Physics 107 HOMEWORK ASSIGNMENT #14


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1 Physics 107 HOMEWORK ASSIGNMENT #14 Cutnell & Johnson, 7 th edition Chapte 17: Poblem 44, 60 Chapte 18: Poblems 14, 18, 8 **44 A tube, open at only one end, is cut into two shote (nonequal) lengths. The piece that is open at both ends has a undamental equency o 45 Hz, while the piece open only at one end has a undamental equency o 675 Hz. What is the undamental equency o the oiginal tube? 60 Concept Questions A coppe block is suspended in ai om a wie in Pat 1 o the dawing. A containe o mecuy is then aised up aound the block as in Pat. (a) The undamental equency o the wie is given by Equation 17.3 with n 1: 1 v/(l). How is the speed v at which individual waves tavel on the wie elated to the tension in the wie? (b) Is the tension in the wie in Pat less than, geate than, o equal to the tension in Pat 1? (c) Is the undamental equency o the wie in Pat less than, geate than, o equal to the undamental equency in Pat 1? Justiy each o you answes. 14 Two tiny conducting sphees ae identical and cay chages o 0 µc and +50 µc. They ae sepaated by a distance o.50 cm. (a) What is the magnitude o the oce that each sphee expeiences, and is the oce attactive o epulsive? (b) The sphees ae bought into contact and then sepaated to a distance o.50 cm. Detemine the magnitude o the oce that each sphee now expeiences, and state whethe the oce is attactive o epulsive. *18 The dawing shows an equilateal tiangle, each side o which has a length o.00 cm. Point chages ae ixed to each cone, as shown. The 4.00 µc chage expeiences a net oce due to the chages q A and q B. This net oce points vetically downwad and has a magnitude o 405 N. Detemine the magnitudes and algebaic signs o the chages q A and q B. 8 Fou point chages have the same magnitude o.4 x 101 C and ae ixed to the cones o a squae that is 4.0 cm on a side. Thee o the chages ae positive and one is negative. Detemine the magnitude o the net electic ield that exists at the cente o the squae.
2 44. REASONING AND SOLUTION The oiginal tube has a undamental given by v/(4l), so that its length is L v/(4 ). The cut tube that has one end closed has a length o L c v/(4 c ), while the cut tube that has both ends open has a length L o v/( o ). We know that L L c + L o. Substituting the expessions o the lengths and solving o gives o c ( 45 Hz)( 675 Hz) 16 Hz Hz + 45 Hz c o 60. CONCEPT QUESTIONS a. The speed v at which individual waves tavel on the wie F elated to the tension F accoding to Equation 16.: v, whee m/l is the mass pe m / L unit length o the wie. b. The tension in the wie in Pat less than the tension in Pat 1. The eason is elated to Achimedes pinciple (Equation 11.6). This pinciple indicates that when an object is immesed in a luid, the luid exets an upwad buoyant oce on the object. In Pat the upwad buoyant oce om the mecuy suppots pat o the block s weight, thus educing the amount o the weight that the wie must suppot. As a esult, the tension in the wie is less than in Pat 1. c. Since the tension F is less in Pat, the speed v is also less. The undamental equency o the wie is given by Equation 17.3 with n 1: 1 v/(l). Since v is less, the undamental equency o the wie is less in Pat than in Pat 1. SOLUTION Using Equations 17.3 and 16., we can obtain the undamental equency o the wie as ollows: v 1 F 1 (1) L L m / L In Pat 1 o the dawing, the tension F balances the weight o the block, keeping it om alling. The weight o the block is its mass times the acceleation due to gavity. The mass, accoding to Equation 11.1 is the density ρ coppe times the volume V o the block. Thus, the tension in Pat 1 is Pat 1 tension F ( mass) g ρcoppevg In Pat o the dawing, the tension is educed om this amount by the amount o the upwad buoyant oce. Accoding to Achimedes pinciple, the buoyant oce is the weight o the liquid mecuy displaced by the block. Since hal o the block s volume is immesed, the volume o mecuy displaced is V/. The weight o this mecuy is the mass times the acceleation due to gavity. Once again, accoding to Equation 11.1, the mass is the density ρ mecuy times the volume, which is V/. Thus, the tension in Pat is Pat tension F ρcoppevg ρmecuy V / g
3 With these two values o the tension we can apply Equation (1) to both pats o the dawing and obtain Pat 1 Pat ρcoppevg L m / L ( / ) 1 ρcoppevg ρmecuy V g L m / L Dividing the Pat by the Pat 1 esult, gives ( / ) 1 ρ Vg ρ V g L m / L 1 ρ Vg L m / L coppe mecuy ρ 1 1, Pat coppe 1, Pat 1 coppe coppe ρ ρ mecuy kg/m kg/m kg/m As expected, the undamental equency is less in Pat than Pat REASONING a. The magnitude o the electostatic oce that acts on each sphee is given by Coulomb s law as F k q 1 q /, whee q 1 and q ae the magnitudes o the chages, and is the distance between the centes o the sphees. b. When the sphees ae bought into contact, the net chage ate contact and sepaation must be equal to the net chage beoe contact. Since the sphees ae identical, the chage on each ate being sepaated is onehal the net chage. Coulomb s law can be applied again to detemine the magnitude o the electostatic oce that each sphee expeiences. SOLUTION a. The magnitude o the oce that each sphee expeiences is given by Coulomb s law as: ( N m /C )( C)( C) ( m) k q1 q 4 F N Because the chages have opposite signs, the oce is attactive.
4 b. The net chage on the sphees is 0.0 µ C µ C µ C. When the sphees ae bought into contact, the net chage ate contact and sepaation must be equal to the net chage beoe contact, o µ C. Since the sphees ae identical, the chage on each ate being sepaated is onehal the net chage, so q q µ C. The electostatic oce 1 that acts on each sphee is now ( N m /C )( C)( C) ( m) k q1 q 3 F N Since the chages now have the same signs, the oce is epulsive. 18. REASONING The unknown chages can be detemined using Coulomb s law to expess the electostatic oce that each unknown chage exets on the 4.00 µc chage. In applying this law, we will use the act that the net oce points downwad in the dawing. This tells us that the unknown chages ae both negative and have the same magnitude, as can be undestood with the help o the eebody diagam o the 4.00 µc chage that is shown at the ight. The diagam shows the attactive oce F om each negative chage diected along the lines between the chages. Only when each oce has the same magnitude (which is the case when both unknown chages have the same magnitude) will the esultant oce point vetically downwad. This occus because the hoizontal components o the oces cancel, one pointing to the ight and the othe to the let (see the diagam). obseved downwad net oce. The vetical components einoce to give the SOLUTION Since we know om the REASONING that the unknown chages have the same magnitude, we can wite Coulomb s law as ollows: ( C) q ( 6 A C) F k k The magnitude o the net oce acting on the 4.00 µc chage, then, is the sum o the magnitudes o the two vetical components F cos 30.0º shown in the eebody diagam: q A µc 30.0º F cos 30.0º q B F F sin 30.0º q B
5 6 6 ( C) qa ( C) qb Σ F k cos k cos ( C) qa k cos 30.0 Solving o the magnitude o the chage gives q ( Σ ) F k C cos 30.0 A 6 ( 405 N)( m) C N m / C C cos Thus, we have 6 qa qb C. 8. REASONING Each chage ceates an electic ield at the cente o the squae, and the ou ields must be added as vectos to obtain the net ield. Since the chages all have the same magnitude and since each cone is equidistant om the cente o the squae, the magnitudes k q o the ou individual ields ae identical. Each is given by Equation 18.3 as E. The diections o the vaious contibutions ae not the same, howeve. The ield ceated by a positive chage points away om the chage, while the ield ceated by a negative chage points towad the chage. SOLUTION The dawing at the ight shows each o the ield contibutions at the cente o the squae (see black dot). Each is diected along a diagonal o the squae. Note that E D and E B point in opposite diections and, theeoe, cancel, since they have the same magnitude. In contast E A and E C point in the same diection towad cone A and, theeoe, combine to give a net ield that is twice the magnitude o E A o E C. In othe wods, the net ield at the cente o the squae is given by the ollowing vecto equation: B A + E A E D E C E B + + C D Σ E EA + EB + EC + ED EA + EB + EC EB EA + EC E A
6 Using Equation 18.3, we ind that the magnitude o the net ield is Σ E EA k q In this esult is the distance om a cone to the cente o the squae, which is one hal o the diagonal distance d. Using L o the length o a side o the squae and taking advantage o the Pythagoean theoem, we have magnitude o the net ield becomes 1 ( L + L ) 1 1 d L + L. With this substitution o, the 9 1 ( )( ) k q 4k q N m / C.4 10 C Σ E 54 N/C L ( m)
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