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Winter 2017 Section A/B/C PHYSICS 115 FINAL EXAM PRACTICE - SOLUTION Last Name (Print): ______________________ Seat No_________ First Name (Print): _____________________ I certify that the work I shall submit is my own creation, not copied from any source, and that I shall abide by the examination procedures outline below. Signature: ______________________________ Student ID Number: _________________________ READ THIS ENTIRE PAGE NOW Do not open the exam until told to do so. You will have 120 minutes to complete the examination. NO CELL PHONES, TEXT MSG, etc. ALLOWED AT ANY TIME. • • • • • • • • • • Before the exam begins: Print and sign your name, and write your student ID number in the spaces above. Write your name, student ID number, and exam version on your bubble sheet and fill in the corresponding “bubbles” using dark pencil marks. During the exam When the exam begins, print your name and student ID number on the top of each page. Do this first when you are told to open your exam. If you are confused about a question, raise your hand and ask for an explanation. If you cannot do one part of a problem, move on to the next part. This is a closed book examination. All equations and constants are provided. You may use a calculator, but not a computer, or other internet connected devices (smart-phones, iPads, etc.). For multiple-choice questions: Clearly circle your answer choice. Make no stray marks. If you must erase, erase completely. Also circle your choices directly on the exam paper for later reference. End of exam: Out of respect to other students, please remain seated for the last 20 minutes of the exam. At the end of the exam, please remain seated until all exams have been collected. Name ____________________________________ Last First Student ID: _______________ Score: _______ 1. [6 pts] A research submarine has a 20-cm-diameter window 8.0 cm thick. The manufacturer says the window can withstand net forces up to 1.0 x 106 N. What is the submarine’s maximum safe depth in seawater? The pressure inside the submarine is maintained at 1.0 atm. (𝜌#$%&%'$( = 1030kg/m2 ) A) 1.3 km B) 1.7 km C) 2.3 N D) 2.8 km E) 3.2 km 𝐹5$' = 𝑃5$' 𝐴 = 𝑃85#89$ − 𝑃;<'#89$ 𝐴 = 𝑃%'= − 𝑃%'= + 𝜌𝑔𝑑 𝐴 𝐹5$' = 𝜌𝑔𝑑𝐴 𝑑= 𝐹5$' 4𝐹5$' = = 3150m ≈ 3.2km 𝜌𝑔𝐴 𝜌𝑔𝜋𝑑 C 2. [6 pts] Two cylindrical containers, 1 and 2, contain different unknown liquids. The heights of the liquids in containers 1 and 2 are equal. The masses of the liquids in containers 1 and 2 are also equal. (Note that the diameter of container 2 is greater than that of container 1.) Is the pressure at point A greater than, less than, or equal to that at point B? A B 1 2 A) Greater than B) Less than C) Equal to D) Not enough information to answer. The masses of the liquids are the same but the volume of container 2 is larger, so the density of the liquid in container 2 must be smaller than the density of the liquid in container 1. Points A and B are at the same depth and the surfaces of the two liquids are at atmospheric pressure. The increase in pressure as depth increases is proportional to the density of the liquid, so the pressure at A must be larger than the pressure at B since liquid 1 is more dense. 3. [6 pts] Block A of mass m and volume V floats in a fluid of density 𝜌 as shown. Half of block A is submerged. Block B of mass 2m and 3V is placed in a container with the same fluid density. Which of the blocks, 1-5 best represents the location of block B in its container? 𝐹H = 𝑊 A) 1 B) 2 C) 3 D) 4 E) 5 ObjectA 𝐹H = 𝑊 𝜌J 𝑔𝑉#<L = 𝜌M 𝑔𝑉M 𝑉M 𝑉#<L = 2 𝑉M 𝜌J 𝑔 = 𝜌M 𝑔𝑉M 2 𝜌J = 2𝜌M Physics 115 – Winter 2017 ObjectB 𝐹H = 𝑊 𝜌J 𝑔𝑉#<L = 𝜌H 𝑔𝑉H 𝜌J = 2𝜌M 2𝜌M 𝑔𝑉#<L 𝑉#<L = C 𝜌H = 𝜌M 2 2 = 𝜌M 𝑔𝑉H 3 1 𝑉 3 H Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 4. [7 pts] The 3.0-cm diameter water line shown at right splits into two 1.0-cm diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa. What is the gauge pressure at point B? A) 12 kPa B) 22 kPa C) 29 kPa D) 38 kPa E) 42 kPa 𝐴M 𝑣M = 2 𝐴H 𝑣H 𝑣H = 𝐴M 𝑣M 2𝐴H 1 1 𝑃M + 𝜌𝑣MC = 𝑃H + 𝜌𝑣HC 2 2 1 1 𝐴MC 𝑣MC 𝑃M + 𝜌𝑣MC = 𝑃H + 𝜌 2 2 4𝐴CH 1 1 𝐴MC 𝑣MC 1 C 𝐴MC 𝑃H = 𝑃M + 𝜌𝑣MC − 𝜌 = 𝑃 + 𝜌𝑣 1 − = 11,500Pa ≈ 12kPa M 2 2 4𝐴CH 2 M 4𝐴CH 5. [6 pts] The viscosity of blood at normal hematocrit is about 4.75 mPa-s, compared to the viscosity of water, which is 1 mPa-s. In an adult human the coronary arteries have a diameter of about 4 mm. Suppose you are able to produce a flow rate Q of water in a tube of a given diameter. Which of the following could you do to produce that same flow rate if you were to use blood instead? A) Increase the diameter of the tube to 19 mm but keep the pressure difference the same. B) Increase the diameter of the tube to 5.9 mm but keep the pressure difference the same. C) Decrease the pressure difference by a factor of 4.75 but keep the diameter the same. D) Increase the pressure difference by a factor of 22.5 but keep the diameter the same. E) I don’t need to change anything; the flow rates will be the same. DpA2 Volume flow rate 𝑄 = Q = , so if hblood = 4.75hwater then DpA2 = 4.75 DpA2 S blood water 8phl to keep the same flow rate. Since A2 µ d 4 , to increase A2 by a factor of 4.75 we must increase the ST ( ) ( diameter of the tube by a factor of (4.75)1/4 = 1.48, to 1.48 x 4 mm = 5.9 mm. Physics 115 – Winter 2017 Practice Final Exam ) Name ____________________________________ Last First Student ID: _______________ Score: _______ 6. [6 pts] At 18.75°C a brass sleeve has a diameter of 2.21988 cm and a steel shaft has a diameter of 2.22258 cm. To what temperature must the sleeve be heated in order for it to slip over the shaft? 𝛼L(%## = 19×10XY K X[ , 𝛼#'$$\ = 12×10XY K X[ A) 64°C B) 83°C C) 140°C D) 101°C E) 120°C ∆𝑙 = 𝛼𝑙_ 𝑇J − 𝑇8 𝑇J = ∆𝑙 + 𝑇8 = 83℃ 𝛼𝑙_ 7. [7 pts] A laboratory technician drops an 85.0 g solid sample of unknown material at a temperature of 100.0°C into a well-insulated copper can. The copper can has a mass of 0.150 kg and contains 0.200 kg of water, and both the can and the water are initially at 19.0°C. The final temperature of the system is measured to be 26.1°C. Compute the specific heat capacity of the sample. (ccopper = 390 J/kg-K, cwater = 4190 J/kg-K) A) 2.51×103 J/kg-K B) 1.01×103 J/kg-K C) 0.912×103 J/kg-K D) 1.62×103 J/kg-K E) 2.11×103 J/kg-K Qnet = 0 Qsample + Qwater + Qcopper = 0 ms cs ΔTs + mw cw ΔTw + mc cc ΔTc = 0 cs = −mw cw ΔTw − mc cc ΔTc −(0.2kg)(4190J/kg-K)(7.1K) − (0.15kg)(390J/kg-K)(7.1K) = ms ΔTs (0.085kg)(−73.9K) cs = 1.01×103 J/kg-K 8. [6 pts] A window has dimensions of 1.40 m x 2.50 m and is made of glass with a thickness d. On a winter day, the outside temperature is -20.0°C, while the inside temperature is a comfortable 19.56°C. The rate of heat loss by conduction is known to be 2.13 x 104 W. Determine the thickness of the window. (kglass = 0.8 W/m-K) A) 3.49 mm 𝑄 = 𝑘𝐴 B) 4.31 mm C) 5.20 mm D) 6.71 mm E) 7.48 mm ∆𝑇 ∆𝑙 ∆𝑙 = 𝑘𝐴 ∆𝑇 = 5.20mm 𝑄 Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ The pV diagram at right illustrates two different processes in which an ideal gas changes from state 1 to state 2. 9. [6 pts] Determine the ratio of the work done on the gas in process A (1-A-2) to the work done on the gas in process B (1-B-2)? Explain. A) 6 B) 4 C) 3 D) 2 E) 1 The work done on the gas is equal to the negative of the area under the P-V curve. Process A – area = (3P)(3V-V) = 6PV Process B – area = (P)(3V-V) = 2PV 𝑊M −6𝑃𝑉 = =3 𝑊H −2𝑃𝑉 10. [7 pts] Suppose that the energy of the gas in state 1 is U. In terms of U, how much heat is added in process B (1-B-2)? A) 6U B) 7.5U C) 9.3U D) 12U E) 14U 2 The first law of thermodynamics states: 𝑊 + 𝑄 = 𝛥𝑈 = 𝑁𝑘H 𝛥𝑇 C Each point on a P-V diagram corresponds to a particular temperature since T = PV/nR, and in these processes n is a constant. In process B, the gas goes from a temperature of PV/nR to 9PV/nR. Thus the absolute temperature increases by a factor of 9. The thermal energy is proportional to the change in temperature. Since the temperature increases by a factor of 9, the final internal energy must be 9 times larger than the initial thermal energy (9U). The change in thermal energy is therefore 8U. The initial thermal energy is equal to 3/2NkBTi, which is equivalent to 3/2PV and U as stated in the question. The work in process B is -2PV from question 9 above. The work is thus equal in magnitude to -1.33U. The heat transferred is equal to DU – W and equal to 8U –(-1.33U) = 9.3U. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 11. [6 pts] An ideal engine does 10 J of work and exhausts 15 J of waste heat. If the cold reservoir temperature is 20°C, what is the temperature in degrees of the hot reservoir? A) 100°C B) 180°C C) 220°C D) 350°C E) 490°C 𝑄h = 𝑊 + 𝑄i = 25J 𝑄h 𝑇h = 𝑄i 𝑇i 𝑇h = 𝑇i 𝑄h 25J = 293𝐾 = 488K = 215℃ 𝑄i 15J 12. [6 pts] How far above a proton would an electron have to be held to exert an electrostatic force that just equals the weight of the proton? (mp = 1.67 x 10-27 kg, me = 9.11 x 10-31 kg) A) 14 mm B) 21 mm C) 58 mm D) 0.12 m E) 8.4 m FCoulomb = Kqe q p r 2 = mp g ® r = Kqe q p mp g = [(9 x 109 N-m2/C2)(1.6 x 10-19 C)(1.6 x 10-19 C)/(1.67 x 10-27 kg)(9.8 m/s2)]1/2 = 0.12 m 13. [6 pts] In case C, two positive point charges +2Q are each a distance s away from a third positive point charge +q. In case D, four positive point charges +Q are each a distance s away from a fifth positive point charge +q. (The angle shown is the same in both cases.) Is the net electric force on the +q charge in case C greater than, less than, or equal to net electric force on the +q charge in case D? A) Greater than B) Less than C) Equal to D) Not enough information given to answer. The net force by the upper and lower +Q charges in case D will be half as large as the net force applied by the two charges in case C. However, the two inner charges in case D will exert a net force on the +q charge that is greater than half of net force in case C. Thus the combined force from all the charges in case D is greater than that in case C. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 14. [6 pts] Consider a system of three point charges on the x-axis. Charge 1 is at x = 0, charge 2 and charge 3 are located to the right of charge 1 at x = r, and at x = 2r, respectively. In addition, the charges have the following values q1 = -q, q2 = q3 = q. The electric field is zero at some point along the x-axis between x = r and x = 2r. Is the point of zero field: A) At x = 1.5r B) To the left of x = 1.5 r C) To the right of x = 1.5 r. D) Not possible to tell. At x = 1.5r the contributions to the electric field from charge 2 (to the right) and 3 (to the left) are equal and opposite and the contribution from charge 1 points to the left. Therefore this is not a valid answer. Moving to the left of x = 1.5r decreases the contribution from charge 3 and increases the contribution from charges 1 and 2. Since E µ 1/ r 2 , the contribution from charge 2 (which is to the right) changes more rapidly than the contributions from the other two charges, and the net change is to make the magnitude of the left-pointing field at x = 1.5r get smaller as you move to the left. To the right of 1.5r, the electric field from charge 3 is greater in magnitude than that from charge 2. The electric field from charge 1 also points to the left. To the right of 1.5r, the net field cannot be zero. 15. [6 pts] The charge q1 in the figure at right is positively charged. When a small test charge, +q0, travels from point A to point B, it is known that the net work done by the electric field on the charge is negative. Determine the sign of q2 and its relative magnitude in comparison to q 1. A) q2 is negative, 𝑞C < 𝑞[ B) q2 is negative, 𝑞C > 𝑞[ C) q2 is negative, 𝑞C = 𝑞[ D) q2 is positive, 𝑞C < 𝑞[ E) q2 is positive, 𝑞C > 𝑞[ The work done on a charge is given as: 𝑊$\$i = -𝛥𝑉𝑞 Since the charge moving through the potential is positive, and the work done is negative, the change in potential must be positive. For the potential at point B to be more positive than at point A, q2 must be positive and have a larger magnitude than q1. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 16. [6 pts] A point charge, +Q, is fixed in place as shown at right. Points a, c, and e are equidistant from the +Q charge, as are points b, d and f. shown in the diagram at right. A small negative test charge, -q0, travels along three different paths, 1 (a to b), 2 (d to a) and 3 (c to e to f). Rank the paths according to the work done by the electric field on the negative test charge. Rank positive work values as larger than negative work values. A) 3 > 2 > 1 B) 3 = 2 = 1 C) 2 > 1 = 3 D) 1 = 2 = 3 E) 3 > 1 = 2 In path 1, the force on the charge is to the left and the displacement is to the right. The work done by the field is negative. In path 2, a component of the force is in the same direction as the displacement of the charge. The work done by the field is positive. In path 3, no work is done when the charge moves from point c to point e, since these two points are at the same potential. In moving from point e to point f, the force opposes the displacement and the work is negative. Since the force and displacement have the same magnitude as that along path 1, the works for path 1 and 3 are both negative and have the same magnitude. 17. [7 pts] A parallel-plate capacitor is constructed with plates of a particular shape. The plates are separated by 0.25 mm, and the space between the plates is filled with a dielectric with dielectric constant of value 4.6. When the charge on the capacitor is 1.2 µC the potential difference between the plates is 750 V. Which of the following choices best describes the shape of the plates. A) Square plates with length 0.056 m. B) Rectangular plates with length 0.056 m and width 0.112 m. C) Circular plates with radius 0.056 m. D) None of the above. 𝐶= 𝜅𝜀_ 𝐴 𝑑 𝐴= 𝐶𝑑 𝑄𝑑 = = 9.83×10X2 mC 𝜅𝜀_ 𝑉𝜅𝜀_ Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 18. [6 pts] Consider the two cases at right. Is the magnitude of the flux through the Gaussian surface in case A greater than, less than, or equal to that through the surface in case B? A) Greater than B) Less than C) Equal to D) Not enough information given to answer. There is no change enclosed by the surface in case A, so the magnitude of the flux is zero. The magnitude of the flux in case B is non-zero since the surface encloses a +4Q charge. 19. [7 pts] Consider the circuit at right, where R1 = R, R2 = ½ R, and R3 = 2R. The battery has a voltage V. What is the voltage across the resistor, R1? A. 5V/7 B. 3V/5 D. 8V/11 E. 11V/13 C. 5V/6 R1 R2 R3 The resistance of the parallel combination is -1 -1 æ 1 1 ö 2R æ2 1 ö R23 = ç The voltage across R1 andt he parallel combination are + ÷ =ç + ÷ = R R R 2 R 5 è ø 3 ø è 2 V R R 5 proportional to their resistances: 1 = 1 = = . V1 + V23 = V, so V1 = 5V/7 V23 R23 2 R 2 5 and V23 = 2V/7. 20. [6 pts] Consider the circuit shown at right. The current flowing through the battery is 0.92 A. Determine the potential difference across the 12-Ω resistor. A. 15 V B. 4.9 V C. 3.2 V D. 4.3 V E. 3.7 V The potential drop across the 11 Ω resistor = (0.92 A)(11 Ω) = 10.1 V Thus the potential drop VBA = 15 V – 10.1 V = 4.9 V. The 6.2 Ω and 12 Ω resistors are in series, so V6.2 + V12 = 4.9 V. The potential drop across each resistor is proportional to its resistance: V6.2 6.2 = = 0.52 . Thus V12 = 4.9 - V6.2 = 4.9 - 0.52V12 ® V12 12 V12 = 3.2 V Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 21. [7 pts] A flash unit in a camera consists of an RC circuit. The capacitor has a capacitance of 1500 µF. What resistance is needed in this RC circuit if the flash is to charge to 90% of its full charge in 21 s? A. 4900 Ω B. 5400 Ω C. 5800 Ω D. 6100 Ω E. 6700 Ω As the flash charges up the voltage across the RC combination changes with time according to V = V0 (1 - e-t / RC ) . We want V/V0 = 0.9 when t = 21 s. 1- æ V ö V t = e-t / RC ® ln ç1 - ÷ = ®R=V0 RC è V0 ø t æ V ö C ln ç1 - ÷ è V0 ø = - (21 s)/(1500 x 10-6 F)ln(0.1) = 6100 Ω 22. [6 pts] The time constant for the RC circuit at right is 3.5 ms. Calculate the current in the circuit 7.0 ms after the switch is closed. Assume that the capacitor is uncharged initially and that the emf of the battery is 9.0 V. A) 11 mA I = I 0e −t / RC = B) 9.2 mA C) 8.4 mA 145 Ω 9.0 V + D) 6.8 mA E) 5.6 mA I 0 −t / RC e R τ = RC 9.0 V −0.007s/0.0035s I= e 145Ω I = 8.4 mA 23. [6 pts] In a mass spectrometer, charged particles pass through a velocity selector with electric and magnetic fields at right angles to each other, as shown. If the electric field has a magnitude of 450 N/C and the magnetic field has a magnitude of 0.18 T, what speed must the particles have to pass through the selector undeflected? A. 1500 m/s 𝑣= B. 1800 m/s C. 2000 m/s D. 2200 m/s E. 2500 m/s 𝐸 450N/C = = 2500m/s 𝐵 0.18T Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 24. [6 pts] An solenoid with 500 turns has a diameter of 10 cm, is 40 cm long, and carries a current of 30 A. What is the magnitude of the magnetic field at the center of this solenoid? A) 7.5 mT B) 19 mT C) 47 mT D) 190 mT B= µ0 NI L = (1.26 x 10 -6 T-m/A)(500)(30 A)/(0.4 m) = 4.7 x 10-2 T 25. [7 pts] Two long straight wires carry current in the positive xdirection. At an instant in time, three charges, q1, q2, and q3 are moving in the positive x-direction at the points shown. Rank the magnitude of the net magnetic force exerted on each charge due to the current-carrying wires. A) 𝐹[ > 𝐹C > 𝐹2 B) 𝐹C > 𝐹[ > 𝐹2 C) 𝐹C > 𝐹[ = 𝐹2 D) 𝐹[ > 𝐹2 > 𝐹C E) 𝐹C = 𝐹[ > 𝐹2 𝐵∝ 𝐼 𝑟 The fields from both wires point in the same direction between the wires (right hand rule). The field will be strongest at this location. The field at q1 will be greater than that at the location of q3 as the field is proportional to the current through the wire. The force exerted on the charges is proportional to the field strength at their location and all charges have the same speed. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ 26. [7 pts] Consider the set-up at right that was presented in the Magnetic Interactions tutorial. The wire is observed to move out of the page when the wire carries a current. Which of the following choices is consistent with this observation? Score: _______ X A) Pole X is a north pole and current travels right to left through the horizontal section of the wire. B) Pole X is a north pole and current travels left to right through the horizontal section of the wire. C) Pole X is a south pole and current travels left to right through the horizontal section of the wire. D) Descriptions A and C are both correct. E) Descriptions B and C are both correct. In choices A and C, the magnetic force on the horizontal segment is into the page. 27. [6 pts] A conducting rectangular loop of wire is near a long straight wire. The long straight wire carries a constant current I in the direction shown in the figure at right. If the loop moves with a speed v in the direction shown, what is the direction of the current in the loop? A) clockwise v I B) counterclockwise C) There is no current. D) There is not enough information to tell. At the loop the magnetic field points into the page, and its magnitude decreases with increasing distance from the wire. Thus the flux through the loop decreases, and a current is induced to keep the flux constant by producing a field pointing into the page. That requires a clockwise current. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 28. [6 pts] An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. The area enclosed by the loop is in the plane of the page, and the loop has a resistance of 0.62 Ω. Find the magnitude of the induced current and its direction, if the change in shape occurs in 4.25 s and the local 0.125-T magnetic field is perpendicular to the plane of the loop and points out of the page. A) 1.2 mA, counterclockwise B) 1.2 mA, clockwise C) 2.4 mA, counterclockwise D) 2.4 mA, clockwise E) 3.1 mA, clockwise Csquare = 4d = 1.22 m Since the area increases, the flux increases. The induced magnetic field opposes this change, and thus points into the page. This is consistent with a clockwise current in the loop. 1.22m = 0.305 m 4 Asquare = d 2 = (0.305 m)2 = 0.093 m 2 d= Ccircle = 2π r = 1.22 m 1.22m = 0.194 m 2π Acircle = π r 2 = π (0.194 m)2 = 0.118 m 2 " 0.118 m 2 − 0.093 m 2 % dΦ dA −4 EMF = =B = (0.125T ) $ ' = 7.4 ×10 V dt dt 4.25s # & r= I= EMF 7.4 ×10−4 V = = 1.2 mA R 0.62Ω vo 29. [6 pts] The observer on the right observes a counter-clockwise current in loop 2. Which of the following statements is consistent with this observation? X Eye A) The pole labeled with an X is the north pole of the magnet. B) The pole labeled with an X is the south pole of the magnet. C) The pole labeled X could be a north or south pole; it depends on the magnet’s velocity. D) There is not enough information to answer. The current in the loop is counter-clockwise from the perspective of the observer, which means the induced magnetic field points to the right. The flux through the loop is increasing, since the magnet is moving toward the loop. The induced magnetic field opposes the change in the flux, therefore the magnetic field of the magnet must point to the left. This means that the pole label with an X is a south pole. Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 30. [6 pts] A conducting stationary loop (copper loop) is placed in a uniform external magnetic field as shown. A clockwise current is induced in the loop. While the current is flowing through the loop, how is the external magnetic field changing? A) The magnetic field is increasing. B) The magnetic field is decreasing. C) The magnetic field remains constant. D) Not enough information to tell. A clockwise current around the loop produces a magnetic field that points into the page at the center of the loop. Lets call the flux positive if the field lines point through the loop into the page. This would make the external flux and the induced flux positive. The induced flux always opposes the change in external flux. Since the induced flux is positive, the change in external flux must be negative. This means that the field must be decreasing in strength, such that the change in external flux is negative. 31. The figure at right shows a zero-resistance rod sliding to the right on two zero-resistance rails separated by the distance L = 0.500 m. The rails are connected by a 10.0 ohm resistor, and the entire system is in a uniform magnetic field with a magnitude of 0.750 T. Find the speed at which the bar must be moved to produce a current of 0.175 A in the resistor? A) 4.67 m/s B) 3.75 m/s C) 3.13 m/s D) 2.29 m/s E) 1.58 m/s ℇ = 𝐵𝑣𝑙 𝑣= ℇ 𝐼𝑅 0.175A (10Ω) = = = 4.67m/s 𝐵𝑙 𝐵𝑙 (0.750T)(0.5m) Physics 115 – Winter 2017 Practice Final Exam Name ____________________________________ Last First Student ID: _______________ Score: _______ 32. [6 pts] An ac generator with a maximum voltage of 12.0 V and a frequency of 60.0 Hz is connected to a resistor with a resistance R =150 Ohm. Determine the average power dissipated in the resistor. A) 0.057 W B) 0.48 W C) 1.1 W D) 4.3 W E) 8.5 W 𝑃%•‚ 𝑉=%ƒ 2 = = 𝑅 𝑅 C 𝑉(=# Physics 115 – Winter 2017 C 12.0V 2 = 150Ω C = 0.48W Practice Final Exam