kangaroo is in contact with the ground. That is, 0 20 s. T =  (b) For a mass on a spring, 2 . m k T  = 2 2 2 2 4 4 (70 kg) 69 kN/m (0 20 s) m k T   = = =  (c) The amplitude would have to increase, but the contact time of each bounce would remain the same since the period does not depend on the amplitude. Assess: It would take a large k to make such a short period. PptIV.21. Reason: It is a logarithmic scale. 2 7 0 0 (dB) 10 log 27 dB 10 500 I I I I   = −  = = Assess: The answer may be surprising to those who don’t realize the decibel scale is logarithmic. PptIV.22. Reason: The air chambers allow the sound to reflect back and forth. With each reflection a bit is absorbed, so the intensity decreases. Assess: This is a reasonable explanation of how to get a series of clicks with decreasing intensity from one loud initial click.

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    V-1 PptV.1. Reason: Certainly a diverging lens diverges parallel rays, but so does a converging lens after the focal point. The answer is C. Assess: In a converging lens parallel rays initially converge, but then after the focus point they diverge again. PptV.2. Reason: The light would be more spread out, making its intensity lower. Further, the image would be blurry because regions near to the focus would also contribute light that could get through the aperture. So the answer is C. Assess: It is important to have high - quality lenses. PptV.3. Reason: Smaller wavelengths can probe smaller features, just as in a regular microscope. The answer is A. Assess: For this reason more information fits on a Blu - ray disk than a CD. PptV.4. Reason: Since the laser source is small and then gets diverged, after the light leaves the object the diverging needs to be undone to return to a small hole in the screen. So the two distances are the same through the same optics. The answer is B. Assess: Symmetry leads us to this conclusion. PptV.5. Reason: A nearsighted person has an eyeball that is too long, but can still focus on near objects. So where the distance from the cornea to the retina in the horse’s eye is the longest is the answer. The answer is A. Assess: Close objects focus farther away. PptV.6. Reason: A farsighted person has an eyeball that is too short, but can still focus on distant objects. So where the distance from the cornea to the retina in the horses’ eye is the shortest is the answer. The answer is B. Assess: One would expect that in general close objects are at the bottom of the field of view and distant objects more often in the upper part of the field of view. PptV.7. Reason: Close things that are at the bottom of the field of vision are in focus, so the feet will be in focus; but in the upper part of the field of vision distant objects are in focus, so the close head of the person will be out of focus. The answer is B. Assess: See the previous two problems. PptV.8. Reason: If all the distances are increased then near objects would be focused at the bottom of the retina (the top of the field of view), but distant objects could not be focused at all, so the horse would be nearsighted. The answer is A. Assess: Having an eyeball that is too long is a cause of nearsightedness in humans as well. O PTICS P A R T V

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    V - 2 Part V PptV.9. Reason: Recall that the refraction at an interface depends on the indices of refraction according to Snell’s law: ( ) ( ) 1 1 2 2 sin sin . n n   = If the indices of refraction of the lens and the surrounding water are very similar, then ( ) ( ) 1 2 2 1 sin 1. sin n n   =  Thus, the angles will be very similar. The correct answer is B. If the indices of refraction are very similar, the refraction at the lens - water interface will be minimal. Assess: The curvature of the lens is important for focusing, but not for the basic refraction that occurs at the interface. PptV.10. Reason: Recall that there is a phase change due to reflection when light is incident from a medium of lower index of refraction onto one of higher index of refraction. So when the light is in the cytoplasm ( 1.34 n = ) and reflects off the guanine ( 1.83 n = ) there will be a phase change. However, at the back of the guanine crystals, where the reflection occurs off the interface with cytoplasm, there is not a phase change. So the correct answer is B. There is one phase change. Assess: This is analogous to waves on strings flipping over when they reflect off an interface with a more massive rope. PptV.11. Reason: In order for there to be a strong reflection, a full number of wavelengths N  plus one half a wavelength / 2  must fit into the path length difference (which is twice the thickness of the crystal). Thus 2 / 2. t N   = + We want the largest possible ,  which corresponds to the smallest possible . N With 0, N = we have ( ) 4 4 80 nm 320 nm. t  = = = This is the wavelength in guanine (since that is where the extra path length is). In air, this light would have a wavelength: ( ) ( ) ( ) guan air guan air 1.83 320 nm 586 nm 1.0 n n   = = = The correct answer is C, since this is closest to 600 nm. Assess: This means the scallops can see at least part of the spectrum that is visible to us, but not all. PptV.12. Reason: A spherical mirror focuses light down to make a real image. Light rays cross the axis during the focusing process, leaving the image inverted. The correct answer is B: the image is real and inverted. Assess: The fact that the image is inverted is not important to the scallop. The brain can still flip the image in processing. PptV.13. Reason: The f number is the ratio of the focal length to the aperture of an optical system. For humans, the focal length is close to the 2.5 cm diameter of the eye, and the aperture is given as 3.0 mm. The ratio is thus about 8.3. The closest answer is D. Assess: This is significantly higher than the f number for scallops. PptV.14. Reason: The incident power depends on the intensity (which we will assume is constant once the contraction begins) and the area of the pupil: 2 . P IA I r  = = The area depends on (half) the diameter squared. So halving the diameter means reducing the power by a factor of four. The correct answer is B. Assess: Of course, your pupil contracts because the intensity increased. In this problem we are only dealing with the effect of the pupil’s contraction once the intensity has already changed. PptV.15. Reason: Spherical aberration occurs when light rays strike the edge of an optical device and don’t quite form the image in the same place as the other rays. Since contracting the pupil removes these aberrant rays formerly at the edge of the pupil, the contraction should reduce the spherical aberration. The correct option is C. Assess: The eye often has a positive spherical aberration. PptV.16. Reason: The Rayleigh criterion is 1.22 D   = for a circular aperture. If the diameter of the pupil decreases, the minimum angular separation that can be resolved goes up. The correct answer is A. Assess: This can be semantically confusing. The “resolution” refers to the minimum angular spread that can be resolved. So when that goes up, it means it is more difficult for you to resolve nearby objects.

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    Optics V - 3 PptV.1 7 . Reason: In bright light the pupil is small and so many off - axis rays do not enter the pupil. Aberrations are reduced. The focus is better with a smaller aperture (pupil). Assess: This is the same in cameras. The smaller the aperture, the greater the depth of field (objects at different distances are all in focus). PptV.18 . Reason: This is a simple ratio problem. The pupil acts like a pinhole. 8 8 mm (24 mm) 35 m 6 1 m h h h h s s s s      =  = = =   Assess: The angle between the outermost rays is the same on both sides of the pupil. PptV.19 . Reason: The focal length is six times longer so the image will be six times taller, or 7.2 mm. The total light hitting the detector is the same, but is now spread out over 6 2 times the area, so the intensity is reduced by 36; the new intensity is 2 0 069W/m .  Assess: As the text says, “if f is large, the image is large and its light is spread out and dim. If f is small, the image is small and the light is concentrated and bright.” PptV.20 . Reason: We need to compute the wavelength of 1200 Hz sound at 20 C.  The speed of sound in air at that temperature is 343 m/s. 343 m/s 0 286 m 1200 Hz v f  = = =  Use the formula for the positions of dark fringes in the single - slit diffraction pattern with 1. p = (1)(0 286 m)(12 m) 3 4 m 1 0 m p p L y a   = = =   Assess: This seems to be a reasonable diffraction pattern width given the dimensions in the problem.

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    VI-1 PptVI.1. Reason: The intensity of sunlight decreases with the square of the distance from the sun. If Mars is 1.5 times farther away then the sunlight is 2 1 1 5   times as intense. 2 2 Mars 2 1400 W/m 620 W/m 1 5 I = =  The correct choice is B. Assess: This is one reason it is quite cold on Mars. The lack of a greenhouse effect is another. PptVI.2. Reason: If the earth received appreciably more or less energy than it emitted then the temperature would increase or decrease. The overall average temperature of the earth is fairly stable over the long haul (global warming notwithstanding). So energy in must equal energy out. The correct choice is C. Assess: The greenhouse effect prevents the earth from radiating as much energy as it would have otherwise, so the equilibrium temperature is higher than without the greenhouse effect. But whatever the temperature is, if it is stable then energy in equals energy out. PptVI.3. Reason: As black bodies get hotter the peak of their emission spectrum shifts toward shorter wavelengths, so the correct choice is C. Assess: The peak of the sun’s radiation curve is at a much shorter wavelength than the earth’s because the surface of the sun is so much hotter. PptVI.4. Reason: The energy of a photon is 15 8 (4 14 10 eV s)(3 0 10 m/s) 0 12 eV 10 m hc E hf   −      = = = =  The correct choice is D. Assess: The units work out correctly. PptVI.5. Reason: A water molecule has no net charge, so the net force on it due to an electric field is zero. However, because of the molecule’s electric dipole moment, an electric field does exert a torque on the molecule. The correct answer is B. Assess: The alternating net torque on the dipole causes the water molecule to rotate back and forth. This causes the molecules to jostle each other and the temperature rises. PptVI.6. Reason: Electrons like to go against the field, so the field is to the left since the electrons are accelerating to the right. The correct choice is A. Assess: Electrons roll up the potential hill. E LECTRICITY AND M AGNETISM P A R T VI

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    VI - 2 Part VI PptVI.7. Reason: Use Newton’s second law. 19 18 2 31 (1 6 10 C)(70 kV/1 0 cm) 1 2 10 m/s 9 11 10 kg V d q F qE a m m m −  −    = = = = =     The correct choice is A. Assess: This is a surprisingly large acceleration, but so were the other choices, so we trust the math. PptVI.8. Reason: Since current is charge per time, then mA s  is equal to mC. The correct choice is C. Assess: This goes back to the fundamental definitions of current and charge in the SI system. PptVI.9. Reason: We are given 10 mA I = and V 70 kV.  = (10 mA)(70 kV) 700 W P I V =  = = The correct choice is B. Assess: This is a decent amount of power, but it isn’t on for very long. PptVI.10. Reason: Multiply the usual energy = power time equation by 1%.  0 01 (700 W)(0 75 s)(0 01) 5 J E P t =    =   = The correct choice is C. Assess: This seems like a reasonable amount of energy for the x - ray pulse. PptVI.11. Reason: Assume the maximum energy of the x - ray photon is the kinetic energy of the electron just before it strikes the target electrode. 2 31 18 2 14 1 1 (2 ) (9 11 10 kg)(1 2 10 m/s )(0 010 m) 1 1 10 J 2 2 E mv m a x ma x − − = =  =  =      =   The correct choice is C. Assess: For a photon this is fairly energetic (68 keV), as we expect for an x - ray. PptVI.12. Reason: Use . P I V =  130 MJ 5910 s 100 min (100 A)(220 V) E E t P I V  = = = =   The correct choice is A. Assess: This seems like a long time at such a large current and voltage, but the point is that gasoline really stores a lot of energy. PptVI.13. Reason: Use . P I V =  270 kW 675 A 400 V P I V = = =  To two significant digits, this would round to 680 A. The correct choice is C. Assess: This is a lot of current, but what is needed to produce good car performance. PptVI.14. Reason: Assume that the transformer is ideal: 1 2 . P P = 1 2 1 2 220 V 100 A 55 A 400 V V I I V = = = The correct choice is C. Assess: This is a step up transformer, so the current in the secondary is less than the current in the primary.

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    Electricity a nd Mag netism V I - 3 PptVI.15. Reason: The energy stored in a capacitor is 2 1 2 ( ) . C V  Halving the voltage cuts the energy stored to 1/4 the original value. The correct choice is D. Assess: This is a challenge for a capacitor - powered car. PptVI.16. Reason: Use the right hand rule and curl your fingers in the clockwise direction as viewed from the secondary coil. Your thumb points to the left. The correct choice is B. Assess: A compass needle at the center of the primary coil would verify this. PptVI.17. Reason: The induced current in the secondary will be in such a direction as to oppose the increase of the field from the primary, so the induced current will create a field to the left. The correct choice is B. Assess: If the right - pointing field from the primary were decreasing, the induced current would be in the other direction, such as to oppose the decrease. PptVI.18. Reason: The frequency of the secondary is supposed to match the frequency in the primary. To double the resonant frequency in the secondary requires a decrease in the capacitance by a factor of 4 because 2 1 . f LC  =  The correct choice is D. Assess: Variable capacitors are used to tune circuits like this. PptVI.19. Reason: Use . P I V =  rms rms 60 W 0 50 A 120 V P I V = = =   The peak current is 2 more than this. The correct choice is C. Assess: We know that the peak current has to be more than the rms current; this eliminates all other choices. PptVI.20. Reason: We need to pull in lots of ideas for this integrated problem. 1 0 L  of water has a mass of water 1 0 kg. 4190 J (kg C). c  =  Use 2 ( ) V E t R P   = = and . E cm T =  Combine and solve for . T  2 2 ( ) (120 V) (60 s) 10 C (20 )(4190 J/(kg C))(1 0 kg) V t T Rcm    = = =   So the final temperature is f i 20 C 10 C 30 C. T T T = +  = + = Assess: The power dissipated by the resistor is 720 W, so it seems reasonable the temperature would go up by 10 C. PptVI.21. Reason: Each side of the wire loop has a mass of 50 g. To hold the loop steady we must have 0.   = Compute the torques around the pivot axis. The downward torque due to the weight of the two sides of the loop perpendicular to the axis plus the torque due to the weight of the side parallel to the axis is 2 2 grav 2(50 g)(9 8 m/s )(5 0 cm) (50 g)(9 8 m/s )(10 cm) 0 098 N m  =   +  =   The upward torque due to the current in the magnetic field must counterbalance that torque. No torque is generated in the two sides perpendicular to the axis because sin 0,  = so all the torque must be generated on the side parallel to the axis. The current will need to be clockwise (looking from above) to produce the upward torque, and in the parallel side sin 1.  = 2 mag mag 2 2 0 098 N m ( ) 19.6 A 20 A (0 10 m) (0 50 T) r F L ILB IL B I L B   ⊥   = = =  = = =    Assess: It seems reasonable that a current of this magnitude could hold up a 200 g loop of wire.

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    VII-1 PptVII.1. Reason: At low speeds the quantum nature becomes more important, and the waves of the atoms can overlap and create new forms of matter. The answer is C. Assess: These new states of matter have only recently been observed, but they are very exciting. PptVII.2. Reason: The momentum of a photon is directly related to its energy , which is directly related to its frequency. The ultraviolet photon has the highest frequency and therefore the greatest momentum, so it would give the highest recoil speed to the atom due to conservation of momentum. The answer is D. Assess: We can assume that the answer has to be one of the extreme answers, not the blue or red photons which are between the others in energy and momentum. PptVII.3. Reason: Due to the Doppler effect the laser photons appear to be blue - shifted which is greater energy. The answer is A. Assess: The greater energy pushed the atoms back to the equilibrium position. PptVII.4. Reason: The faster moving atoms see the photons as Doppler shifted so the photons no longer match the absorption energy as well. The answer is B. Assess: The warming of the gas does not change the atomic energy levels. PptVII.5. Reason: Temperature is related to the kinetic energy of the atoms, which is proportional to the speed squared. Halving the speed reduces the temperature to 2 (1 2) 1 4  =  the previous temperature. The answer is D. Assess: These nanokelvin temperatures are quite impressive. PptVII.6. Reason: Use the equation for the de Broglie wavelength. 34 6 27 h 6 63 10 J s 4 6 10 m 5000 nm mv (87)(1 67 10 kg)(0 0010 m/s)  − − −    = = =       The answer is D. Assess: This is not the wavelength of a photon, it is the wavelength of the rubidium atoms themselves. With this long wavelength they can overlap to form the Bose - Einstein condensates. M ODERN P HYSICS P A R T V I I

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    VII - 2 Part VII PptVII.7. Reason: The same temperature means the same average kinetic energy. 2 2 p ( ) 2 m 2 m h T K    = = This shows that for the same temperature a more massive atom has a shorter wavelength. Sodium, with less mass, has a longer wavelength. The answer is A. Assess: Redoing the previous problem for sodium would allow a direct comparison. PptVII.8. Reason: Add up the mass numbers of the daughter nuclei and subtract from the mass number of the original uranium and the incident neutron . (235 1) (87 147) 2 + − + = The answer is B . Assess: Because some neutrons escape and don’t cause further fissions, the average number of released neutrons must be greater than one to sustain a chain reaction. PptVII.9. Reason: The two fragments fly off at high speeds, and their kinetic energy is what is eventually converted to electric energy by a turbine and generator. The answer is D. Assess: The released neutrons also carry away some energy, but not as much as the two large fragments. PptVII.10. Reason: The law of conservation of momentum applies here, and since we are told to ignore the momentum of the neutrons, the two fragments must have momenta that are equal in magnitude and opposite in direction. The answer is C. Assess: The speeds will be different because the masses are different, but mv will be the same for each. PptVII.11. Reason: Because of conservation of momentum (see previous problem) we know the Br nucleus is going faster than the La nucleus (because the Ba is lighter). In the formula for kinetic energy the speed is squared, so the nucleus with the higher speed has the greatest kinetic energy when the momenta have the same magnitude. The answer is A. Assess: We saw this idea back in the chapter on conservation of momentum. PptVII.12. Reason: The energy of an electron (or a positron) is just barely over half an MeV, so a pair would be about one MeV. So 200 MeV can create about 200 pairs. The answer is C. Assess: This is quite a bit of energy! PptVII.13. Reason: In beta decay it is like a neutron decaying into a proton and an electron. This reduces the neutron number while increasing the proton number and thus makes a more stable isotope. The answer is B. Assess: Gamma decay doesn’t change the neutron number nor the proton number. Alpha decay carries away just as many protons as neutrons. PptVII.14. Reason: The magnetic field will bend the path of charged particles. Because those that go in Detector 4 are not bent very much, those particles must have a relatively high mass. The alpha particles have a mass of 4 u and are positively charged. So the answer is A. Assess: Particles of opposite charges will be bent in different directions. PptVII.15. Reason: The radiation detected by Detector 5 was not bent by the magnetic field; this means it isn’t charged. Gamma radiation is electromagnetic radiation (photons) and not charged. The correct choice is D. Assess: Charged particles are deviated by a magnetic field, but gamma rays (and all other electromagnetic radiation is not. PptV II.16. Reason: The magnetic field will bend the path of charged particles. Because those that go in Detector 10 are bent quite a bit, those particles must have a relatively low mass. The beta - minus particles have a small mass and are negatively charged. So the answer is B. Assess: Particles of opposite charges will be bent in different directions.

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    Modern Physics VI I - 3 PptVII.17. Reason: (a) The equation for relativistic kinetic energy is 0 ( 1) . K E  = − Solve for v . The mass of an electron is 2 0 511 MeV . c   2 2 2 1 1 1 v c K mc  + = =  − 2 2 2 7 19 keV 511 keV 1 1 1 1 0 265 8 0 10 m/s 1 1 K mc v c c c         = − = − =  =       + +     (b) 19 8 8 1 decay/s 1 6 10 J (15 MBq) (19 keV decay) 4 56 10 W 4 6 10 W 1 Bq 1 eV − − −        =              (c) The activity decays exponentially. 1 2 5 12 0 1 1 15 MBq 11 MBq 2 2 t t y y R R        = = =         Assess: The 11 MBq in the last part seems reasonable after less than half of a half - life. PptVII.18. Reason: 1% of 700 W is 7.0 W. The energy of a 0.030 nm photon is 34 8 15 (6 63 10 J s)(3 0 10 m/s) 6 63 10 J (0 030 nm) hc  − −      = =    (a) The number of photons emitted per second is 15 15 15 7 0 W 1 056 10 photons s 1 1 10 photons s 6 63 10 J photon N −  = =           (b) The dose in grays is the energy absorbed (7.0 W for one second i s 7.0 J) per kilogram of tissue . 80%, or 0.80, of the x - ray energy is absorbed. 7.0 J dose (0.80) 0.075 Gy 75 kg = = The dose equivalent is the dose multiplied by the RBE which is 1.0 for x - rays. dose equivalent 0 075 Gy 1 0 75 mSv =    = Assess: This dose equivalent is quite high relative to the values in Table 30.5. PptVII.19. Reason: First express the speed of the ship in m/s. 8 7 (0 12)(3 0 10 m/s) 3 6 10 m/s.    =   (a) For a non - relativistic calculation we use 2 6 7 2 21 1 1 (1 7 10 kg)(3 6 10 m/s) 1 1 10 J 2 2 K mv = =     =  

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    VII - 4 Part VII (b) The number of reactions to produce twice this kinetic energy (since only half of the fusion energy ends up as kinetic energy of the ship) is 21 32 32 19 (2)(1 1 10 J) 1 eV 7 65 10 reactions 7 7 10 reactions 18 MeV 1 6 10 J N −     = =            (c) We need 32 7 65 10   of each kind of nucleus. The mass of a 2 1 H nucleus is 27 27 (2)(1 67 10 kg) 3 34 10 kg − −   =   and the mass of a 3 2 He nucleus is 27 27 (3)(1 67 10 kg) 5 01 10 kg. − −   =   32 27 6 2 1 32 27 6 3 2 (7 65 10 )(3 34 10 kg) 2 6 10 kg of H (7 65 10 )(5 01 10 kg) 3 8 10 kg of H − −     =       =   Assess: We would need to haul around (or collect from space) a few million kilograms of hydrogen and helium. PptVII.20. Reason: Start with 0 ( 1) K E  = − where 2 2 1 1 v c  =  −  and solve for v. ( ) ( ) 0 0 2 2 2 2 0 1 1 1 1 1 1 1 K K E E K v v c E c  + =  = −  = − + + Call the later speed , v  which corresponds to 4 0 GeV. K  =  The ratio of the new speed to the original speed is (after canceling c ) ( ) ( ) ( ) ( ) 2 2 4 0 GeV 105 MeV 0 2 2 6 0 GeV 105 MeV 0 1 1 1 1 1 1 1 1 1 1 0 999821 1 1 K E K E v v    + + + + − −  = = =  − − The fraction by which the speed decreases is 1 1 0 999821 0 00018 v v  − = −  =  Assess: Even though the muon lost a third of its kinetic energy its speed is only a tiny fraction slower. PptVII.21. Reason: If we follow the discussion in the text and assume the muons are traveling at a speed close to the speed of light then the time needed to reach the ground in the earth’s reference frame is 8 (120 km) (3 0 10 m/s) 400 s. t   =    = Next we find  for the 10 GeV muons. 0 10 GeV 1 1 96 24 105 MeV K E  = + = + =  In the muons’ reference frame the time interval (proper time) is 400 4 16 s 96 24 t s       = = =   This is (4 16 s) (1 5 s) 2 77      =  half - lives, so the number reaching the ground is 2 77 0 1 15000 2 N N    = =     Assess: Many more muons reach the ground than expected without time dilation. From the other reference frame we could say the atmosphere is length contracted and arrive at the same answer.

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    1-1 Q1. 1 . Reason: The softball player starts with an initial velocity but as he slides he moves slower and slower until coming to rest at the base. The distance he travels in successive times will become smaller and smaller until he comes to a stop. See the figure below. Assess: Compare to Figure 1.10 in the text. Q1. 2 . Reason: Assess: The dots are equally spaced until the brakes are applied to the car. Equidistant dots indicate constant average speed. On braking, the dots get closer as the average speed decreases. Q1. 3 . Reason: R EPRESENTING M OTION 1

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    1 - 2 Chapter 1 Assess: The spacing between dots is initially large, since the initial speed with which the bush baby leaves the ground is large. As the bush baby rises and gravity slows the ascent, the speed decreases and therefore the spacing between adjacent dots in the motion diagram decreases. Q1.4 . Reason: As the ball drops from the tall building t 02. he ball will go faster and faster the farther it falls under the pull of gravity. The motion diagram should show the displacements for later times to be getting larger and larger. The successive displacements in the diagram given in the text get smaller and smaller. So the diagram given in the problem is incorrect. The correct diagram is below. Assess: Compare to Figure 1.5 in the text, which shows a motion diagram for two objects falling under the influence of gravity. The displacements increase during the fall of the object as we reasoned. Q1.5 . Reason: Position refers to a location in a coordinate frame. A displacement is the difference between two positions. In general, displacement is a vector and requires a direction. But in this one - dimensional case, we ignore that subtlety. The four miles Mark and Sofia walked definitely refer to a difference between their starting and ending positions. There is no information given about a reference frame. So it is more reasonable to associate the four miles with a displacement (magnitude) than with an absolute position. Assess: Mark and Sofia’s position could be specified as (for example) 100 m west of a particular intersection. But what is described is a difference between two positions. This is more like a displacement magnitude. Note that if their starting point had been the origin of a coordinate system, then both Mark and Sofia would be correct. Q1.6. Reason: The distance you travel will be recorded on the odometer. As you travel, the distance you travel accumulates, is recorded by the odometer, and is independent of the direction of travel. Your displacement is the difference between your final position and your initial position. If you travel around a 440 m track and end up where you started, you have traveled 440 m; however, since you ended up where you started, your change in position and hence displacement is zero. Assess: If you watch a track meet, you will observe the 440 - m race. As you watch the race, it is obvious that the runners travel a distance of 440 m (assuming they complete the race). Yet since they end up where they started, their final position is the same as their initial position and hence their displacement is zero. Q1.7 . Reason: Both speed and velocity are ratios with a time interval in the denominator, but speed is a scalar because it is the ratio of the scalar distance over the time interval while velocity is a vector because it is the ratio of the vector displacement over the time interval. Speed and velocity have the same SI units, but one must specify the direction when giving a velocity. An example of speed would be that your hair grows (the end of a strand of hair moves relative to your scalp) at a speed of about 0.75 in/month. An example of velocity (where direction matters) would be when you spring off a diving board. Your velocity could initially be 2.0 m/s up, while later it could be 2.0 m/s down.

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