Physics 212 Spring 2009 Final Exam Version B (872413)

Transcription

1 Physics 212 Spring 2009 Final Exam Version B (872413) 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 DetailsElectric Field Lines (3) [ ] Which of the following is true about the charges q A for particle A and q B for particle B in the figure below? x*07*q A = 32 and q B = -4 y*03*q A = -32 and q B = 4 v*10*q A = - 4 q B w*06*q A = + 4 q B z*01*nothing can be said about them without more information. A has four times as many field lines as B and the field lines point out of A and into B. Note that A is a positive charge and B is a negative charge. 2. Question DetailsMagnetic Field (3) [ ] What is the direction of the magnetic field in each section of the figure?

2 z*01*none of these. y*03*1 left, 2 up, 3 left, 4 down v*10*1 toward you, 2 away from you, 3 away from you, 4 toward you x*04*1 up, 2 left, 3 down, 4 left w*07*1 away from you, 2 toward you, 3 toward you, 4 away from you Using the right hand, we point our fingers in the direction of the initial velocity (left, up, left, down) and curl them in the direction of the magnetic field (toward you, away from you, away from you, toward you). Our thumb then points in the direction of the force or acceleration (up, left, down, left). 3. Question DetailsApparent Depth (1) [750391] A silver medallion is actually 1.33 cm beneath the surface of a calm pool. How far below the surface of the pool would someone see the medallion if they were viewing it from above the pool? (In other words, what is its apparent depth?) Note: The refractive index of water is y*03*0.75 cm w*01*1.77 cm z*00*1.33 cm x*04*1.00 cm v*10*1.00 cm The apparent depth is d' = d (1) / n water = 1.00 cm. 4. Question DetailsRay Trace or Thin-Lens Equation (1) [ ] Do a ray trace for the object and lens shown here or use the number of blocks to do the thin-lens equation. Which of the following is closest to the magnification of the image?

3 y*02*-1/4 w*07*-4 x*03*1/4 v*10*4 z*01*1 The ray trace is shown below. The image is about 4 times bigger than the object and both are right-side up.

4 5. Question DetailsElectric Potential (1) [ ] A charge of 3 nc is placed at the orgin. A second - 21 nc charge is placed on the x-axis 3 m to the right of the origin. What is the potential at a point on the x-axis 10 m to the left of the origin? w*06*17.22 V y*05*-0.85 V z*01*4.36 V x*04*-9.89 V v*10* V The potential for each point charge is given by the equation. V = k q / r Note that this is a scalar and thus we do NOT take the absolute value of the charge. For the first charge we have a distance of 10 m between the charge and where the potential is measured. For the second charge that distance is 13 m. Thus, V 1 = 2.70 V and V 2 = V. We then just add these together. 6. Question DetailsLenz's Law (4) [ ]

5 The figure below shows an external magnetic field and the induced current caused by that field. What must be true about this external magnetic field? y*02*it is negative. v*10*it is decreasing. z*01*it points out of the page. x*02*it is positive. w*07*it is increasing. If the field is decreasing, it generates an induced magnetic field that points in the same direction to counteract the change. This induced magnetic field comes from the current in the wire. We can see the direction of the induced magnetic field by wrapping our right hand around the wire in the direction of the current and the induced field points in the direction of our thumb. 7. Question DetailsComplex Circuits (2) [ ] What is the equivalent resistance for the circuit below? (If it helps, assume that there is a battery attached between points A and B.) v*10*17.5 ohms x*07*72 ohms z*01*70 ohms w*03*48.75 ohms y*01*110 ohms The 20Ω and the 30Ω resitors in the upper-right of the diagram are in parallel. Their equivalent resistance is R eq = 1 / (1/20Ω + 1/30Ω) = 12Ω. This equivalent resistor would then be in series with the remaining 30Ω resistor in the top of the figure to give us R eq =30Ω + 12Ω = 42Ω. Finally, this is in parallel with the bottom 30Ω resistor. To give us the final answer R eq = 1 / (1/42Ω + 1/30Ω) = 17.5Ω. 8. Question DetailsTotal Internal Reflection (1) [ ]

6 The refractive index of a material is What is the critical angle for a light ray traveling in that material toward a plane layer of air above that material? y*03*0.02 o z*01*90.00 o w*07*0.01 o v*10*23.78 o x*04*66.22 o Use the equation θ c = sin -1 (n 2 / n 1 ) where n 1 is the refractive index of the material and n 2 is the refractive index of air. 9. Question DetailsRay Trace or Thin-Lens Equation (1) [ ] Do a ray trace for the object and lens shown here or use the number of blocks to do the thin-lens equation. Which of the following is closest to the magnification of the image? w*07*-1/2 x*03*2 z*01*1 y*02*2 v*10*1/2

7 The ray trace is shown below. The image is about 2 times smaller than the object and both are right-side up. 10. Question DetailsWhat do we see? (3) [ ] (For this problem, ignore the top blank box in the figure.) What would the box containing the letters A and B look like to the eye in the figure? (Be careful. The answers may not be in order from I to IV.)

8 w*05*figure II v*10*figure I z*00*none of these x*03*figure III y*03*figure IV

9 Trace a ray from the left and right side of the A, as well as from the left and right side of the B, and you will see that the mirror image of a mirror image looks just like the object. By the way, I was able to find at least one ray from every point on the A and B that hit the eye. Assignment Details Name (AID): Physics 212 Spring 2009 Final Exam Version B (872413) Submissions Allowed: 100 Category: Exam Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsu.edu ) Last Saved: May 6, :40 PM MDT Permission: Protected Randomization: Person Which graded: Last Feedback Settings Before due date Nothing After due date Nothing