Equip yourself to ace the SAT Subject Test in Physics with The Princeton Review's comprehensive study guide—including 2 full-length practice tests, thorough reviews of key physics topics, and targeted strategies for every question type. Physics can be a tough subject to get a good handle on—and scoring well on the SAT Subject Test isn't easy to do. Techniques That Actually Work. Equip yourself to ace the SAT Physics Subject Test with The Princeton Review's comprehensive study guide—including 2 full-length practice tests, thorough reviews of key physics topics, and targeted strategies for every question type.
This eBook edition has been specially formatted for on-screen reading with cross-linked questions, answers, and explanations. Author : Steven A. This eBook version of the edition of Cracking the SAT Physics Subject Test has been optimized for on-screen viewing with cross-linked questions, answers, and explanations.
We don't try to teach you everything there is to know about physics-only what you'll need to score higher on the exam. There's a big difference. Equip yourself to ace the SAT Physics Subject Test with The Princeton Review's comprehensive study guide--including 2 full-length practice tests, thorough reviews of key physics topics, and targeted strategies for every question type.
Physics can be a tough subject to get a good handle on--and scoring well on the SAT Subject Test isn't easy to do. Both the feather and the hammer hit the lunar surface at the same time, verifying the fact—first stated by Galileo—that under conditions of no air resistance, all objects fall with the same acceleration, regardless of their mass. This result tells us that the mass of the satellite is irrelevant; only the mass of the earth, M, remains in the formula. Since kinetic energy is conserved in an elastic collision, the total kinetic energy after the collision must also be J.
Now, multiplying this rate by the time of flight, T, gives the total horizontal distance covered. The mass of the cannonball does not change, eliminating E. The answer is B. As the cannonball falls, its height decreases, so its gravitational potential energy decreases.
C is false since it generally requires much more energy to break the intermolecular bonds of a liquid to change its state to vapor than to loosen the intermolecular bonds of a solid to change its state to liquid. And D and E are false: While a substance undergoes a phase change, its temperature remains constant. The answer must be A. These four forces could not give the object an acceleration greater than this.
Therefore 35 C If both the source and detector travel in the same direction and at the same speed, there will be no relative motion and hence no Doppler shift. Ultraviolet light has a higher energy and shorter wavelength than visible light.
To see that the hole does indeed get bigger, imagine that it was filled with a flat circular plug of metal. This plug would get bigger as the entire plate expanded, so if the plug were removed, it would leave behind a bigger hole.
So, the answer must be either C or D. This eliminates B and C. This eliminates A and E. During these phase transitions, the temperature remains constant.
To produce photoelectrons, each photon of the incident light must have an energy at least as great as the work function of the metal.
A, 17 eV, is equal to the energy emitted by the photon when an electron drops from the —21 eV level to the —38 eV level. C, 64 eV, is equal to the energy emitted by the photon when an electron drops from the —21 eV level to the —85 eV level. D, eV, is equal to the energy emitted by the photon when an electron drops from the —85 eV level to the — eV level. And E, eV, is equal to the energy emitted by the photon when an electron drops from the —38 eV level to the — eV level.
However, no electron transition in this atom could give rise to a 42 eV photon. In the region between the wires, the individual magnetic field vectors due to the wires are both directed into the plane of the page use the right-hand rule with your right hand wrapped around the wire and your right thumb pointing in the direction of the current , so they could not cancel in this region.
Therefore, the total magnetic field could not be zero at either Point 2 or Point 3. Because the magnetic field created by a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire, the fact that Point 1 is in a region where the individual magnetic field vectors created by the wires point in opposite directions and that Point 1 is twice as far from Wire 2 as from Wire 1 imply that the total magnetic field there will be zero.
Since the induced negative charge is closer than the induced positive charge to the charged sphere, there will be a net electrostatic attraction between the spheres. Decreasing d will cause C to increase. Since Wire B has the greatest length and smallest cross-sectional area, it has the greatest resistance. Since the distance between the maximum positive displacement and the maximum negative displacement is 0.
The drawing given with the question shows three such consecutive humps having a total length of 0. Since this is half a wavelength, the full wavelength must be 0.
So, we find that 58 A Because each half-life is 1. After each half-life elapses, the mass of the sample is cut in half, so after 4 half-lives, the mass of the sample decreases from 2 grams to 1 gram to 0. Of the five diagrams given, the charges are closest together in diagram E.
The first diagram below shows the directions of the individual electric forces that each of the other three charges exerts separately on the lower right-hand charge.
This eliminates C, D, and E. However, just swinging the handle over as described in choice A changes neither the area presented to the magnetic field lines from the bottom coil nor the density of the field lines at the position of the loop.
No change in magnetic flux means no induced emf and no induced current. The electrostatic force on Charge 1 due to Charge 2 is a repulsive force, of magnitude , and the electrostatic force on Charge 1 due to Charge 3 is a weaker, attractive force, of magnitude.
So, the total electric force on Charge 1 has magnitude 67 A In order for total internal reflection to occur, the beam must be incident in the medium with the higher index of refraction and strike the interface at an angle of incidence greater than the critical angle. Since k is known, all we need is f in order to calculate m. If we know the quantity given in E, we can double it to get the period, then take its reciprocal to obtain f.
At the moment the block passes through the equilibrium position, O, this energy 2 2 has been converted entirely to kinetic. By setting mv equal to kd , we find that 70 D Because the particle travels in a circular path with constant kinetic energy which implies constant speed , the net force on the particle is the centripetal force.
The time it takes the light pulse to travel a 8 distance of 4.
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