Practice Test: Physics (69)

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Question Number	Your Response	Correct Response (show answershide answers)	Related Objectives and Rationale (show objectives & rationaleshide objectives & rationales)
1	Answer to question 1	C	Objective 001 The number of radioactive atoms remaining after a given period of time, t, can be modeled with the equation N of t = N naught e to the minus lambda t , where N of t is the number of radioactive atoms remaining after time t, N naught is the initial number of atoms, and lambda is equal to (natural log of 2) divided by the half-life ( T half ) of the isotope. Solving the equation for t, t = minus [natural log of N of t divided by N naught ] divided by lambda = minus natural log of [10 to the 3 divided by 10 to the 6] divided by [natural log of 2 divided by 27.7 years] = 276 y, which equals 280 years when rounded to two significant figures (Correct Response C). Incorrect Response A is about 1 half-life. After 30 years, there would be 4.72 times 10 to the 5 strontium-90 atoms. Incorrect Response B is about 4 half-lives. After 110 years, there would be 6.39 times 10 to the 4 strontium-90 atoms. Incorrect Response D is about 26 half-lives. After 730 years, there would most likely be 0 strontium-90 atoms since N(730) = 0.0119.
2	Answer to question 2	D	Objective 001 An emitted photon results from the energy level change of an electron from a high energy orbital to a low energy orbital. The frequency of the emitted photon is directly proportional to this energy change (Correct Response D): h new = delta E = (one over n low squared minus one over n high squared) times 13.6 e V . Responses A, B, and C are incorrect because the orbital radius of the electron about the nucleus (Incorrect Response A), the acceleration of the electron during the transition (Incorrect Response B), and the electromagnetic force acting on the orbiting electron (Incorrect Response C) are not directly related to the frequency of the emitted photon.
3	Answer to question 3	A	Objective 001 The energetic alpha particles can ionize matter and hence cause cell damage or death; however, because alpha particles are massive, large, and highly charged (+2e) they readily interact with matter. Clothing and dead skin layers easily stop alpha particles (Correct Response A), resulting in no damage to interior living cells; however, if they are inhaled or ingested they can damage or kill living cells. Response B is incorrect because the energetic electrons or positrons that are emitted during beta decay are much smaller in mass and size and have a single unit of charge e. Beta particles can pass through cloth and dead skin layers to living cells, causing damage or death to a living cell. Responses C and D are incorrect because these types of energetic electromagnetic radiation, which have no mass or charge, can travel through the exterior of a living organism with ease and interact with molecules in a living cell, causing cellular damage or death.
4	Answer to question 4	C	Objective 001 The binding energy is given by the rest mass energy of the individual components in the nucleus minus the rest mass energy of the bound nucleus or (938.27 M e V + 939.57 M e V minus 1,875.62 M e V) = 2.22 M e V (Correct Response C). Responses A, B, and D are incorrect. The value 0.56 MeV (Incorrect Response A) is one quarter the correct answer, while 1.30 MeV (Incorrect Response B) is approximately half the correct value and is close to the binding energy per nucleon. The value 3.86 MeV (Incorrect Response D) is close to the deuterium's rest energy minus twice the rest mass energy of a neutron plus about 0.3 MeV.
5	Answer to question 5	D	Objective 001 The scattering of the alpha particles that were used to bombard a thin sheet of gold foil demonstrated the existence of a small, dense, positively charged atomic nucleus (Correct Response D). Although Rutherford did develop the concept of half-life of a radioactive element (Incorrect Response A), the alpha-scattering experiments did not contribute to the development of that concept. Creation of new radioactive isotopes by nuclear bombardment (Incorrect Response B) is incorrect because Rutherford's alpha-scattering experiments had nothing to do with creating new isotopes. Changing one atomic element into another through nuclear reactions (Incorrect Response C) is incorrect because Rutherford's alpha-scattering experiments had nothing to do with changing one atomic element into another.
6	Answer to question 6	C	Objective 001 The Pauli exclusion principle (Correct Response C) states that no two electrons of an atom can occupy the same quantum states, which means they cannot have the same set of quantum numbers. If the three electrons were to simultaneously occupy the energy level n = 1, two would have the same spin, which is not permitted within energy level n = 1. Incorrect Response A pertains to the scattering of X-rays or gamma rays as they interact with electrons, resulting in ionization and scattered photons of reduced energy. Incorrect Response B pertains to superconductors that expel magnetic flux when cooled to the point of superconductivity. Incorrect Response D asserts that there is a fundamental limit to the precision with which certain pairs of measurements can be made, such as position and velocity. The more precisely one is measured, the less precisely the other can be known.
7	Answer to question 7	B	Objective 001 Radioactive tracers are used for the medical diagnoses of organs. The tracer can be tracked as it passes through the organ by detecting the gamma radiation emitted by the radioisotope. Radioisotopes with short half-lives are used so that they do not remain in the body for long periods of time (Correct Response B). Incorrect Response A is a true statement that describes the basis for medical radiography commonly known as X-rays and is most commonly performed with X-rays, not gamma rays. Response C is incorrect, because radioactive tracers use smaller quantities and their purpose is not for destroying tissue but for the diagnosis of imaging of organs. Response D is incorrect because radioactive tracers emit gamma radiation, which is ionizing radiation.
8	Answer to question 8	D	Objective 002 The two most important forces at play in the nucleus are the two largest forces, the strong force and the electromagnetic force. The electromagnetic force is repulsive and only about 100th the strength of the strong force. The electromagnetic force must be smaller than the strong force (Correct Response D) for the nucleus to stay bound. Response A is incorrect because the gravitational force is the weakest force, of the order of 10 to the minus 38 times the strength of the strong force and 10 to the minus 36 that of the electromagnetic force. Response B is incorrect because the weak nuclear force is of the order of 10 to the minus 13 that of the strong force or 10 to the minus 11 that of the Coulomb force. Response C is incorrect because neutrons do not have a charge, so the electromagnetic force is not affected by their presence.
9	Answer to question 9	C	Objective 002 The photoelectric effect is one of the primary experiments that provided evidence of energy quantization. In the photoelectric effect, when an electron within some material absorbs the energy of a photon and acquires more energy than its energy, the electron will be ejected. If the photon energy is too low, the electron is unable to escape the material. The energy of the emitted electrons will not depend on the intensity of the incoming light of a given frequency but only on the energy of the individual photons. This shows that there is a specific amount, or quantum, of energy needed to eject the electrons (Correct Response C). Wave interference (Incorrect Response A), electrostatic force (Incorrect Response B), and many-body theory (Incorrect Response D) have no such requirements related to energy quantization.
10	Answer to question 10	C	Objective 002 Classical physics predicted that at short wavelengths, a blackbody in thermal equilibrium would emit an infinite amount of energy, a situation sometimes referred to as the ultraviolet catastrophe. Planck's analysis of the spectrum emitted by a blackbody showed that the situation could be resolved by postulating that light is quantized (Correct Response C). Einstein applied this postulate to explain the photoelectric effect, setting the stage for further acceptance of the idea of quantization. Michelson's and Morley's effort to detect the luminiferous ether (Incorrect Response A) influenced the development of the theory of relativity. Röntgen's discovery of the emission of X-rays from a vacuum tube (Incorrect Response B) represented the discovery of a new type of radiation in the form of electromagnetic waves, but it did not directly lead to the quantum theory of light. Hertz's detection of electromagnetic radiation (Incorrect Response D) demonstrated the existence of electromagnetic waves, lending support to Maxwell's electrodynamic theory. This led to the understanding of light as a wave, not as a photon, which is the electromagnetic particle associated with the quantum theory of light.
11	Answer to question 11	D	Objective 002 The circuit would be based primarily on transistors because transistors can be used to amplify weak signals by outputting higher power to a circuit, given a low power input (Correct Response D). Capacitors (Incorrect Response A) help regulate current in a circuit. Diodes (Incorrect Response B) help control the direction of current flow. Transformers (Incorrect Response C) will adjust the current and voltage values while keeping the same power output. It is the transistor that performs the amplification of the weak signal from the transducer.
12	Answer to question 12	B	Objective 002 The elementary particle moving at 0.99c with respect to the laboratory frame will appear to have a longer half-life to an observer in the laboratory's frame of reference compared with the half-life observed by an observer in the particle's frame of reference. The half-life observed in the particle's frame of reference can be found using the equation delta t naught = delta t times the square root of (1 minus v squared over c squared) = (16 microseconds) times the square root of (1 minus 0.99 squared) = 2.3 microseconds (Correct Response B). Incorrect Response A would be the half-life observed in the particle's frame of reference if the 16 μs half-life had been observed when the relative speed of the particle was 0.9995c. Incorrect Response C would be the half-life observed in the particle's frame of reference if the 16 μs half-life had been observed when the relative speed of the particle was 0.97c. Incorrect Response D would be the half-life observed in the particle's frame of reference if the 16 μs half-life had been observed when the relative speed of the particle was 0.35c.
13	Answer to question 13	A	Objective 002 The de Broglie wavelength of the electron can be determined with the equation lambda = h over p , where h = Planck's constant and p is the momentum of the electron, which is equal to the mass of the electron times its velocity (p = m v ). The kinetic energy of the electron traveling at speeds much less than the speed of light can be determined with the equation E = half m v squared , where m is the mass of the electron and v is its speed. Substituting m v with p in the kinetic energy equation produces the equation E = half p v = half p m v over m = p squared over 2 m . Solving this equation for momentum yields p = root (2 m E) . Substituting momentum with root (2 m E) in the equation for the de Broglie wavelength produces lambda = h over root (2 m E) (Correct Response A). Incorrect Response B shows the denominator of the correct answer being multiplied by its numerator. Incorrect Response C is the reciprocal of the correct answer, except that m appears in the denominator rather than under the radical sign in the numerator. Incorrect Response D shows the denominator of the correct answer being multiplied by its numerator but with m appearing as a factor rather than under the radical sign.
14	Answer to question 14	B	Objective 002 The wave function sigh of x describes the state of a particle, in this case the electron. absolute value of sigh of x squared describes the probability density. To find the relative probability that an electron is in a region delta , the absolute value of sigh of x squared is integrated over delta , so the relative probability equals the integral , where x sub 1 and x sub 2 represent the endpoints of delta . The absolute probability can then be determined by normalizing the relative probability density function absolute value of sigh of x squared so that the total probability over all possible values of x is set to equal one (Correct Response B). Response A is incorrect because it takes the square root of the wave function, not the square of absolute value of sigh of x . Incorrect Response C is the integral of sigh of x not absolute value of sigh of x squared . Incorrect Response D is the momentum operator and is not directly related to probability density.
15	Answer to question 15	B	Objective 003 This problem can be solved with the formula delta x = v average times t = [(v naught + v final) divided by 2] times t = [(23 Meters per second + 29 Meters per second divided by 2] times [10 seconds] = 260 meters (Correct Response B). Incorrect Response A can be obtained using the incorrect formula delta x = v naught t = 23 Meters per second times 10 seconds = 230 meters . Incorrect Response C can be obtained using the incorrect average speed of 27.5 meters per second in the formula (27.5 Meters per second) times (10 seconds) = 275 meters . Incorrect Response D can be obtained using the incorrect formula delta x = v final times t = (29 Meters per second) times (10 seconds) = 290 meters .
16	Answer to question 16	B	Objective 003 This problem can be solved with the formula: v squared = v naught squared + 2 a delta x , where v = final velocity, v naught = initial velocity, a = acceleration, and delta x is the displacement. Solving for a, the equation is rewritten as a = (v squared minus v naught squared) divided by (2 delta x) = [0 minus (12 meters per second) squared] divided by (2) times (20 meters) = negative 3.6 meters per second squared (Correct Response B). Incorrect Response A would be obtained if the factor 2 were omitted from the equation. This would make the solution twice as large, since the solution includes 2 in the denominator of the fraction. Incorrect Response C would be obtained if the factor 2 were omitted from the solution and the velocities were not squared. Incorrect Response D would be obtained by failing to square the velocities in the solution.
17	Answer to question 17	C	Objective 003 To calculate the distance traversed any of the three kinematic relationships that contain the displacement delta x could be applied since the acceleration is constant (a requirement for these equations). The acceleration can be found using the relationship a x = v x minus v naught x divided by t=0.4 meters per second squared , which is just the slope of the line in the graph. For the kinematic relationship delta x = v naught x t + one half a x t squared , the initial velocity at time t = 0 s is found from the graph provided, v naught x = 2 meters per second . Therefore, delta x = v naught x t + one half a x t squared = 2 meters per second times 5 seconds + one half times (0.4 meters per second squared) times 5 seconds squared = 15 meters (Correct Response C). Response A incorrectly uses delta x = one half a x t squared , ignoring initial velocity. Response B incorrectly uses delta x = a x t squared . Response D incorrectly uses delta x = vt, with v equal to the maximum velocity of 4 meters per second .
18	Answer to question 18	D	Objective 003 Resistive forces are assumed to be negligible. To solve for the horizontal distance of travel the following kinematic equation relating position as a function of time is used: a_xt² = 10 ^m/_s (3.0 s) x minus x naught = v naught x times t + one half a x t squared = 10 meters per second times 3.0 seconds + 0 = 30.0 meters The horizontal acceleration is zero (Correct Response D). Responses A and C are incorrect because either an incorrect equation and/or one or more variables were incorrectly assigned. Incorrect Response B uses the wrong component of velocity.
19	Answer to question 19	B	Objective 003 Vector addition is commutative (2 + 5 = 5 + 2) and therefore the steps can be taken out of order (Correct Response B). Response A is incorrect because many random vectors may add up to be zero when added tip to tail but, for this scenario, we do not require the vectors to add up to zero because there should be a net displacement from the start line to the finish line; further, vectors are not added tail to tail. Response C is incorrect because the net displacement cannot be zero for this scenario. Response D is incorrect because the students will have a net displacement (the finish line is not at the location of the start line) and it will take them some time to reach the finish line, hence the average velocity cannot be zero.
20	Answer to question 20	A	Objective 003 The sum of the vectors equals the resultant, Vector V R . The x-component of the resultant equals the sum of the x-components of the vectors being added, so V R x = V 1 x + V 2 x + V 3 x = 4 i hat minus 2 i hat minus 3 I hat = negative 1 I hat . The y-component of the resultant equals the sum of the y-components of the vectors being added, so V R y = V 1 y + V 2 y + V 3 y = 2 j hat plus 1 j hat minus 4 j hat = minus 1 j hat, and vector V R = negative 1 I hat minus 1 j hat (Correct Response A). Incorrect Response B has an incorrect x-component due to a negative sign error. Incorrect Response C has a y-component with a negative sign error. Incorrect Response D has a negative sign error for both components of the vector.
21	Answer to question 21	C	Objective 003 The graph gives v as a function of t. Since d x over d t = v, then d x = v d t and the displacement of the particle is given by the area under the curve. During the first second, the particle moves at a constant positive velocity of 2 m/s, starting from position x = 0. This means the position-versus-time graph increases with a constant, positive slope until the end of the first second, when t = 1 s and x = 2 meters per second times 1 second = 2 m. This position is equal to the area under the velocity graph for this interval: a rectangle of height = 2.0 m/s and width = 1.0 s. During the next 2 seconds (from t = 1 s to t = 3 s), the particle accelerates at a constant rate from 2 m/s to negative 2 m/s. It momentarily comes to rest at t = 2 s as it reverses direction from positive to negative. The position-versus-time graph between t = 1 s and t = 3 s must therefore be initially increasing, peaking at t = 2 s, and then decreasing as the particle travels in the negative direction between t = 2 s and t = 3 s. More specifically, the shape of the position-versus-time curve between 1 and 3 seconds must be parabolic, concave downward, and with its vertex at t = 2 s, making Response C the correct answer while eliminating Incorrect Response B. Note that the slope of the line tangent to the graph also matches the graph given in the question. Response A is incorrect because it does not begin at the origin, it incorrectly indicates the particle is at rest for the first second, and it incorrectly indicates that the position of the particle is continuously decreasing between t = 1 s and t = 3 s. Response D is incorrect because it does not begin at the origin, incorrectly indicates that the particle is moving in the negative direction for the first 2 seconds, and incorrectly indicates that the position of the particle is increasing between t = 2 s and t = 3 s.
22	Answer to question 22	B	Objective 003 The velocity of the boat relative to the shore ( V B S ) equals the vector sum of the velocity of the boat relative to the water ( V B W ) + the velocity of the water relative to the shore ( V W S ). Since the vectors being added form a right angle, the Pythagorean theorem can be used to calculate the magnitude of the resultant. The magnitude of V B S is the square root of (2.5 meters per second squared + 1.5 meters per second squared) = 2.9 m/s, and its direction relative to the east, theta = inverse tangent of (2.5 divided by 1.5) = 59 degrees north of east, so V B S = 2.9 m/s, 59° north of east (Correct Response B). Response A has an incorrect direction that is the complement of theta . This direction can be obtained by making the mistake of using theta = inverse tangent of (1.5 divided by 2.5) rather than inverse tangent of (2.5 divided by 1.5) . Incorrect Response C has the wrong magnitude equal to the scalar sum, V B W + V W S = 2.5 m/s + 1.5 m/s = 4.0 m/s. Incorrect Response D has the wrong magnitude equal to the scalar sum, V B W + V W S = 2.5 m/s + 1.5 m/s = 4.0 m/s, and the wrong direction, obtained using theta = inverse tangent of (1.5 divided by 2.5) rather than inverse tangent of (2.5 divided by 1.5) .
23	Answer to question 23	D	Objective 004 The magnitude of the acceleration can be calculated with the equation a = F net divided by m , where a is the acceleration of the object, F net is the net force on the object, and m is the mass of the object. The net force can be determined from its x-component with the equation F net x = F net cosine theta . Solving the equation for F net results in F net = F net x divided by cosine theta = 30 Newtons divided by cosine 50 degrees = 46.7 Newtons, so a = F net divided by m = (46.7 Newtons) divided by (10 kilograms) = 4.7 m/s² (Correct Response D). Response A is obtained by incorrectly using F net x cosine theta in place of F net x divided by cosine theta . Response B is obtained by incorrectly using F net x sine theta in place of F net x divided by cosine theta . Response C is obtained by incorrectly using F net x divided by sine theta in place of F net x divided by cosine theta .
24	Answer to question 24	C	Objective 004 An approach to solving this problem is to draw a free body diagram and then apply Newton's second law, summing the forces in each direction. A force diagram of the box with F N pointed up, F g pointed down, F friction pointed left and F Horizontal pointed right with a value of 75 Newtons. First equation: The sum of all forces in the y = 0 = F N minus m g yields F N = mg = 20 N Second equation: The maximum static friction force = mew static F N = 0.40 times 20 N) = 8.0 N Third equation: The kinetic friction force = mew kinetic F N = 0.38 times 20 N) = 7.6 N Since the horizontal applied is greater than the opposing maximum static friction, the box will move with a net force equal to the applied horizontal force minus the kinetic frictional force, or The sum of all forces in the x = m a = F Horizontal minus the kinetic friction force yields m a = 75 N minus 7.6 N = 67.4 N A non-zero net force will cause the velocity of the box to increase (Correct Response C). Response A is incorrect because the maximum static frictional force is not larger than the applied 75 N horizontal force. Response B is incorrect because the kinetic frictional force is not equal to the applied 75 N horizontal force. Response D is incorrect because to slow down, the net force would need to be opposite to the direction of motion—for example, if the applied horizontal force was removed after the box was in motion.
25	Answer to question 25	D	Objective 004 The normal force results from the incline's surface acting on the mass and points perpendicular to the incline's surface toward the mass, as indicated in the free body diagram. The sum of the forces perpendicular to the incline is zero (Correct Response D): The sum of the forces perpendicular to the incline = zero = F N minus m g cosine theta yields F N = m g cosine theta. A force diagram of the mass on an incline at angle theta from horizontal. The gravity force points down, the static friction force points up along the incline, and the F N points up perpendicular to the surface of the incline. A dashed line directed down and perpendicular to the incline makes an angle, theta, between the dashed line and the gravity force. Incorrect Responses A, B, and C improperly apply the angle of the incline. Incorrect Response A may indicate a misunderstanding of what the normal force is as it is often incorrectly taken to be equal to the weight, not accounting for the incline angle at all. Incorrect Responses B and C apply the incorrect trigonometric function when calculating the appropriate component of gravity perpendicular to the incline.
26	Answer to question 26	B	Objective 004 This scenario emphasizes Newton's first law. The car places two forces on the cup: the normal force pointing upward on the cup and a frictional force between the surfaces. Since the cup is not moving with the car, static friction was not great enough to move the cup with the car; kinetic friction between the surfaces acts on the cup until the car is done passing underneath it (Correct Response B). Response A is incorrect because the frictional force the car places on the cup is in the same direction as the car's motion. Response C is incorrect as there are only two forces that the car is applying to the cup: the normal force and the frictional force. Response D is incorrect because inertia is a property of matter that resists motion and is not an agent that provides or causes forces.
27	Answer to question 27	B	Objective 004 The drag force always points opposite to the direction of motion (Correct Response B); hence, while the projectile moves upward it has an overall acceleration on it downward that is greater than the acceleration due to gravity because the drag force also points downward, and while the projectile moves downward it has an overall acceleration on it that is less than the acceleration due to gravity because the drag force points upward. These differences result in shorter rise times than fall times. Response A is incorrect because although the force of gravity diminishes the momentum as it travels in time upward, it increases the momentum in a symmetrical way on the way downward, so that the overall result would be equal rise and fall times between the same two vertical points. Response C is incorrect because the acceleration due to gravity results in equal changes in speed and times of flight for the upward and downward motions between the same two vertical points. Response D is not valid because the cannon is not acting on the projectiles after they are launched.
28	Answer to question 28	C	Objective 004 The weight of the space station and the gravitational force of the space station on Earth is the only answer choice that represents an action-reaction pair. The weight of the space station is the force of Earth's gravity on the space station. This force is equal in magnitude and opposite in direction to the gravitational force of the space station on Earth (Correct Response C). The weight of the space station and the centripetal force on the space station (Incorrect Response A) are not an action-reaction pair. The weight of the space station is the force of Earth's gravity on the space station. This is the same as the centripetal force on the space station, which is also the force of Earth's gravity on the space station. The weight of the astronauts and the centripetal force on the space station (Incorrect Response B) are not an action-reaction pair. The weight of the astronauts is the force of Earth's gravity on the astronauts. Its action-reaction pair is the gravitational force of the astronauts on Earth. It is not the centripetal force on the space station, which equals the force of Earth's gravity on the space station. The weight of the astronauts and the gravitational force of the space station on the astronauts (Incorrect Response D) are not an action-reaction pair. The weight of the astronauts is the force of Earth's gravity on the astronauts. This is independent of the gravitational force of the space station on the astronauts.
29	Answer to question 29	D	Objective 004 The x-component of the resultant is equal to the sum of the x-components of the vectors being added, so Sigma F x = negative 4 cosine 60 degrees + 6 cosine 40 degrees (Correct Response D). Response A is incorrect because the applications of the sine and cosine functions are both incorrectly applied. Response B is incorrect because it uses sine 40 degrees where cosine 40 degrees should be used. Response C is incorrect because it uses cosine 60 degrees where sine 60 degrees should be used.
30	Answer to question 30	D	Objective 005 Momentum is conserved, so the total momentum before the collision equals the total momentum after the collision. Furthermore, the total momentum must be set to zero to model the case described in the problem, where the two masses come to rest after colliding. Expressed mathematically, m v one + m v two = m v one prime + m v two prime = 0. Therefore, (6.0 kilograms) times (10 meters per second) + (0.25 kg)v 2 = 0, and v 2 = negative (6.0 kg) times (10 meters per second) divided by (0.25 kilograms) = negative 240 meters per second , so the speed of the 0.25 kg mass must be 240 m/s (Correct Response D). Response A is obtained by incorrectly assuming the mass of each object is identical or that velocity is conserved instead of momentum. Response B is obtained by using the incorrect equation v 2 = negative (6.0 kg) times (10 meters per second) times (0.25 kg) = negative 15 meters per second . Response C is obtained by using the incorrect equation v 2 = negative (6.0 kg) times (10 meters per second) times (0.25 kg) = negative 15 meters per second
31	Answer to question 31	B	Objective 005 Acceleration is defined as velocity change divided by time. Solving for time: t = delta omega divided by alpha = (15 radians per second) divided by (2.5 radians per second squared) = 6.0 s (Correct Response B). Response A incorrectly divides velocity by acceleration squared. Response C incorrectly multiplies velocity and acceleration. Response D incorrectly multiplies velocity and doubles the acceleration.
32	Answer to question 32	A	Objective 005 Torque is force multiplied by the lever arm distance from the rotation axis, which in this case is the axle. A consistent force applied to the string will produce the most torque when applied at the greatest distance from the axle, which is at the wheel (Correct Response A). Response B is incorrect because inertia only depends on the mass distribution of the object rotating and is independent of applied forces. Response C is incorrect because friction is not affected by the string. Response D is incorrect because the number of revolutions depends on how much force is applied and how much time it is applied to the system.
33	Answer to question 33	C	Objective 005 The rod-mass system is in equilibrium when supported at its center of mass. Let x be the distance from M 1 to the system's center of mass. The sum of the torques about x equals zero, so M 1 x = m 2 times (L minus x), (M 1 + M 2) times x = M 2 L, and x = M 2 L divided by (M 1+ M 2) (Correct Response C). Incorrect Response A is obtained by selecting the midpoint of the rod, which is at the distance One half L from m 1 . Incorrect Response B is the correct distance from m2 , not from m1 . Incorrect Response D is actually equal to Response A but is expressed in a more complicated form.
34	Answer to question 34	C	Objective 005 The change in momentum of the mass can be calculated using the impulse momentum theorem: F average times delta t = delta p = m delta v. Thus delta v = F average delta t divided by m = (area under the force versus time curve) divided by m = (4.5 newton seconds) divided by (2.00 kilograms) = (4.5 kilogram meters per second) divided by (2.00 kilograms) = 2.25 meters per second; speed at 5.00 seconds = v initial + delta v = 3.00 m/s + 2.25 m/s = 5.25 m/s (Correct Response C). Response A (2.25 m/s) is incorrect because it is the change in speed, not the final speed. Response B is incorrect because it is twice the change in speed. Response D is incorrect because it is twice the change in speed added to twice the initial speed.
35	Answer to question 35	A	Objective 005 In this situation, there are no external torques on the star so the magnitude of the angular momentum, L = I I omega , will remain constant. Since the angular speed ( omega ) increases, the moment of inertia (I) must decrease (Correct Response A). Angular momentum (Incorrect Response B) does not decrease; it remains constant because no external torques are applied. Rotational kinetic energy (Incorrect Response C) does not decrease; it increases due to the work done by the gravitational force that pulls matter toward the center of the star. The net external torque (Incorrect Response D) is zero and would remain zero as the moment of inertia decreases and the angular speed increases.
36	Answer to question 36	A	Objective 006 The gravitational force is an inverse square force law. If the distance between two masses is doubled, then the force will decrease by a factor of 4. Since the distance between the two masses doubles, then the force will be 80 N divided by 4 = 20 N (Correct Response A). Incorrect Response B would be obtained if the force were assumed to be inversely proportional to the distance, so that doubling the distance would reduce the force to half its original value. Incorrect Response C would be obtained if the force were assumed to be proportional to the distance, so that doubling the distance would double the force. Incorrect Response D would be obtained if the force were assumed to be proportional to the square of the distance, so that doubling the distance would quadruple the force.
37	Answer to question 37	B	Objective 006 For an object in uniform circular motion, the centripetal acceleration is a c = v squared divided by r . In the case of a satellite, a c is the gravitational acceleration due to Earth's gravity. When r is doubled, a c is reduced to a c divided by 4 , so at the altitude 2r, a c divided by 4 = v prime squared divided by 2 r , where v prime is the speed of the satellite at the altitude 2r. Substituting a c with v squared divided by r in the equation for v prime : v squared divided by 4 r = v prime squared divided by 2 r , so v prime squared = (v squared divided by 4 r) times 2 r = v squared divided by 2 , and v prime = v divided by root 2 = 0.7v (Correct Response B). Incorrect Response A would result if one divided by 2 in place of root 2 . Incorrect Response C would result if one multiplied v by root 2 instead of dividing. Incorrect Response D would result if one multiplied v by 2 instead of dividing by root 2 .
38	Answer to question 38	C	Objective 006 Gravity forces in classical physics can only be attractive due to mass always being positive. Coulomb's law is different than the universal law of gravitation because it deals with charge instead of mass and charge can be negative, leading to both attraction and repulsion (Correct Response C). Response A is incorrect because both forces are conservative and have scalar potential functions. Response B is incorrect because both law's equations have 1 over r squared dependence. Response D is incorrect because both laws are applied to all three dimensions of spatial direction.
39	Answer to question 39	A	Objective 006 Both light intensity and gravity force decrease with distance, r, according to the inverse square law: 1 over r squared (Correct Response A). Response B is incorrect because escape speed depends on 1 over r to the one half power . Response C is incorrect because gravitational potential depends on 1 over r . Response D is incorrect because orbital period depends on r to the 3 halves power .
40	Answer to question 40	D	Objective 006 The hanging mass is in equilibrium, therefore big M g = T, the tension in the string. Tension will be proportional to the centripetal force on the swinging mass: T = small m v squared divided by R . Tension must be equal throughout the string so, big M g = small m v squared divided by R . The terms big M g and v are the variables being investigated, holding everything else constant. Since big M g is directly proportional to v squared , only big M g vs v squared will produce a linear graph (Correct Response D). Response A is incorrect because big M g is not directly proportional to v. Response B is incorrect because big Mg squared is not directly proportional to v. Response C is incorrect because the square root of big M g is directly proportional to v, not v squared .
41	Answer to question 41	A	Objective 006 The direction of the acceleration vector for a mass in uniform circular motion is toward the center of its circular path. Therefore, the vector that represents the acceleration of the person riding the Ferris wheel must point straight up while the person is passing through point Y (Correct Response A). The vector pointing to the right (Incorrect Response B) represents the direction of the velocity vector for the person passing through point Y. The vector pointing straight down (Incorrect Response C) represents the direction opposite to the acceleration vector. The vector pointing upward and toward the right (Incorrect Response D) represents a direction between the directions of the velocity and acceleration vectors.
42	Answer to question 42	A	Objective 007 The situation describes charging by induction, in which excess negative charge on the balloon causes the neutral wall to have a charge separation induced within the wall. Free negative charges in the wall are repelled from the excess negative charge in the balloon, leaving a positive charge in the wall closer to the balloon and negative charge in the wall farther from the balloon (Correct Response A). Responses B and C are incorrect because they incorrectly imply that the wall has a net positive charge, not a charge separation in the wall induced by the balloon. Response D is incorrect because it shows negative charges attracted to, and positive charges repelled from, the negative charges in the balloon.
43	Answer to question 43	B	Objective 007 Electric force is a vector. Due to symmetry, the vertical components of the force vectors will balance each other to have zero net force in the vertical direction. Since all charges are positive and the test charge is positive there will be a repulsion to the right only (Correct Response B). Response A is incorrect because all charges are positive and must repel each other, so the net force cannot be zero. Responses C and D are incorrect because the symmetry relative to the test charge is in the y direction so there can be no force in the y direction.
44	Answer to question 44	C	Objective 007 The situation describes charging by induction, in which excess positive charge on the object causes the neutral electroscope to have a charge separation induced within it. Electrons will be attracted to the positively charged object and will move up to the top of the copper wire, leaving positive net charge on the leaves and causing them to repel each other (Correct Response C). Response A is incorrect because it reverses the direction of the flow of electrons. Response B is incorrect because it does not explain any charge separation induced by the object. Response D is incorrect because opposite charges would attract, not repel, each other.
45	Answer to question 45	D	Objective 007 The energy required, equivalent to the work done, to move a charge, q, across a potential difference, V, is: E = absolute value of q times V = (5.0 Coulombs) times (5.0 Volts) = 25 J (Correct Response D). Responses A and B are incorrect because they divide potential by charge. Response C is incorrect because it adds charge and potential.
46	Answer to question 46	B	Objective 007 By convention, electric field lines are shown leaving positive charges and entering negative charges. Based on this convention, Q 1 must be positive, and Q 2 must be negative (Correct Response B). Response A incorrectly indicates that Q 2 is positive. Response C incorrectly indicates that Q 1 is negative and Q 2 is positive. Response D incorrectly indicates that Q 1 is negative.
47	Answer to question 47	A	Objective 007 The electrostatic potential between points A and B is equal to the product of the electric field and the component of displacement that is parallel to the field, V = E x = (10 Newtons per coulomb) times (0.3 meters) = 3 V (Correct Response A). Response B is obtained by using the incorrect component of the displacement that is perpendicular to the field, V = (10 Newtons per coulomb) times (0.4 meters) = 4 V. Response C is obtained by incorrectly using the magnitude of the displacement in place of its component that is parallel to the field, V = (10 N/C)(0.5 m) = 5 V. Response D is obtained by incorrectly using the scalar sum of the components of the displacement in place of its component that is parallel to the field, V = (10 Newtons per coulomb) times (0.3 meters + 0.4 meters) = 7 V.
48	Answer to question 48	D	Objective 007 The magnitude of the electric field can be determined with the equation E = V over d , where E is the magnitude of the electric field, V is the voltage across the plates of the capacitor, and d is the distance between the plates. In this case, E = (36 V) divided by (0.20 m) = 180 V/m = 180 J/C/m = 180 N/C. The direction of the field is from the positively charged plate to the negatively charged plate, which is to the right. Therefore, the electric field vector E = 180 N/C to the right (Correct Response D). Response A is obtained by using the incorrect equation E = 36 times 0.20 = 7.2 and selecting the opposite of the correct direction. Response B is obtained by using the incorrect equation E = 36 times 0.20 = 7.2 and selecting the correct direction. Response C is the correct magnitude, but the direction is opposite of the correct direction.
49	Answer to question 49	B	Objective 008 Changing the values of two factors at the same time is the most significant problem with this experimental design (Correct Response B). Limiting the change to one variable at a time is important because it isolates the effect of each variable. Not accounting for the resistance of the additional length of wire (Incorrect Response D) would most likely not present a problem for this experiment, since the field strength would increase as the number of turns increases, even though the resistance of the coil also increases. Using paper clips to measure the strength of the electromagnet (Incorrect Response C) is not a problem, since a stronger magnetic field can lift more paper clips than can a weaker magnetic field. Using increments of 2.0 V instead of 1.0 V (Incorrect Response A) does not present a problem as long as the wire can withstand the higher currents and as long as the voltage is manipulated while the number of turns and any other variables are held constant.
50	Answer to question 50	D	Objective 008 Magnetized ferrous materials are magnetic because their dipoles are all aligned in the same direction. Physically striking the magnet will cause vibrations that change the dipoles' orientation to more random direction, thus reducing the strength of the overall magnet (Correct Response D). Response A is incorrect because temperature does not increase significantly due to an impact, and resistance does not affect magnetic dipole direction. Response B is incorrect because work does not affect surface currents, nor do surface currents exist in this scenario or affect dipole direction. Response C is incorrect because impact forces do not create magnetic fields.
51	Answer to question 51	B	Objective 008 According to the right-hand rule applied to magnetic fields, the magnetic field created from a loop of current will face out of the page in this scenario (Correct Response B). Using your right hand, bend your fingers in the direction of the current in the loop and your right thumb points in the direction of the magnetic field interior to the loop. Response A has the incorrect direction, opposite to the result of the right-hand rule. Responses C and D have the incorrect directions perpendicular to the result of the right-hand rule.
52	Answer to question 52	C	Objective 008 According to the right-hand rule applied to force on a current-carrying wire in an external magnetic field, the force on a current-carrying wire will be to the top of the page in this scenario (Correct Response C). Using your right hand, point your fingers in the direction of the current (right) and bend them toward the direction of the magnetic field (down, into the page). Your right thumb points in the direction of the force on the loop (up toward the top of the page). Responses A and B are incorrect because they describe force directions parallel to the magnetic field. Response D incorrectly describes the force direction opposite to the result of the right-hand rule.
53	Answer to question 53	A	Objective 008 Current-carrying wires will feel a force that is inversely proportional to the distance between them. Wires with current in the same direction will attract each other; wires in opposite directions will repel each other. The vertical sections of the frame will have attraction and repulsion; however, the force on the right vertical section of the frame is greater than on the left section because the right section is closer to the wire. Thus, the net force from these two sections will be rightward (Correct Response A). The net forces from the horizontal sections will balance to zero since they carry the same current and are equidistant from the long wire. Responses B, C, and D incorrectly assign the direction of the net force on the frame.
54	Answer to question 54	B	Objective 008 According to the right-hand rule applied to force on a current-carrying wire in an external magnetic field, the force on the top part of the current-carrying loop will be into the page (Correct Response B). Using your right hand, point your fingers in the direction of the current at the top of the loop (left) and bend them toward the direction of the magnetic field (up, to the top of the page). Your right thumb points in the direction of the force on the loop (into the page). Similar analysis will cause the bottom of the loop to feel an upward force. The right and left sections of the loop feel minimal force since the current is closest to parallel to the field at those locations. Responses A and C are incorrect because they describe force directions not consistent with the right-hand rule. Response D incorrectly assumes no motion since the forces balance to zero, but they do cause a net torque and therefore rotation.
55	Answer to question 55	D	Objective 008 By convention, magnetic lines of flux are shown leaving through the north pole and entering through the south pole. Based on this convention, the right side of magnet 1 is a south pole, and the left side of magnet 2 is a north pole (Correct Response D). Response A incorrectly indicates that the left side of magnet 2 is a south pole. Response B incorrectly indicates that the right side of magnet 1 is a north pole. Response C incorrectly indicates that the right side of magnet 1 is a north pole and the left side of magnet 2 is a south pole.
56	Answer to question 56	A	Objective 008 The situation can be explained via Lenz's law, which states that a changing magnetic flux through a closed surface will induce a magnetic field and therefore a current in the surrounding wires to resist the change in magnetic flux. When the switch is closed, the current in the outer coil creates a magnetic flux on the interior. Via Lenz's law, the inner coil will create a magnetic flux that resists this increase. Once the field is created by the outer coil, the magnetic flux is constant and no magnetic field is inducted by the inner coil and therefore the current in the inner coil drops to zero. Therefore, when the switch is closed, the current through the resistor will spike to some value but will then quickly drop to zero as the magnetic field reaches a steady state (Correct Response A). Response B is incorrect because no voltage will be induced in the inner coil once the current in the outer coil stops changing. Response C is incorrect because the inductive properties of the outer coil prevent the battery current from increasing instantaneously from zero to its steady state value. Response D is incorrect because it describes the superposition of an alternating current ( A C ) and a direct current (DC). Without the use of an A C power supply, the A C could be produced only by alternately and repeatedly opening and closing the switch.
57	Answer to question 57	C	Objective 009 The current through the battery equals the sum of the currents through the resistors, which equals 1.2 A + the current through the 40 ohm resistor. The current through the 40 ohm resistor can be found using Ohm's law: I = V divided by R , where I is the current through the resistor, V is the voltage across the resistor, and R is the resistance. In this case, I = (12 V) divided by (40 ohm ) = 0.3 A. The total current through the battery is therefore 1.2 A + 0.3 A = 1.5 A (Correct Response C). Incorrect Response A (0.3 A) is the current only through the 40 ohm resistor. Incorrect Response B (1.2 A) is the current only through the unknown resistor. Incorrect Response D (2.4 A) is twice the current through the unknown resistor.
58	Answer to question 58	C	Objective 009 Since power is the rate at which work is done, the energy can be found using the equation E = P T = V I t , where P is power, t is time, V is voltage, and I is current. Substituting each variable with its given value in units that are consistent with the answer choices, the equation becomes P = (115 Volts) times (20 milliamps) times (0.001 Amps per milliamp) times (24 hours) times (0.001 kilowatts per watt) = 5.5 times 10 to the minus 2 kilowatt hours (Correct Response C). Response A can be obtained by incorrectly manipulating the variables, expressing the time in seconds, and using an incorrect conversion factor when converting watts to kilowatts: P = (115 Volts) times (20 milliamps) times (0.001 Amps per milliamp) times (24 hours) times (0.001 kilowatts per watt) = 5.5 times 10 to the minus 2 kilowatt hours . Response B is incorrect and not directly related to the information provided. Response D is an order of magnitude greater than the correct answer; it can be obtained incorrectly by misplacement of the decimal point in one or more of the variables in a manner that results in an answer that is ten times greater than the correct answer.
59	Answer to question 59	A	Objective 009 Since a live voltage source is exposed to a person, the person is certainly in danger. Although the appliance is not grounded, if someone touches it, that will allow for grounded contact via the person. Thus, a potentially dangerous current would flow through the person (Correct Response A). Response B is incorrect because the person would themselves potentially create a closed circuit. Response C is incorrect because charge buildup itself is not dangerous; rather, potential current flow is the danger. Response D is incorrect because voltage will continue to stay on the case as long as it is plugged in to a live voltage. Moreover, current flow, not charge, is the danger source.
60	Answer to question 60	A	Objective 009 Using Ohm's law, V = I R , one can reason that identical resistors, with identical batteries attached, will draw identical current (Correct Response A). Response B is incorrect because it contradicts Ohm's law. Responses C and D are incorrect because open circuits do not allow current to flow. Thus, they must have identical currents of zero amps.
61	Answer to question 61	D	Objective 009 Using Ohm's law, V = I R , this graph has a slope described as the reciprocal of resistance. The slope = rise divided by run = 6 divided by 4 = three halves , so the resistance is two halves ohms (Correct Response D). Response A is incorrect because it fails to identify resistance as the reciprocal of the slope. Responses B and C are incorrect because they calculate slope incorrectly or misapply Ohm's law.
62	Answer to question 62	A	Objective 009 Resistors in parallel add in a reciprocal fashion, while resistors in series add arithmetically. Reduce the 20 Ω resistors first to 1 over R equivalent = 1 over 20 plus 1 over 20 = 2 over 20 = 1 over 10 ohms, so R equivalent = 10 ohms . That equivalent resistance is in series with the 10 Ω resistor: R equivalent = 10 + 10 = 20 Ω. That equivalent resistor is in parallel with the 5 Ω resistor: 1 over R equivalent = 1 over 20 plus 1 over 5 = 1 over 20 plus 4 over 20 = 5 over 20 , so the final R equivalent =20 over 5 = 4 Ω (Correct Response A). Response B is incorrect because it incorrectly divides 20 over 5 = 5. Response C is incorrect because it judges all resistors in parallel with each other, adding the reciprocals of all resistors. Response D is incorrect because it judges all resistors in series with each other, adding all the resistors arithmetically.
63	Answer to question 63	C	Objective 009 Standard A C current oscillates between plus V zero and minus V zero according to a sinusoidal wave. In a sinusoidal wave signal the instantaneous value varies, so we cannot use the instantaneous value to calculate average values related to the voltage, like the average power. The average voltage value is not useful because that wave spends equal times positive and negative, so the simple average is zero. The R M S squares the sinusoidal wave to remove the effect of the negative voltage and then square roots the value. For sinusoidal waves, this is mathematically equivalent to V zero divided by root 2 (Correct Response C). Responses A and B are incorrect because they falsely describe the R M S voltage as directly related to the sinusoidal functions. Response D is incorrect because it multiplies by root 2 instead of dividing by root 2.
64	Answer to question 64	C	Objective 009 Electric power is calculated by multiplying voltage by current. A voltmeter and an ammeter should be selected to determine the power used since they measure voltage and current, respectively (Correct Response C). The galvanometer and ammeter (Incorrect Response A) should not be chosen. A galvanometer is essentially a sensitive ammeter, and while it can be converted to a voltmeter by adding a suitable resistor, that information is not included in the response. The galvanometer and ohmmeter (Incorrect Response B) will not measure voltage and current since an ohmmeter is used for measuring resistance. The ohmmeter and voltmeter (Incorrect Response D) will measure voltage and resistance, respectively.
65	Answer to question 65	C	Objective 010 Power is the rate at which work is done. The power used to lift the mass is therefore given by P = W divided by t = m g h divided by t = (800 kilograms) times (9.80 meters per second squared) times (20.0 meters) divided by (10.0 seconds) = 1.6 times 10 to the 4 Watts (Correct Response C). Incorrect Response A is off by a factor of 0.01 by ignoring the time and gravity in the power expression. Incorrect Responses B and D are off by a factor of 10 and would be obtained if mistakenly using the incorrect equation P = m g h or m h divided by g .
66	Answer to question 66	D	Objective 010 Since the machine is 100% efficient, the work done by the machine, W out , is equal to the work done by the person, W in , where W out = F out times X out and W in = F in times X . Since W in = (490 Newtons) times x = (100 kilograms) times (9.8 meters per seconds squared) times (6.0 meters), x = 12 meters (Correct Response D). Incorrect Response A halves the distance the box travels instead of doubling the distance. Incorrect Response B incorrectly equates the distances. Incorrect Response C misapplies the work equation or assumes the machine is not 100% efficient.
67	Answer to question 67	A	Objective 010 Since work is distance traveled multiplied by the amount of force parallel to the motion force, the gravity force does zero work (Correct Response A). Gravity is down, into the page in this view and is always perpendicular to the motion of the mass. Responses B, C, and D are incorrect because they do not account for the direction of the gravity force and the displacement of the mass being perpendicular to each other.
68	Answer to question 68	D	Objective 010 Since the ball is moving translationally and rotationally, some of its initial potential energy is converted into rotational energy and therefore rotational motion. The brick does not rotate, so all its initial potential energy is converted into translational kinetic energy. The ball then has less translational energy at the bottom of the hill and less speed (Correct Response D). Response A is incorrect because the ball must have less translational kinetic energy and translational speed than the equal-mass brick. Responses B and C are incorrect because both objects start at the same height and have the same mass and thus the same potential energy. All the potential energy must be converted into kinetic energy if there is negligible friction, so they have the same total kinetic energy at the bottom.
69	Answer to question 69	B	Objective 010 Energy can be calculated as E = P t . Power is given by P = V I for electric appliances. Thus, E = V I t = (9 volts) times (3 amps) times (1800 seconds) = 48.6 kJ for the fast charger. E = V I t = (2.4 volts) times (5.2 amps) times (3000 seconds) = 37.4 kJ for the fast charger. Thus the fast charger uses approximately 10 kJ more energy to charge (Correct Response B). Response A is incorrect because it says the fast charger uses less energy. Responses C and D are incorrect because they say the fast charger uses an incorrect amount of energy compared to the slow charger.
70	Answer to question 70	C	Objective 010 The work performed by the force on the mass is equal to the area under the curve, which equals 50 J. Using the work-kinetic energy theorem, Delta K E = work = 50 J (Correct Response C). Response A would apply only if no work were performed, so that the kinetic energy would remain constant. Response B is the maximum magnitude of the force that performed the work. It does not represent the change in kinetic energy of the mass. Response D is the maximum magnitude of the force that performed the work multiplied by the total displacement. It represents neither the area under the curve nor the change in kinetic energy of the mass.
71	Answer to question 71	C	Objective 010 The maximum speed attainable occurs when friction is negligible so mechanical energy is conserved. Let the potential energy be zero at the bottom of the swing when the ropes are vertical. Then mgh = one half m v squared, or v = root 2 g h . Letting h = 2.00 meters minus (2.00 meters) times cosine 45.0 degrees = 0.586 meters, so v = root (2 times 9.8 meters per second squared times 0.585 meters) = 3.39 m/s (Correct Response C). Responses A and B are incorrect because they do not apply energy conservation. Response A incorrectly calculates the vertical component of the rope length initially. Response B incorrectly equates speed to the length of each rope, 2.00. Response D can be obtained by incorrectly using 4 m rather than 0.586 m as the maximum height, h.
72	Answer to question 72	A	Objective 010 The average amount of electrical power per square meter available from the solar panel is (12 percent) times (250 watts per meter squared) = 30 W/m². The area of the solar panel required for providing 1.2 kW of power can be determined with the equation A = P divided by (30 watts per meter squared) = (1.2 kilowatts) times (1000 watts per kilowatt) divided by (30 watts per meter squared) = 40 m². The 5 m by 8 m panel will meet the requirement because it has an area of 40 m² (Correct Response A). The area in Incorrect Response B would provide (1.2 kilowatts) times (60 meters squared) divided by (40 meters squared) = 1.8 kW. The area in Incorrect Response C would provide (1.2 kW)(169 m²)/(40 m²) = 5.1 kW. The area in Incorrect Response D would provide (1.2 kilowatts) times (400 meters squared) divided by (40 meters squared) = 12 kW.
73	Answer to question 73	B	Objective 011 The average or r m s speed of a particle in an ideal gas at constant volume is related to the temperature by v R M S = square root of the quantity 3 R T over M . When the temperature doubles, the average increases by root 2 (Correct Response B). Responses A, C, and D incorrectly apply the relationship between average molecular velocity and temperature for ideal gasses at constant volume.
74	Answer to question 74	A	Objective 011 The second law of thermodynamics says that the entropy of any closed system must increase. Closed systems imply no external work is done or energy added to the system. Reversible processes must therefore have no entropy increase or decrease (Correct Response A). Response B incorrectly assumes work is done on the closed system and is related to reversibility. Response C incorrectly assumes entropy increases, which means entropy would decrease when the process is reversed, violating the second law. Response D incorrectly assumes work is related to reversibility.
75	Answer to question 75	C	Objective 011 Determine the heat required to convert 0°C ice to 0°C water using the heat of fusion: q 1 = m delta H f = (2.00 kilograms) times (333 kilojoules per kilogram) = 666 kJ. Determine the heat required to raise the temperature of the water from 0°C to 55°C: q 1 = m c delta T = (2.00 kilograms) times (4.19 kilojoules per kilogram degree Celsius) times (55 degrees Celsius) = 461 kJ. The total energy is 1130 kJ (Correct Response C). Response A incorrectly uses a mass of 1 kg of water and only accounts for the temperature change, not the phase change. Response B incorrectly accounts for the temperature change, not the phase change. Response D incorrectly uses the heat of vaporization instead of the heat of fusion.
76	Answer to question 76	A	Objective 011 The thermal expansion coefficient is first multiplied by the temperature change and then used to multiply by the initial length of the rod to get the change in length: delta L aluminum = (25 times 10 to the minus 6) times (250 degrees Celsius) times (25 centimeters) = 0.15625 centimeters and delta L copper = (17 times 10 to the minus 6) times (250 degrees Celsius) times (25 centimeters) = 0.10625 centimeters The difference between these is 0.05 cm (Correct Response A). Response B incorrectly accounts for only the temperature change of the aluminum. Response C incorrectly accounts for only the temperature change of the copper. Response D incorrectly adds the length changes.
77	Answer to question 77	A	Objective 011 For an ideal gas, P V = n R T . Since volume is cut in half, the pressure must double. Since work is done to compress the cylinder, energy is added to the system and the temperature will increase slightly. Taking both into account, the pressure must then increase by more than double its initial value (Correct Response A). Response B incorrectly assumes the gas escapes the closed cylinder. Response C incorrectly determines that internal energy increases, but internal energy does not include the kinetic energy of motion of the system, nor any external energies from surrounding forces or work. Response D incorrectly determines the pressure will increase less than double.
78	Answer to question 78	D	Objective 011 Conduction (Correct Response D) is the thermal-energy-transfer process that involves the transfer of kinetic energy between molecules as they collide. Radiation (Incorrect Response A) is the thermal-energy-transfer process that involves the transfer of energy through electromagnetic waves. Latent heat (Incorrect Response B) is not an energy-transfer process. It is the energy per unit mass absorbed or released by a substance during a change in phase. Convection (Incorrect Response C) is the thermal-energy-transfer process that involves the transfer of energy through currents in fluids caused by variations in density due to uneven heating or cooling.
79	Answer to question 79	B	Objective 011 The amount of randomness or disorder in a system is a good definition of entropy (Correct Response B). Incorrect Response A describes the average kinetic energy of the molecules. Incorrect Response C describes equilibrium states and time, which have no direct relation to the definition of entropy. Incorrect Response D describes quantum systems, also not directly related to entropy.
80	Answer to question 80	D	Objective 012 From the universal law of gravitation, the acceleration of a rock dropped is given by a = G M divided by r squared . M is the mass of the Earth interior to the radial position of the object. When the rock is at the center of the Earth, no mass is radially interior to the rock so M = 0 and thus, the acceleration is zero when the radial position of the rock is zero (Correct Response D). Response A is incorrect because the symmetry of the rock's motion and the symmetry of the spherical Earth is assumed. The motion of the rock in the first and second halves of the motion will then be identical in time and speed. Response B is incorrect because it assumes as r approaches 0, the acceleration increases, but this does not account for the mass interior to the rock's radial position being zero at the center of the Earth. Response C is incorrect because the motion of the rock in the first and second halves of the motion will be identical in time and speed. If the rock is dropped from rest, at its symmetrically opposite position the rock will have the same zero velocity as when it entered.
81	Answer to question 81	D	Objective 012 The standing wave shows one half a wavelength, so the full wavelength is lambda = 2 times (0.625 meters) = 1.25. The speed, v = lambda f = 125 m/s (Correct Response D). Responses A, B, and C are incorrect because they do not account for the full wavelength.
82	Answer to question 82	B	Objective 012 A wave's energy is proportional to the square of the amplitude (Correct Response B). Response A is incorrect because decreasing density would not change the total energy, due to energy conservation. Response C is incorrect because decreasing frequency would decrease the energy in the wave, as they are directly proportional. Response D is incorrect because increasing tension would not change the total energy, due to energy conservation.
83	Answer to question 83	B	Objective 012 A M is an acronym for amplitude modulation, meaning the amplitude of the received radio waves changes with a regular frequency when one changes the radio station to a different broadcast frequency (Correct Response B). Responses A, C, and D have the incorrect definition of A and M.
84	Answer to question 84	C	Objective 012 Receivers of FM radio waves vary the frequency of resonance in the electronic system to tune to the correct signal. When the resonant frequency of the receiver matches the frequency of the radio wave precisely, the radio signal will be received clearly (Correct Response C). Response A is incorrect because resonance does not occur when phase shifts match. Response B is incorrect because resonance does not occur when amplitudes match. Response D is incorrect because resonance does not occur when speeds match, and all radio waves travel at the same speed.
85	Answer to question 85	D	Objective 012 The wavelength of sound is measured when the frequency of the fork matches resonantly with the natural frequency of the tube. This allows a standing wave to occur in the tube that will be in phase with the wavelength sound, causing an increase in volume at resonance. The speed of sound will then be equal to the wavelength of the standing wave in the tube, lambda , times the frequency of the fork, f: v equals lambda f (Correct Response D). Response A is incorrect because the frequencies must be the same at resonance, not change to different values. Response B is incorrect because tone is what we hear as frequency, so the tone would not become inaudible. Response C is incorrect because beats occur when sound waves interfere and are not related to resonance.
86	Answer to question 86	C	Objective 012 The relationship between the period of a pendulum and its length is represented by the equation T = 2 pie times the square root of (L over g) . Solving the equation for L produces L = g T squared over (4 pie squared) . According to the graph, one cycle is completed in 3 s, so the period of the pendulum is 3 s. Setting T = 3 s in the equation produces the answer L = 2.2 m (Correct Response C). Incorrect Response A is not related to the length of the pendulum. Incorrect Response B is numerically equal to one-half the period of the pendulum, not the length of the pendulum. Incorrect Response D is numerically equal to the period of the pendulum, not the length of the pendulum.
87	Answer to question 87	B	Objective 012 The speed of a wave on a string can be calculated with the equation v = Based on this equation, the speed of the wave is highest when mew is minimized and T is maximized (Correct Response B). Incorrect Response A includes the minimum values of both mew and T, rather than the minimum value of mew and the maximum value of T. Incorrect Response C includes the maximum value of mew and the minimum value of T, rather than the minimum value of mew and the maximum value of T. Incorrect Response D includes the maximum value of both mew and T, rather than the minimum value of mew and the maximum value of T.
88	Answer to question 88	D	Objective 013 According to Snell's law, n 1 sine theta 1 = n 2 sine theta 2 so n 2 = n 1 sine theta 1 divided by sine theta 2 = 1.0 sine 40 degrees divided by sine 25 degrees = 1.5 = 1.5 (Correct Response D). Incorrect Response A would be obtained if the two angles were interchanged— sine 25 degrees divided by sine 40 degrees = 0.66. Incorrect Response B would be obtained if the cosines of the two angles were mistakenly used in place of their sines— cosine 40 degrees divided by cosine 25 degrees = 0.85. Incorrect Response C would be obtained if the cosines of the two angles were mistakenly used in place of their sines after interchanging the angles— cosine 25 degrees divided by cosine 40 degrees = 1.2.
89	Answer to question 89	C	Objective 013 According to Snell's law, n 1 sine theta 1 = n 2 sine theta 2 so n 2 = n 1 sine theta 1 divided by sine theta 2 = 1.0 sine 40 degrees divided by sine 25 degrees = 1.48 (Correct Response C). Incorrect Response A would be obtained if Snell's law were incorrectly applied by having the two angles interchanged— sine 19.7 degrees divided by sine 30 degrees = 0.674. Responses B and D also incorrectly apply Snell's law.
90	Answer to question 90	A	Objective 013 Polarized glasses are simply properly oriented polarization filters that block light that is polarized perpendicular to the filter orientation. This is particularly helpful in blocking reflected light from surfaces, which is mostly horizontally polarized (Correct Response A). Response B is incorrect because Mie scattering does not contribute to polarization of light in the atmosphere. Response C is incorrect because polarized lenses do not allow one to choose the percentage of blocked light from the same lens. Response D is incorrect because the frequency of the light does not have such a significant effect on polarization as to perceptibly affect brightness.
91	Answer to question 91	A	Objective 013 The wavelength of the sound produced by each speaker can be determined with the equation Lambda = v over f = 340 meters per second over 680 hertz = 0.5 m. At point P, the sound waves from the two speakers are destructively interfering so that at point P the waves from the speakers are 180 degrees out of phase. Therefore, the difference in the distances traveled by the waves must be an odd multiple of one half lambda = 0.25 m. Stated mathematically, destructive interference will occur whenever this equation is satisfied: D 2 minus d 1 = n times 0.25 meters where n is any odd whole number (Correct Response A). Incorrect Response B describes a point where constructive interference takes place, because D 2 minus d 1 = lambda . The sound at this point would be at a maximum rather than silent. Incorrect Response C describes a point where constructive interference takes place, since 2 m is an even multiple of one half lambda , which means it is also a multiple of lambda . Incorrect Response D describes a point where constructive interference takes place, since 3 m is an even multiple of one half lambda , which means it is also a multiple of lambda .
92	Answer to question 92	B	Objective 013 The wavelength of the third harmonic is one-third the wavelength of the fundamental. For the string oscillating in its first harmonic, one-half the wavelength of the fundamental is equal to the length of the string, so the wavelength of the fundamental is Lambda 1 = 2 times 0.30 Meters = 0.60 m. The wavelength of the third harmonic is therefore Lambda 3 = 0.60 Meters divided by 3 = 0.20 m (Correct Response B). Incorrect Response A is One half times (Lambda 1 over 3) = 0.10 m. Incorrect Response C is One quarter times Lambda 1 times 3 = 0.45 m. Incorrect Response D is One half times Lambda 1 times 3 = 0.90 m.
93	Answer to question 93	D	Objective 013 Compared with light from the upper slit, light from the lower slit travels a greater distance before reaching the screen at the point shown in the diagram. The difference in the distances traveled by the light from the slits equals the length of segment A B . Therefore, the light from the slits arrives at the point on the screen with a phase difference that depends directly on the length of segment A B (Correct Response D). Response A would be correct only in the case where theta is selected, such that the length of segment A B is an integral multiple of the wavelength. Response B is incorrect: The intensity of the bright fringes does depend on the length of segment A B , but it is not proportional to this length. Response C is incorrect: The length of segment A B is not equal to the distance between any two adjacent bright fringes on the screen.
94	Answer to question 94	D	Objective 013 As the second filter is rotated, the angle at which the intensity of the transmitted light will be a minimum is 90 degrees either clockwise or counterclockwise because the light coming from the first filter is vertically polarized up and down, along the 0–180-degree line. The second filter will block all the light coming from the first filter when the polarization directions are perpendicular, or 90 degrees , to each other. This corresponds to the angles plus or minus 90 degrees , which are the same as 90 degrees , 270 degrees (Correct Response D). Incorrect Responses A and C each include one of the two correct angles, and Incorrect Response B includes neither of the correct angles.
95	Answer to question 95	A	Objective 014 From smallest wavelength to greatest: gamma rays, X-rays, ultraviolet, visible, infrared, microwaves, and finally radio waves (Correct Response A). Response B incorrectly lists ultraviolet as having longer wavelength than microwaves. Response C incorrectly lists gamma rays as having longer wavelength than radio and infrared. Response D incorrectly lists X-rays as having longer wavelength than ultraviolet.
96	Answer to question 96	C	Objective 014 Constructive interference occurs when the wavelengths are completely in phase. The size of the film, d, will determine the path length difference between light that passes through and light that is reflected at the surface of the film. These two waves will be in phase and thus constructively interfere when the path length difference is a whole integer, n, times the thickness of the film, d, is equal to one wavelength, lambda = n d . Using each answer to calculate n, 480 nm is the largest that produces a whole integer: n = 4 (Correct Response C). Response A incorrectly halves the thickness. Incorrect Response B will have constructive interference but is not the largest value that produces an integer for n. Incorrect Response D will not produce an integer value for n.
97	Answer to question 97	C	Objective 014 This experiment is trying to determine the relationship between the image distance and the object distance for a single lens. Both independent and dependent variables need to change to investigate their relationship (Correct Response C). The control variable is the focal length of the lens because that stays the same for a single lens. Response A is incorrect because the focal length cannot be a variable for a single lens since focal length is constant. Response B is incorrect because the focal length and radius of curvature cannot be variables for a single lens since they both remain constant for the whole experiment. Response D is incorrect because radius of curvature cannot be a variable for a single lens since it remains constant.
98	Answer to question 98	A	Objective 014 An inductor-capacitor (LC) circuit will oscillate at the frequency f = Since accelerating charges produce electromagnetic waves, the circuit will oscillate waves of the same frequency (Correct Response A). Incorrect Response B is a resistor-inductor (RL) circuit and exhibits exponential decay, to resonance, and will not produce consistent oscillations, nor accelerating charges. Incorrect Response C is a resistor connected to a diode and will not produce consistent oscillations, nor accelerating charges. It does not have a resonant frequency; nor does Incorrect Response D, a capacitor connected to a diode.
99	Answer to question 99	C	Objective 014 X-rays are commonly used to produce medical images. The approximate wavelength of the X-rays is Lambda = c over f, where lambda is the wavelength of X-rays, c is the speed of light in a vacuum, and f is the frequency of the X-rays. In this case, Lambda = (3.0 times 10 to the 8 meters per second) divided by 10 to the 18 hertz = 3.0 times 10 to the negative 10 meters which is closest to 10 to the negative 9 meters (Correct Response C). Incorrect Response A corresponds to a frequency of (3.0 times 10 to the 8 meters per second) divided by 10 to the negative 3 meters = 3 times 10 to the 11 hertz This is closest to infrared, or heat radiation, which is absorbed in the skin and felt as heat. Incorrect Response B corresponds to a frequency of (3.0 times 10 to the 8 meters per second) divided by 10 to the negative 3 meters = 3 times 10 to the 11 hertz This is closest to ultraviolet, which may only penetrate the skin a few millimeters and be absorbed. Incorrect Response D corresponds to a frequency of (3.0 times 10 to the 8 meters per second) divided by 10 to the negative 12 meters = 3 times 10 to the 20 hertz. This is between X-rays and gamma rays and will likely not interact with biological material much, so its use in imaging is limited.
100	Answer to question 100	C	Objective 014 According to the law of reflection, the angle of reflection from mirror 1 equals the angle of incidence, theta . The angle formed by mirror 1 and its reflected ray is 90 degrees minus theta , so the angle formed by this same ray and mirror 2 must be theta . Therefore, the angle of incidence for mirror 2 is 90 degrees minus theta , and by the law of reflection, the angle of reflection must also be 90 degrees minus theta (Correct Response C). Incorrect Response A is equal to half of the incident angle for mirror 1. Incorrect Response B ( theta ) is the incident angle for mirror 1. Incorrect Response D 180 degrees minus 2 theta is twice the reflected angle for mirror 2.
Total Correct:	Total Correct	Review your results against the test objectives.

Open Responses, Sample Responses, and Analyses

Question Number	Your Response Read about how your responses are scored and how to evaluate your practice responses
101	Answer to 101 Open Response Item Assignment #1 For each assignment, you may type your written response on the assigned topic in the box provided. Note: The actual test allows you to handwrite your responses on separate response sheets to be scanned for upload to the test. For this practice test, you may handwrite each response on 1–2 sheets of paper. Essay Response Field
First Sample Weak Response (show sample responsehide sample response)	First Sample Weak Response to Open-Response Item Assignment #1 Within a system you always have conservation of energy. Energy is neither created nor destroyed. It can always be transferred from one form to another. In this problem we have Electric Energy going to Mechanical Energy to Kinetic Energy to Heat Energy. Heat energy occurs at various stages and accounts for the energy that is transferred to the environment (see below). Because the car is 60% efficient, we can determine that 135 KW of power was generated, but a percentage of that was converted to thermal energy which went to heating up the environment. Because of the efficiency of the electric car, we can determine that this is the transportation of the future, and we can also show that energy is neither created nor destroyed, just transformed from one type to another. K E equals one half m times v squared which in this case is 729 kJ P equals w divided by t which in this case is 81 kW
First Weak Response Analysis (show response analysishide response analysis)	Analysis of First Weak Response to Open-Response Item Assignment #1 This is an example of a weak response because it is characterized by the following: Purpose: The purpose is limited because all the bullets are not covered, namely the discussion of how to use the model in the classroom. The second bullet requiring calculations is not fully developed due to limited support. All formulas and calculations are not shown. Subject Matter Knowledge: The origin of the stated numbers is unknown and not all formulas are shown, reflecting limited knowledge. Support: The lack of calculations limits the evidence of understanding of what is happening in the system. Rationale: There is no demonstration that the subject matter is fully understood. The knowledge expressed is limited as is the soundness of the argument. There is no application of the problem.
Second Sample Weak Response (show sample responsehide sample response)	Second Sample Weak Response to Open-Response Item Assignment #1 A universal law of nature is conservation of energy. Energy cannot be created or destroyed. It can only be transformed from one form to another. In the example of electric cars, chemical energy of the battery is transformed into electric energy that is transformed into mechanical energy and partially dissipated as heat. Energy is conserved in the system as input energy and is equal to output energy. Since the efficiency of this system is 60%, then 40% of the system energy is lost. Some losses are always occurring, and not all system energy is transformed into useful work (see below). Students could compare the efficiency of electric cars to gasoline powered vehicles. Kinetic energy of the car is K E equals one half m times v squared K E equals one half times open paren 2,0000 kilograms close paren times open paren 27 meters per second close paren squared = 729,000 J = 729 kJ 729 kJ times 291.6 kJ Energy loss
Second Weak Response Analysis (show response analysishide response analysis)	Analysis of Second Weak Response to Open-Response Item Assignment #1 This is an example of a weak response because it is characterized by the following: Purpose: The purpose of the assignment is mostly limited to the first and second bullets, and so it is partially developed. The required calculation is limited to kinetic energy. Although the last sentence attempts to address the third bullet, it falls short because it makes no connection to the idea of conservation of energy. Subject Matter Knowledge: Calculations are limited to kinetic energy only, showing limited subject matter knowledge. Support: The limited number of calculations shows lack of understanding of energy transformations in the system. Rationale: Lack of full understanding of the problem shows limited knowledge. There is no other example to illustrate the application of energy conservation.
First Sample Strong Response (show sample responsehide sample response)	First Sample Strong Response to Open-Response Item Assignment #1 The law of conservation of energy states that the total energy of an isolated system is constant. Energy, as the ability to do work, is neither created nor destroyed, it can only be transformed from one form to another or from one system to another. The total energy of a system, kinetic, gravitational potential, heat, might change forms, but if energy is conserved then the total will remain the same. An electric car is a machine that converts electric potential energy of a battery into mechanical energy of the moving car. In this process part of the energy is dissipated as heat because of friction, and part as drag since the car is moving on the road surrounded by air. The car accelerates to the constant speed and the kinetic energy of the system can be calculated. Not all potential energy of the battery can be transformed to kinetic energy, and losses are reflected in the efficiency of the system. Due to conservation of energy, the initial energy of the system is the same as the final energy. The energy transformations in this system are shown below: CHEMICAL becomes ELECTRICAL becomes MECHANICAL becomes KINETIC becomes THERMAL K E = one half M V squared = one half times (2,000 kilograms) times (27 meters per second) squared = 729,000 kilogram meters squared per second squared = 729 kJ K E = 729 kJ a = ;(V final minus V initial) divided by delta t , a = (27 meters per second minus 0 meters per second) divided by 9.0 seconds = 3 meters per second squared The distance the car reaches in 9 seconds is determined by: d = a t squared divided by 2 d = 3 meters per second squared times (9 seconds) squared divided by 2 = 121.5 m W = F times d or W = m times a times d W = 2000 kilograms times 3 meters per second squared times 121.5 meters = 729,000 J or 729 kJ P = delta W divided by delta t P = 729,000 joules / 9 seconds equals 81,000 or 81 kilowatts P = 81 kW Efficiency = W output divided by W input , 60 percent = 81 kilowatts divided by x , x = 81 kilowatts divided by 60 percent , x = 135 kW The work input is calculated to be 135 kW. A 40% loss of input power dissipated as heat would be 135 kW times 0.40 = 54 kW Thermal energy heat loss = 54 kW Kinetic energy is the energy of motion observable as the movement of an object, in this example, the electric car. It is defined as the energy needed to accelerate a car of a given mass from rest to its stated velocity. This energy is provided by an electric battery through the conversion of chemical energy into electrical and then to the mechanical energy of the car. This example shows that the initial potential energy of a battery can't be used without some losses due to friction and drag in the system, car/road/air. The law of conservation of energy enables us to calculate the initial potential energy of the system and to calculate any energy dissipated. In the classroom, students will research the average mileage range for several electric vehicles ( E Veez ) and compare the M P G E ratings (a calculation of E V efficiency) of combined/city/highway. Students will then select one model to calculate how many times a recharge is necessary on a trip, determine the location of charging stations, calculate charging time, and determine charging cost.
First Strong Response Analysis (show response analysishide response analysis)	Analysis of First Strong Response to Open-Response Item Assignment #1 This is an example of a strong response because it is characterized by the following: Purpose: The response demonstrates a strong 4 score because the assignment is fully achieved. The three bullets are fully addressed and answered with appropriate subject matter knowledge, relevant vocabulary and algebraic expressions, and accurate calculations. Subject Matter Knowledge: The concept on which the prompt is based, conservation of energy, is fully explained in the opening paragraphs and appropriately applied at the end to justify the energy used by the system. The energy transformations are identified and support the idea about energy changing forms, but always being conserved and accounted for. Support: The three calculations—kinetic energy, power of the car, and thermal energy dissipated—are all shown. The formulas for these calculations are given, and all the work is shown. The final answer for each is labeled, underlined, and accurate. Rationale: The fourth paragraph applies the idea of energy conservation to the electric car in the problem. It summarizes some of the points mentioned earlier in the response. It points out where the energy to run the car comes from, how it gets transformed to energy that moves the car, and how some of the energy is dissipated due to friction and drag. The final calculations support the idea that the energy of a system is conserved. At 60% efficiency, not all the initial energy actually powers the car, but what gets dissipated can be quantified and accounted for. The classroom activity adds relevancy to the response. As people begin to transition to electric vehicles, many questions come to mind about their use: how far can an E V go, what is involved in charging, where does that occur, how much does it cost? Searching out answers to these questions would be helpful and practical in preparing for the transition and in understanding energy conservation. The response is thoroughly answered and demonstrates comprehensive understanding of the subject matter.
Second Sample Strong Response (show sample responsehide sample response)	Second Sample Strong Response to Open-Response Item Assignment #1 An electric car is powered by electricity produced by a battery. Energy is transferred from the battery to make the engine work and move the car. As this all happens, CHEMICAL energy in the battery gets converted to ELECTRICAL energy to power the engine, and then this MECHANICAL energy causes the car to move giving it KINETIC energy. Throughout all these energy conversions, some energy is dissipated as HEAT. The THERMAL energy dissipated is due to friction of the internal parts and externally between the tires and the road and the resistance of the air. The law of conservation of energy states that energy can neither be created nor destroyed, but it can change forms. The total amount of energy put into this system can be accounted for as it is transferred from one form to another (see below). Efficiency is defined as the ratio of work output to work input. The efficiency of all machines is always less than 100% because the work output is always less than the work input. This is due to friction and thermal energy being dissipated. Efficiency = work output divided by work input In this case the E f f = 60% as stated in the problem. Having calculated the work output, 81 kW, we can solve for the work input and then take 40% of that to determine the amount of thermal energy lost or dissipated. 60% = 81 kW divided by work input work input = 81 kW/.60 work input = 135 kW 40% of that would be 54 kW of Thermal Energy Dissipated Electric vehicles ( E Veez ) are more efficient than gasoline-powered vehicles. Students could research and compare the energy transfers of each to better understand why the energy loss in an E V is less. The total energy of the system can always be accounted for because energy is conserved; however, if less energy is dissipated by the system, the system will work with greater efficiency. K E = one half m v squared m = 2,000 kg v = 27 m/s Kinetic Energy = 729,000 J = 729 kJ P = ^Work/_t = 729,000 J divided by 9s Power = 81,000 W = 81 kW
Second Strong Response Analysis (show response analysishide response analysis)	Analysis of Second Strong Response to Open-Response Item Assignment #1 This is an example of a strong response because it is characterized by the following: Purpose: The response demonstrates a strong 4 because the purpose of the assignment is fully achieved. All three bullets are answered with the information asked for in the prompt. The information is accurate, the required calculations are done, and the work is shown. Subject Matter Knowledge: There is strong evidence of subject matter knowledge with the use of appropriate scientific vocabulary. Relevant formulas are used, and the calculations are accurate and complete. Support: The answer is well supported with the explanation of the concept, conservation of energy, and a description of the appropriate energy transfers in the electric car. Rationale: The overall response shows good understanding of the subject matter. In the third bullet, by applying the concept of energy conservation and energy accountability to both electric and gas-powered vehicles, the students should develop a greater understanding of and appreciation for the current trend to promote both hybrid and electric vehicles. A basic physics concept would be applied to a contemporary real-world situation, making it relevant to the student.
102	Answer to 102 Open-Response Item Assignment #2 For each assignment, you may type your written response on the assigned topic in the box provided. Note: The actual test allows you to handwrite your responses on separate response sheets to be scanned for upload to the test. For this practice test, you may handwrite each response on 1–2 sheets of paper. Essay Response Field
First Sample Weak Response (show sample responsehide sample response)	First Sample Weak Response to Open-Response Item Assignment #2 If there is a magnet moving within a coil of copper wire, then it should induce a current within that wire that can be read by an ammeter. Have a coil of wire connected to an ammeter. Using bar magnets of varying sizes, move the magnets within the coil of wire to see if there is a reading in the ammeter. Repeat the procedure by using different sized magnets and at different speeds to see how this would affect the readings. The variables would be the size of the magnets and the speed with which they move within the coil of wire, and the current is what is being collected in the data for this test. The collected data should not only show that the changing magnetic field will produce current in the wire, but also to see how the different variables (speed and size) would affect that result. This experiment should show the students that electricity and current are two sides of the same coin and should deepen their understanding of why it is called the electromagnetic spectrum.
First Weak Response Analysis (show response analysishide response analysis)	Analysis of First Weak Response to Open-Response Item Assignment #2 This is an example of a weak response because it is characterized by the following: Purpose: The purpose of the assignment is only partially achieved because the response lacks the key variable, the number of loops in the wire coil, to effectively complete the demonstration. Also, it does not address the final bullet requiring how a magnetic field and electric current are related. Subject Matter Knowledge: The response does use scientific vocabulary, but the number of coils is not varied to show how much electric current is generated. The different sizes of the magnets will show different currents, but not enough of a changing current. Support: The response is attempting to describe an investigation to support the idea that a magnetic field can induce a current but does not go far enough to prove how changing a magnetic field will produce a current. Rationale: The response is limited not only in the design of the investigation, but it also falls short on the last bullet.
Second Sample Weak Response (show sample responsehide sample response)	Second Sample Weak Response to Open-Response Item Assignment #2 When a wire is wound around an F e core, and current moves through the wire, it creates a magnetic field. This apparatus is called an electromagnet. Testing this can be done by placing a compass near the wire. A stronger electromagnet can be made by increasing the number of wire coils. A changing magnetic field can produce current in a conductor that is placed in this field. This is called electromagnetic induction. Electro motors are an example of electromagnetism. Due to magnetic field interaction, forces are induced, and motor blades are rotating and converting electrical energy into mechanical. Another example of application of the phenomenon of electromagnetic induction is the electric generator. The changing magnetic field produces current. In the demonstration, students should notice that the conductor is stationary, and the magnet is rotating.
Second Weak Response Analysis (show response analysishide response analysis)	Analysis of Second Weak Response to Open-Response Item Assignment #2 This is an example of a weak response because it is characterized by the following: Purpose: The purpose of the assignment is only partially achieved because part of the response is related to an electromagnet that is not addressed in the prompt. The response begins by describing how a magnetic field can be produced rather than how a magnetic field can produce a current. Subject Matter Knowledge: The response uses some scientific vocabulary, but no procedure is described to show how current can be induced in a conductor in a changing magnetic field. No relevant variables were discussed. Support: Real-life applications of electromagnetic induction are given but without adequate explanation of the concept of induction. Rationale: There is some basic understanding of the working principles of the equipment mentioned, but not enough to help students understand how a changing magnetic field produces an electric current.
First Sample Strong Response (show sample responsehide sample response)	First Sample Strong Response to Open-Response Item Assignment #2 Electromagnetic induction is a phenomenon described by Faraday's law. When an electrical current is flowing in a conductor, there is an associated magnetic field created around the wire. Current is produced in a conductor when it is moved through a magnetic field. This process of generating current in a conductor by placing it in a changing magnetic field is called induction. There is no physical connection between the conductor and the magnet, and the current is said to be induced in the conductor by the magnetic field. If we move a bar magnet near or inside of a conductor loop, a current will be produced. A conductor loop could be made by wrapping copper wire around a cylindrical form such as a cardboard tube then removing the tube. The ends of the wire coil would be connected to an ammeter. A bar magnet could be moved through the coil of copper wire. As the magnet passes through the coil, a current will be induced in the wire, and the ammeter will indicate the value/strength of the current. Increasing the number of coils in the loop will cause a stronger current to be produced. The number of loops in the wire coil could be varied in different trials -- 5, 10, 15, 20, etc. -- and the strength of the current could be recorded on a data table. Data should support the finding that increasing the number of loops in the coil results in a stronger current. A hand crank generator with a light bulb could be used as a demonstration to show students that the motion of the magnet relative to the wire coil can produce current to light the bulb. This concept could then be expanded upon to understand how electric generators are used to produce electricity. An electric generator is an everyday device used to convert mechanical energy or chemical energy into electrical energy by the movement of a loop of wire between the poles of a magnet. The conductor coil is wound on a metal core and horseshoe magnets are used in electric generators.
First Strong Response Analysis (show response analysishide response analysis)	Analysis of First Strong Response to Open-Response Item Assignment #2 This is an example of a strong response because it is characterized by the following: Purpose: The response demonstrates a strong 4 score because the assignment is fully achieved. The four bullets are thoroughly addressed and answered with relevant subject matter knowledge and vocabulary. The investigation described uses appropriate equipment and a logical experimental design. The final paragraph makes the connection between the concept of electromagnetic induction demonstrated by the experiment and its use in generators, helping to broaden the understanding. Subject Matter Knowledge: The concept of electromagnetic induction, Faraday's law, is fully explained in the opening paragraph. This is followed by a statement of a testable scientific claim that addresses the concept of electromagnetic induction. Support: An investigation is proposed, appropriate equipment mentioned, and a doable procedure described. Although there are many ways to investigate this concept, this plan is simple and straightforward and incorporates the basic equipment needed to show that a current is in fact induced in the conductor and that it will vary directly in strength with the number of loops in the coil of wire. Variables (number of loops) and Data (measurement of current) are noted. A data table is mentioned as an organized means of recording the data. The expected outcome that would support the claim is pointed out. Rationale: The final paragraph describes a piece of equipment that would provide a hands-on activity that would allow students to produce electricity. The mechanical action of turning the crank in order to rotate a wire coil within magnets would light a bulb. Students would be able to observe that the brightness or dimness of the bulb relates to the speed at which the crank is turned. By applying the concept of electromagnetic induction, portable and commercial generators could be introduced to deepen the student's understanding of how generators work. The response is thoroughly answered and demonstrates comprehensive understanding of the subject matter and experimental design.
Second Sample Strong Response (show sample responsehide sample response)	Second Sample Strong Response to Open-Response Item Assignment #2 The discovery of electromagnetic induction was made by Faraday. He asked the question: is it possible to produce electric current without a battery? He hypothesized that a changing magnetic field is necessary to induce a current in a nearby circuit. To test his hypothesis, he designed an experiment like the one described below. Electromagnetic induction is defined as electric current generated in a closed circuit because of the electromotive force induced by a changing magnetic field. Or in other words, electromagnetic induction is the process by which a changing magnetic field induces a current in another conductor. This happens either when the conductor is placed in a moving magnetic field or when a conductor is constantly moving in a stationary magnetic field. Students will make two circuits, a primary circuit and a secondary circuit that will be next to each other. The primary circuit will consist of a coil, battery, and switch. The secondary circuit will have a coil and a galvanometer. When the current in the primary circuit is passed by closing the switch, the galvanometer in the secondary circuit will deflect in one direction and then go to zero as the constant magnetic field is established. When the current is cut off by the switch in the primary circuit, the galvanometer in the secondary circuit will deflect in the other direction. When the secondary circuit is moved toward or away from the primary circuit, the galvanometer needle will deflect. The faster the movement, the greater the deflection. The variable in this experiment is the speed of the movement of the secondary circuit. Students will also notice that the change in direction of the secondary circuit's movement will correlate with the direction of deflection of the galvanometer. Based on this experiment, students will be able to support the tested claim that the induced current in the secondary circuit is generated only by changing magnetic fields. The deflection direction of the galvanometer depends on the direction of movement of the secondary circuit. Also, the magnitude of deflection depends upon the speed at which the secondary circuit is moved. The faster the movement, the greater the deflection. There are numerous applications of the electromagnetic induction phenomenon in everyday life: electric generators, induction cooking, induction motors, magnetic card readers, and digital cameras. For example, an induction stove produces heat by inducing a current in the cooking vessel. When current passes through a coil of copper wire located below the cooktop, the resulting magnetic field induces an electrical current in the cooking vessel. It uses a rapidly alternating magnetic field to generate current that, because of the electrical resistance of the vessel, is transformed into heat.
Second Strong Response Analysis (show response analysishide response analysis)	Analysis of Second Strong Response to Open-Response Item Assignment #2 This is an example of a strong response because it is characterized by the following: Purpose: The response demonstrates a 4 score because it shows thorough knowledge and the assignment is fully achieved. The four bullets are answered with appropriate subject matter knowledge and relevant vocabulary. The concept of electromagnetic induction is introduced through a testable hypothesis and investigation described in the response. Subject Matter Knowledge: The experimental design and investigation procedure are precisely explained. The variables and data to be collected are clearly identified. Support: The collected data should support the hypothesis that a changing magnetic field produces current in the circuit without a battery. Students will observe that both the speed and the direction of the movement of the magnetic field will influence the current induced. Rationale: The final paragraph provides examples as well as an explanation of an everyday application of the electromagnetic induction principle. The response is thoroughly answered and demonstrates full understanding of the subject matter.
Review the Performance Characteristics and Score Scale for Written Performance Assignments.

Multiple Choice Question
Practice Test Evaluation Chart

In the evaluation chart that follows, the multiple-choice questions are arranged in numerical order and by test objective. Check your responses against the correct responses provided to determine how many questions within each objective you answered correctly.

Subarea 1 : Matter and Its Interactions

Objective 0001: Apply knowledge of atomic and nuclear physics.

Question Number	Your Response	Correct Response
1		C
2		D
3		A
4		C
5		D
6		C
7		B

Total Correct for Objective 1 out of 7

Objective 0002: Demonstrate knowledge of the basic principles of modern physics.

Question Number	Your Response	Correct Response
8		D
9		C
10		C
11		D
12		B
13		A
14		B

Total Correct for Objective 2 out of 7

Subarea 1 (Objectives 0001–0002) Total Total Correct for Subarea 1 out of 14

Subarea 2 : Motion and Stability: Forces and Interactions

Objective 0003: Apply knowledge of kinematics to interpret motion.

Question Number	Your Response	Correct Response
15		B
16		B
17		C
18		D
19		B
20		A
21		C
22		B

Total Correct for Objective 3 out of 8

Objective 0004: Apply knowledge of forces and Newton's laws.

Question Number	Your Response	Correct Response
23		D
24		C
25		D
26		B
27		B
28		C
29		D

Total Correct for Objective 4 out of 7

Objective 0005: Apply knowledge of linear momentum, angular momentum, and rotational dynamics.

Question Number	Your Response	Correct Response
30		D
31		B
32		A
33		C
34		C
35		A

Total Correct for Objective 5 out of 6

Subarea 2 (Objectives 0003–0005) Total Total Correct for Subarea 2 out of 21

Subarea 3 : Motion and Stability: Forces and Interactions in Fields and Circuits

Objective 0006: Apply knowledge of gravitational forces and circular motion.

Question Number	Your Response	Correct Response
36		A
37		B
38		C
39		A
40		D
41		A

Total Correct for Objective 6 out of 6

Objective 0007: Apply knowledge of electrostatics, electric fields, and electric potential.

Question Number	Your Response	Correct Response
42		A
43		B
44		C
45		D
46		B
47		A
48		D

Total Correct for Objective 7 out of 7

Objective 0008: Apply knowledge of magnetic fields and electromagnetism.

Question Number	Your Response	Correct Response
49		B
50		D
51		B
52		C
53		A
54		B
55		D
56		A

Total Correct for Objective 8 out of 8

Objective 0009: Apply knowledge of electric circuits.

Question Number	Your Response	Correct Response
57		C
58		C
59		A
60		A
61		D
62		A
63		C
64		C

Total Correct for Objective 9 out of 8

Subarea 3 (Objectives 0006–0009) Total Total Correct for Subarea 3 out of 29

Subarea 4 : Energy

Objective 0010: Apply knowledge of energy, power, and the conservation of energy.

Question Number	Your Response	Correct Response
65		C
66		D
67		A
68		D
69		B
70		C
71		C
72		A

Total Correct for Objective 10 out of 8

Objective 0011: Apply knowledge of the basic laws of thermodynamics and the kinetic molecular theory.

Question Number	Your Response	Correct Response
73		B
74		A
75		C
76		A
77		A
78		D
79		B

Total Correct for Objective 11 out of 7

Subarea 4 (Objectives 0010–0011) Total Total Correct for Subarea 4 out of 15

Subarea 5 : Waves and Their Applications in Technologies for Information Transfer

Objective 0012: Apply knowledge of simple harmonic motion and wave properties and characteristics.

Question Number	Your Response	Correct Response
80		D
81		D
82		B
83		B
84		C
85		D
86		C
87		B

Total Correct for Objective 12 out of 8

Objective 0013: Apply knowledge of wave interactions and phenomena.

Question Number	Your Response	Correct Response
88		D
89		C
90		A
91		A
92		B
93		D
94		D

Total Correct for Objective 13 out of 7

Objective 0014: Apply knowledge of electromagnetic waves and the electromagnetic spectrum.

Question Number	Your Response	Correct Response
95		A
96		C
97		C
98		A
99		C
100		C

Total Correct for Objective 14 out of 6

Subarea 5 (Objectives 0012–0014) Total Total Correct for Subarea 5 out of 21

Practice Test Score Calculation

The practice test score calculation is provided so that you may better gauge your performance and degree of readiness to take an MTEL test at an operational administration. Although the results of this practice test may be used as one indicator of potential strengths and weaknesses in your knowledge of the content on the official test, it is not possible to predict precisely how you might score on an official MTEL test.

The Sample Responses and Analyses for the open-response items may help you determine whether your responses are more similar to the strong or weak samples. The Scoring Rubric can also assist in estimating a score for your open responses. You may also wish to ask a mentor or teacher to help evaluate your responses to the open-response questions prior to calculating your total estimated score.

How to Calculate Your Practice Test Score

Review the directions in the sample below and then use the blank practice test score calculation worksheet to calculate your estimated score.

Multiple-Choice Section

Enter the total number of multiple-choice questions you answered correctly:

Use Table 1 below to convert that number to the score and write your score in Box A:

197

Open-Response Section

Enter the number of points (1 to 4) for your first open-response question:

Enter the number of points (1 to 4) for your second open-response question:

Add those two numbers (Number of open-response question points):

Use Table 2 below to convert that number to the score and write your score in Box B:

Total Practice Test Score (Estimated MTEL Score)

Add the numbers in Boxes A and B for an estimate of your MTEL score:

A + B =

247

Practice Test Score Calculation Worksheet: Physics (69)

Table 1:

Number of Multiple-Choice Questions Correct	Estimated MTEL Score
0 to 25	117
26 to 30	125
31 to 35	133
36 to 40	141
41 to 45	149
46 to 50	157
51 to 55	165
56 to 60	173
61 to 65	181
66 to 70	189
71 to 75	197
76 to 80	205
81 to 85	213
86 to 90	221
91 to 95	229
96 to 100	237

Table 2:

Number of Open-Response Question Points	Estimated MTEL Score
2	31
3	36
4	41
5	46
6	50
7	55
8	60

Use the form below to calculate your estimated practice test score.