Physics 212 Spring 2009 Makeup Exam Version A...

Physics 212 Spring 2009 Makeup Exam Version A (866539)

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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.

Circuits (5) [1102756] 1. Question Details

Given the figure below, how do the voltages of the capacitors relate?

Solution or Explanation For both resistors and capacitors in parallel. Veq = V1 = V2 = ...

v*10*V1 = V2

w*05*V1 = 1/2 V2

y*01*V1 = C1 / Q2

x*04*V1 = 2 V2

z*02*V1 = C2 / Q2

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Lenses (2) [1102765] 2. Question Details

An object is place 20 cm in front of a lens. The image is found 30 cm on the other side of the lens. What is the magnification of the lens?

Solution or Explanation The magnification equation gives us, m = - di / do. Note that di and do are both positive in this case.

w*07*1.50 z*00*I would need to know the heights to figure this out.

v*10*-1.50 x*04*-0.67 y*05*0.67

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Apparent Depth (1) [1102759]3. Question Details

For the figure below, what is the apparent depth D of the light source?

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Solution or Explanation For this case, D = nair / nwater d = (1 / 1.33) 4 m.

z*02*3.99 m y*02*2.26 m x*01*4.00 m w*05*5.32 m

v*10*3.01 m

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Mirrors (2) [1102760] 4. Question Details

For a concave spherical mirror, how far from the mirror must the object be placed so that height of a real image is equal to the negative of the height of the object?

Solution or Explanation In this case, hi = - ho. Thus, m = -1 and di = + do. This gives us the mirror equation 1/f = 1/do + 1/do. Solving for do gives the correct answer (r = 2f).

z*02*There is no place where this occurs. v*10*r

y*02*1/2 f w*05*f x*01*2r

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Coulomb's Law (3) [1102718]5. Question Details

For which of the following do the particles have the greatest magnitude of force acting on them.

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Solution or Explanation The formula is F = k |q1| |q2| / r2

Note that the distance is squared and so has much more to do with the magnitude than the charges.

v*02*a) w*03*b) x*05*c)

y*10*d) z*02*e)

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Ray Trace or Thin-Lens Equation (1) [1092933]6. Question Details

Do a ray trace for the object and mirror shown here or use the number of blocks to do the thin-lens equation. Determine which of the following describes the image.

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Solution or Explanation The ray trace is shown below.

x*04*It is virtual, upright, and smaller than the object. z*02*It is virtual, inverted and smaller than the object. y*02*It is real, upright, and smaller than the object.

v*10*It is virtual, upright, and larger than the object. w*04*It is real, inverted, and larger than the object.

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Mass Spectrometer (1) [1102757]7. Question Details

The figure below shows the paths for four particles in a mass spectrometer. Which of the following is true?

z*01*m1 = m2 = m3 = m4

w*07*m1 < m2 < m3 < m4

x*05*m1 = m2 > m3 > m4

v*10*m1 = m2 < m3 < m4

y*04*m2 > m3 > m4 > m1

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Solution or Explanation The larger the radius of the circle, the larger the mass of the particle. You can see this from the equation R = (m v) / (q B).

Ray Trace or Thin-Lens Equation (1) [1093275]8. Question Details

Do a ray trace for the object and mirror shown here or use the number of blocks to do the thin-lens equation. Determine which of the following describes the image.

Solution or Explanation

v*10*It is virtual, upright, and smaller than the object. w*04*It is real, inverted, and smaller than the object. x*04*It is virtual, upright, and larger than the object. y*02*It is real, upright, and larger than the object. z*02*It is virtual, inverted and larger than the object.

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Gauss's Law (2) [1102755] 9. Question Details

A charge of 5 nC is found inside an imaginary sphere of radius 25 nm. What is the electric flux through the sphere?

Solution or Explanation The Gauss' Law formula for electric flux is Φ = 4 π k Qinside Note that if the charge had been negative, then the flux would have been negative as well.

x*02*39269.91 Nm2/C z*00*125.00 Nm2/C

v*10*564.86 Nm2/C w*05*62.83 Nm2/C y*01*2454.37 Nm2/C

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Faraday's Law (3) [1102758]10. Question Details

A single square loop of wire with side length 7 cm is placed inside a decreasing magnetic field. The angle between the field and the wire loop area is 90o. The initial magnetic field magnitude is 9 T and the final, after 10 seconds, is 4 T. What is the absolute value of the EMF of the coil?

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Solution or Explanation For this case, | | = |∆B / ∆t| A = |∆B / ∆t| 2. And don't forget to convert cm2 to m2.

v*10*2.45 10-3 T m2 w*05*2.45 101 T m2 z*02*1.02 10-2 T m2 x*03*2.45 10-1 T m2 y*03*1.02 102 T m2

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Assignment Details

Name (AID): Physics 212 Spring 2009 Makeup Exam Version A (866539)

Submissions Allowed: 100

Category: Exam

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Locked: No

Author: DeAntonio, Michael ( [email protected] )

Last Saved: Apr 28, 2009 01:04 AM MDT

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