Physics section test (12)

17 172 0
Physics section test (12)

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

Thông tin tài liệu

MCAT Section Tests Dear Future Doctor, The following Section Test and explanations should be used to practice and to assess your mastery of critical thinking in each of the section areas Topics are confluent and are not necessarily in any specific order or fixed proportion This is the level of integration in your preparation that collects what you have learned in the Kaplan classroom and synthesizes your knowledge with your critical thinking Simply completing the tests is inadequate; a solid understanding of your performance through your Score Reports and the explanations is necessary to diagnose your specific weaknesses and address them before Test Day All rights are reserved pursuant to the copyright laws and the contract clause in your enrollment agreement and as printed below Misdemeanor and felony infractions can severely limit your ability to be accepted to a medical program and a conviction can result in the removal of a medical license We offer this material for your practice in your own home as a courtesy and privilege Practice today so that you can perform on test day; this material was designed to give you every advantage on the MCAT and we wish you the best of luck in your preparation Sincerely, Albert Chen Executive Director, Pre-Health Research and Development Kaplan Test Prep © 2003 Kaplan, Inc All rights reserved No part of this book may be reproduced in any form, by Photostat, microfilm, xerography or any other means, or incorporated into any information retrieval system, electronic or mechanical without the written permission of Kaplan, Inc This book may not be duplicated, distributed or resold, pursuant to the terms of your Kaplan Enrollment Agreement PHYSICAL SCIENCES TEST EXPLANATIONS Passage I (Questions 1–7) C This question deals with the stoichiometry of Reaction Looking at the balanced reaction given, we see that when mol AgBr react with mol hydroquinone, two moles of metallic silver are generated mol AgBr would therefore be expected to react with 0.5 mol hydroquinone to produce mol Ag In this case, however, hydroquinone is a limiting reagent The 0.25 moles of hydroquinone we have will react with 0.5 mol AgBr to produce 0.5 moles of metallic silver A The quickest way to get the answer is to remember that the passage mentions that hydroquinone is a mild reducing agent at the beginning of the second paragraph That means that the hydroquinone itself is oxidized in the reaction It loses electrons in the process The only choice that shows hydroquinone as the reactant of an oxidative half-reaction is choice A Choice B shows the other half-reaction from Reaction It shows Ag+ gaining an electron and going to solid silver Since electrons are added in this half-reaction, it is a reduction Choice C can also be eliminated because it is a reduction half-reaction also, but since we know that we are looking for a half-reaction of reaction 1, the half-reaction must conform to the direction of reaction In choice C, we see hydroquinone as a product, so we know that this choice cannot be right Looking at choice D, we see that while it is an oxidative half-reaction that releases one electron, it is the oxidation of solid silver to form silver ion We can eliminate this choice because the product of reaction is solid silver, not silver ion So, if you forgot the definition of a reducing agent, you could have gotten the answer anyway by eliminating choices B and C as reductions and choice D since it wasn't a part of reaction D Question is really a reading comprehension question To answer it, you have to think through what the passage has told you about photographic negatives and the process involved in developing them, especially the fixer The third paragraph describes what a negative really is and the second describes how it is made The second paragraph also describes the role of the fixer solution in the development of film If a developing piece of film were left long enough, the unactivated silver bromide would eventually react with the hydroquinone and deposit silver over the entire film The fixer solution prevents this by stopping the reaction The fixer removes the unreacted silver bromide by ion-exchange solvation That means that any activated silver halide that hadn't reacted yet would also be removed from the film If the film is never treated with the fixer and we assume an excess of hydroquinone, we know that Reaction would proceed until all the silver bromide on the film had reacted, leaving metallic silver all over the film The film would then appear black because the metallic silver is black That lets us know that choices A and B are wrong because they say that the negative would be completely white If we look at the rest of the answers, we see that C says that the negative is black because the unactivated silver bromide would bond with the hydroquinone Be careful! We just finished saying that all the silver bromide will be reduced to metallic silver, but in the reaction, the silver bromide and the hydroquinone not bind together The unreacted silver bromide will reduce to metallic silver as in choice D So choice D is right because it describes the correct appearance of the negative and also states the correct reasons why Notice that choice A also explains that silver will be deposited all over the film It is imperative in this question, therefore, to know not only that the entire film is covered in silver but how this makes the film look The only choice with both parts correct is choice D C In AgBr, silver has an oxidation number of +1: the oxidation numbers of Ag and Br have to add up to zero since AgBr is neutral, and Br, as a halogen, will typically have an oxidation number of –1 On the product side, silver is associated with the thiosulfate ion Note that since the thiosulfate ion is already present on the reactant side, and so instead of assigning oxidation numbers to sulfur and oxygen separately, we can just look at the ion as a whole which has an oxidation number of –2, since that is its charge In silver thiosulfate, there are two Kaplan MCAT Physical Sciences Test Explanations thiosulfate ions, which together will give an oxidation state of -4 This, added to the oxidation number of silver, must equal –3, the charge of the complex ion The oxidation number of silver must therefore be +1 A Essentially, all this question asks is which halogen has the greatest electronegativity Electronegativity measures the attractive force that an element has for electrons in a bond and is often used to help determine whether a bond between two atoms will be covalent or ionic It increases as we move right across a period (except for the noble gases) and as we move up a column That means that the halogens are the most electronegative elements in their periods and that fluorine, at the top of the halogen group, has the strongest electronegativity of the halogens C From paragraph one, we know that what the question really asks is where the most light struck the film The most light strikes the film where the image of the brightest object is on the print That would mean the white ball, or answer choice C Remember that we were given a print to work backward from The white ball on the print would appear black on the negative, which means that the most metallic silver was deposited on the film there The black background, choice A, would have appeared white on the negative meaning that hardly any silver was deposited on the area that showed its image The gray ball, choice D, would be an intermediate between these two, having more light than the black background but not as much as the white ball Choice D, the boundary between the white ball and the background, is more of a distraction than anything else The boundary defines a part of the film between two concentrations of activated silver bromide and doesn't represent any particular value C Molar solubility is the number of moles of a compound that will dissolve in a solution That means that we're looking for the number of moles of silver bromide that will dissolve in a liter of water that already has 0.10 moles of silver nitrate in it From the definition of the solubility product constant, Ksp, we know that: [Ag+][Br –] = 5.0 ∞ 10–13 (in the saturated solution) So, the maximum concentrations of silver and bromide ions dissolved cannot have a product greater than 5.0 ∞ 10−13 x moles of silver halide produces x moles of Ag+ and x moles of Br − in a solution which means the product is x2 We have, however, already got 0.1 moles of silver in the solution from silver nitrate, so the amount of silver ions will be x + 0.10, the amount dissolved from AgBr plus the amount already in solution So the relationship that needs to be satisfied is: 5.0 ∞ 10–13 = x (x + 0.10) Since the molar solubility of silver bromide is very small compared to the 0.10 moles of silver already in the solution, whether we actually add the x to the 0.10 in our product or not won't make much difference So, if we just treat our product as 0.10x, all we have to to find the molar solubility is divide 5.0 ∞ 10–13 by 0.10 The answer is 5.0 ∞ 10-12, which is choice C Passage II (Questions 8–12) A The last paragraph of the passage gives the definition of mass defect: mass defect = sum of masses of nucleons – actual mass of nucleus What is the sum of the masses of the constituent nucleons? There are two types of nucleons: protons and neutrons Their masses are given as mp and mn respectively The nucleus under consideration has atomic number Z and mass number A: the nucleus has Z protons and (A - Z) neutrons The sum of the masses of its constituent nucleons, then, is: Kaplan MCAT Physical Sciences Test Explanations Z ∞ mp + (A – Z) ∞ mn This is what one would expect the mass of the nucleus to be It actually has a smaller mass, measured to be M The difference between the two is the mass defect; i.e.: mass defect = Z ∞ mp + (A Z) ∞ mn – M C The mass condition referred to in the question is simply a requirement of energy conservation A decay reaction is special in that it will happen spontaneously That is, the initial particle will decay even if it is not given any energy So the energy condition for a spontaneous decay dictates that the rest energy of the reactants must be greater than the rest energy of the products The difference is converted into kinetic energy for the emitted electron, for example We know that rest energies are related to masses by E = mc2, and, since c is a constant, we may conclude, that the mass of the initial particle must be greater than the mass of the decay products Some of the mass of the initial particle is converted to energy to make the reaction occur, and therefore the remaining mass, which is the mass of the product, must be smaller than the mass of the initial particle The initial particle in beta decay is called the parent nucleus, and the decay products are the daughter nucleus and an electron (Remember that the electron emitted in ordinary beta-decay comes from the nucleus itself, not from the cloud of electrons around the nucleus.) So, we conclude that choice C is correct The mass of the parent nucleus must be greater than the sum of the masses of the daughter nucleus and an electron, in order for beta decay to occur Choice A can be ruled out because the mass defect is related to the binding energy of the nucleus, which in turn is an indication of its stability: the higher the binding energy, the more stable the nucleus since the more energy needs to be supplied to break it apart If the daughter nucleus has a lower mass defect than the parent nucleus, that would mean it is less stable than the parent nucleus Together with the kinetic energy of the emitted beta particle, this would imply that energy needs to be supplied, which contradicts our knowledge of the spontaneity of such reactions Answer choice B is wrong as well; although the statement it makes is true, it is not pertinent The mass of the parent particle plus the mass of an electron will be greater than the mass of the daughter particle, but this fact is not enough to ensure that there is enough energy for the decay to occur Answer choice D is again true, but it is a statement that has no bearing on the beta decay of the parent nucleus It simply restates the concept of mass defect and is true of every nucleus 10 C The second paragraph identifies the two forces acting within the nucleus There is the familiar electrostatic force which acts to repel protons from one another because they are similarly charged There is also the strong force which is stronger than electrostatic repulsion and works to hold the nucleus together Both forces decrease as distance increases—the electrostatic force falls off as the distance squared: F=k q1q2 r2 The passage does not tell us precisely how the strong force depends on the distance Since a large binding energy means a stable (and hence tightly bound) nucleus, and the strong force is what keeps the nucleons together, stronger strong forces means a higher binding energy The electrostatic force, on the other hand, works to destabilize the nucleus: the stronger the electrostatic forces, the smaller the binding energy If the binding energy per nucleon is decreasing, then, either the strong force is decreasing or the electrostatic force is increasing (per nucleon) This alone helps us eliminate choice A Choice B proposes that larger nuclei have stronger electrostatic forces because of their larger size Although stronger electrostatic forces would decrease binding energy, electrostatic forces decrease with larger distances, not increase Kaplan MCAT Physical Sciences Test Explanations Choice C states that the strong nuclear forces decrease in magnitude as the size of the nucleus increases This is consistent with what we know about the strong force and also successfully accounts for the trend Choice C is the correct answer Choice D is incorrect because greater total charges increase the electrostatic repulsion between protons 11 C This is the only question based on the graph and it requires the ability to interpret information presented in a graphical format This is a difficult question How we decide which reactions give out the most energy? The reaction that gives off the most energy is that one with the greatest energy difference between the reactant particles and the product particles We know from the passage that the total binding energy of a nucleus is the amount of energy needed to completely break it apart into its constituents Look at each reaction given in the answer choices The number of protons and neutrons on each side of the reaction balances, which means that the sum of the energies of the nucleons on each side of the reaction is the same Therefore: Ereactants – Eproducts = binding E (products) – binding E (reactants) So, to chose the reaction that gives out the most energy, we need to chose the reaction for which the product particles have the highest total binding energy compared to the reactant particles The total binding energy of a nucleus is the binding energy per nucleon, shown on the graph, multiplied by the number of nucleons, which is just the mass number, A It looks like we are going to have to a lot of math To make our calculations as short as possible, let's make integer estimates of the binding energies per nucleon from the graph This estimation should allow us to differentiate answer choices from each other Take answer choice A The binding energy of N-14 is approximately MeV (from the graph) multiplied by 14, which is the number of nucleons The binding energy of C-13 is approximately MeV ∞ 13 The binding energy of H-1 is zero Therefore, the energy released for a single reaction is approximately: (14 ∞ – 13 ∞ 8) = (14 – 13) ∞ = MeV Now look at answer choice B The binding energy of He-4 is about MeV ∞ The binding energy of C12 is about MeV ∞ 12 The binding energy of N-15 is about MeV ∞ 15 The binding energy of H-1 is zero Therefore the energy released for a single reaction is approximately ∞ + ∞ 12 – ∞ 15 = ∞ – ∞ = MeV Next consider answer choice C Following the reasoning we used for the first two answer choices, the energy given off by one occurrence of this reaction is given by: about MeV ∞ ∞ – about MeV ∞ – about MeV ∞ 6, which is 24 MeV Thus, the energy released in this reaction is much greater than that released in answer choices A and B In fact, answer C turns out to be the correct answer, as we can check by considering final reaction, given in answer choice D The energy emitted in answer choice D is given by: about MeV ∞ – about MeV ∞ 2, which is about MeV The energy given off in reaction C, over 20 MeV is much greater than the energy given off in the other three reactions Answer choice C is correct 12 C This question requires some interpretation and a calculation using the formula given To answer this question, we must figure out what causes the neutron to lose mass in a γ decay We are told in the question stem that a photon is emitted So the mass that the nucleus loses during its decay must correspond to the energy that it loses in emitting the gamma photon The energy taken away by the photon is related to its wavelength by the hc Since we are told to assume that the nucleus does not recoil, we know that the nucleus does not formula E = λ carry any kinetic energy So, the difference between the rest energy of the nucleus before the decay, and the rest energy of the nucleus after the decay, is equal to the energy of the photon Kaplan MCAT Physical Sciences Test Explanations We can eliminate choices A and B immediately because they are negative numbers The initial rest mass of the nucleus must be greater, not less, than the final mass of the nucleus This must be so, since the emission of the gamma photon takes away some of the energy, and therefore some of the mass, from the nucleus Now we are going to have to some calculating to choose between answers C and D We know that the difference in rest energies of the initial and final nuclei, E, has an associated rest mass, m, where E = mc2 Coupling this equation with the photon energy, we find that: hc λ hc h m= 2= λc λc E = mc2 = m= (6.63 ∞ 10–34) (3.83 ∞ 10–12 ∞ ∞ 108) Since the remaining answer choices are pretty far apart, it is a good idea to save time by estimating the answer: m ∞ 10(–34 + 12 – 8) 4∞3 ♠ 0.5 ∞ 10–30 = ∞ 10–31 kg ♠ This is closest to choice C so it must be the correct answer Passage III (Questions 13–18) 13 A The final paragraph of the passage states that a sacrificial anode for iron is a metal that is more easily oxidized than iron In other words, a sacrificial anode for iron has a reduction potential that is more negative than that of iron Remember that a positive potential corresponds to a negative free energy, and so a more negative reduction potential would mean a lower tendency to be reduced, or equivalently, a higher tendency to be oxidized Looking at table one, you can see that iron, with a reduction potential of –0.44 volts, would be located between zinc and nickel Since choices B and D, tin and palladium, have reductions potentials that are less negative than that of iron, they can be eliminated So, those metals that are more easily oxidized than iron are zinc, aluminum, and magnesium Looking at the answer choices you can see that you have two possibilities: magnesium, with a reduction potential of –2.37 volts, and zinc, with a reduction potential of –0.76 volts Both of these metals can be used as sacrificial anodes for iron, but since magnesium has the more negative potential of the two, it is the best choice 14 D Corrosion is an electrochemical process, and in any electrochemical process, charged conductors, ions, are needed in order for charge to flow So, let's start eliminating some answer choices Choice C, salt water, is out As is well known, salt water contains sodium and chlorine ions, and these ions definitely increase the rate of corrosion Choice A, dissolved carbon dioxide gas, also increases the rate of corrosion Dissolved CO2 gas produces carbonic acid and hence protons according to the following equilibria: CO2 + H2O H2CO3 H+ + HCO3– You can see that H+ is one of the reactants in Reaction 1; increasing its concentration will drive this reaction forward increasing the rate of corrosion So, choice A is out Choice B, air pollution, increases the rate of corrosion as well The sulfur oxides in air pollution react with water to form dissolved sulfuric acid; this speeds up corrosion much in the same way as dissolved CO2 does The only choice left is choice D, making it the correct response Nitrogen gas is relatively inert and in no way assists in the corrosion of iron Kaplan MCAT Physical Sciences Test Explanations 15 A Recall that oxidation occurs at the anode, and upon oxidation iron will acquire a positive charge to become cations In a typical galvanic cell, because the anode and cathode are separated, the ions themselves cannot migrate, and so charge is balanced by the movement of the ions in the salt bridge In this case, however, no such separation is performed, and so the positive Fe ions will be attracted to the cathode Looking at Equation 2, which is the reaction for the formation of rust, you can see that oxygen and water are required Since the exposed areas of iron are rich in oxygen and water, they serve as the cathodes and the unexposed areas serve as the anodes 16 C To answer this question you need to know how to interpret cell schematics or cell diagrams First, the anode process is represented on the left with a vertical line drawn to represent phase boundaries So in this case, we have a zinc anode where zinc metal is oxidized to Zn 2+ Next, two vertical lines are drawn to represent the separation between anode and cathode, and the cathode is represented on the right side of the schematic In this example we have an iron cathode that has Fe2+ being reduced to metallic iron Now that we know how to construct the cell schematic, all we have to is determine the cell potential The cell potential is determined by subtracting the emf of the cathode from the emf of the anode when they are written as reductions: E cell = E cathode – E anode Again, it is important to realize that subtraction is valid only when both potentials are given as reduction potentials If the anode reaction is already written as an oxidation reaction, and the emf given accordingly, then we can simply add the two half-cell potentials Looking at Table we can see that the reduction potential of zinc is – 0.76, and the passage tells us that the reduction potential of iron is –0.44 So the cell potential is (–0.44) – (–0.76) = 0.32 volts 17 D The standard hydrogen electrode is the common reference electrode to which all half-cell potentials are measured The standard hydrogen electrode is a hydrogen-gas electrode specifying hydrogen ion concentration at one molar and the partial pressure of hydrogen gas at atmosphere By convention, the SHE is assigned a value of zero volts Choice A says the SHE “uses hydrogen gas at atmosphere (760 mm Hg)” — this is a true statement, so we can rule out this choice Statement II says the SHE “potential is assigned a value of zero volts” — again this is a true statement which eliminates choice B Choice C is also a true statement: if a half-cell has a negative reduction potential, it has a lower tendency to be reduced than the SHE, and so when the two are coupled it will undergo oxidation instead, yielding a positive emf overall which corresponds to a spontaneous net reaction The statement in choice D is untrue and so this is the correct choice 18 D A galvanic cell is an electrochemical cell where a chemical reaction produces electrical energy An electrolytic cell is an electrochemical cell that requires an input of electrical energy to produce a chemical change Since the corrosion of iron does not require an input of electrical energy, but rather occurs spontaneously, it is considered a galvanic cell Since this is the case, choice A and B can be eliminated You are told in the passage that Fe2+ is further oxidized to Fe3+ to form rust As discussed above in the explanation to #15, the positive ions migrate toward the cathode This piece of information doesn’t really help us in this question since the remaining two choices both contain refer to cathodes What you should realize, however, is that these iron cations go on to react with oxygen and water to form rust Where is there an abundance of these two reactants? In the atmosphere Choice D is the correct response to question 18 Discrete Questions 19 B The acceleration of gravity near the surface of the earth is the constant g = 9.8 m/s2 The fact that this is a constant means that all objects, regardless of their mass, are accelerated towards the earth at the same rate If their acceleration is the same, their velocity changes the same amount in any given time So, if they fall from the same height with similar initial velocities, in this case zero, they will hit the ground with the same speed Since distance Kaplan MCAT Physical Sciences Test Explanations = (1/2)at2 and they fall the same distance with the same acceleration, the time it takes to reach the ground is also the same 20 B A Brønsted acid is defined as a substance that is capable of donating a proton, and a Brønsted base is a substance capable of accepting a proton In this question HCl donates a proton to form Cl–: HCl is the Brønsted acid and Cl– is the Brønsted base In addition, water accepts a proton to form the hydronium ion water accepts a proton making it a Brønsted base and H3O+ the Brønsted acid Two substances that differ from each other only by a proton are called an acid-base conjugate pair This question is asking for the conjugate base of the hydronium ion If we remove a proton from H3O+ we have water, making choice B the correct response 21 C Radiation is the transfer of energy by electromagnetic waves In this manner energy is transferred from the sun to the Earth Convection and conduction both require the presence of a material substance Conduction is the transfer of heat energy from molecule to molecule in a material substance, and convection is the transfer of heat energy as a result of the large scale motion of a material substance In a vacuum neither of these can occur 22 D A reducing agent loses electrons and gets oxidized and an oxidizing agent gains electrons and gets reduced Since a reducing agent loses electrons, its oxidation number increases, and since an oxidizing agent gains electrons, its oxidation number decreases 23 A The original weight is 180 newtons Weight is the force on a mass due to gravity, and is given by the equation W = mg, where g is the acceleration due to gravity Since the mass, m, doesn't change, the acceleration due to gravity, g, is what changes with W Now the weight force is also given by Newton's law of gravity which is given by the equation: F= GM1M2 r2 where G is the universal gravitational constant, M1 and M2 are the two masses, and r is the distance between the centers of the two masses If we let the mass of the Earth be Me, the mass of an object be m, and the distance that separates them be r, then we get that F = GMem/r2 Here r is the distance from the center of the Earth, which at the Earth's surface is the radius of the Earth Now, when the object's distance from the center of the Earth is tripled, then F= GMem GMem = (3r)2 9r2 The force is hence only 1/9 that of its magnitude on the surface At the surface of the Earth the object weighs 180 newtons So at the distance under consideration W = 180/9 = 20 newtons, answer choice A Passage IV (Questions 24–29) 24 A In the second paragraph we are told that the pressure in the chamber is lowered This has the effect of lowering the temperature at which the hydrogen in the chamber boils So by lowering the pressure in the chamber, the hydrogen may remain a liquid even though its actual temperature is above the boiling temperature of the liquid at the new lower pressure By quickly looking over the answer choices, we see there is only one choice involving pressure Answer choice A states that the pressure is decreased in an isothermal process An isothermal process is one in which the temperature remains constant We are told in the passage that the temperature of the liquid Kaplan MCAT Physical Sciences Test Explanations remains constant So answer choice A corresponds to what happens when the piston is lowered, and is probably the correct answer However let's look over the other possibilities Choice B states that the temperature of the liquid is increased in an adiabatic process We don't even need to know that an adiabatic process is one in which no heat flow occurs We are told in the passage that the temperature of the liquid in the chamber remains constant so this choice must be incorrect Choice C states that the volume of the liquid is decreased in an isothermal process Again, just by looking at the first part of this answer choice, we can determine that this is the opposite of what happens In the first paragraph we are told that the piston is lowered This creates a bigger space inside the chamber and increases the volume So choice C is also incorrect Choice D states that the density of the liquid is increased in an adiabatic process You should know that liquids are not very compressible so the variation in density of a liquid is very small We know that the volume of the chamber containing the liquid increases If anything this might cause a small decrease in the density of the liquid, so answer choice D is also incorrect 25 B The passage states that when the pressure drops slightly below the liquid-gas coexistence line and no phase change occurs, it is considered an unstable superheated liquid Looking at the answer choices we see they all involve pressure and temperature so we need to determine what the defining pressure and temperature is of a superheated liquid In the passage we are told that if we drop the pressure below the liquid-gas coexistence line while keeping the temperature constant the temperature of the liquid will be above the boiling temperature of the liquid at the new lower pressure Now let's look over the answer choices to see if any match this condition Choice A states that a superheated liquid is a liquid at a temperature above its boiling temperature at atmospheric pressure Boiling points are defined by specific pressures and temperatures So the temperature at which the liquid boils at atmospheric pressure will be different than the temperature at which the liquid boils at its present pressure Therefore, choice A is incorrect We are looking for a liquid that is above the boiling temperature that corresponds with its present pressure Choice B states that a superheated liquid is a liquid that is at a temperature above its boiling point at its present pressure Therefore it is the correct answer Let's quickly review the other answer choices to make sure we are not missing anything Look at choice C It states that a superheated liquid is normally a gas at room temperature and atmospheric pressure This isn't mentioned anywhere in the passage and it doesn't really make sense that the term superheated liquid would be used to refer to a substance which is a gas at room temperature Choice D states that a superheated liquid is a liquid which remains a liquid at 1000 °C and atmospheric pressure Nowhere in the passage is this quantitative description of a superheated liquid given 26 A This question requires some outside physics knowledge because it deals with a situation that is not discussed in the passage Let's examine what is going on when the liquid hydrogen boils The liquid becomes a gas and expands Therefore, the vapor pressure inside the chamber increases In the question stem we are told we need to keep the pressure of the system constant So to compensate for the increasing pressure inside the chamber, the volume must expand The hydrogen gas will exert a force on the piston moving it farther away from the interior of the chamber Now let's look over the answer choices to see if any agree with our analysis of the situation In looking over the answer choices, we realize they have two parts Both parts of the correct answer choice must be correct So if we can rule out one part of an answer choice, we can eliminate that choice Choice A states that the piston moves away from the interior We determined earlier that this is true The second part of this choice states that work is done by the system in order to move the piston away We decided that the piston moves out because the expanding gas exerts a force on it, so the gas is in fact doing work Both parts of choice A seem to be correct so A is probably the correct answer Let's go over the other answer choices to make sure The first part of answer choice B is correct, the piston does move away from the interior of the chamber What about the second part of choice B? It states that work is done on the system We decided that this is not true, and indeed the piston does not work on the gas when the gas expands The work is done by the liquid-gas system when the hydrogen absorbs heat and becomes a gas So choice B is incorrect The first part of answer choice C states that the piston moves toward the interior of the chamber; this is incorrect So choice C must not be the right choice, because both parts of the correct answer must be correct Answer choice D is incorrect for the same reason; the first part states that the piston moves toward the interior of the chamber So answer choice A is the correct answer Kaplan MCAT Physical Sciences Test Explanations 27 B The answer to this question depends on the information given in Figure and requires a bit of common sense Let's look at the diagram and try to figure out what is going on in the tank and what interaction is indicated by the tracks labeled “a” We note that the particles enter the tank from the left and exit from the right, so the initial particle or particles must be indicated by the tracks on the left side of the tank So looking at the track labeled “a”, we see that there is one line which starts from the left and splits into two lines as we move towards the right side of the tank So this indicates that one moving particle experiences some event in which the single line becomes two lines, which represent two particles Now we have to look over the answer choices to decide which one best describes the situation Answer choice A suggests a particle decays into three charged particles If this were the case, we would observe one line which starts from the left and then splits into three lines at some point in the chamber which then exit on the right Well, this is not the case so answer choice A must be incorrect Choice B states that an elastic collision between two particles occurs You should know that in an elastic collision two particles collide and then go off in different directions If this type of collision occurred between two moving particles, we would observe two tracks which intersect and then split again We don't observe this However, one of the particles in the elastic collision can initially be at rest somewhere in the chamber In this case we would not see a track for it before the collision We would only observe one line branching into two lines as we look from left to right The particle at rest is sitting at the point of intersection and only starts to move after the other particle collides into it So choice B is correct if we consider this case There is nothing in the question stem or answer choice excluding the case where one particle is at rest so choice B is probably correct Let's look over the other answer choices for completeness Choice C states that a completely inelastic collision between two particles occurs In a completely inelastic collision two particles collide and then stick together So we would have either two tracks intersecting and merging into one track or one track if one of the particles in the collision were initially at rest Clearly, we don't have one track, but let's make sure the tracks labeled “a” don't indicate an inelastic collision between two moving particles We must be careful and pay close attention to the direction from which the particles are entering the chamber In the fourth paragraph and in Figure 2, we are told that the particles enter from the left and exit from the right So an inelastic collision between two moving particles would be observed as two lines starting from the left and becoming one line as we move toward the right side of the chamber This is the opposite of the case shown, and therefore choice C is incorrect Answer choice D suggests that an inelastic collision of three particles occurs If the particles were all moving initially, this would show up as three lines starting from the left which intersect, merge into one line, and leave the chamber on the right Clearly the tracks labeled “a” not show three lines intersecting and forming a single line An inelastic collision between three particles might also be represented by two tracks merging into one if one particle were initially at rest, but we've already seen that this is not what the tracks labeled “a” indicate So choice D must be incorrect Therefore answer choice B best describes the diagram and is the correct choice Note that, of course, the tracks may also indicate the decay of one particle into two particles That, however, is not one of the answer choices 28 B To answer this question we need to figure out what feature of the particle tracks in a bubble chamber indicate its charge We are told in the passage that charged particles curve due to the magnetic field in the chamber and that the direction in which they curve indicates their charge The direction of the magnetic force can be figured out by applying the right hand rule The magnetic field points into the page, and that is the direction our fingers should point For track “b,” the particle is experiencing a force pointing upwards Our palm should therefore be facing upwards Our thumb is now pointing to the right, and this is the direction of qv Since all particles enter the chamber from the left and exit at the right, our thumb is pointing in the same direction as the velocity of the particle This implies that q must be positive for the particle that left track “b.” As for track “d,” our fingers still point into the page, but now our palm faces down Our thumb thus points opposite to the direction of the paricle’s velocity, and so q must be negative Kaplan MCAT Physical Sciences Test Explanations F v qv qv (also direction of v) B X B X F d b 29 C To answer this question we need to know what feature of a particle track relates to the momentum of the particle responsible for that track The logical place to go for this information is the passage Sure enough in the third paragraph we are told that the curve of the track is related to the momentum of the particle that left it We are given a formula involving the radius of the circle that the particle travels in, the momentum of the particle, the charge of the particle, and the magnetic field In the question stem we are told to assume that all particles in the chamber have the same magnitude of charge From the equation we see that the radius of the circle is directly proportional to the momentum of the particle So we can gauge the momentum of the particles in this bubble chamber solely by the curve of their tracks Therefore, a particle traveling in a small circle will have a small momentum Now we need to relate the size of the circle in which a particle travels to the curved paths we see in the chamber The curved paths are just sections of a circle: track track R larger R: higher momentum R smaller R, lower momentum So the track that is the most curved must come from the smallest circle If we are looking for the particle with the smallest momentum, we must look for the particle that left the most curved track Looking at the diagram, the upper branch of track c is definitely the most curved Therefore it must be the track of the particle with the lowest momentum in the chamber The spiral occurs because the particle is losing energy, and therefore the momentum and radius are getting smaller Passage V (Questions 30–35) 30 C The kinetic theory of gases attempts to explain the macroscopic properties of gases from a microscopic perspective The basic assumptions are: (1) the gas particles are separated by distances much greater than their size; (2) the only intermolecular forces occur during collisions, which are totally elastic; (3) the gas particles are moving in a random fashion with a distribution of speeds; and (4) the average kinetic energy of a gas is directly proportional to the absolute temperature You should be aware that since the AVERAGE kinetic energy is constant at a particular temperature so is the AVERAGE speed This does not imply that the speed of a particular molecule remains constant, it changes with each elastic collision It is the average speed that remains constant Just one more aside about elastic collisions, an elastic collision is one in which the sum of the kinetic energies of the colliding particles does not change Anyway, it is the third and fourth assumptions that allow us to pick C as the correct response to this Kaplan MCAT Physical Sciences Test Explanations question As just stated, it is the average speed of the collection of gas molecules that remains constant, not the speed of the individual particles 31 B Ideal gases obey the ideal gas law at all temperatures and pressures ideal gases exist only in theory Real gases, on the other hand, follow the ideal gas law only at high temperatures and low pressures At low temperatures real gases exhibit a lower volume than Charles' law would predict This is because attractive forces exist between the molecules of a real gas; whereas, it is assumed that the molecules of an ideal gas not interact in any way We know that these attractive forces exist because if we continue to lower the temperature of a gas, the molecules associate with each other, which eventually leads to condensation Ideal gases never condense; they just continue to follow Charles' law Another way in which real gases differ from ideal gases, is that at high pressures, real gases occupy a larger volume than that predicted by Boyle’s law This is because the ideal gas law does not take into consideration the volume occupied by the molecules themselves It assumes that the volume predicted is the volume that the gas would occupy if the molecules themselves occupied no space You should remember from the passage that the Van der Waals equation of state attempts to account for these deviations Anyway, looking at the answer choices we can see that choice B, attractive forces, is the correct response But what if you didn't remember this is there any way that the passage could help? The answer is yes The answer to this question can be found directly in the passage Equation 2, the Van der Waals equation, predicts the behavior of nonideal, or real, gases How does this equation differ from the ideal gas equation? It has two additional constants that account for the intermolecular forces of the gas molecules and for the "excluded volume" of the gaseous molecules Looking at the answer choices you can see that choice B, attractive forces, is the correct response 32 D The Gay-Lussac law states that at constant volume the pressure of an ideal gas is directly related to its temperature (This has been subsumed into the ideal gas law so there is no reason to memorize these separately.) So, since pressure is directly related to temperature, whenever the temperature increases so goes the temperature With this in mind you can eliminate choices A and B since they state that the pressure decreases Answer choices C and D both state that the pressure increases, but choice C says that the density increases as well If the volume is constant, this can't be true! Choice D is, therefore, the correct response If you didn't realize this, you could have used your knowledge of the kinetic theory in arriving at choice D as the correct response Choice D states that the pressure increases because the frequency of collisions between gas particles and the container increase This is absolutely true 33 B This has been discussed above in question 31 34 D This question is testing you on the physical meaning of the constant b in the van der Waals equation, which is Equation in the passage The passage tells you that the purpose of this constant is to account for the volume that the gas molecules occupy, something that the ideal gas doesn't take into consideration How can you approximate the magnitude of this constant for a particular molecule? If you need to figure out how much space the molecules take up in the gas phase, a good place to look is at the molar volume of the liquid state In this state the molecules are in extremely close contact The molar volume can be found from dividing the molecular weight by the density: molecular weight ÷ density = mass mass volume volume mass ÷ = ∞ = mass mole mole volume mole You can see from the table that the densities are all pretty close to each other so it is the molecular weight that predominates Ethanol, choice D, is by far the heaviest so it is the correct response 35 C 11 Kaplan MCAT Physical Sciences Test Explanations In a nutshell, the kinetic energy of a gas depends only on the absolute temperature, and the average speed of a gas is inversely proportional to the square root of its molecular weight: vave = k mw According to the kinetic molecular theory, since there are no intermolecular interactions, the energy of a gas is all kinetic The higher the temperature, the more energetic is the gas, and at a specific temperature, the lighter the gas, the faster is its average speed The question stem tells you that gases X and Y are in a closed container at a constant temperature You are also told the gas Y is heavier than gas X Let's look for the true statement in the answer choices Choice A states that both gases have the same average speed This isn't true: since Y is heavier than X, X has the greater average speed Choice B states that gas X has a greater average kinetic energy than Y Not true, according to the kinetic molecular theory: all gases at the same temperature have the same average kinetic energy Choice C states that gas X has a greater average speed than that of gas Y This is a true statement If two gases are maintained at the same temperature, the lighter of the gases will have the higher average speed Choice C is the correct response Choice D is another choice that does not equate the kinetic energies, so it is also incorrect Passage VI (Question 36–41) 36 B The question stem indicates that we only need to look at one experiment First we have to figure out how power is related to the parameters described in the passage We are told that a constant force is applied to the spaceship for the first 200 kilometers in Experiment The formula power = force ∞ velocity may have escaped your mind, but you should be able to see why it makes sense and ideally be able to drive it on your own Power is defined as energy per unit time, and energy in this case is the change in kinetic energy which is also the work done by the engine Work = force ∞ distance, and so: power = (force ∞ distance) ÷ time = force ∞ distance = force ∞ velocity time So for the first 100 km of Experiment 1, in which the ship experiences a constant force, the power is directly proportional to the velocity Also, a constant force results in a constant acceleration, and thus the velocity is directly proportional to the time as seen in the kinematic equation v = at Therefore, the power is also directly proportional to the time So the power also must increase linearly with time Now we must look for a graph with a line that tilts upward as we go farther away from zero on the positive side of the x-axis Therefore choice B is the correct answer choice 37 D In this question we are asked to determine in which of the three experiments the spaceship achieves the greatest maximum velocity We can assume that the greatest maximum velocity in Experiments and will occur before the spaceship hits the wall because the collision only serves to slow the spaceship down So let's figure out what determines the velocity We are told that in each experiment there are stretches over which the spaceship is accelerated The velocity must increase during these stretches We are also told that there are stretches over which the engines remain off and that the velocity is constant over these stretches So the stretches we should be concerned with for this question are only those over which the ship is accelerating Now to relate acceleration, velocity, and distance, we can use the equation vf2 – vi2 = 2ad So to find the maximum velocity in each experiment, we need to consider the velocity at the beginning and end of each stretch over which the spaceship is accelerated for each experiment Since the spaceship starts at Kaplan MCAT Physical Sciences Test Explanations rest, the velocity at the beginning of the first stretch is zero The velocity at the end of the first stretch can be determined from the equation v12 = 2ad1 where v1 is the velocity at the end of the first stretch, a is the acceleration of the spaceship over the stretch and d1 is the length of the first stretch Now we need to find the velocity at the end of the second stretch What is the velocity at the beginning of the second stretch over which the spaceship is accelerated? It must be the final velocity at the end of the first stretch because we are told in the passage that between the two accelerated stretches in each experiment the spaceship travels at a constant velocity So we can use v1 for the velocity at the beginning of the second stretch So the velocity at the end of the second stretch v2 can be determined by the following formula v22 = v12 + 2ad2 where d2 is the length of the second stretch If we plug in the relation for v12 which we determined earlier we get v22 = 2a(d1 + d2) Now v22 is the velocity at the end of the second stretch so it is also the maximum velocity d1 + d2 is the total distance over which the spaceship was accelerated in each experiment We know that the acceleration a is the same in each experiment because F = ma, the same force F is applied in each experiment, and the same spaceship is used so m is the same in each experiment Therefore the maximum velocity the spaceship reaches in each experiment will be the same if the total distance over which the spaceship is accelerated is the same in each experiment In Experiment the spaceship is accelerated over two 200-kilometer stretches, so the total distance over which the spaceship is accelerated is 400 kilometers In Experiment the spaceship is accelerated over a stretch of 100 kilometers and then 300 kilometers, so the total distance over which the spaceship is accelerated is also 400 kilometers In the third experiment the distance which the spaceship travels with its engines shut off is increased and the wall is removed Removing the wall just increases the distance the spaceship can travel without meeting any resistance The distance over which the spaceship is accelerated is still 400 kilometers So the maximum velocity of the spaceship in all three cases is the same Therefore, answer choice D is correct 38 B The question asks us how we can determine if the fuel company charged the correct amount for the fuel used in the experiments So we need to find out from the passage when the fuel tank was filled We also need to figure out when fuel is used in each experiment Looking to the passage, we see that the tank is refilled at the end of each experiment So we need to calculate how much fuel was used in each experiment Now ask yourself when fuel is used by the engines Fuel is only consumed when the engines are on So we only need to consider the stretches in each experiment when the engines are on Now we know which part of each experiment to consider, but we still need to determine which parameters of each experiment to consider From looking over the answer choices, we see they involve time, force times distance, and mass times average speed We know that fuel provides the energy the engines need to propel the spaceship Also, we are told in the passage that the fuel used is proportional to the square root of the work done This information is crucial Now we can determine how much fuel was used by considering the work done in each experiment The work done in each case is the force times the distance over which the force is applied So by multiplying the force applied and the distance over which the spaceship was accelerated, we can determine the energy used in the form of fuel Thus, the correct answer is choice B Let's look over the other choices Choice A suggest that we should consider the entire distance traveled The engines aren't running over the entire distance, and fuel is not used when the engines aren't running So choice A is incorrect because we are only concerned with the distances over which fuel is being consumed Choice C claims all we need to know is the time each experiment took But we can't figure out the work done from the time alone, we must know the magnitude of the applied force in addition to the time Choice D is incorrect because the work done cannot be determined from the mass times the average speed over the entire distance traveled Again we are only concerned with the distance over which the spaceship is accelerated, and we are looking for a quantity which has units of work Mass times average speed has units of momentum 39 C 13 Kaplan MCAT Physical Sciences Test Explanations We are asked to compare the time for which the spaceship was in contact with the wall in each experiment The fact that we are given average force and that we are looking for the time for which the wall and spaceship are in contact should bring to mind impulse, because impulse equals the average force times time The impulse is the change in momentum the spaceship experiences after a collision with the wall So the change in momentum of the ship due to its collision with the wall equals the average force multiplied by the time In the question stem we are told that the average force imparted by the wall is the same in both experiments So the change in momentum of the spaceship must be directly proportional to the time of contact of the spaceship and wall in the collision So we need to compare the change in momentum of the spaceship in each experiment Momentum is mass times velocity It is a vector quantity Since mass has no direction associated with it, the momentum of an object will have the same direction as the velocity of the object To figure out the initial momentum, we must first determine the velocity of the spaceship right before it hits the wall in each experiment We know the mass is the same in each experiment because the same ship is used To find the velocity before the ship hits the wall, we can use the formula vf2 = vi2 + 2ad, where a is the acceleration and d is the distance over which the ship is accelerated The spaceship experiences the same constant acceleration in each experiment because acceleration equals force over mass, and the force and mass associated with the spaceship are the same in both experiments Let d1 be the first distance over which the spaceship is accelerated in either experiment In each experiment the spaceship starts from rest, so the initial velocity is zero Therefore the speed at the end of the first distance is given by v12 = 2ad1 Let d2 be the second distance over which the spaceship is accelerated in either experiment The speed at the end of the second distance is given by v22 = v12 + 2ad2 Now plugging in the previously determined value of 2ad1 for v12, we determine that v22 = 2ad1 + 2ad2, or v22 = 2a (d1 + d2) In Experiment 1, d1 + d2 = 200 + 200 or 400 kilometers In Experiment 2, d1 + d2 = 100 + 300 or 400 kilometers So d1 + d2 = 400 kilometers in both experiments Therefore the velocity of the spaceship when it reaches the wall must be the same in each experiment So the momentum of the spaceship before it hits the wall is the same in each experiment Now we must consider the momentum of the spaceship after the collision with the wall So we need to determine the velocity of the spaceship after the collision with the wall in each experiment Well looking back to the passage, we find that we are given information about the spaceship after the collision We are told that it takes less time for the spaceship to coast back to the starting point in the second experiment Since the distance is the same in each experiment, this implies that the velocity and therefore the speed with which the spaceship travels after the collision is greater in the second experiment So the momentum of the spaceship after the collision is greater in the second experiment Now to figure out the change in momentum we subtract the momentum of the spaceship before it hits the wall from the momentum of the spaceship after it hits the wall Don't forget to take into consideration the direction of the velocity It changes after the ship hits the wall So if we take the direction the ship travels when it is moving towards the wall as positive, then the direction it travels in after it hits the wall is negative But when we subtract this negative vector, it becomes positive So the change in momentum is the mass times the initial speed plus the mass times the final speed Therefore the change in momentum is greater in the second experiment because the mass times the speed after the collision in this experiment is greater than in the first experiment Since the change in momentum equals average force times time and the average force is the same in each experiment, the change in momentum is proportional to time Therefore the spaceship must be in contact with the wall for a greater amount of time in Experiment because the change in momentum due to the collision is greater in Experiment 40 A The stem of this question might seem a little vague So it is important to quickly read over the answer choices to get a sense of what factors must be considered From reading the answer choices, one can determine that getting the spaceship to stop at the starting line will involve the application of a force So to answer this question we must go back to the passage and find out what is the last thing mentioned about Experiment The last thing we are told about experiment one is that the spaceship collides into the wall and drifts back to the starting line We need to have the spaceship come to a stop at the starting line This will happen if the spaceship experiences a deceleration on its way back A deceleration must be caused by some external force Now we have to figure out if there is any force that the spaceship experiences on its way back to the starting point Since it is in outer space, we know there is no air resistance The term outer space also implies that there are no planets or stars nearby, therefore there will be no gravity due to nearby stars or planets In the first paragraph of the passage, we are told that the force of gravity between the spaceship and wall is negligible So the Kaplan MCAT Physical Sciences Test Explanations spaceship will not experience an external force when it coasts back to the starting line But, what about the collision with the wall? Didn't that cause the spaceship to decelerate? Initially it did, but once the spaceship left the wall, it was traveling at a velocity which was no longer affected by the wall Therefore the spaceship must be coasting back toward the starting line at a constant velocity This means that the spaceship will drift back right past the starting line So in order to have the spaceship come to rest at the starting line, we must apply a force In order to slow the spaceship down, a force must be applied opposite to the direction in which the spaceship is traveling Therefore the force must be applied in the direction pointing from the starting line toward the wall Now that we've analyzed the situation, let's look over the answer choices to see if any concur with our reasoning Answer choice A states that a force must be applied in the direction pointing from the starting line to the wall This agrees with our analysis so it is probably the correct answer But to make sure we are not overlooking anything, let's look at the other choices Choice B states that a force must be applied in the direction pointing from the wall toward the starting line Well this means a force is applied in the same direction in which the spaceship is traveling This will accelerate the spaceship causing it to go past the starting line at a greater speed So choice B is incorrect Answer choice C states that no force need be applied Well, we figured out that the spaceship is moving at a constant velocity, and according to Newton's first law it will keep moving until an outside force is applied So this choice is also incorrect Choice D states that a force perpendicular to the direction the spaceship is traveling in must be applied Well this will only serve to change the direction in which the spaceship is traveling A force must be applied in the opposite direction to that in which an object is traveling in order to slow it down So answer choice D is incorrect Therefore, choice A is the correct answer choice 41 A All you need to to answer this question is rearrange the given equation solving for Mi/Mf and relate it to one of the graphs in the answer choices First divide each side by u to obtain: ln Mi (vf – vi) = u Mf Now to get rid of the ln, you must raise each side to be a power of e (Remember elnx = x.) So we get Mi/Mf = e to (v – v ) the [(vf – vi)/u] power This implies that Mi/Mf increases exponentially as the quantity f i increases Now look u at the answer choices to determine the axes of the graphs We see that all of the choices have Mi/Mf as the y(vf – vi) as the x-coordinate So we are looking for a graph that shows exponential growth of the ycoordinate and u coordinate as the x-coordinate increases The correct answer choice is therefore choice A Choice B implies an exponential decay of the y-coordinate as the x-coordinate increases Choice C implies a linear increase of the ycoordinate with respect to the x-coordinate, and choice D implies that the y-coordinate increases and then decreases as the x-coordinate increases Discrete Questions 42 D The key word here is “melting” When ice is melting the temperature remains constant until all of the ice has melted, after which the temperature of the melted water starts to rise The question stem does not tell us how much ice there is and whether the heat is enough to melt all of it Yet choice D is the only possibility of the four choices: it shows that the temperature is constant as the ice is undergoing the phase change Choices A and B cannot be correct since it shows the temperature falling in at least one portion of the graph Choice C is also incorrect because at least at the beginning, the temperature should be flat as the ice finishes melting 43 B The capacitance of a parallel-plate capacitor is given by 15 Kaplan MCAT Physical Sciences Test Explanations C= ε0A d where ε0 is the permittivity of free space and is a constant, A is the area of overlap of the two plates, and d is the separation of the two plates From this we can see that the capacitance is proportional to the area of the plates and inversely proportional to the separation of the plates Now in the question we are told that the separation of the plates increases which means that the capacitance must decrease assuming that all the other variables remain constant This eliminates Roman numeral I and therefore answer choices A and C That leaves us choices B and D So we know that Roman numeral III must be part of the answer, and we only need to look at Roman numeral II Now, we are also told in the question stem that the capacitor is isolated Since it is isolated, charge cannot flow or be generated This tells us that the charge on the plates must remain constant This eliminates Roman numeral II, and therefore answer choice D This leaves us with answer choice B which is indeed the correct answer Let's go on and see why Roman numeral III is part of the answer The capacitance of a capacitor is related to the voltage across the plates and the charge stored on the plates by the equation C = Q/V, where C is the capacitance, Q is the charge stored, and V is the voltage across the plates We have already established that the charge on the capacitor remains constant, and we can therefore say that the capacitance of the capacitor is inversely proportional to the voltage across the plates In other words as the capacitance decreases, the voltage across the plates increases So Roman numeral III is correct, and choice B is the answer 44 A The force of a spring is a conservative force which means that the total energy inherent in a spring system is equal to the sum of its potential and kinetic energies Remember, potential energy is the energy that's stored, while kinetic energy is the energy of motion Now, the potential energy of a spring can be expressed by the formula U= kx where U is the potential energy, k is the spring constant specific for that spring, and x is the distance that the spring is either stretched or compressed from its equilibrium length As the distance of compression or stretching increases, the x2 term increases so the potential energy must increase Let's say we have a compressed spring with a certain amount of stored potential energy When this spring is released, its potential energy will be converted to kinetic energy until the spring has reached its equilibrium length It will then continue to stretch, converting the kinetic energy back to potential energy The spring will stretch until its displacement from the equilibrium length is equal to the compressed spring's displacement from the equilibrium length Looking at the energies involved, it's easy to see that the potential energy would be a maximum at the maximum compression, or the maximum stretch, because at either point the displacement x would be at a maximum Furthermore, potential energy would be a minimum at the equilibrium length because at this point x would be equal to Since we're dealing in a conservative system, it follows that the sum of the potential and kinetic energies will be constant Therefore, the maximum kinetic energy would have to occur at the x = point, or at the equilibrium length, and the minimum kinetic energy will occur at the maximum stretch and maximum compression points In this question we're told that the spring is compressed to its minimum length and that it is not allowed to spring back, so this would be a point of maximum potential energy and minimum kinetic energy (= 0), and choice A is correct 45 A This question tests how well you understand vector addition If the two forces act in the same direction, they will reinforce each other and then the resultant force will be 12 + or 17 newtons If they act in opposite directions, they will detract from each other in which case the resultant force is 12 – or newtons These represent the maximum and minimum resultant forces that can occur But any other value between and 17 is also possible these will occur when the forces are at an angle to each other So we are looking for an answer between and 17 newtons, inclusive The only Roman numeral that satisfies this condition is Roman numeral II, 13 newtons ... determine in which of the three experiments the spaceship achieves the greatest maximum velocity We can assume that the greatest maximum velocity in Experiments and will occur before the spaceship... –2, since that is its charge In silver thiosulfate, there are two Kaplan MCAT Physical Sciences Test Explanations thiosulfate ions, which together will give an oxidation state of -4 This, added... silver must therefore be +1 A Essentially, all this question asks is which halogen has the greatest electronegativity Electronegativity measures the attractive force that an element has for electrons

Ngày đăng: 04/05/2017, 09:04

Tài liệu cùng người dùng

  • Đang cập nhật ...

Tài liệu liên quan