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# final

Course: PHY 303L, Spring 2007

School: University of Texas

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Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A static uniform magnetic field is directed into the page. A charged particle moves in the plane of the page following a counterclockwise spiral of...

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Matthew Daniels, Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire This print-out should have 46 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A static uniform magnetic field is directed into the page. A charged particle moves in the plane of the page following a counterclockwise spiral of decreasing radius as shown. B 1 We know that when a charged particle moves in a uniform magnetic field with a constant speed, it undergoes a circular motion with the centripetal force provided by the magnetic force, namely m v2 = qvB, r so we know that the radius is in fact proportional to the speed, r= m v. qB B Neglect the effect due to gravity. What is a reasonable explanation? 1. The charge is neutral and speeding up. 2. The charge is positive and with a constant speed. 3. None of these 4. The charge is neutral and with a constant speed. 5. The charge is negative and with a constant speed. 6. The charge is positive and slowing down. correct 7. The charge is negative and slowing down. 8. The charge is neutral and slowing down. 9. The charge is positive and speeding up. 10. The charge is negative and speeding up. Explanation: Since the particle follows a spiral of decreasing radius, we can judge that it is slowing down. The magnetic force F = q v B must be in the direction for the centripetal force -^ r (pointed inward) of this particle in counterclockwise circular motion. Since v B is in the negative ^ direction, the particle has a r positive charge. keywords: 002 (part 1 of 1) 10 points Which travels most slowly in glass? 1. Red light 2. Orange light 3. Yellow light 4. Violet light correct 5. Brown light 6. Green light 7. Blue light Explanation: c v Since violet light is refracted more than the light of other frequencies its index of refraction is the highest, making its speed the smallest. In other words, violet light travels slowest in glass. n= Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire keywords: 003 (part 1 of 1) 10 points When a potential difference of 65 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge density of 38 nC/cm2 . The permittivity of a vacuum is 8.85419 -12 2 10 C /N m2 . What is the spacing between the plates? Correct answer: 1.51453 m. Explanation: Let : = 38 nC/cm2 = 0.00038 C/m2 , V = 65 V , and -12 2 C /N m2 . 0 = 8.85419 10 104 mA? Correct answer: 1.58587 ms. Explanation: Let : R = 7.29 , L = 81 mH = 0.081 H , E = 5.7 V . 2 and R E L S b a The time constant of an RL circuit is = L 0.081 H = = 0.0111111 s . R 7.29 s= 0V (8.85419 10-12 C2 /N m2 ) (65 V) = (0.00038 C/m2 ) = 1.51453 10-6 m = 1.51453 m . The final current reached in the circuit is I0 = 5.7 V E = = 0.781893 A . R 7.29 The switch is in position a in an RL circuit connected to a battery at t = 0 when I = 0. Then the current vs time is I = I0 1 - e-t / . keywords: 004 (part 1 of 2) 10 points An inductor and a resistor are connected with a double pole switch to a battery as shown in the figure. The switch has been in position b for a long period of time. Solving the above expression for t, when I = I1 gives t1 = - ln 1 - I1 I0 0.104 A 0.781893 A = -(0.0111111 s) ln 1 - = 1.58587 ms . 7.29 81 mH S b a 005 (part 2 of 2) 10 points What is the maximum current in the inductor a long time after the switch is in position a? Correct answer: 0.781893 A. Explanation: After a long time compared to , we have a d.c. circuit with a battery supplying an emf E, 5.7 V If the switch is thrown from position b to position a (connecting the battery), how much time elapses before the current reaches Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire which is equal to the voltage drop I R across the resistor. Thus I= E 5.7 V = = 0.781893 A . R 7.29 3 The Coulomb constant is 8.98755 109 N m2 /C2 . Find the electrical potential at point p. Correct answer: 81570.9 V. Explanation: Let : = 1.7 C/m = 1.7 10-6 C/m , R = 1.3 m , 2 R = 2.6 m , and ke = 8.98755 109 N m2 /C2 . keywords: 006 (part 1 of 1) 10 points A concave mirror has a focal length of 30.3 cm. Determine the object position for which the resulting image is upright and five times the size of the subject. Correct answer: 24.24 cm. Explanation: Given : f = 30.3 cm M = 5. and R 2R p 2R Let p be the origin. Consider the potential due to the line of charge to the right of p. Vright = = ke = ke R The magnified, virtual images formed by a concave mirror are upright, so M > 0. Thus q M =- p q = -M p The mirror equation gives 1 1 1 1 1 M -1 = + = - = f p q p Mp Mp p= (M - 1) f M (5 - 1) (30.3 cm) = 5 = 24.24 cm . dV dq r 3R dx x 3R = ke ln x R = ke ln 3 . By symmetry, the contribution from the line of charge to the left of p is the same. The contribution from the semicircle is Rd Vsemi = ke R 0 = ke 0 0 d = ke keywords: 007 (part 1 of 1) 10 points A wire that has a uniform linear charge density of 1.7 C/m is bent into the shape as shown below, with radius 1.3 m. 1.3 m 2.6 m p 2.6 m = ke . Hence the electric potential at p is Vp = Vright + Vlef t + Vsemi = 2 ke ln 3 + ke = ke (2 ln 3 + ) = (8.98755 109 N m2 /C2 ) (1.7 10-6 C/m) (2 ln 3 + ) = 81570.9 V . Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire keywords: 008 (part 1 of 1) 10 points An infinite line charge is located along the z-axis. A particle of mass m that carries a charge q whose sign is opposite to that of is in a circular orbit in the xy plane about the line charge. Obtain an expression for the period T of the orbit in terms of m, q, R, and , where R is the radius of the orbit. 1. T = 2. T = 3. T = 4. T = 5. T = 6. T = 7. T = 8. T = 9. T = 10. T = (2 )3 m (2 )3 m 0 4 Because of symmetry, the electric field produced by the line charge is perpendicular to the line. We can choose a closed cylinderical surface and apply Gauss' Law to obtain the electric field: En dA = S Qnet 0 E 2RL = L 0 R2 R4 | q | 0 correct E= 2 0R . 2 | q | 2 m 0 R2 | q | 2m 0R | q | (2 )3 m m 0 R2 | q | (2 )3 m 0 R2 2 | q | (2 )3 m 0 R4 | q | 2m 0R 4 0 Applying uniform circular motion, 2 T 2 T 2 | qE | = m R4 | q | | q | =m 2 0R T = R 2 R (2 )3 m 0 R2 | q | 2m o . | q | = 2R | q | keywords: m 0R | q | z m,q r y x Explanation: 009 (part 1 of 1) 10 points A solenoid with circular cross section produces a steadily increasing magnetic flux through its cross section. There is an octagonally shaped circuit surrounding the solenoid as shown below. The increasing magnetic flux gives rise to a counterclockwise induced emf E per loop. Initial Case: The circuit consists of two identical light bulbs of equal resistance, R, connected in series, leading to a loop equation 2 E - 2 i R = 0. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire Figure 1: 5. 5 B B X B B Y i The corresponding electrical power consumed by bulb X and bulb Y are PX and PY , respectively. Primed Case: Each bulb is on a separated loop with the two loops connected at one point, (see the figure below). Figure 2: PX PY 1 = =2 and PX 2 PY P PY 6. X = 8 and =8 PX PY P PY 7. X = 4 and =4 PX PY P PY 1 1 8. X = and = PX 4 PY 4 P PY 1 1 = 9. X = and PX 8 PY 8 PY P =0 10. X = 0 and PX PY Explanation: Basic Concepts: Induced emf. Solution: Let E and R be the induced emf and resistance of the light bulbs, respectively. For the first case, since the two bulbs are in series, the equivalent resistance is simply Req = R + R = 2 R and the current through the bulbs is i= 2E E = . 2R R B B X B B Y Hence, for the first case, the power consumed by bulb X is PX = E R E2 R 2 R i The corresponding electrical power consumed by bulb X and bulb Y are PX and PY , respectively. P P The ratios X and Y are respectively PX PY given by PX PX P 2. X PX P 3. X PX P 4. X PX 1. = 1 2 and PY PY PY PY PY PY PY PY = 1 2 = For the second (primed) case, since there is only one bulb per loop, current through bulb X is now E i = , R and the power consumed by bulb X is PX = = Hence the ratio is PX PX E2 = R E2 R = 1. E R E2 . R 2 R = 1 and = 2 and = 2 and = 1 correct = 1 2 =2 Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire For the second (primed) case, bulb Y has the same conditions as bulb X; i.e., PY PY E2 = R E2 R 6 Hint: Use a small angle approximation; e.g., sin = tan . S1 a 10 S2 keywords: 010 (part 1 of 1) 10 points Consider the diagram a 2.74 c 3.97 A E L Determine the height y3 , where the third minimum occurs. 1. y3 = 2. y3 = 3. y3 = 5 L a 7 L 2a L a 3 L 2a 9 L 2a 2 L a 4 L a 5 L 2a 3 L correct a L 2a 1.67 b Find the potential difference Va - Vb . Correct answer: 17.5077 V. Explanation: R1 R2 E 4. y3 = 5. y3 = 6. y3 = 7. y3 = I Let : I = 3.97 A R1 = 2.74 , R2 = 1.67 . and 8. y3 = 9. y3 = 10. y3 = Currents through resistor elements in a series combination are the same. We can find the potential difference by finding the equivalent resistance R = R1 + R2 and multiplying the equivalent resistance by the current: V = (R1 + R2 ) I = (2.74 + 1.67 ) (3.97 A) = 17.5077 V . Explanation: The third minimum occurs at = 6 , which corresponds to a path difference between two end rays: b3 = = k keywords: 011 (part 1 of 1) 10 points Consider the setup of a single slit experiment. 6 2 = 3 b3 = a viewing screen y3 = 1. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire y3 L b3 y3 = L a 3 = L. a = keywords: 012 (part 1 of 3) 10 points Consider two concentric spherical conducting shells. The inner shell has radius a and charge q1 on it, while the outer shell has radius 3 a and charge q2 on it. 7 Set up a Gaussian surface of radius r = 2 a between the shells: q1 q2 O a 3a p Due to symmetry, the electric field is constant over the surface of the sphere, so the flux is simply = E A, and the enclosed charge is q1 . From Gauss's Law, = E 4 r2 = 4 E (2 a)2 = E= q1 0 r q1 q2 O a 3a p q1 0 Assume the electric potential V at is zero. Determine the electric field E at p, where the distance Op = 2 a. q1 + q 2 1. E = 16 0 a2 q1 2. E = 18 0 a2 q1 3. E = 4 0 a2 q1 4. E = 8 0 a2 q1 5. E = 12 0 a2 q1 6. E = 6 0 a2 q1 7. E = correct 16 0 a2 q1 8. E = 14 0 a2 q2 9. E = 4 0 a2 q1 10. E = 20 0 a2 Explanation: q1 q1 = . 2 4 0 (2 a) 16 0 a2 013 (part 2 of 3) 10 points Find the electric potential V at point p. 1 4 1 2. V = 4 1 3. V = 4 1. V = 4. V = 0 5. V = 1 4 1 6. V = 4 q1 + q 2 3a q2 q1 + 3a 2a q1 + q 2 a q2 q1 + a 3a q1 q2 + correct 2a 3a 0 0 0 0 0 7. V = 8. V = 1 4 q2 q1 + 3a a 0 Explanation: The potential due to a spherical charge distribution is the same as the potential due to a point charge at its center, if V at is zero Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire (otherwise we would add a constant). Due to q1 , k q1 1 q1 V1 = = . r 4 0 2a The outer sphere requires more thought. Just outside the surface, the potential due to the outer shell is like a point charge at the center; i.e., at a distance 3 a. As we go inside the shell, Gauss's Law tells us that since we enclose no charge (we are only considering the outer shell right now), there is no electric field due to the outer shell inside the outer shell. If there is no electric field, the potential stays the same as we go inwards. Therefore the potential due to the outer shell at any point inside the outer shell is V2 = and V = V 1 + V2 = 1 4 0 8 same as that at the surface of the inner shell, so k q1 1 q1 V1 = = . a 4 0 a The potential V2 is the same everywhere inside the outer shell, so the total potential at the center is V = V 1 + V2 = 1 4 0 q2 q1 + a 3a . keywords: 015 (part 1 of 3) 10 points Consider the application of two potential differences across the same metallic ohmic conductor where the conductor is maintained at a constant temperature of 300 K. k q2 1 = 3a 4 0 q2 3a q1 q2 + 2a 3a J1 V1 . J2 V2 014 (part 3 of 3) 10 points Find the electric potential V at the point O. 1. V = 1 4 q1 + q 2 a L L 0 2. V = 3. V = 0 1 q1 + q 2 4 0 3a q1 1 q2 5. V = + 4 0 3a a 1 q2 q1 6. V = + 4 0 3a 2a q1 1 q2 7. V = + correct 4 0 a 3a q1 q2 1 + 8. V = 4 0 2a 3a Explanation: Now we are inside both shells, so a reasoning similar to the one for V2 above leads us to conclude that the contribution to the potential at the center from the inner shell is the 4. V = R R Case 1: The potential difference is V1 , the current density is J1 , and the average thermal speed of the free electron within the conductor is v1 . Case 2: The corresponding quantities are V2 , J2 , and v2 . J2 If V2 = 2 V1 , the corresponding ratio of J1 current densities is given by 1. 2. 3. 4. 5. 6. J2 J1 J2 J1 J2 J1 J2 J1 J2 J1 J2 J1 = 1. = 1 . 8 = 16 . = 8. = 4. = 2 . correct Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire J2 = 0. 7. J1 J2 1 8. . = J1 16 J2 1 9. = . J1 2 J2 1 10. = . J1 4 Explanation: Using the relation between current density for an ohmic material E = J, we find E2 J2 = E1 J1 E2 = E1 V2 = L V1 L V2 = = 2 . V1 016 (part 2 of 3) 10 points v2 is approximately The ratio v1 v2 1 1. . = v1 16 v2 1 = . 2. v1 4 v2 = 1 . correct 3. v1 v2 1 4. = . v1 8 v2 = 0. 5. v1 v2 = 2. 6. v1 v2 1 7. = . v1 2 v2 8. = 4. v1 v2 9. = 16 . v1 v2 = 8. 10. v1 9 Explanation: When two different voltages are applied across the conductor, there will be two different drift velocities. However, we recall that drift velocity is negligible compared with the average thermal velocity of free electrons. We denote the average velocity of free electrons in the absence of applied voltage to be vabsent . Thus, the ratio of the magnitude of the average velocities (the vector sum of vabsent and average drift velocities) is |vabsent + vdrif t2 | v2 = v1 |vabsent + vdrif t1 | vabsent vabsent 1 . 017 (part 3 of 3) 10 points Consider the following effects due to the increase of the temperature. A. Variations on the average collision time: (i) the average collision time increases (ii) the average collision time remains the same (iii) the average collision time decreases B. Variations on the resistivity: (a) the resistivity increases (b) the resistivity remains the same (c) the resistivity decreases Choose the correct pair of statements which represent the variations as the temperature increases. 1. (i) and (b) 2. (ii) and (c) 3. (ii) and (b) 4. (iii) and (a) correct 5. None of them. 6. (i) and (a) 7. (ii) and (a) Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire keywords: 8. (iii) and (b) 9. (i) and (c) 10. (iii) and (c) Explanation: As temperature increases, thermal motion of particles in the conductor increases, so the collision time decreases (i.e., the electrons 1 collide more frequently). Since , we find that increases. keywords: 018 (part 1 of 1) 10 points A converging lens with a diameter of 40.6 cm forms an image of a satellite passing overhead. The satellite has two green lights (wavelength 497 nm) spaced 1.1 m apart. If the lights can just be resolved according to the Rayleigh criterion, what is the altitude of the satellite? Correct answer: 736.55 km. Explanation: Given : d = 1.1 m , D = 40.6 cm = 0.406 m , = 497 nm = 4.97 10 The angular resolution is = m = 1.22 D 4.97 10-7 m = 1.22 0.406 m = 1.49345 10-6 rad . -7 10 019 (part 1 of 1) 10 points Four resistors are connected as shown in the figure. c a 66 25 b 47 81 96 V d S1 Find the resistance between points a and b. Correct answer: 16.3411 . Explanation: c R1 a R3 b R2 and m. EB R4 d S1 Thus the altitude is h= = d 1 km 1.1 m -6 rad 1 103 m 1.49345 10 Let : R1 R2 R3 R4 E = 25 , = 47 , = 66 , = 81 , = 96 V . and = 736.55 km . Ohm's law is V = I R . A good rule of thumb is to eliminate junctions connected by zero resistance. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire R1 0.61 a d R4 EB The series connection of R2 and R3 gives the equivalent resistance R23 = R2 + R3 = 47 + 66 = 113 . The total resistance Rab between a and b can be obtained by calculating the resistance in the parallel combination of the resistors R1 , R4 , and R23 ; i.e., 1 1 1 1 + + = Rab R1 R2 + R 3 R4 R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 ) = R1 R4 (R2 + R3 ) R1 R4 (R2 + R3 ) Rab = R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 ) The denominator is R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 ) = (81 )[47 + 66 ] + (25 ) (81 ) + (25 ) [47 + 66 ] = 14003 2 , so the equivalent resistance is Rab = (25 ) (81 ) [47 + 66 ] (14003 2 ) E R2 R3 8.48 1.37 4.46 24.5 11 b c 13.36 V 3.56 A Correct answer: 38.3338 V. Explanation: r2 r3 r1 r4 E I E1 A I1 r5 I2 Let : r1 r2 r3 r4 r5 I E1 = 0.61 , = 8.48 , = 4.46 , = 1.37 , = 24.5 , = 3.56 A , and = 13.36 V . Applying Kirchhoff's junction rule at point A, I1 + I 2 = I . Applying Kirchhoff's loop rule counterclockwise around the right loop, E1 + r5 I2 - (r3 + r4 ) I1 = 0 E1 - r5 (I - I1 ) - (r3 + r4 ) I1 = 0 E1 + r 5 I I1 = r3 + r 4 + r 5 13.36 V + (24.5 ) (3.56 A) = 4.46 + 1.37 + 24.5 = 3.31619 A . = 16.3411 . keywords: 020 (part 1 of 1) 10 points What is the emf E of the battery at the lower left in the figure? Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire Applying Kirchhoff's loop rule to the left loop, E = (r1 + r2 ) I + (r3 + r4 ) I1 - E1 = (0.61 + 8.48 ) (3.56 A) + (4.46 + 1.37 ) (3.31619 A) - 13.36 V = 38.3338 V . Let : C = 42 F = 4.2 10-5 F , E = 20 V , R = 4 k = 4000 , and I = 1 mA = 0.001 A . 12 Applying Kirchhoff's law keywords: 021 (part 1 of 1) 10 points A ray of light has an angle of incidence of 30 on a block of quartz and an angle of refraction of 20.7 . What is the index of refraction for this block of quartz? Correct answer: 1.41453 . Explanation: By Snell's Law, nair sin i = n sin r Thus n= since nair = 1. keywords: 022 (part 1 of 1) 10 points At t=0 the switch S is closed with the capacitor is uncharged. 42 F 4 k 20 V sin i sin r so Q = C (E - I R) = (4.2 10-5 F) [20 V - (0.001 A) (4000 )] = 0.000672 C . E- Q - I R = 0, C keywords: 023 (part 1 of 1) 10 points A proton moves at a speed of 0.53 107 m/s at right angles to a magnetic field with a magnitude of 0.37 T. Find the magnitude of the acceleration of the proton. Correct answer: 1.87543 1014 m/s2 . Explanation: Let : v = 0.53 107 m/s , B = 0.37 T , m = 1.673 10-27 kg , q = 1.60 10-19 C . and S i Fnet = Fmag ma = qvB qvB a= m (1.6 10-19 C) (5.3 106 m/s) = 1.673 10-27 kg (0.37 T) = 1.87543 1014 m/s2 . What is the charge on the capacitor when I = 1 mA? Correct answer: 0.000672 C. Explanation: C R E S i Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire keywords: 024 (part 1 of 2) 10 points A metal bar with length OA = L is rotating in a counter-clockwise manner about the point O with a constant angular velocity . There is a constant magnetic field B directed into the paper. Explanation: 13 A v +q FB B v B A O From the figure above we can see that the force on +q is directed radially inward . The sign of the charge can reverse the direction of the force FB . This may be obtained from the equation F = +q v B . Because of this magnetic force, the positive charges begin to accumulate at O (or normally, the negative charges begin to accumulate at A), producing an electric field that points from O to A. Hence, VO > VA . Second of eight versions. 025 (part 2 of 2) 10 points The length of the bar is 10 m and the magnitude of the magnetic field is 9 T . The speed of the bar at point C is 25 m/s , and the length of OC = 3 m . Determine the magnitude of the potential difference |VO - VA | . 1. |VO - VA | = 1500 V 2. |VO - VA | = 750 V 3. |VO - VA | = 7500 V 4. |VO - VA | = 15000 V 5. |VO - VA | = 135 V 4 405 6. |VO - VA | = V 2 7. |VO - VA | = 3750 V correct C +q L O B B The direction of the magnetic force on the positive charge +q at a fixed point C due to the magnetic field, and the relationship VO between and VA are respectively given by 1. out of the plane, VO = VA . 2. in the plane, VO = VA . 3. opposite to the direction of v, VO = VA . 4. radially outward, VA > VO . 5. radially inward, VO = VA . 6. radially outward, VO > VA . 7. radially inward, VO > VA . correct 8. radially outward, VO = VA . 9. radially inward, VA > VO . 10. in the direction of v, VO = VA . Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire 750 8. |VO - VA | = V 7 135 V 9. |VO - VA | = 2 10. |VO - VA | = 375 V Explanation: Let : OC = R = 3 m , L = m, v = 25 m/s , and B = 9 T. 14 Polarizers #1 and #3 are "crossed" such that their transmission axes are perpendicular to each other. Polarizer #2 is placed between the polarizers #1 and #3 with its transmission axis at 60 with respect to the transmission axis of the polarizer #1 (see the sketch). #1 60 #2 #3 Recall that the induced electric field at a point r is given by Eind = |v B| = v B = r B , where in this problem, the angular velocity is = v (25 m/s) 25 = = rad/s . R (3 m) 3 R After passing through polarizer #2 the intensity I2 (in terms of the intermediate intensity I1 ) is 1. None of these. 2. I2 = I1 3. I2 = 4. I2 = 5. I2 = 1 2 1 8 3 4 1 3 1 4 I1 I1 I1 I1 I1 correct Hence the magnitude of the induced E is Eind = 0 r B dr 1 = B R2 2 1 25 rad/s = 2 3 = 3750 V . (9 T) (10 m) 2 6. I2 = 7. I2 = keywords: 026 (part 1 of 2) 10 points Consider 3 polarizers #1, #2, and #3 ordered sequentially. The incident light is unpolarized with intensity I0 . The intensities after the light passes through the subsequent polarizers are labeled as I1 , I2 , and I3 , respectively (see the sketch). I0 I1 I2 I3 Explanation: When the light passes through the polarizer #1 it is polarized vertically. Thus the angle between its polarization and the orientation of polarizer #2 is = 60 . Thus the transmitted intensity is I2 = I1 cos2 = I1 cos2 (60 ) 1 = I1 . 4 When polarized light passes through a polarizer, the transmitted intensity is I2 = #1 #2 #3 Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire I1 cos2 , where is the angle between the polarization of the light (of I1 ) and the orienI1 tation of the polarizer #2. Thus I2 = . 4 027 (part 2 of 2) 10 points What is the final intensity I3 ? 3 I0 16 1 2. I3 = I0 4 1. I3 = 3. I3 = 0 1 I0 8 5 5. I3 = I0 32 1 I0 6. I3 = 16 1 7. I3 = I0 2 3 I0 correct 8. I3 = 32 Explanation: After the polarizer #1 4. I3 = I1 = 1 2 Circuit B I0 . keywords: 028 (part 1 of 2) 10 points 15 Four identical light bulbs are connected either in series (circuit A) or parallel (circuit B) to a constant voltage battery with negligible internal resistance, as shown. Circuit A E E After the polarizer #2 I2 = I1 cos (60 ) = After the polarizer #3 I3 = I2 cos (90 - 60 ) 3 I2 = 4 3 1 = I1 4 4 3 1 1 = I0 4 4 2 = 3 I0 . 32 2 2 Compared to the individual bulbs in circuit A, the individual bulbs in circuit B are 1 4 1. not lit up at all. I1 . 2. 16 times brighter. correct 3. the same brightness. 4. 4 times brighter. 5. 1 as bright. 8 6. 8 times brighter. 1 as bright. 4 1 8. as bright. 2 7. 9. 2 times brighter. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire 1 10. as bright. 16 Explanation: In circuit A, the voltage across each light bulb is V =IR= E E R= , 4R 4 keywords: 16 Since the bulbs are parallel, after one of the bulbs is unscrewed, the voltage across each remaining bulb is unchanged, and the brightness is unaffected. so the power of each bulb in circuit A is V2 E2 PA = = . R 16 R In circuit B, the voltage across each bulb is identical; namely E. Hence the power of each bulb in circuit B is PB = E2 = 16 PA . R 030 (part 1 of 1) 10 points The second-order bright fringe (m = 2) is 4.45 cm from the center line. 4.45 cm P viewing screen O -1 0.0307 mm S2 029 (part 2 of 2) 10 points If one of the bulbs in circuit B is unscrewed and removed from its socket, the remaining 3 bulbs 1. turn red, white and blue. 2. become dimmer. 3. become brighter. 4. begin to flash. 5. burn out the battery twice as fast. 6. pop. 7. are unaffected. correct 8. one become brighter, one became dimmer and one unaffected. 9. go out. 1.11 m Determine the wavelength of the light. Be sure to use the small angle approximation, sin( ) Correct answer: 615.383 nm. Explanation: Let : y = 4.45 cm , L = 1.11 m , and d = 0.0307 mm , S2 Q S1 90 r1 S1 d 10. begin to blink in Morse code "physics sucks". Explanation: S2 an r2 = t Q y L d sin r2 - r1 L y viewing screen We can see that the bulbs in circuit B are 16 times brighter than the bulbs in circuit A. S1 Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire 17 r1 S1 Determine the magnitude of the magnetic field at the center of the loop. Correct answer: 3.31327 10-5 T. Explanation: d = tan S2 Q -1 y L 90 S1 - r1 S2 Q r2 sin d r2 Let : 0 = 1.25664 10-6 N/A2 , I = 3.42 A , and R = 0.0855 m . We can think of the total magnetic field as the superposition of the field due to the long straight wire having magnitude Bstraight = 0 I 2R For constructive interference L m, (1) ybright = d with m = 2, y2 = 0.0445 m, L = 1.11 m, and d = 3.07 10-5 m d y2 = mL (3.07 10-5 m) (0.0445 m) = (2) (1.11 m) = 6.15383 10-7 m = 615.383 nm . keywords: 031 (part 1 of 1) 10 points A conductor consists of a circular loop of radius 0.0855 m and straight long sections, as in the figure. The wire lies in the plane of the paper and carries a current of 3.42 A. The permeability of free space is 1.25664 10-6 N/A2 . and directed into the page and the field due to the circular loop having magnitude Bloop = 0 I 2R and directed into the page. Using appropriate right hand rules we find both B fields are in the same direction. Therefore the resultant magnetic field is B = Bstraight + Bloop 1 0 I = 1+ 2R 1 (1.25664 10-6 N/A2 ) (3.42 A) = 1+ 2 (0.0855 m) = 3.31327 10-5 T . keywords: 032 (part 1 of 2) 10 points Two converging lenses, each of focal length 14.7 cm, are placed 39.3 cm apart, and an object is placed 29.4 cm in front of the first. How far from the first lens is the final image formed? Correct answer: 8.98125 cm. Explanation: Given : f1 = f2 = 14.7 cm , L = 39.3 cm , and p1 = 29.4 cm . I R Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire The thin lens equation gives the image position of the first lens as q1 = p 1 f1 p1 - f1 (29.4 cm) (14.7 cm) = 29.4 cm - 14.7 cm = 29.4 cm . 18 In the arrangement shown in the figure, the resistor is 9 and a 6 T magnetic field is directed into the paper. The separation between the rails is 6 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 7 m/s . 6T m 1g V . R 6m I 9 7 m/s The real image formed by the first lens serves as the object for the second lens, so p2 = L - q1 = 39.3 cm - 29.4 cm = 9.9 cm. Then, the thin lens equation gives q2 = p 2 f2 p2 - f2 (9.9 cm) (14.7 cm) = 9.9 cm - 14.7 cm = -30.3187 cm 6T At what rate is energy dissipated in the resistor? Correct answer: 7056 W. Explanation: Basic Concept: Motional E E = B v. Ohm's Law I= Thus, the final image is located 8.98125 cm front of the second lens, and 8.98125 cm behind the first lens. 033 (part 2 of 2) 10 points What is the magnification of the system? Correct answer: -3.0625 . Explanation: The magnification of the lenses are q1 M1 = - p1 q2 M2 = - p2 and Solution: The motional E induced in the circuit is E =B v = (6 T) (6 m) (7 m/s) = 252 V . From Ohm's law, the current flowing through the resistor is I= E R B v = R (6 T) (6 m) (7 m/s) = R = 28 A . so the total magnification is M = M 1 M2 -q1 -q2 = p1 p2 -29.4 cm -(-30.3187 cm) = 29.4 cm 9.9 cm = -3.0625 . keywords: 034 (part 1 of 1) 10 points Given: Assume the bar and rails have negligible resistance and friction. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire The power dissipated in the resistor is P = I2 R B 2 2 v2 = R R2 B 2 2 v2 = R (6 T)2 (6 m)2 (7 m/s)2 = (9 ) = 7056 W . Note: Second of four versions. keywords: 035 (part 1 of 1) 10 points The large loop shown has a diameter of 87 cm and carries a constant 7 A current. The loop rotates about an axis in the plane of the loop and passes through the center with a constant angular velocity of 48 rad/s , as shown. A small, fixed coil with 20 turns is concentric with the large loop. The coil has a total resistance 0.11 and a 4 cm diameter. Note: The diameter of the coil is much less than the diameter of the loop. The permeability of free space is 4 -7 10 T m/A. 48 rad/s 0 I ds ^ r d = 2 cm , 2 D = 87 cm , I = 7 A, N = 20 , R = 0.11 , = 48 rad/s , and 0 = 4 10-7 T m/A . 19 d I D Let : r0 = In order to calculate the strength of the field at the center of the outer loop (prior to effects upon the inner loop), use the Biot-Savart law 4 cm I 87 cm , 4 r2 where r is the distance from the differential D point of charge generating the field . 2 In this case, however, d s ^ = r d z . r ^ Therefore, since we are integrating around a circle, all factors not involving theta can be considered constants, so B= B= 0 I 4 r d z . ^ What is the amplitude of the induced current in the small coil? Correct answer: 0.000110887 A. Explanation: The integral over gives 2 , and since r D= , 2 0 I B= z ^ D Now, let the outer loop initially have its magnetic moment pointing in the z-direction. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire Then, taking B constant near the center of the inner wire, B = B dA = 0 I d 2 cos( t) . 4D keywords: 20 Then, applying Faraday's Law, V =- d B dt 0 I r 0 2 sin( t) . = D 037 (part 1 of 1) 10 points The speed of light is 3 108 m/s. What is the frequency of an X-ray with a wavelength of 0.4 nm? Correct answer: 7.5 1017 Hz. Explanation: Let : c = 3 108 m/s and = 0.4 nm = 4 10-10 m . The frequency of this X-ray is f= c 3 108 m/s = 4 10-10 m V Therefore, by Ohm's Law, we get I = , R which has amplitude I= 0 N I r 0 DR (4 10-7 T m/A) (20) (7 A) = (87 cm) (0.11 ) (2 cm)2 (48 rad/s) 2 = 7.5 1017 Hz . = 0.000110887 A . keywords: 038 (part 1 of 1) 10 points A point light source delivers a power P . It radiates light isotropically. A mirror is placed at point B that is at distance r from the source. The mirror has a cross section A. Make the approximation that all the light hitting the mirror comes in perpendicular to it. The mirror is a totally reflecting surface. B keywords: 036 (part 1 of 1) 10 points A "slinky" toy spring has a radius of 2.7 cm and an inductance of 175 H when extended to a length of 1.4 m. How many turns are in the spring? Correct answer: 291.769 turns. Explanation: Let : r = 2.7 cm = 0.027 m , L = 175 H = 0.000175 H , l = 1.4 m . and r 0 N 2 A From L = , we find l N= = Ll 0 A (0.000175 H)(1.4 m) (1.25664 10-6 n/A2 ) (0.027 m)2 Find the total force on the mirror. 2 r2 P c A2 2AP 2. F = c/, r 2 1. F = = 291.769 turns . Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire A2 P 2 c r2 2P c P c P c2r 4 r2 P cA AP correct c 2 r2 2P c r2 21 3. F = 4. F = 5. F = 6. F = 7. F = 8. F = 9. F = Find the maximum value of for which the refracted ray will undergo total internal reflection at the left vertical face of the block. (Assume the refractive index of polystyrene is 1.5.) Correct answer: 31.0645 . Explanation: Given : np = 1.5 and nw = 1.333 . 1 3 2 10. F = r 2 P c Explanation: Since the mirror is small enough (or far enough away) that we may approximate the incoming light as perpendicular, the force is simply F = pA, where p is the radiation pressure. The mirror is totally reflecting, so p= 2I . c Intensity is defined by I= Power hitting the mirror . Area of mirror A P P , 4 r2 2I A AP = c c 2 r2 For polystyrene surrounded by water, total internal reflection at the left vertical face requires that 3 c = sin-1 nw np 1.333 = sin-1 1.5 = 62.7062 . A 4r2 I= A leading to a force of = F = pA = on the mirror. keywords: From the geometry shown in the figure, 2 = 90.0 - 3 90.0 - 62.7062 = 27.2938 . Thus, use of Snell's law at the upper surface gives 1 = sin-1 np sin 2 nw 039 (part 1 of 1) 10 points A light ray of wavelength 553 nm is incident at an angle on the top surface of a block of polystyrene surrounded by water. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire sin-1 1.5 sin 27.2938 1.333 7. into the plane and counter-clockwise. 8. out of the plane and clockwise. 22 = 31.0645 . keywords: 040 (part 1 of 1) 10 points A pendulum consists of a supporting rod and a metal plate (see figure). The rod is pivoted at O. The metal plate swings through a region of magnetic field (directed out of the paper). Consider the case where the pendulum's metallic plate enters the magnetic field region from left to right. O 9. along the rod away from the pivot point and counter-clockwise. 10. along the direction of swing and counterclockwise. Explanation: A conduction electron in the pendulum will experience a magnetic force opposite the direction of v B, so using the right hand rule, we can determine that the motion of the electrons will be in the counter-clockwise direction, which produces a clockwise current. Because the magnetic field is pointing out of the paper, the magnetic flux through the pendulum is increasing, so by Lenz's law we know that the induced magnetic field is in the opposite direction, or into the paper. O e entering fi ld B B B B Fw The direction of the induced magnetic field at the center of the circulating eddy current and the direction of the eddy currect are 1. into the plane and clockwise. correct 2. out of the plane and counter-clockwise. 3. opposite to the direction of swing and counter-clockwise. 4. along the rod toward the pivot point and clockwise. 5. opposite to the direction of swing and clockwise. 6. along the direction of swing and clockwise. B entering fi el d i B B B Alternative Solution: Use the result of Part 1, and the right hand rule on the current flow to determine that the induced magnetic field must be directed into the paper. keywords: 041 (part 1 of 1) 10 points In a Young's double-slit experiment using 153 nm light, a thin piece of plexiglass of refractive index 1.63 covers one of the slits. Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire If the center point on the screen is a dark spot instead of a bright spot, what is the minimum thickness of the plexiglass? Correct answer: 121.429 nm. Explanation: Passage through the plastic delays light by one-half wavelength. Let the minimal thickness of the plexiglass sheet be t. For the bright spot to become dark spot we must have: 1) The number of wavelengths through t thickness t for the light in air to be , and 2) The number of wavelengths through thickness t for the light in the plexiglass to t be . /n To have destructive interference the number of wavelengths must differ by one-half wavelength, so 1 t t + = 2 /n or, equivalently, t+ = tn 2 2 23 B 9 mm 8m v A 5.3 If the wires make an angle of 56 with the vertical when in equilibrium (v = 0 m/s), determine the strength of the magnetic field. Correct answer: 0.219581 T. Explanation: B r v I t (n - 1) = Solving for t gives: t= 2 (n - 1) 153 nm = 2 (1.63 - 1) = 121.429 nm . Let : g = 9.8 m/s2 , v = 0 m/s , r = 9 mm , = 8 m, I = 5.3 A , and = 0.0801 kg/m . Since the rod is in equilibrium, the total forces acting on it must be zero: Fx = T sin - I |L B| = 0 Fy = T cos - mg = 0 keywords: 042 (part 1 of 1) 10 points A metal rod having a mass per unit length of 0.0801 kg/m carries a current of 5.3 A . The rod hangs from two wires (in the same plane as the rod) in a uniform vertical magnetic field as in the figure. The acceleration of gravity is 9.8 m/s2 . or T sin - I L B sin 90 = 0 T cos - L g = 0 Dividing these equations, we have tan = IB , g so B g B 9.8 m/s2 Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire B= g tan I (0.0801 kg/m) (9.8 m/s2 ) tan (56 ) = (5.3 A) = 0.219581 T . keywords: 043 (part 1 of 1) 10 points Consider the capacitor circuit 24 F 33 F 51 F 6 F and the individual voltages are the same. C1 and C2 are in parallel, so C12 = C1 + C2 = 24 F + 33 F = 57 F . C3 and C4 are in parallel, so C34 = C3 + C4 = 51 F + 6 F = 57 F . 24 C12 EB C34 65 V What is the effective capacitance of the circuit? Correct answer: 28.5 F. Explanation: Let : C1 C2 C3 C4 EB = 24 F , = 33 F , = 51 F , = 6 F , and = 65 V . C12 and C34 (Note: C12 = C34 ) are in series with the battery, so C1234 = 1 1 1 + C12 C34 C12 C34 = C12 + C34 (57 F) (57 F) = 57 F + 57 F = 28.5 F . keywords: C1 C2 C3 C4 044 (part 1 of 1) 10 points A thin layer of liquid methylene iodide (n = 1.756) is sandwiched between two flat parallel plates of glass (n = 1.4). What must be the minimum thickness of the liquid layer if normally incident light of wavelength 598 nm in air is to be strongly reflected? Correct answer: 85.1367 nm. Explanation: Given nf ilm = 1.756 , nglass = 1.4 , and = 598 nm . EB For capacitors in series, 1 Cseries = 1 Ci Vi , Vseries = and the individual charges are the same. For parallel capacitors, Cparallel = Qparallel = Ci Qi , Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire A phase change of occurs upon reflection at the first interface, but not at the second. The condition for strong reflection is then 2t = m+ 1 2 n = 2m + 1 , 2 nf ilm m = 0, 1, 2, ... 25 L rb ra Thus t= 4 nf ilm 598 nm = 4(1.756) = 85.1367 nm . The current I = 10 mA . Since the cylindrical shell is infinitely long, and has cylindrical symmetry, Ampere's Law gives the easiest solution. Consider a circle of radius r1 centered around the center of the shell. To use Ampere's law we need the amount of current that cuts through this circle of radius r1 . To get this, we first need to compute the current density, for the current flowing through the shell. J= = I A 2 rb keywords: 045 (part 1 of 1) 10 points A long cylindrical shell has a uniform current density. The total current flowing through the shell is 10 mA. The permeability of free space is 1.25664 10-6 T m/A . 7 cm 3 cm 16 k m I 2 - ra (10 mA) = [(0.07 m)2 - (0.03 m)2 ] = 0.795775 A/m2 . The current enclosed within the circle is 2 2 Ienc = [r1 - ra ] J = [(0.042 m)2 - (0.03 m)2 ] (0.795775 A/m2 ) = 0.00216 A . The current is 10 mA . Find the magnitude of the magnetic field at a point r1 = 4.2 cm from the cylindrical axis. Correct answer: 10.2857 nT. Explanation: Let : L = 16 km , ra = 3 cm = 0.03 m , rb = 7 cm = 0.07 m , r1 = 4.2 cm = 0.042 m , I = 10 mA , and b = 1.25664 10-6 T m/A . Ampere's Law, B ds = 0 Ienc B 2 r1 = 0 Ienc 0 Ienc B= 2 r1 1.25664 10-6 T m/A = 2 (0.042 m) (0.00216 A) = 10.2857 nT . Daniels, Matthew Final 1 Due: Dec 16 2006, 6:00 pm Inst: Ditmire keywords: 046 (part 1 of 1) 10 points A uniformly charged insulating rod of length 12 cm is bent into the shape of a semicircle as in the figure. The value of the Coulomb constant is 8.98755 109 N m2 /C2 . Ex = - 26 ke r 3/2 cos d /2 3/2 /2 ke =- (sin ) r =2 where = ke , r -8.71 C L q and r = . Therefore, L O Ex = 2 ke q L2 2 (8.98755 109 N m2 /C2 ) = (0.12 m)2 (-8.71 10-6 C) = -3.41568 107 N/C . If the rod has a total charge of -8.71 C, find the horizontal component of the electric field at O, the center of the semicircle. Define right as positive. Correct answer: -3.41568 107 N/C. Explanation: Let : L = 12 cm = 0.12 m and q = -8.71 C = -8.71 10-6 C . 12 cm Since the rod has negative charge, the field is pointing to the left (towards the charge distribution). A positive test charge at O would feel an attractive force from the semicircle, pointing to the left. keywords: Call the length of the rod L and its charge q. Due to symmetry Ey = and Ex = dE sin = ke dq sin , r2 dEy = 0 where dq = dx = r d, so that y x

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University of Texas - PHY - 303L
Daniels, Matthew Quiz 1 Due: Sep 21 2006, 11:00 pm Inst: Ditmire This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Daniels, Matthew Quiz 2 Due: Oct 19 2006, 11:00 pm Inst: Ditmire This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Daniels, Matthew Quiz 3 Due: Nov 16 2006, 9:00 pm Inst: Ditmire This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - PHY - 303L
Daniels, Matthew Quiz 4 Due: Dec 7 2006, midnight Inst: Ditmire This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
University of Texas - CH - 302
1A11Periodic Table of the Elements2A28A181.0079H3A134A145A156A167A1724.0026He36.941Li49.0122Be510.811B612.011C714.0067N P815.9994O S918.9984F1020.1797Ne Ar1122.989
University of Texas - CH - 302
1A11Periodic Table of the Elements2A28A181.0079H3A134A145A156A167A1724.0026He36.941Li49.0122Be510.811B612.011C714.0067N P815.9994O S918.9984F1020.1797Ne Ar1122.989
University of Texas - CH - 302
1A11Periodic Table of the Elements2A28A181.0079H3A134A145A156A167A1724.0026He36.941Li49.0122Be510.811B612.011C714.0067N P815.9994O S918.9984F1020.1797Ne Ar1122.989
University of Texas - CH - 302
1A11Periodic Table of the Elements2A28A181.0079H3A134A145A156A167A1724.0026He36.941Li49.0122Be510.811B612.011C714.0067N P815.9994O S918.9984F1020.1797Ne Ar1122.989
University of Texas - CH - 302
1A11Periodic Table of the Elements2A28A181.0079H3A134A145A156A167A1724.0026He36.941Li49.0122Be510.811B612.011C714.0067N P815.9994O S918.9984F1020.1797Ne Ar1122.989
University of Texas - CH - 302
Elbel, Brittany Homework 1 Due: Jan 23 2006, noon Inst: J A Holcombe This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (
University of Texas - CH - 302
Elbel, Brittany Homework 2 Due: Jan 30 2006, noon Inst: J A Holcombe This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. The g
University of Iowa - BUSINESS - 06a
IMPORTANT: This list in not all inclusive! I don't know what is on the test; this is just an overview of the keytopics in each chapter. REVIEW FOR MIDTERM #1 *Review Chapters *Review Notes *Review HomeworkCHAPTER ONE: Do not spend much time on this
University of Texas - CH - 302
Elbel, Brittany Homework 3 Due: Feb 6 2006, noon Inst: J A Holcombe This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
University of Iowa - BUSINESS - 06a
IMPORTANT: This list in not all inclusive! We don't know what is on the test; this is just an overview of thekey topics in each chapter. REVIEW FOR MIDTERM #2:*Review Chapters *Review Notes *Review HomeworkCHAPTER EIGHT:Inventory Relationship: B
University of Texas - CH - 302
Elbel, Brittany Homework 4 Due: Feb 13 2006, noon Inst: J A Holcombe This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (
University of Iowa - BUSINESS - 06a
Practice Questions for Midterm I 1) The Balance Sheet: i. Lists a firm's resources and obligations at a specific point in time ii. Is designed to measure performance over a given period of time iii. Contains all economic assets relevant to the value
University of Iowa - BUSINESS - 06a
Practice Questions for Midterm II 1) A preliminary count of items in XYZ Company's warehouse as of 12/31/04 indicates an inventory balance of \$100,000. Included in this figure is \$5,000 worth of inventory on consignment from ABC Co. (XYZ is the consi
University of Texas - CH - 302
Elbel, Brittany Homework 5 Due: Feb 20 2006, noon Inst: J A Holcombe This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (
University of Iowa - BUSINESS - 06a
6A:120 Financial Accounting and Reporting Homework 9 Part 1: Financial Statement Application: 1. How many shares of common stock and preferred stock is Lowe's authorized to issue? Lowe's authorized to issue \$5.6 billion shares of common stocks and \$5
University of Texas - CH - 302
Elbel, Brittany Homework 6 Due: Feb 27 2006, noon Inst: J A Holcombe This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (
University of Texas - CH - 302
Elbel, Brittany Homework 7 Due: Mar 6 2006, noon Inst: J A Holcombe This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
Columbia Basin College - MTH - 155
Section 5.1 Angles and Their MeasureOBJECTIVE 1RadiansOBJECTIVE 2Find the length of the arc of a circle of radius 4 meters subtended by a central angle of 0.5 radian.OBJECTIVE 3OBJECTIVE 4Find the area of the sector of a circle of radiu
University of Texas - CH - 302
Elbel, Brittany Homework 8 Due: Mar 29 2006, noon Inst: J A Holcombe This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (
University of Texas - CH - 302
Elbel, Brittany Homework 9 Due: Apr 3 2006, noon Inst: J A Holcombe This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (p
University of Texas - CH - 302
Elbel, Brittany Homework 10 Due: Apr 10 2006, noon Inst: J A Holcombe This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
University of Texas - CH - 302
Elbel, Brittany Homework 11 Due: Apr 20 2006, noon Inst: J A Holcombe This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
Columbia Basin College - MTH - 155
Section 5.2 Right Triangle TrigonometryOBJECTIVE 1Trigonometric Functions of Acute AnglesFind the value of each of the six trigonometric functions of the angle .OBJECTIVE 210 3 10 Given sin and cos , 10 10 find the value of each of the four
University of Texas - CH - 302
Elbel, Brittany Homework 12 Due: Apr 28 2006, noon Inst: J A Holcombe This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
Columbia Basin College - MTH - 155
Section 5.3 Computing the Values of Trigonometric Functions of Acute AnglesOBJECTIVE 1OBJECTIVE 2 2 (a) sin tan 3 42Find the exact value of each expression. (b) sin 60 cos 45 (c) cos cot 3 4OBJECTIVE 3OBJECTIVE 4
University of Texas - CH - 301
Columbia Basin College - MTH - 155
Section 5.4 Trigonometric Functions of General AnglesOBJECTIVE 1OBJECTIVE 2Two angles in standard position are said to be coterminal if they have the same terminal side.OBJECTIVE 3If sin &gt; 0 and cos angle lies.&lt; 0, name the quadrant in wh
University of Texas - CH - 301
University of Texas - CH - 301
1A 1 H 1.008 3 Li 6.94 11 Na 22.99 19 K 39.10 37 Rb 85.5 55 Cs 132.9 87 Fr (223)2A 3A 4 5 Be B 9.01 10.81 12 13 Mg Al 24.30 3B 4B 5B 6B 7B -8B- 1B 2B 26.98 20 21 22 23 24 25 26 27 28 29 30 31 Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga 40.08 44.96 47.90 50.
Columbia Basin College - MTH - 155
Section 5.5 Unit Circle Approach; Properties of the Trigonometric FunctionsOBJECTIVE 1OBJECTIVE 2OBJECTIVE 3Find the exact value of:(a) cos 48011 (b) tan 4 7 (c) sin 3OBJECTIVE 4Find the exact value of:(a)cos (-60 ) (b) sin (-390) (c)
University of Texas - CH - 301
Pickering, Tracey Exam 2 Due: Oct 19 2005, 3:00 pm Inst: McCord This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Results wi
Columbia Basin College - MTH - 155
Section 5.6 Graphs of the Sine and Cosine FunctionsOBJECTIVE 1OBJECTIVE 202 3OBJECTIVE 3Determine the amplitude and period of y = - 4 cos (3x)OBJECTIVE 4OBJECTIVE 5
Columbia Basin College - MTH - 155
Section 5.7 Graphs of the Tangent, Cotangent, Cosecant, and Secant FunctionsOBJECTIVE 1OBJECTIVE 2
University of Texas - CH - 301
1A 1 H 1.008 3 Li 6.94 11 Na 22.99 19 K 39.10 37 Rb 85.5 55 Cs 132.9 87 Fr (223)2A 3A 4 5 Be B 9.01 10.81 12 13 Mg Al 24.30 3B 4B 5B 6B 7B -8B- 1B 2B 26.98 20 21 22 23 24 25 26 27 28 29 30 31 Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga 40.08 44.96 47.90 50.
Columbia Basin College - MTH - 155
Section 5.8 Phase Shifts; Sinusoidal Curve FittingOBJECTIVE 1OBJECTIVE 2
University of Texas - CH - 301
1A 1 H 1.008 3 Li 6.94 11 Na 22.99 19 K 39.10 37 Rb 85.5 55 Cs 132.9 87 Fr (223)2A 3A 4 5 Be B 9.01 10.81 12 13 Mg Al 24.30 3B 4B 5B 6B 7B -8B- 1B 2B 26.98 20 21 22 23 24 25 26 27 28 29 30 31 Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga 40.08 44.96 47.90 50.
Columbia Basin College - MTH - 155
Section 6.1 The Inverse Sine, Cosine, and Tangent FunctionsReview of Properties of Functions and Their InversesOBJECTIVE 1OBJECTIVE 2ytan 1 xytan 1 xOBJECTIVE 3
University of Texas - CH - 301
1 1A 1 H 1.008 3 Li 6.94 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223)Periodic Table of the Elements2 2A 4 Be 9.01 12 Mg 24.30 20 Ca 40.08 38 Sr 87.62 56 Ba 137.3 88 Ra (226) 13 3A 5 B 10.81 13 4 5 6 7 8 9 10 11 12 Al 4B 5B 6B 7B -8B-
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Section 6.2 The Inverse Trigonometric Functions (Continued)OBJECTIVE 1OBJECTIVE 2OBJECTIVE 3
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Pickering, Tracey Prequiz 1 Due: Sep 11 2005, 6:00 pm Inst: McCord This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (pa
University of Texas - CH - 301
Pickering, Tracey Prequiz 2 Due: Sep 18 2005, 6:00 pm Inst: McCord This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (pa
University of Texas - CH - 301
Pickering, Tracey Prequiz 3 Due: Sep 25 2005, 6:00 pm Inst: McCord This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (par
University of Texas - CH - 301
Pickering, Tracey Prequiz 4 Due: Oct 2 2005, 6:00 pm Inst: McCord This print-out should have 8 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part
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Section 6.3 Trigonometric IdentitiesOBJECTIVE 1OBJECTIVE 2
University of Texas - CH - 301
Pickering, Tracey Quiz 5 Due: Apr 18 2006, 1:00 pm Inst: John McDevitt This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. The
Columbia Basin College - MTH - 155
Section 6.4 Sum and Difference FormulasOBJECTIVE 1OBJECTIVE 2OBJECTIVE 3
University of Texas - CH - 301
Pickering, Tracey Quiz 6 Due: Apr 27 2006, 10:00 am Inst: John McDevitt This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
University of Texas - CH - 301
Pickering, Tracey Quiz 7 Due: Apr 25 2006, 10:00 am Inst: John McDevitt This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001
Columbia Basin College - MTH - 155
Section 6.5 Double-angle and Half-angle FormulasOBJECTIVE 1OBJECTIVE 2OBJECTIVE 3
Columbia Basin College - MTH - 155
Section 6.6 Product-to-Sum and Sum-to-Product FormulasOBJECTIVE 1OBJECTIVE 2
Columbia Basin College - MTH - 155
Section 6.7 Trigonometric Equations (I)OBJECTIVE 1
Columbia Basin College - MTH - 155
Section 6.8 Trigonometric Equations (II)OBJECTIVE 1OBJECTIVE 2OBJECTIVE 3OBJECTIVE 4
Columbia Basin College - MTH - 155
Section 7.1 Applications Involving Right TrianglesOBJECTIVE 1Find the value of each of the six trigonometric functions of the angle .OBJECTIVE 2OBJECTIVE 3OBJECTIVE 4
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Section 7.2 The Law of SinesOblique Triangle(None of the angles is a right angle)OBJECTIVE 1OBJECTIVE 2SSA - The Ambiguous CaseOBJECTIVE 3
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Section 7.3 The Law of CosinesOBJECTIVE 1OBJECTIVE 2OBJECTIVE 3
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Section 7.4 Area of a TriangleOBJECTIVE 1OBJECTIVE 2
Columbia Basin College - MTH - 155
Section 7.5 Simple Harmonic Motion; Damped Motion; Combining WavesOBJECTIVE 1FrequencySuppose that an object attached to a coiled spring is pulled down a distance of 5 inches from its rest position and then released. If the time for one oscilla