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Two point charges –Q and +2Q are placed distance R apart. Where should a third charge be placed so that it is in equilibrium. [a] at a point on the right of +2Q [b] at a point on the left of charge –Q [c] between –Q and +2Q [d] at a point on a line perpendicular to line joining –Q and +2Q

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A 22-gauge cylindrical Nichrome wire has a radius of 0.321 mm. The resistivity of Nichrome is ρ= 1.5x10 -6 Ωm. If a potential of 10 V is maintained across a 1.0 m long Nichrome wire, what is the current in the wire? A) 22 mA B) 280 mA C) 460 mA D) 1.5 A E) 2.2 A

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The magnitude of the charge on an electron is approximately: A) 10 23 C B) 10 –23 C C) 10 19 C D) 10 –19 C E) 10 9 C

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A 5.0-C charge is 10 m from a –2.0-C charge. The electrostatic force is on the positive charge is: A) 9.0 × 10 8 N toward the negative charge B) 9.0 × 10 8 N away from the negative charge C) 9.0 × 10 9 N toward the negative charge D) 9.0 × 10 9 N away from the negative charge E) none of these

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Two identical charges, 2.0 m apart, exert forces of magnitude 4.0 N on each other. The value of either charge is: A) 1.8 × 10 –9 C B) 2.1 × 10 –5 C C) 4.2 × 10 –5 C D) 1.9 × 10 5 C E) 3.8 × 10 5 C An electron traveling north enters a region where the electric field is uniform and points north. The electron: A) speeds up B) slows down C) veers east D) veers west E) continues with the same speed in the same direction

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If 500 J of work are required to carry a 40-C charge from one point to another, the potential difference between these two points is: A) 12.5 V B) 20,000 V C) 0.08 V D) depends on the path E) none of these

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A parallel-plate capacitor has a plate area of 0.2 m 2 and a plate separation of 0.1 mm. To obtain an electric field of 2.0 × 10 6 V/m between the plates, the magnitude of the charge on each plate should be: A) 8.9 × 10 ¯7 C B) 1.8 × 10 ¯6 C C) 3.5 × 10 ¯6 C D) 7.1 × 10 ¯6 C E) 1.4 × 10 ¯5 C

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A ball is held 50 cm in front of a plane mirror. The distance between the ball and its image is: A) 100 cm B) 150 cm C) 200 cm D) zero E) 50 cm

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An object is 30 cm in front of a converging lens of focal length 10 cm. The image is: A) real and larger than the object B) real and the same size than the object C) real and smaller than the object D) virtual and the same size than the object E) virtual and smaller than the object

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An erect object is 2f in front of a convex lens of focal length f. The image is: A) real, inverted, magnified B) real, erect, same size C) real, inverted, same size D) virtual, inverted, reduced E) real, inverted, reduced

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A 3-cm high object is in front of a thin lens. The object distance is 4 cm and the image distance is –8 cm. The image height is: A) 0.5 cm B) 1 cm C) 1.5 cm D) 6 cm E) 24 cm

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A convex lens has a focal length f. An object is placed between infinity and 2f from the lens on its axis. The image formed is located: A. at 2f. B. between f and 2f. C. at f. D. between the lens and f.

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Which diagram best represents image I, which is formed by placing object O in front of a plane mirror? A. B. C. D.

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The diagram below shows an arrow placed in front of a converging lens. The lens forms an image of the arrow that is A. real and inverted B. real and erect C. virtual and inverted D. virtual and erect

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A parallel-plate capacitor has a capacitance of C. If the area of the plates is doubled and the distance between the plates is halved, what is the new capacitance? C/2 C/4 2C 4C

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10) In the figure. If R=11Ω, what is the equivalent resistance between points a and p? 1 A)26 Ω B) 31 Ω C)4.6 Ω D) 36 Ω

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Water flows through a pipe. The diameter of the pipe at point B is larger than at point A. Where is the speed of the water greater? a.Point A b.Point B c.Same at both A and B d.Cannot be determined from the information given.

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Oil is flowing at a speed of 1.22 m/s through a pipeline with a radius of 0.305 m. How much oil flows in 1 day?

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The volume rate of flow in an artery supplying the brain is 3.6x10 -6 m 3 /s. If the radius of the artery is 5.2 mm, determine the average blood speed. Find the average blood speed if a constriction reduces the radius of the artery by a factor of 3 (without reducing the flow rate).

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Examples A diverging lens with f = -20 cm h = 2 cm, s = 30 cm The image is virtual and upright A converging lens with f = 10 cm (a) s = 30 cm (b) s = 10 cm (c) s = 5 cm The image is real and inverted The image is virtual and upright The image is at infinity

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46 Review Problem: A ball is dropped from rest and falls towards the earth due to gravity. After a distance d, the velocity of the ball is proportional to 1) d 1/2 2) d 3) d 2 4) d 4 5) I have no idea

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A particle with a mass of 3.5 g and charge of +0.045 C is released from rest at point A. The electric field strength is E=1200 V/m. (a) In which direction will this charge move? (b) What speed will it have after moving through a distance of 5.0 cm? F = qE q=+0.045·10 C > 0, Force parallel to E Charge accelerates to left Work done on charge = qEd = Change in Kinetic Energy W = (0.045·10 C)(1200 V/m)(0.05m) = 54. ·10 V ·C =0.054J (K f -K i ) = W (1/2) mv f 2 – 0 = 0.054 J v f 2 = 2 (0.054 J) / ( 3.5 ·10 kg)= 30.8 m 2 /s 2 v f =5.6 m/s F=qE

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48 Two test charges are brought separately into the vicinity of a charge +Q. First test charge +q is brought to point A a distance r from +Q. Then +q is removed and test charge +2q is brought to a point B a distance 2r from +Q. The electrostatic potential felt by which test charge is greater: 1)+q 2)+2q 3)It is the same for both +Q r 2r2r +q+q +2q

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49 Two test charges are brought separately into the vicinity of a charge +Q. First test charge +q is brought to a point a distance r from +Q. Then this charge is removed and test charge -q is brought to the same point. The electrostatic potential energy of which test charge is greater: 1)+q 2)-q 3)It is the same for both +Q r r +q+q -q-q

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50 Two test charges are brought separately into the vicinity of a charge +Q. First test charge +q is brought to point A a distance r from +Q. Then +q is removed and test charge +2q is brought to a point B a distance 2r from +Q. The electrostatic potential felt by which test charge is greater: 1)+q 2)+2q 3)It is the same for both +Q r2r2r +q+q +2q

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51 Rank in order, from largest to smallest, the potential energies U a to U d of these four pairs of charges. Each + symbol represents the same amount of charge. 1. U a = U b > U c = U d 2. U a = U c > U b = U d 3. U b = U d > U a = U c 4. U d > U b = U c > U a 5. U d > U c > U b > U a

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52 An electron is accelerated from rest through a potential difference V. Its final speed is proportional to 1. V 2. V 2 3. V 1/2 4. 1/V 5. 1/V ½ 6. You need to know the x dependence of V(x)

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53 A proton is released from rest at point B, where the potential is 0 V. Afterward, the proton 1. moves toward A with an increasing speed. 2. moves toward A with a steady speed. 3. remains at rest at B. 4. moves toward C with a steady speed. 5. moves toward C with an increasing speed.

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54 E d i f b Consider a charge q in a uniform field E. How much work does the electric field do in moving the charge from i to b ? 1)qEdcos 2)qEdsin 3)qEd 4) -qEdcos 5) -qEdsin 6) -qEd 7) Zero 8) Impossible to determine

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55 E d i f b Consider a charge q in a uniform field E. How much work does the electric field do in moving the charge from b to f ? 1)qEdcos 2)qEdsin 3)qEd 4) -qEdcos 5) -qEdsin 6) -qEd 7) Zero 8) Impossible to determine

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56 The electric potential inside a capacitor 1. is constant. 2. increases linearly from the negative to the positive plate. 3. decreases linearly from the negative to the positive plate. 4. decreases inversely with distance from the negative plate. 5. decreases inversely with the square of the distance from the negative plate.

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57 If 500 J of work is required to carry a 40 C charge from one point to another, the potential difference between the two points is 1. 12.5 V 2. 20,000 V 3. 0.08 V 4. depends upon the path 5. none of these From U = qV (potential energy = charge potential) we get U = q V. Using potential differences:

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58 During a lightning discharge, 30 C of charge move through a potential difference of 1.0 10 8 V in 0.02 s. The energy released by this lightning bolt is 1. 1.5 10 11 J 2. 3.0 10 9 J 3. 6.0 10 7 J 4. 3.3 10 6 J 5. 1500 J U = q V.

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59 U = q V. Sample Problem: In a cathode ray tube (CRT, like in your TV or computer monitor) electrons initially at rest are accelerated through a potential difference of 10 kV. What is their final speed? Solution. The final kinetic energy equals the change in potential energy, m e v 2 /2 = e V v = (2e V/m e ) 1/2 = [2(1.6e-19 C)(10e3 V)/(9.11e-31kg)] 1/2 = 5.93e7 m/s.

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60 Suppose that a beam of 0.2-MeV photon is scattered by the electrons in a carbon target. What is the wavelength of those photon scattered through an angle of 90 o ? A. 0.00620 nm B. 0.00863 nm C. 0.01106 nm D. 0.00243 nm E. Non of the above Example

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61 Solution First calculate the wavelength of a 0.2 MeV photon: E = hc/ = 1240 eV nm/ = 0.2 MeV =1240 nm / 0.2 x 10 6 = 0.062 nm From Compton scattering formula, the shift is = ’ = e (1 – cos 90 ) = e Hence, the final wavelength is simply ’ = + = e + 0.00243nm + 0.062 nm = 0.00863 nm ANS: B,

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62 X-rays of wavelength 0.2400 nm are Compton scattered and the scattered beam is observed at an angle of 60 degree relative to the incident beam. Find (a) the wave length of the scattered x-rays, (b) the energy of the scattered x-ray photons, (c) the kinetic energy of the scattered electrons, and (d) the direction of travel of the scattered electrons Example

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63 ’= + e (1 - cos ) = 0.2400nm+0.00243nm(1–cos60 o ) = 0.2412 nm ’ = hc/ ’ = 1240 eV nm /0.2412 nm = 5141 eV solution

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64 EE K E’ < E meme Initial photon kinetic energy gained by the scattered electron = energy transferred by the incident photon during the scattering: K.E = hc/ - hc/ ’ =(5167–5141)eV = 26 eV pp p’ pepe

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65 (c)A 0.0016-nm photon scatters from a free electron. For what scattering angle of the photon do the recoiling electron and the scattered photon have the same kinetic energy? Example

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66 Solution The energy of the incoming photon is E i = hc/ = 0.775 MeV Since the outgoing photon and the electron each have half of this energy in kinetic form, E f = hc/ ’ = 0.775 MeV / 2 = 0.388 MeV and ’ = hc/E f = 1240 eV nm / 0.388 MeV = 0.0032 nm The Compton shift is = ’ - = (0.0032 – 0.0016) nm = 0.0016 nm By = c (1 – cos ) = (h/m e c) (1 – cos ) 0.0016 nm = 0.00243 nm (1 – cos ) = 70 o

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67 Which of the following statement(s) is (are) true? I.Photoelectric effect arises due to the absorption of electrons by photons II.Compton effect arises due to the scattering of photons by free electrons III. In the photoelectric effect, only part of the energy of the incident photon is lost in the process IV.In the Compton effect, the photon completely disappears and all of its energy is given to the Compton electron A. I,IIB. II,III,IVC. I, II, III D. III,IVAns: E [I = false; II = true; III = false; IV = false] Example

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68 To produce an x-ray quantum energy of 10 -15 J electrons must be accelerated through a potential difference of about A. 4 kV B. 6 kV C. 8 kV D. 9 kV E. 10 kV ANS: B

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69 Which of the following statement(s) is (are) true? I. -rays have much shorter wavelength than x-rays II.The wavelength of x-rays in a x-ray tube can be controlled by varying the accelerating potential III. x-rays are electromagnetic waves IV. x-rays show diffraction pattern when passing through crystals A. I,IIB. I,II,III,IVC. I, II, III D. III.IVE. Non of the above Ans: B

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more electronsmore photonsgreater intensity Each photon can only knock out one electron. So to increase the current, we would have to knock out more electrons, which means we need more photons, which means we need a greater intensity! frequencywavelength energy number of electrons Changing the frequency or wavelength will change the energy of each electron, but we are interested in the number of electrons in this case. ConcepTest 27.2ePhotoelectric Effect V A photocell is illuminated with light with a frequency above the cutoff frequency. The magnitude of the current produced depends on: 1) wavelength of the light 2) intensity of the light 3) frequency of the light 4) all of the above 5) none of the above

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A metal surface is struck with light of = 400 nm, releasing a stream of electrons. If the light intensity is increased (without changing ), what is the result? 1) more electrons are emitted in a given time interval 2) fewer electrons are emitted in a given time interval 3) emitted electrons are more energetic 4) emitted electrons are less energetic 5) none of the above higher intensitymore photons more electrons. A higher intensity means a more photons, which in turn means more electrons. On average, each photon knocks out one electron. ConcepTest 27.2dPhotoelectric Effect IV

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reducedwavelengthhigherfrequency higher energymore energetic A reduced wavelength means a higher frequency, which in turn means a higher energy. So the emitted electrons will be more energetic, since they are now being hit with higher energy photons. ConcepTest 27.2cPhotoelectric Effect III A metal surface is struck with light of = 400 nm, releasing a stream of electrons. If the 400 nm light is replaced by = 300 nm light of the same intensity, what is the result? 1) more electrons are emitted in a given time interval 2) fewer electrons are emitted in a given time interval 3) emitted electrons are more energetic 4) emitted electrons are less energetic 5) none of the above c = f E = h f Remember that c = f and that E = h f

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reducedwavelengthhigherfrequency more energetic A reduced wavelength means a higher frequency, which in turn means a higher energy. So the emitted electrons will be more energetic, since they are now being hit with higher energy photons. ConcepTest 27.2cPhotoelectric Effect III A metal surface is struck with light of = 400 nm, releasing a stream of electrons. If the 400 nm light is replaced by = 300 nm light of the same intensity, what is the result? 1) more electrons are emitted in a given time interval 2) fewer electrons are emitted in a given time interval 3) emitted electrons are more energetic 4) emitted electrons are less energetic 5) none of the above c = f E = h f Remember that c = f and that E = h f

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greatercutoff frequencyhigherenergy work function is greater A greater cutoff frequency means a higher energy is needed to knock out the electron. But this implies that the work function is greater, since the work function is defined as the minimum amount of energy needed to eject an electron. ConcepTest 27.2aPhotoelectric Effect I If the cutoff frequency for light in the photoelectric effect for metal B is greater than that of metal A. Which metal has a greater work function? 1) metal A 2) metal B 3) same for both W 0 4) W 0 must be zero for one of the metals f KE f0f0 Follow-up: What would you expect to happen to the work function of a metal if the metal was heated up? of a metal if the metal was heated up?

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greatercutoff frequencyhigherenergy work function is greater A greater cutoff frequency means a higher energy is needed to knock out the electron. But this implies that the work function is greater, since the work function is defined as the minimum amount of energy needed to eject an electron. ConcepTest 27.2aPhotoelectric Effect I If the cutoff frequency for light in the photoelectric effect for metal B is greater than that of metal A. Which metal has a greater work function? 1) metal A 2) metal B 3) same for both W 0 4) W 0 must be zero for one of the metals f KE f0f0 Follow-up: What would you expect to happen to the work function of a metal if the metal was heated up? of a metal if the metal was heated up?

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RADIOACTIVITY Types of Radioactive decay: – Beta emission: Converts neutron into a proton by emission of energetic electron; atomic # increases: E.g. Determine product for following reaction: – Alpha emission: emits He particle. E.g. Determine product: – Positron emission: Converts proton to neutron: E.g. Determine product of Gamma emission: no change in mass or charge but usually part of some other decay process. E.g. Electron capture: electron from electron orbitals captured to convert proton to neutron. E.g. Determine product:

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Problem: A sample contains 4.5g of 1 3 H (tritium), which decays by a - decay to 2 3 He with a half-life of 12.26 yr. (a) what is the activity of the sample? (b) About how long would it take the number of tritium atoms to decrease by a factor of a million? First, we need to find N. The Avogadro number of tritium atoms (the number of atoms in 1 mole) have a mass of 3g. Avogadro’s number N A 6.022045 10 23 Activity of the sample: Decay constant:

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