Physics 212 Spring 2009 Exam 2 Version C (832725)

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1 Physics 212 Spring 2009 Exam 2 Version C (832725) Question Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly. 1. Question Details Electric Current and Resistance (2) [742620] Five resistors are made of a material with the same value of resistivity. Which has the most resitance? *v*10*a cylindrical resistor with length 3 cm and radius 0.01 cm. *w*05*a cylindrical resistor with length 1 cm and radius 0.01 cm. *y*02*a cubic resistor with side length 4 cm. *z*02*a cubic resistor with side length 2 cm. *x*05*a cylindrical resistor with length 2 cm and radius 0.02 cm. Use the equation R = L/A where L is the length of the resistor, A is its area and is its resistivity. 2. Question Details Magnetic Force (4) [623658] A proton is moving through a magnetic field whose magnitude is T. The proton experiences only a magnetic force and has an acceleration of magnitude m/s 2. At a certain instant, it has a speed of m/s. Determine the angle (less than 90 ) between the proton's velocity and the magnetic field. y*02*37.20 x*02*40.63 v*10*49.37 w*05*24.46 z*06*0.86 The angle can be found from the equation, F = q vb sin, where q = 1.602x10-19 C, v = m/s and B = 10-4 T. The force is equal to F = m a = (1.67x10-27 kg) ( m/s 2 ). 3. Question Details Equivalent Circuits (4) [567357]

2 In the figure below, the resistors all have a value of 5 is the current across the resistor R 1? except R 1, which has a value of 15. The battery is a 9 V battery. What y*03*0.26 A w*04*1.80 A v*10*0.60 A z*01*0.00 A x*06*1.50 A Because the resistor, R 1, is in parallel with the battery, we see that V 1 = V = 9 V. Then, Ohm's law gives us I 1 = V 1 / R 1 = (9 V) / (15 ) = 0.60 A. 4. Question Details Capacitance (3) [742619] How many electrons are on the negative plate of a 37 F capacitor with a 8 V potential difference across the plates? v*10* electrons x*04* electrons z*01* electrons y*03* electrons w*04* electrons By definition, the capitance of a capacitor is C = Q/ V. Thus, Q = - C V = - (37 F) (8 V) = C. Each electron has x C. Thus, n = Q / -e = ( C) / ( x C) = electrons.

3 5. Question Details Electric Current and Resistance (3) [742622] How many electrons per second must a 1.5 V battery produce in order to generate a current of 41 A? *v*10* electrons/second *y*02* electrons/second *w*04* electrons/second *x*03* electrons/second *z*01*i would need to know the resistance in the circuit to calculate this. Remember that by definition, the charge per unit time that passes through a given cross-sectional area is the current. Thus, the amount of chrage produced by the battery in one second is q = - I * (1 sec) = - 41 C. The charge of a single electron is e = x C. Thus, we have the number of electrons produced by the battery. n = q / -e = - 41 C / ( x C) = electrons/second. 6. Question Details Mass Spectrometer (3) [750382] A negative ion with only one extra electron moves through the magnetic field (B = 0.36 mt) of a mass spectrometer. The ion had an initial speed of 200 m/s. If the radius of the circle is 1.94 cm, what is the mass of the ion? x*02* kg y*01* kg w*05* kg v*10* kg z*01*i would need to know what kind of ion it is to determine this. The mass can be found from the equation, R = (m v) / ( q B), where R = 1.94 cm, v = 200 m/s, B = 0.36 mt, and q = x C. Remember to convert your units. 7. Question Details Capacitors (4) [742609] An isolated spherical conductor of radius 15 cm is charged to 2 kv and has a capacitance of pf. What would the capacitance be if the sphere is charged to 7 kv? z*02*1.19 pf y*02* pf w*05*58.42 pf x*05*4.77 pf v*10*16.69 pf The capacitance does not change unless the shape, size or dielectric of the capacitor changes. In this case it was the voltage that changed. Since the charge changes proportional to the voltage, the capacitance deos not change.

4 8. Question Details Capacitors (3) [742612] A 4.0-µF and an 8.0-µF capacitor are connected in series across a 25-V battery. Find the total charge stored on the two capacitors. z*05*9.38 µc v*10*66.67 µc y*04*0.11 µc x*02*0.48 µc w*05* µc The equivalent capacitance for two capacitors in SERIES is given by the equation 1 / C eq = 1 / C / C 2 and the total charge on both capacitors is then Q eq = C eq V. 9. Question Details Resistors (4) [742618] Which of the following equivalent resistances CANNOT be made with one 1 and two 3 resistors? (Hint: Try ALL of the combinations of parallel and series. There are 4 unique combinations each having a different equivalent resistance.) v*10*1.67 z*03*0.60 y*05*3.75 w*01*7.00 x*05*2.50 Put the resistors is series to get Put the 3 resistors in parallel and this combination in series with the 1 resistor to get Put one of the 3 resistors in parallel with the 1 resistor and put this combination in series with the other 3 resistor to get Put all of the resistors in parallel to get Question Details Equivalent Circuits (3) [567355] A 28 F and a 69 F capacitor are connected in series with each other and a 5 V battery. What is the voltage across each capacitor? v*10*3.56 V and 1.44 V w*04*5.00 V and 5.00 V y*02*17.32 V and 7.03 V x*02* V and V z*05*2.53 V and 2.47 V For resistors in series, the charge is the same and the voltages add. Thus,

5 Q 1 = Q 2 = Q eq = C eq V eq = (1 / C / C 2 ) -1 V eq = C. Thus, V 1 = Q eq / C 1 = 3.56 V and V 2 = Q eq / C 2 = 1.44 V Assignment Details Name (AID): Physics 212 Spring 2009 Exam 2 Version C (832725) Submissions Allowed: 100 Category: Exam Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsu.edu ) Last Saved: Mar 10, :00 AM MDT Permission: Protected Randomization: Person Which graded: Last Feedback Settings Before due date Nothing After due date Nothing

Physics 222 Spring 2009 Exam 1 Version D (808898)

Physics 222 Spring 2009 Exam 1 Version D (808898) Question 1 2 3 4 5 6 7 8 9 10 Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form.