# Physics 251 August 1, 2002 n Questions from yesterday? n Review Practice Final n Additional Problems n Bonus Points for Formal Lab Reports…see homepage.

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Physics 251 August 1, 2002 n Questions from yesterday? n Review Practice Final n Additional Problems n Bonus Points for Formal Lab Reports…see homepage

These notes are available on-line Http://webphysics.iupui.edu/251/251su02/Final_review.ppt

Question 1 n Q1 An ideal 10 V battery connected across a resistor generates 1000 J in 20 seconds. The value of the resistor is n a) 1 Ohm. n b) 2 Ohms. n c) 50 Ohms. n d) 100 Ohms.

Answer n V=10 V, P=1000J/20s=50 W n b) 2 Ohms

Q2 n A wire carries current from north to south. An electron is located directly below the wire and is traveling straight up. The force on the electron is n a) To the North n b) To the South n c) To the East n d) To the West n e) Zero n f) None of the above n

Answer n I north to south n v cross B points to the North, the force on a negative electron points opposite, or South. n b) To the South N S EW I v B

Q3 n A series RLC circuit has X L < X C. Which of the following is true? n a) The phase is zero. n b) The phase is negative. n c) The power factor is zero. n d) The power factor is positive. n e) Two of the above. n f) None of the above.

answer n X L < X C so the circuit is more capacitor like…V lags I, negative phase n The power factor is always positive n e) Two of the above

Q4 n The image you see when you look at yourself in the mirror is n a) real and erect n b) real and inverted n c) virtual and erect n d) virtual and inverted

Answer n c) virtual and erect

P1 n A charge distribution consists of three charges located at the corners of a triangle. The charges are q1=q2 = 2 microcoulombs at the points (-1,-1) and (1,-1) and a third charge q3 at the point (0,1). n (a) Find the charge q3 that gives a zero electric field at the origin. n (b) Find the total potential energy of this system.

Answer n A) By superposition, the field at the origin is E 1 +E 2 +E 3. So we want E 3 =-(E 1 +E 2 ). By symmetry, (E 1 +E 2 ) must be in the j direction and must have magnitude |(E 1 +E 2 )|=2E 1,y. If q 3 is positive E 2 will automatically be in the -j direction and thus opposite to (E 1 +E 2 ). So, we need only set |E 3 |=2E 1,y.

n B) The potential energy for a pair of charges is U=kq 1 q 2 /r. The potential for the system must take into account all three pairs

P2 n Three moles of a monatomic ideal gas begin at volume V = 0.1 m 3 and 1 atmosphere. The gas is taken through the following cycle. n 1.Volume doubled at constant pressure (ab) n 2.Pressure reduced by a factor of two at constant volume (bc) n 3.Volume reduced by a factor of two at constant pressure (cd) n 4.Pressure doubled at constant volume (da) n a) Draw a P-V diagram for this cycle n b) Calculate Q, W, and the change in internal energy for each of the four processes.

Answer n A) The pressure vs. volume diagram n V = 0.1 m 3 and p=1 atmosphere V=0.1m 3 V=0.2m 3 p=0.5 atm p=1 atm ab cd

Answer b) n Two isobaric processes, two isochoric processes (1 atm =1e5 Pa)

Cont. n To get the q’s, frst calculate all the T’s using the ideal gas law:

Cont. n Now, use the T’s in the equation for heat. Also use Cp=5R/2 and CV=3R/2 for monotomic ideal gas

Cont. n Finally, use the first law of thermodynamics: n Notice the total change in internal energy is zero!

P3 n A solid insulating sphere of radius a carries a uniform charge density and has total charge Q. Concentric with this sphere is a conducting spherical shell with inner radius b and outer radius c. The conducting shell carries a total charge 2Q. n a) Find the electric field in each of the regions: r less than a, r between a and b, r between b and c, and r greater than c. n b) Find the surface charge density on the inner and outer surfaces of the spherical shell.

Answer n A) By symmetry, the electric field must be radial, so we can use Gauss’ Law. Since the field is radial, the flux integral simplifies and we get

Cont. n To get the surface charge density on the inner surface, we require that a Gaussian surface drawn inside the conductor contain no net charge, this means that the charge at r=b is -Q. So, at r=b,  =-Q/4  b 2. The sum of the charge at r=c and r=b must be 2Q, so there must be 3Q at r=c. This means at r=c,  =3Q/4  c 2.

P4 n Consider a series RLC circuit. The source supplies a voltage of V=150cos(120  t+  ) volts, maintaining a current I = I0cos(120  t). The values of the circuit elements are R = 30 Ohms, L = 175 mH, and C = 75 microfarads. Find n a) The frequency n b) The impedance n c) the phase angle n d) The average power delivered by the source. n e) The current amplitude.

Answer n The frequency n The impedance

Cont. n c) the phase angle

Cont. n d) The average power delivered by the source

Cont. n e) The current amplitude.

P5 n An object is placed 30 cm to the left of lens #1, which has focal length f1 =10 cm. Ten centimeters to the right of lens #1 is lens #2, which has of focal length f2 = -10. n a.Is lens #1 converging or diverging? How about Lens #2? n b.Find the location of the final image produced by these two lenses together n c.Find the total magnification of the system. n d.Is the final image real or vitual? Erect or inverted?

Answer n A) Lens #1 is converging (f>0). Lens #2 is diverging (f<0). n B) first image due to lens #1 n with new object (virtual) distance, we have

cont n c.Find the total magnification of the system

Cont. n d.Is the final image real or vitual? Erect or inverted? n The final image is real (s 2 ’>0) and inverted (m T <0)

P6 n The figure to the right shows an apparatus consisting of a conducting rail that is bent into a circular arc of radius a =0.5 m, and a resistor of value R = 10 Ohms that is connected to the arc at one end, and to a pivot point at the otherend. A conducting rod is connected to the resistor at the pivot, and slides along the rail near its other end. The entire apparatus is in a uniform magnetic field of magnitude B0 = 0.2 T that is perpendicular to the plane of the loop (into the page). The rod rotates around the pivot at constant angular velocity of 2 rad/sec. Please find n a.The direction of the induced current through the resistor. n b.The magnitude of the induced current.

Answer n A) As the rod swings, the area of the “pie slice” defined by the rod, the resistor, and the curved wire gets larger, and thus the flux (into the plane) goes too. Lenz’s law says the induced current opposes this, so the induced current will flow in the direction required to produce a field out of the plane. Using the right hand rule, this direction must be from left to right in the resistor

n B) The area of the “pie slice” is (  /2)a 2 where  is the angle between the rod and the resistor. Since the field is perpendicular to the plane, the flux is BA=B (  /2)a 2. Faraday’s Law gives

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