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A current-carrying wire and coordinate system is shown above. Initially, the wire carries a current I toward the top of the page. The amount of current is steadily decreased until it is 0 A, then steadily increased until it reaches a value of I in the downward direction. This change in current takes time t. The magnetic field versus time at the observation point P from t = 0 until t is graphed. Which statement is true concerning the graph if +z is above the horizontal axis and -z is below the horizontal axis?
A. The graph has a constant slope.
B. The graph is piecewise linear with a negative slope for the first half and a positive slope for the second half.
C. The graph is piecewise linear with a positive slope for the first half and a negative slope for the second half.
D. The graph is a curve showing an inverse relationship.
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B As the current decreases, the magnetic field observed at P will decrease in magnitude. When the current reaches 0, the observed field will be 0. As the current increases, the field magnitude will again increase.
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Physics/176
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Physics
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Radiothorium-228 decays into Radon-220 through two alpha decays, as shown below.Radiothorium-228 decays into Radon-220 through two alpha decays, as shown below.
Radiothorium-228 decays into Radon-220 through two alpha decays, as shown below.Which equation correctly describes the energy released during this process?
A. (mTh - mRn)c2
B. (mTh + 2mHe - mRn)c2
C. (mTh - 2mHe - mRn)c2
D. (mRn + 2mHe - mTh)c2
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D Rest-mass energy of a particle is given by E = mc2. The released energy is given by the mass that remains after the decay less the mass that existed before the decay.
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Physics/177
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The wave function of a quantum object vs. position is graphed above. Which of the following correctly ranks the probabilities of observing the particle at the listed positions?
A. P(x = 0.5 m) > P(x = 2.25 m) > P(x = 1.75 m)
B. P(x = 0.5 m) > P(x = 1.75 m) > P(x = 1.25 m)
C. P(x = 3.0 m) > P(x = 1.75 m) > P(x = 2.0 m)
D. P(x = 3.5 m) > P(x = 3.0 m) > P(x = 1.75 m)
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B The square of the wave function of a quantum object can be interpreted as the likelihood of an observation occurring at the stated position.
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Physics/178
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Physics
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An experiment is conducted to determine the power output from a circuit as various voltages are supplied to the circuit. The circuit is set up so that it draws a constant current. What is the value of the current in the circuit that produced the graph above?
A. 0.00 A
B. 0.25 A
C. 1.0 A
D. 6.0 A
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B Power dissipated in a circuit is directly proportional to the product of the current and voltage. The slope of the line of a power vs. voltage graph gives the current.
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Physics/179
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A positive charge of 2Q and a negative charge of 4Q are arranged at positions as shown above. What is the correct ranking of the electric potential at points along the x-axis?
A. V(x = 4 m) = V(x = 2 m) > V(x = 1 m)
B. V(x = 4 m) > V(x = 1 m) > V(x = 2 m)
C. V(x = 1 m) > V(x = 2 m) > V(x = 4 m)
D. V(x = 4 m) > V(x = 2 m) > V(x = 1 m)
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C In the presence of a positive and a negative point charge, the electric potential will be higher at locations closer to the positive charge and farther from the negative charge.
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Physics/180
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Physics
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An insulating rod separates two conducting spheres as shown above. Point A is midway between the spheres. Point B lies on the axis of the rod. Which of the following arrangements of charges would result in a net torque on the rod?
A. Charge both spheres negatively and place a positive charge at point A.
B. Charge one sphere negatively and the other positively and place a positive charge at point A.
C. Charge both spheres negatively and place a positive charge at point B.
D. Charge one sphere negatively and the other positively and place a positive charge at point B.
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B By the process of elimination, placing charges at B will result in the lever arm being parallel to the force and will not cause torque, eliminating (C) and (D). The sign of the charges on the two spheres must be different to have a net torque.
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Physics/181
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The tube shown above carries water. At some point, the tube splits, as shown above. Three points within the tube are labeled A, B, and C. The greatest diameter is at A and the smallest diameter is at B. How do the pressures in the pipes compare at the three points?
A. PA > PC > PB
B. PA = PB = PC
C. PB = PC > PA
D. PB > PC > PA
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D Pressure is highest when speed is slowest. At the widest point in the tube, position A, the speed will be lowest. Because B is narrower than C, the speed at B will be greater than at C, making the pressure highest at point B.
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Physics/182
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A ray of light enters the pair of tanks as shown above. The light rays in the water and glass tank are shown at the left, with the light exiting the pair of tanks at a point Y. The light ray in the air tank is identical to the ray in the water tank, entering at the same height and same angle. Both sets of tanks use the same glass. How will the ray exiting the second glass tank compare to height Y?
A. The ray will not exit the second tank due to total internal reflection.
B. The exiting ray will be closer to the top of the tank than Y.
C. The exiting ray will be at the same height as Y.
D. The exiting ray will be farther from the top of the tank than Y.
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B The index of refraction of air is less than the index of refraction of water. According to Snell's Law, n1sin(θ1) = n2sin(θ2), a smaller n1 results in a smaller θ2. The ray in the glass will be at an angle closer to the normal than the ray in glass from the water/glass apparatus.
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Physics/183
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Three neutral conducting spheres sit on insulating bases. The spheres are separated by a very large distance. The sphere in the center is given a positive charge. The spheres are brought close together, as shown above, but not allowed to come into contact with one another. Which is the correct description of the net charge on each sphere?
A. All three spheres are positively charged.
B. The center sphere is positively charged and the two outer spheres are negatively charged.
C. The center sphere is positively charged and the other two are neutral.
D. All three spheres are neutral.
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C The spheres do not touch, so there cannot be any charging by contact. While the outside sphere will have non-symmetric charge distributions, each will still have a net charge of 0 C.
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Physics/184
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The instrument in an aircraft to measure airspeed is known as the pitot tube, shown in the figure above. The opening facing the incoming air (with the small aperture) is the part meant to capture the air at rest. The opening perpendicular to the flow of air (with the large aperture) is meant to capture air at speed. If h = 1 m and the fluid within the manometer is water, what is the airspeed? Take the density of air to be Ïair = 1.2 kg/m3.
A. 27 m/s
B. 68 m/s
C. 95 m/s
D. 128 m/s
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D First, use Bernoulli's Equation. Call the point with the air at rest B and the point with the speeding air A. You have vB = 0 and yA = vB, so the equation reduces to The pressure difference, PB = PA arises from the fluid depth Ïfgh. Solving for va and using the given values
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Physics/185
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A beam of light goes from water to air. Depending on the actual angle that the light strikes the surface, which of the following rays are possible outcomes?
A. A only
B. B only
C. A or D
D. C or D
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D Water has a higher index of refraction than air, so if light refracts through the water the light will bend away from the normal-so C is possible. Another possibility is D, total internal reflection when the angle in water is greater than the critical angle.
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Physics/186
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A tube with two T branches that has an open end is inserted in a liquid. However, the section of the tube above part B is hidden from view. The hidden section may be wider or narrower. Air is blown through the tube and the water levels rise as shown. You can conclude which of the following?
A. The picture as drawn is impossible-A and B must be at equal heights.
B. The tube is narrower and the air speed is greater above section B.
C. The tube is narrower and the air speed is less above section B.
D. The tube is wider and the air speed is greater above section B.
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B Because the fluid is higher in column B, the air speed must be greater above column B. This must be due to a narrower tube.
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Physics/187
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In the figure above, a ray of light hits an object and travels parallel to the principal axis as shown by the dotted line. Which line shows the correct continuation of the ray after it hits the concave lens?
A. a
B. b
C. c
D. d
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A The rules for ray-tracing diagrams state that a line parallel to the principle axis bends away from the focal point as shown.
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Physics/188
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A machine shoots a proton, a neutron, or an electron into a magnetic field at various locations. The paths of two particles are shown above. Assume they are far enough apart so that they do not intersect and the magnetic field is going out of the page, as shown. What can you say about the paths that represent each particle?
A. a is the proton and b is the electron.
B. b is the proton and a is the electron.
C. Either may be a neutron.
D. You cannot make any conclusions without knowing the velocities.
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A Path a follows the right-hand rule so must be positively charged. Path b is opposite the right-hand rule so must be negatively charged. Neutrons would follow a straight line path, so it is impossible for either a or b to be a neutron.
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Physics/189
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Which statement correctly characterizes the work done by the gas during the ABCDA cycle shown in the above P-V diagram?
A. There is no work done by the gas because the system both starts and concludes in state A.
B. There is no work done because the work done during the transition from AâB cancels out the work done in transition from CâD.
C. The work done by the gas is positive because the work done during the transition from AâB is greater than the work done in transition from CâD.
D. The work done by the gas is positive because the work done during the transition from BâC is greater than the work done in transition from DâA.
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C Work is done when there is a change in volume. At constant pressure, the equation W = -PÎV tells you that when P is higher (as it is at path AâB), a greater amount of negative work is done on the gas than at lower P. Thus, the work done by the gas during the entire cycle is positive, and the constant volume paths have no influence on the amount of work.
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Physics/190
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A neutral conducting sphere is hung from a thin insulating string. A positively charged object is brought to point P. The two objects are not allowed to touch. What is true about the string when the positively charged object is present at point P?
A. The tension is the same as before the object was present at point P.
B. The tension is greater than when the object was not at point P, and the string stretches to the left of its original orientation.
C. The tension is greater than when the object was not at point P, and the string stretches to the right of its original orientation.
D. The tension is less than when the object was not at point P, and the string stretches to the left of its original orientation.
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C Negative charges will be drawn to the right side of the sphere, causing a force on the sphere to the right. The vertical tension will still have to balance with gravity, but there will also be a component of tension to the right.
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Physics/191
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Points P and Q lie between the plates of a fully charged parallel plate capacitor as shown above. The lower plate is negatively charged and the upper plate is positively charged. How do the magnitudes of the electric fields at points P and Q compare?
A. The field is 0 N/C at both points.
B. The field is the same at both points, but not 0 N/C.
C. EP > EQ
D. EQ > EP
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B Between the plates of a parallel plate capacitor, the field will be constant and non-zero.
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Physics/192
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Physics
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A pipe with a diameter of D splits into two smaller, identical pipes with diameter d. If the speed of the water in the small pipes is v, what is the speed of the water in the large pipe?
A. dv/D
B. 2dv/D
C. d2v/D2
D. 2d2v/D2
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D The continuity equation, A1v1 = A2v2 applies here as A1v1 = 2A2v2 where position 1 is in the large pipe. Solving for > https://img.apstudy.net/ap/physics-2/br20/Page_031_Image_0001.jpg .
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Physics/193
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What's the missing particle in the following nuclear reaction?What's the missing particle in the following nuclear reaction?
A. Proton
B. Neutron
C. Electron
D. Positron
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B In order to balance the mass number (the superscripts), you must have 2 + 63 = 64 + A, so A = 1. In order to balance the charge (the subscripts), you need 1 + 29 = 30 + Z, so Z = 0. A particle with a mass number of 1 and no charge is a neutron, https://img.apstudy.net/ap/physics-2/br20/Page_328_Image_0001.jpg .
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Physics/194
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A partial energy-level diagram for an atom is shown below. What photon energies could this atom emit if it begins in the n = 3 state?A partial energy-level diagram for an atom is shown below. What photon energies could this atom emit if it begins in the n = 3 state?
A partial energy-level diagram for an atom is shown below. What photon energies could this atom emit if it begins in the n = 3 state?
A. 5 eV only
B. 3 eV or 7 eV only
C. 2 eV, 3 eV, or 7 eV
D. 3 eV, 4 eV, or 7 eV
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D If the atom begins in the n = 3 state, it could lose energy by making any of the following transitions: 3 â 2, 3 â 1, or 3 â 2 â 1. The 3 â 2 transition would result in the emission of -5 eV - (-8 eV) = 3 eV; the 3 â 1 transition would emit a -5 eV - (-12 eV) = 7 eV photon; and the 2 â 1 transition would result in the emission of -8 eV - (-12 eV) = 4 eV. Therefore, if the atom is initially in the n = 3 state, it could emit photons of energy 3 eV, 4 eV, or 7 eV.
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Physics/195
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The above picture shows a converging mirror and an object, as well as the image, which is formed from standard ray tracing. After the image is formed, an opaque block, also shown in the picture, is inserted to block the top half of the mirror. What changes, if any, will be observed in the image?
A. The image will remain complete but will be diminished in brightness.
B. Part of the image will be absent, but the rest will be as bright was it was before the insertion of the block.
C. Part of the image will be absent and the part which remains will be diminished in brightness.
D. No change will be observed.
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A All of the rays which originate at the tip of the image and travel along paths that remain above the optic axis will be blocked. However, all of the rays which originate at the tip of the image and travel along paths that are below the optic axis at the plane of the block will be unaltered. As a result, fewer rays will converge at the image location, resulting in a less bright image.
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Physics/196
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The above picture shows a converging mirror and an object, as well as the image, which is formed from standard ray tracing. Also shown, a dotted vertical line indicates the image place for this configuration. Also shown in a ray that travels from the top of the image to the point P on the curved mirror. The point P lies above the line which is parallel to the optic axis from the tip of the image. Where will the reflected ray from point P intersect the image plane?
A. On the optic axis
B. Below the optic axis but above the tip of the image
C. At the tip of the image
D. Farther from the optic axis than the tip of the image
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C Every ray, regardless of whether it is a principle ray or not, which originates at the tip of an object and strikes mirror, will converge at the tip of the image.
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Physics/197
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A beam of light in air is incident upon the smooth surface of a piece of flint glass, as shown:A beam of light in air is incident upon the smooth surface of a piece of flint glass, as shown:
A beam of light in air is incident upon the smooth surface of a piece of flint glass, as shown:As the incident angle is increased toward θ = 90°, what observation is made of the refracted ray? All angle references are relative to the surface as shown for both rays.
A. The refracted ray angle increases as the incident angle increases, but the value of the refracted angle is always smaller than the incident angle.
B. The refracted ray angle increases as the incident angle increases, but the value of the refracted angle is always larger than the incident angle.
C. The refracted ray angle increases as the incident angle increases until at some angle total internal reflection begins to occur.
D. The refracted ray angle decreases as the incident angle increases, but the value of the refracted angle is always smaller than the incident angle.
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B As light travels into an optically dense medium, it will refract in toward the normal and away from the surface. So the light in the glass will always be at a greater angle from the surface than the light in the air, so (B) is correct. Note that total internal reflection, (C), will not occur in this situation because it only happens with the initial medium being more optically dense. https://img.apstudy.net/ap/physics-2/br20/Page_323_Image_0001.jpg
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Physics/198
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As shown in the figures below, a bar magnet is moved as a constant speed through a loop of wire. Figure A shows the bar magnet when it is as a position below the loop of wire and figure B shows the loop of wire after it has passed completely through the loop.As shown in the figures below, a bar magnet is moved as a constant speed through a loop of wire. Figure A shows the bar magnet when it is as a position below the loop of wire and figure B shows the loop of wire after it has passed completely through the loop.
As shown in the figures below, a bar magnet is moved as a constant speed through a loop of wire. Figure A shows the bar magnet when it is as a position below the loop of wire and figure B shows the loop of wire after it has passed completely through the loop.Which of the following best describes the direction or directions of the current induced in the loop when the loop is looked at from above? Note that when looking at the loop from above, the bar magnet will be moving toward the viewer.
A. Always clockwise
B. Always counterclockwise
C. First clockwise, then counterclockwise
D. First counterclockwise, then clockwise
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C First, you can eliminate (A) and (B). By definition, magnetic field lines emerge from the north pole and enter at the south pole. The magnetic field from the bar magnet will always point toward a viewer who is looking down at the loop from above. As the north pole gets closer to the loop, the field at the loop grows in strength, but as the south pole recedes from the loop, the magnetic field strength shrinks. Because of this, the flux changes from growing to shrinking as the magnet passes through the loop and the induced current must also change directions. Therefore, (A) and (B) are wrong. To determine whether (C) or (D) is correct, look at the first half of the motion. As the north pole get closer to the loop, the magnetic flux increases. To oppose an increasing flux, the direction of the magnetic field generated by induced current must be downward. Looking from above, a clockwise-induced current generates a downward magnetic field. Therefore, (C) is correct.
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Physics/199
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A conducting rod of length 0.2 m and resistance 10 ohms between its endpoints slides without friction along a U-shaped conductor in a uniform magnetic field B of magnitude 0.5 T perpendicular to the plane of the conductor, as shown in the diagram below.A conducting rod of length 0.2 m and resistance 10 ohms between its endpoints slides without friction along a U-shaped conductor in a uniform magnetic field B of magnitude 0.5 T perpendicular to the plane of the conductor, as shown in the diagram below.
A conducting rod of length 0.2 m and resistance 10 ohms between its endpoints slides without friction along a U-shaped conductor in a uniform magnetic field B of magnitude 0.5 T perpendicular to the plane of the conductor, as shown in the diagram below.If the rod is moving with velocity v = 3 m/s to the left, what is the magnitude and direction of the current induced in the rod?
A conducting rod of length 0.2 m and resistance 10 ohms between its endpoints slides without friction along a U-shaped conductor in a uniform magnetic field B of magnitude 0.5 T perpendicular to the plane of the conductor, as shown in the diagram below.Current Direction
A. 0.03 A down
B. 0.03 A up
C. 0.3 A down
D. 0.3 A up
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A As we discussed in this chapter, the magnitude of the emf induced between the ends of the rod is ε = BLv = (0.5 T)(0.2 m)(3 m/s) = 0.3 V. Since the resistance is 10 Ω, the current induced will be I = V/R = (0.3 V)/(10 Ω) = 0.03 A. To determine the direction of the current, note that since positive charges in the rod are moving to the left and the magnetic field points into the plane of the page, the right-hand rule states that the magnetic force, qv à B, points downward. Since the resulting force on the positive charges in the rod is downward, so is the direction of the induced current.
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Physics/200
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Physics
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What is the direction of force acting on the current-carrying wire as shown below?What is the direction of force acting on the current-carrying wire as shown below?
A. To the bottom of the page
B. Into the page
C. Out of the page
D. To the right of the page
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C Magnetic fields point from north to south. Therefore, the magnetic field between the two magnets is toward the right of the page. Use the right-hand rule. Because the B field is to the right and the charges through the wire flow to the bottom of the page, the force must be out of the page.
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Physics/201
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Here is a section of a wire with a current moving to the right. Where is the magnetic field strongest and pointing INTO the page?Here is a section of a wire with a current moving to the right. Where is the magnetic field strongest and pointing INTO the page?
A. A
B. B
C. C
D. D
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C Use the right-hand rule for wires. If you point your thumb to the right and wrap your fingers along the wire, you will note that the magnetic field goes into the page when you are below the wire and comes out of the page above the wire. This allows you to eliminate (A) and (B). Because https://img.apstudy.net/ap/physics-2/br20/Page_319_Image_0003.jpg , the closer you are to the wire, the stronger the magnetic field. Choice (C) is closer, so it is the correct answer.
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Physics/202
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In the figure below, what is the magnetic field at the Point P, which is midway between the two wires?In the figure below, what is the magnetic field at the Point P, which is midway between the two wires?
A. 2μ0I/(Ïd), into the plane of the page
B. μ0I/(2Ïd), out of the plane of the page
C. μ0I/(2Ïd), into the plane of the page
D. Zero
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D The strength of the magnetic field at a distance r from a long, straight wire carrying a current I is given by the equation B = (μ0/2Ï)(I/r). Therefore, the strength of the magnetic field at Point P due to either wire is B = (μ0/2Ï)(I/ https://img.apstudy.net/ap/physics-2/br20/1by2.jpg d). By the right-hand rule, the direction of the magnetic field at P due to the top wire is into the plane of the page and the direction of the magnetic field at P due to the bottom wire is out of the plane of the page. Since the two magnetic field vectors at P have the same magnitude and opposite directions, the net magnetic field at Point P is zero.
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Physics/203
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In the figure below, what must be the direction of the particle's velocity, v?In the figure below, what must be the direction of the particle's velocity, v?
A. Downward, in the plane of the page
B. Upward, in the plane of the page
C. Out of the plane of the page
D. Into the plane of the page
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C Since FB is always perpendicular to v, v cannot be upward or downward in the plane of the page; this eliminates (A) and (B). Because the charge is positive, the direction of FB will be the same as the direction of v à B. In order for v à B to be downward in the plane of the page, the right-hand rule implies that v must be out of the plane of the page.
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Physics/204
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A particle of charge -q moves to the left at speed v through a uniform magnetic field B which points into the second quadrant, as shown below.A particle of charge -q moves to the left at speed v through a uniform magnetic field B which points into the second quadrant, as shown below.
A particle of charge -q moves to the left at speed v through a uniform magnetic field B which points into the second quadrant, as shown below.A second experiment is performed which results in the same magnitude and same direction of the magnetic force on the particle. Which of the following could NOT be the conditions for the new experiment?
A. B is unchanged, but q is now positive and v is directed to the right.
B. The charge is still negative, but v and B are rotated 90° into the plane of the paper so that v is directed along the -z-axis.
C. The charge is still negative, but v and B are rotated 90° counterclockwise so that v is directed along the -y-axis.
D. V is unchanged, but q is now positive and B is rotated 180° to point into the fourth quadrant.
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B All of the new experimental set-ups have the same magnitude of force as the original. In order to have the new experiment result in the same magnitude of the force, FB = |q||v||B| sin(θ) must be unchanged. Changing the signs of q, v, or B do not change the strength of the force. Rotating v and B by the same amount leaves θ unchanged, and rotating B by 180° results in the same value for sin(θ). Thus, the invalid experiment must have a different direction of the magnetic force. Using the right-hand rule, rotating your thumb and fingers by 90° into the plane of the paper causes the magnetic force to rotate from the original direction of into the plane (recalling q is negative) to finally point to the right.
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Physics/205
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If each of the capacitors in the array shown above is C, what is the capacitance of the entire combination?
A. C/2
B. 2C/3
C. 5C/6
D. 2C
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D https://img.apstudy.net/ap/physics-2/br20/Page_316_Image_0001.jpg Capacitors 1 and 2 are in series, so their equivalent capacitance is C1-2 = C/2. (This is obtained from the equation 1/C1-2 = 1/C1 + 1/C2 = 1/C + 1/C = 2/C.) Capacitors 4 and 5 are also in series, so their equivalent capacitance is C4-5 = C/2. The capacitances C1-2, C3, and C4-5 are in parallel, so the overall equivalent capacitance is (C/2) + C + (C/2) = 2C.
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Physics/206
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What is the voltage drop across the 12 Ω resistor in the portion of the circuit shown above?
A. 24 V
B. 36 V
C. 48 V
D. 72 V
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B The three parallel resistors are equivalent to a single 2 Ω resistor, because https://img.apstudy.net/ap/physics-2/br20/Page_315_Image_0003.jpg . This 2 Ω resistance is in series with the given 2 Ω resistor, so their equivalent resistance is 2 Ω + 2 Ω = 4 Ω. Therefore, three times as much current will flow through this equivalent 4 Ω resistance in the top branch as through the parallel 12 Ω resistor in the bottom branch, which implies that the current through the bottom branch is 3 A, and the current through the top branch is 9 A. The voltage drop across the 12 Ω resistor is therefore V = IR = (3 A)(12 Ω) = 36 V.
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Physics/207
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Physics
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Three identical light bulbs are connected to a source of emf, as shown in the diagram above. What will happen if the middle bulb burns out?
A. The light intensity of the other two bulbs will decrease (but they won't go out).
B. The light intensity of the other two bulbs will increase.
C. The light intensity of the other two bulbs will remain the same.
D. More current will be drawn from the source of emf.
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C If each of the identical bulbs has resistance R, then the current through each bulb is ε/R. This is unchanged if the middle branch is taken out of the parallel circuit. (What will change is the total amount of current provided by the battery.)
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Physics/208
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The circuit shown above has a constant voltage between points a and b. The voltage has been flowing for a long time. What is the resistance of the circuit?
A. 0.47 Ω
B. 1.5 Ω
C. 2.1 Ω
D. 10 Ω
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C Because the voltage has been established for a long time, the capacitor will be full, so it behaves like an open switch. This puts 4 Ω resistor and the top 3 Ω resistor in series with one another, making an equivalent resistor of 7 Ω. That combination in parallel with the other 3 Ω resistor, so https://img.apstudy.net/ap/physics-2/br20/Page_315_Image_0002.jpg .
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Physics/209
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Physics
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A charge Q creates an electric field through which a second charge q moves, as shown below. q is initially at point B and is moved to point A. The potential from Q at position A is VA = 100 V and at B is VB = 200 V. The charge on q is negative. What is the sign of Q and the sign of the work done by the electric field of Q as q is moved from B to A?A charge Q creates an electric field through which a second charge q moves, as shown below. q is initially at point B and is moved to point A. The potential from Q at position A is VA = 100 V and at B is VB = 200 V. The charge on q is negative. What is the sign of Q and the sign of the work done by the electric field of Q as q is moved from B to A?
A. Q is positive and the work done by the electric field is positive.
B. Q is positive and the work done by the electric field is negative.
C. Q is negative and the work done by the electric field is positive.
D. Q is negative and the work done by the electric field is negative.
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B The charge Q must be positive because the potential decreases as the distance from the charge increases. A negative charge will be attracted to a positive charge. This means that the electric field of Q would do positive work if q were brought closer, and it would do negative work if q were moved farther away.
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Physics/210
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Physics
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Which points in this uniform electric field (between the plates of the capacitor) shown above lie on the same equipotential?
A. 1 and 3 only
B. 2 and 4 only
C. None lie on the same equipotential.
D. 1, 2, 3, and 4 all lie on the same equipotential since the electric field is uniform.
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B Because E is uniform, the potential varies linearly with distance from either plate (ÎV = Ed). Since Points 2 and 4 are at the same distance from the plates, they lie on the same equipotential. (The equipotentials in this case are planes parallel to the capacitor plates.)
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Physics/211
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Physics
| null |
Below is shown a section near the center of a parallel plate capacitor. There are 4 labeled positions between the plates shown as A, B, C, and D. Relative to A, which point has the largest potential difference and why?Below is shown a section near the center of a parallel plate capacitor. There are 4 labeled positions between the plates shown as A, B, C, and D. Relative to A, which point has the largest potential difference and why?
A. Point A because the potential difference is infinite when the position between points is 0 m.
B. Point B because it is the same distance from the -Q plate as A.
C. Point C because it is farther in distance from Point A.
D. Point D because it is closest to the +Q Plate.
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D Equipotential curves for a parallel plate capacitor are horizontal lines. Both (A) and (B) are at the same potential as (A), so their potential difference is 0 V. As one moves closer to the +Q plate, the potential will increase, so (D) is correct.
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Physics/212
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Physics
| null |
What is the total work performed on the gas as it is transformed from state a to state c, along the path indicated?What is the total work performed on the gas as it is transformed from state a to state c, along the path indicated?
A. 1,500 J
B. 3,000 J
C. 4,500 J
D. 9,500 J
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C No work is done during the step from state a to state b because the volume doesn't change. Therefore, the work done from a to c is equal to the work done from b to c. Since the pressure remains constant (this step is isobaric), find that W = -PÎV = -(3.0 Ã 105 Pa)[(10 - 25) Ã 10-3 m3] = 4,500 J
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Physics/213
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Physics
| null |
The figure above shows a portion of a conduit for water, one with rectangular cross sections. If the flow speed at the top is v, what is the flow speed at the bottom?
A. 4v
B. 8v
C. 12v
D. 16v
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D Each side of the rectangle at the bottom of the conduit is 1/4 the length of the corresponding side at the top. Therefore, the cross-sectional area at the bottom is (1/4)2 = 1/16 the cross-sectional area at the top. The Continuity Equation states that the flow speed, v, is inversely proportional to the cross-sectional area, A. So, if A at the bottom is 1/16 the value of A at the top, then the flow speed at the bottom is 16 times the flow speed at the top.
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Physics/214
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Physics
| null |
A ball is tied to a string and placed in a container of water as shown. The water is slowly drained from the container until the water level is below the position of the ball shown, but the draining stops while there is still water in the container. The tension in the string is measured while this occurs. Which describes why the tension measurements decrease?
A. The tension decreases because there is less pressure on the ball when the water level drops because of the change in the water depth.
B. The tension decreases because there is less pressure on the ball when the water level drops because the water is moving at a faster speed.
C. The tension decreases because the force of gravity on the ball decreases.
D. The tension decreases because the buoyant force on the ball decreases.
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D The upward force on the ball is the buoyant force. The downward forces are gravity and tension. Gravity will remain constant, so you can eliminate (C). The change in tension results from a change in the buoyant force and not because of pressure, so you can eliminate (A) and (B).
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Physics/215
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Physics
| null |
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