Science Invitational A 2019 PDF

UNIVERSITY INTERSCHOLASTIC LEAGUE Science Invitational A 2019 GENERAL DIRECTIONS: DO NOT OPEN EXAM UNTIL TOLD TO DO SO. Ninety minutes should be ample time to complete this contest, but since it is not a race, contestants...

UNIVERSITY INTERSCHOLASTIC LEAGUE Science Invitational A 2019 GENERAL DIRECTIONS: DO NOT OPEN EXAM UNTIL TOLD TO DO SO. Ninety minutes should be ample time to complete this contest, but since it is not a race, contestants may take up to two hours. If you are in the process of actually writing an answer when the signal to stop is given, you may finish writing that answer. Papers may not be turned in until 30 minutes have elapsed. If you finish the test in less than 30 minutes, remain at your seat and retain your paper until told to do otherwise. You may use this time to check your answers. All answers must be written on the answer sheet provided. Indicate your answers in the appropriate blanks provided on the answer sheet. You may place as many notations as you desire anywhere on the test paper except on the answer sheet, which is reserved for answers only. You may use additional scratch paper provided by the contest director. All questions have ONE and only ONE correct (BEST) answer. There is a penalty for all incorrect answers. If a question is omitted, no points are given or subtracted. On the back of this page is printed a copy of the periodic table of the elements. You may wish to refer to this table in answering the questions, and if needed, you may use the atomic weights and atomic numbers from the table. Other scientific relationships are listed also. Silent hand-held calculators that do not need external wall plugs may be used. Graphing calculators that do not have built-in or stored functionality that provides additional scientific information are allowed. Small hand-held computers are not permitted. Calculators that accept memory cards or memory sticks are not permitted. Each contestant may bring one spare calculator. All memory must be cleared. SCORING: All questions will receive 6 points if answered correctly; no points will be given or subtracted if unanswered; 2 points will be deducted for an incorrect answer.  Science ! Invitational A ! 2019 Biology B01.! All organisms within the same Order belong to the B06.! Which organ system has no known function in same defense from microbial invaders? A)! Domain. A)! Integumentary B)! Genus. B)! Lymphatic C)! Family. C)! Immune D)! Species. D)! Nervous E)! All of the above are correct. E)! All of the above have some defense function. B02.! If an organism is homozygous recessive for a trait, B07.! A Mendelian monohybrid cross would yield a which of the following is not a possible genotype of genotypic ratio of. the organism’s progeny? A)! 3:1 A)! AA B)! 1:1 B)! Aa C)! 1:2:1 C)! aa D)! 9:3:3:1 D)! Unable to determine. E)! none of the above B03.! The light-harvesting pigments of photosynthesis in B08.! Which process specifically generates a protein’s eukaryotes are typically found within primary structure from an mRNA sequence? A)! nuclear membranes. A)! Gene expression B)! the endoplasmic reticula. B)! Translation C)! the plasma membrane. C)! Transcription D)! thylakoid membranes. D)! Replication E)! mitochondria. E)! Transformation B04.! Which of the following would you not find B09.! Which of the following is the correct order of phases associated with the plasma membrane of a cell? during the nuclear division portion of the eukaryotic A)! Transmembrane proteins cell cycle for somatic cells? B)! Phospholipids A)! Prophase, Anaphase, Metaphase, Telophase, C)! Integral proteins Cytokinesis D)! Selective permeability B)! Prophase I, Metaphase I, Anaphase I, Telophase E)! Ribosomes I, Cytokinesis, Prophase II, Metaphase II, Anaphase II, Telophase II, Cytokinesis C)! Telophase, Anaphase, Metaphase, Prophase, Cytokinesis B05.! With the extinction of dinosaurs nearly 65.5 million D)! Prophase, Metaphase, Anaphase, Telophase, years ago, which group of animals adapted and Cytokinesis evolved very quickly afterwards? E)! G1, S, G2, mitosis A)! Arthropods B)! Mammals C)! Amphibians D)! Reptiles E)! Aquatic vertebrates University Interscholastic League ! page 2  Science ! Invitational A ! 2019 Biology B10.! DNA replication in Domain Bacteria occurs B14.! If a mutation occurred that changed a cytosine to an A)! within the nucleus. adenine in the DNA sequence, this would be called a B)! prior to mitosis. A)! deletion. C)! immediately prior to cell division. B)! base insertion. D)! during gene expression. C)! base substitution. E)! only for one of the DNA strands in the double D)! frameshift mutation. helix. E)! silent mutation. B11.! The field of biology that studies characteristics of B15.! Organisms within the same geographic area that organisms and attempts to classify them into specific can mate and produce fertile offspring are part groups is called of the same A)! population biology. A)! community. B)! phylogeny. B)! population. C)! biological diversity. C)! ecosystem. D)! ecology. D)! biosphere. E)! taxonomy. E)! phylum. B12.! Consider the cell biology of eukaryotic cells. Which B16.! In the water cycle, the process of carrying water of the following statements is incorrect? through plants and exiting leaves as vapor into the A)! Cell walls are present in all eukaryotes. atmosphere is termed B)! The nucleus contains chromatin. A)! condensation. C)! The mitochondria are responsible for glucose B)! precipitation. oxidation to generate ATP. C)! infiltration. D)! Lysosomes are digestive organelles. D)! evaporation. E)! The rough endoplasmic reticulum contains E)! transpiration. ribosomes. B13.! From August 2014 through September 2018, 386 B17.! According to natural selection, which would occur if cases of acute flaccid myelitis (AFM) have been a phenotype in 1 out of 100 organisms provided an confirmed by the Centers for Disease Control and advantage for acquiring food and reproducing? Prevention (CDC), mostly in children. The cause of A)! Organisms with that gene would die out. the disease has yet to be determined. What disease B)! Organisms with that phenotype would thrive and does AFM mimic? the allele(s) for that phenotype would increase in A)! chickenpox the population over time. B)! measles C)! The phenotype in the population would remain C)! influenza relatively stable. D)! food poisoning D)! After several generations, the phenotype would E)! polio still represent 1% in the population. E)! Since it is only 1% of the population, organisms with that phenotype would fail to pass on any useful genes. University Interscholastic League ! page 3  Science ! Invitational A ! 2019 Biology B18.! A viral disease that causes the parotid salivary B20.! Large polymers made of amino acids that catalyze a glands to swell is called majority of chemical reactions are called A)! measles. A)! enzymes. B)! roseola. B)! ribozymes. C)! mumps. C)! inorganic catalysts. D)! rubella. D)! nucleic acids. E)! German measles. E)! transmembrane proteins. B19.! The type of tissue found lining the internal surface of the stomach is A)! muscle. B)! nervous. C)! connective. D)! epithelial. University Interscholastic League ! page 4  Science · Invitational A · 2019 Chemistry C01. What is the molar mass of the mineral C06. In a mercury thermometer, why does the magnetite, Fe3O4, in grams per mole? mercury level in the tube go up as the temperature increases? A) 231.55 B) 34 A) Because that's how thermometers work. C) 71.85 B) Because heat rises, so as the mercury D) 271.4 warms up it rises up the tube. E) 37.00 C) Atoms expand in size as they are heated, increasing the volume of the mercury in C02. What is the sum of the coefficients in the the tube. balanced equation for the complete combustion D) The mercury atoms move farther apart as of butane, C4H10? the temperature increases, increasing the volume of the mercury in the tube. A) 4 E) Some mercury at the bottom of the tube B) 9 begins to boil, pushing the mercury above C) 16.5 it higher up the tube. D) 22 E) 33 C07. Under which of these conditions does the ideal gas law begin to fail? C03. Which of these is the best ground state electron configuration for strontium? A) large volume A) 2 2 6 2 6 2 1s 2s 2p 3s 3p 4s 3d 4p 5s 10 6 2 B) high temperature B) 1s22s22p63s23p63d104s24p64d05s2 C) high pressure C) 1s2 2s2p6 3s2p6d10 4s2p6 5s2 D) large number of moles D) [Kr]4p65s2 E) when R is not in units of L·atm/mol·K E) [Kr]4p65s0 C08. Which chemical equation below would have C04. Which of the following compounds contains the equilibrium expression K = [A]2[B] exactly one unshared pair of valence electrons? A) 2A(aq) ⇌ B(aq) A) BF3 B) B(aq) ⇌ 2A(aq) B) PH3 C) A2B(aq) ⇌ 2A(aq) + B(aq) C) H2S D) 2A(aq) + B(aq) ⇌ A2B(s) D) NO2 E) A2B(s) ⇌ 2A(aq) + B(aq) E) CCl4 C09. The salt concentration in the Dead Sea is C05. What is absolute zero (0 K) measured in 34.2%, almost ten times saltier than the open degrees Fahrenheit? ocean. What is the boiling point elevation of Dead Sea water due to the NaCl concentration? A) 0 B) 32 A) 9.11°C C) –273.15 B) 4.55°C D) –459.67 C) 2.99°C E) –218.52 D) 5.99°C E) 16.5°C University Interscholastic League · page 5  Science · Invitational A · 2019 Chemistry C10. Which of the following are all members of the C13. According to the Arrhenius definition of acids same group on the periodic table? and bases, is carbon dioxide an acid, a base, or neither? A) Copper, silver, gold, and platinum B) Nitrogen, oxygen, fluorine, and neon A) Neither, because CO2 does not contain H C) Uranium, neptunium, plutonium, and or OH in its chemical formula. mercury B) Neither, because CO2 does not have the D) Carbon, silicon, tin, and lead words "acid" or "base" in its name. E) Adamantium, vibranium, dilithium, and C) Neither, because CO2 is a covalent dalekanium compound. D) It is a base, because when you dissolve it C11. The reaction in water the pH goes up. 2 NO(g) + 2 H2(g) → N2(g) + H2O(g) E) It is an acid because when you dissolve it is second order with respect to NO and first in water the pH goes down. order with respect to H2. What is the rate law for the reaction? C14. The number of atoms in one formula unit of [)*], [-, ], potassium dichromate is A) rate = ' [), ][-, *] A) 3 B) rate = k [NO]2[H2][N2][H2O] B) 8 C) [N2][H2O] = k [NO]2[H2] C) 11 D) rate = k [NO]2[H2] D) 12 E) k = rate [NO]2[H2] E) 14 C12. 1026 kJ of heat increases the temperature of a sample of water at sea level from 40°C to C15. What is the difference between an electrolytic 80°C. If 1026 kJ more heat is added, what cell and a galvanic cell? will the final temperature of the water be? A) An electrolytic cell contains electrolytes A) 120°C because the water will continue to and a galvanic cell does not. heat up at the same rate. B) Galvanic cell is an older term for B) More than 100°C but less than 120°C electrolytic cell. because some of the heat will be used to C) A galvanic cell generates a positive voltage boil the water. and an electrolytic cell does not. C) Less than 100°C because as the water gets D) Electrolytic cells give off light as well as hotter, it takes a greater amount of heat to voltage. raise the temperature by one degree. E) Electrolytic cells have a cell wall and D) 100°C because that's the highest galvanic cells have a cell membrane. temperature water can reach at 1 atm pressure. E) The water will all boil away before that much heat can be added. University Interscholastic League · page 6  Science · Invitational A · 2019 Chemistry C16. Which of these is not a valid empirical C20. Our current periodic table lists 118 elements. formula? Is it possible that there are still any undiscovered elements in between the A) CH elements listed in the periodic table? B) CH2 A) Yes, but to avoid detection for this long C) CH3 they would have to exist in very small D) CH4 quantities or not occur on Earth at all. E) C2H4 B) Yes, but they would have to be totally non-reactive reactive so they are never found in C17. What is the silver ion concentration in a compounds. saturated solution of silver oxalate, Ag2C2O4? C) Yes, but they would have to be radioactive Ksp = 5.4 × 10–12. with extremely short half-lives half so they decay before they can be detected. A) 1.11 × 10–4 M D) No, because all the boxes in the periodic B) 2.21 × 10–4 M table are already filled. C) 1.16 × 10–6 M E) No, because the atomic number is the D) 1.64 × 10–6 M number of protons rotons in the nucleus and we E) 2.32 × 10–6 M have accounted for all possibilities po from 1 through 118. C18. What is the molar mass of caffeine, shown below, in grams per mole? A) 194.22 B) 193.21 C) 169.19 D) 133.16 E) 121.14 C19. How many π bonds are in the caffeine molecule shown in question C1 C18? A) 3 B) 4 C) 6 D) 8 E) 12 University Interscholastic League · page 7  Science Invitational A 2019 Physics P01. According to Natarajan, the Babylonians recorded P06. A tennis ball, initially travelling at 42.0m/s West, is the motion of a celestial object that, after two years returned by Serena Williams. After being hit by of normal motion, would reverse its motion for Serena’s racquet, the ball has a velocity of 30.0m/s about 90 days. Which celestial object were they East. If the racquet was in contact with the ball for a observing? total time of 0.650sec, what was the average A) Mars acceleration of the ball caused by the racquet? B) Venus A) 18.5 m/s2 C) The Moon B) 46.2 m/s2 D) Jupiter C) 64.6 m/s2 E) Mercury D) 72.0 m/s2 E) 111 m/s2 P02. According to Natarajan, Cepheid variable stars can be used to measure astronomical distances out to P07. While cleaning the library, Belle pushes a 21.0kg box about five million light years. However, Edwin of books across a frictionless floor. She pushes with a Hubble used a different “standard candle” star to force of 35.0N angled downward at 20.0° below the measure distances to galaxies that were even farther horizontal. What is the acceleration of the box of away. What did he use? books? A) Supernovae A) 1.77 m/s2 B) Red supergiant stars B) 1.67 m/s2 C) Bright O and B stars C) 1.57 m/s2 D) White dwarf stars D) 0.639 m/s2 E) T-Tauri variable stars E) 0.570 m/s2 P03. According to Natarajan, a modern technology that P08. To maintain a constant running speed, Gaston exerts a relies heavily on Einstein’s theories of special and constant force of 320.0 N. If Gaston runs a total general relativity is… distance of 5.00km in 30.0minutes, what is Gaston’s A) The International Space Station. average power output? B) Medical X-Ray devices. A) 1600 W C) Magnetic Resonance Imaging. B) 889 W D) The Global Positioning System. E) Atomic clocks. C) 470 W D) 167 W E) 53.3 W P04. Which planet in our solar system has an axis of rotation that is nearly perpendicular to the axis of its P09. A meter stick is hung horizontally with a pivot orbital plane? exactly at the stick’s center of mass (the 50.0cm A) Venus mark). A 150.0g object is attached to the meter stick B) Mars at the 30.0cm mark and is balanced by another object C) Jupiter (of unknown mass) attached at the 90.0cm mark. D) Saturn What is the mass of the unknown object? E) Uranus A) 300 g B) 150 g P05. An object accelerates at a rate of 285 inches/minute2. C) 113 g What is this acceleration in cm/s2? D) 75.0 g A) 0.0312 cm/s2 E) 50.0 g B) 0.0792 cm/s2 C) 0.201 cm/s2 D) 1.87 cm/s2 E) 4.75 cm/s2 University Interscholastic League page 8  Science Invitational A 2019 Physics P10. A 500.0g mass is hung from a spring. When pulled P15. A light ray travelling in air enters an unknown liquid and released, the mass on the spring oscillates with a at an angle of 45.0° with respect to the liquid’s period of 0.390sec. What is the spring constant of the surface (as shown). The refracted ray makes an spring? angle of 56.0° with respect to the liquid’s surface. A) 130 N/m What is the index of refraction of the unknown B) 83.3 N/m liquid? C) 32.2 N/m D) 8.06 N/m E) 3.29 N/m P11. You dive into a vat of cooking oil (which has a density of 0.91g/cm3) to retrieve your dropped cell phone. The depth of the oil is 3.80m. What is the gauge pressure at the bottom of the vat? A) 0.237 atm B) 0.335 atm A) 1.79 C) 0.368 atm B) 1.41 D) 0.404 atm C) 1.26 E) 0.473 atm D) 1.17 P12. Two resistors, a 27.0Ω and a 47.0Ω, are placed in E) 0.85 series and connected to a 12.0V battery. Once P16. A converging lens is used to form an image of a connected, what is the current passing through the neon sign. The sign is 2.20m from the lens, and the 27.0Ω resistor? focal length of the lens is 80.0cm. At what distance A) 700 mA from the lens is the image formed? B) 600 mA A) 3.46 m C) 444 mA B) 2.26 m D) 255 mA C) 1.40 m E) 162 mA D) 1.26 m E) 0.59 m P13. A balloon holding a charge of -12.0nC is placed with its center 45.0cm from the center of a Van de P17. A hydrogen atom is excited into the 𝑛 = 5 state. It Graff generator with a charge of +90.0µC. What is subsequently transitions down to the 𝑛 = 3 state. the magnitude of the electric force on the balloon? What is the wavelength of the photon emitted in this A) 0.022 N transition? B) 0.040 N A) 2280 nm C) 0.048 N B) 1282 nm D) 0.053 N C) 821 nm E) 0.097 N D) 683 nm E) 603 nm P14. A beam of electrons moving at a velocity of 4.50 × 105 m/s is directed into a region with a P18. A subatomic particle classified as a meson, such as perpendicular magnetic field. If the electron beam the pion, is composed of … traces out a circle with a radius of 22.0cm, then A) three quarks what is the magnitude of the magnetic field? B) two quarks A) 8.62 µT C) a quark and an anti-quark B) 11.6 µT D) a quark and a lepton C) 14.7 µT D) 21.3 µT E) and anti-quark and a lepton E) 27.8 µT University Interscholastic League page 9  Science Invitational A 2019 Physics P19. You do an experiment in which you drop an object P20. The resistance of a type of wire is measured as a from different heights and measure the final velocity function of its length. Based on the graph of the data, of the object when it reaches the ground. The object and knowing that the diameter of the wire is 0.40mm, is always at rest when it is released. You then graph determine the resistivity of the wire metal. the final velocity, v, as a function of the initial height, h. What would you expect the graph to look like? A) 5.8 × 10−7 Ωm B) 4.4 × 10−7 Ωm C) 2.9 × 10−7 Ωm D) 1.4 × 10−7 Ωm E) 1.1 × 10−7 Ωm University Interscholastic League page 10  Science · Invitational A · 2018 Water Data Tmp = 0°C Tbp = 100°C cice = 2.09 J/g·K cwater = 4.184 J/g·K csteam = 2.03 J/g·K ∆Hfus = 334 J/g ∆Hvap = 2260 J/g Kf = 1.86 °C/m Kb = 0.512 °C/m Constants R = 0.08206 L·atm/mol·K R = 8.314 J/mol·K R = 62.36 L·torr/mol·K e = 1.602 × 10‒19 C NA = 6.022 × 1023 mol‒1 k = 1.38 × 10‒23 J/K h = 6.626 × 10‒34 J·s c = 3.00 × 108 m/s RH = 2.178 × 10‒18 J me = 9.11 × 10‒31 kg University Interscholastic League · page 11  Science Invitational A 2019 Physics Physics Useful Constants quantity symbol value Free-fall acceleration g 9.80 𝑚/𝑠 2 Permittivity of Free Space ε0 8.854 × 10−12 𝐶 2 ⁄𝑁𝑚2 Permeability of Free Space μ0 4𝜋 × 10−7 𝑇𝑚/𝐴 Coulomb constant k 8.99 × 109 𝑁𝑚2 /𝐶 2 Speed of light in a vacuum c 3.00 × 108 𝑚/𝑠 Fundamental charge e 1.602 × 10−19 𝐶 Planck’s constant h 6.626 × 10−34 𝐽𝑠 Electron mass me 9.11 × 10−31 𝑘𝑔 Proton mass mp 1.67265 × 10−27 𝑘𝑔 Neutron mass mn 1.67495 × 10−27 𝑘𝑔 Atomic Mass Unit u 1.66 × 10−27 𝑘𝑔 Gravitational constant G 6.67 × 10−11 𝑁𝑚2⁄𝑘𝑔2 Stefan-Boltzmann constant σ 5.67 × 10−8 𝑊 ⁄𝑚2 𝐾 4 Universal gas constant R 8.314 𝐽⁄𝑚𝑜𝑙 · 𝐾 Boltzmann’s constant kB 1.38 × 10−23 𝐽⁄𝐾 Speed of Sound (at 20ºC) v 343 m/s Avogadro’s number NA 6.022 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙 Electron Volts eV 1.602 × 10−19 𝐽/𝑒𝑉 Distance Conversion Meters → inches 1.00 meter = 39.37 inches Rydberg Constant R∞ 1.097 × 107 𝑚−1 Standard Atmospheric Pressure 1 atm 1.013 × 105 𝑃𝑎 University Interscholastic League page 12  DO NOT DISTRIBUTE TO STUDENTS BEFORE OR DURING THE CONTEST! UIL HIGH SCHOOL SCIENCE CONTEST ANSWER KEY 2019 INVITATIONAL A Biology Chemistry Physics B01. A C01. A P01. A B02. A C02. E P02. C B03. D C03. A P03. D B04. E C04. B P04. E B05. B C05. D P05. C B06. D C06. D P06. E B07. C C07. C P07. C B08. B C08. E P08. B B09. D C09. A P09. D B10. C C10. D P10. A B11. E C11. D P11. B B12. A C12. D P12. E B13. E C13. E P13. C B14. C C14. C P14. B B15. B C15. C P15. C B16. E C16. E P16. D B17. B C17. B P17. B B18. C C18. A P18. C B19. D C19. B P19. A B20. A C20. E P20. D CHEMISTRY SOLUTIONS – UIL INVITATIONAL A 2019 C01. (A) The molar mass = (3 × 55.85) + (4 × 16.00) = 231.55 g/mol C02. (E) 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O C03. (A) C04. (B) C05. (D) C06. (D) C07. (C) At high pressures and at low temperatures intermolecular forces become significant and a gas behaves less like an ideal gas. C08. (E) A2B(s) is not included in the equilibrium expression because it is a solid. C09. (A) ∆Tb = ikbm. i for NaCl = 2, kb = 0.512 °C/m, m = moles NaCl/kg solvent. Assume 100.0 g of Dead Sea water. At 34.2% NaCl, 100 g of Dead Sea water contains 34.2 g NaCl and 65.8 g H2O. Moles of NaCl = (34.2 g)/(58.44 g/mol) = 0.5852 moles. m = (0.5852 mol NaCl)/(0.0658 kg H2O) = 8.894 mol/kg ∆Tb = ikbm. = (2)(0.512°C/m)(8.894 m) = 9.11°C. C10. (D) Carbon, silicon, tin, and lead are all in Group 14 (IVA) in the periodic table. C11. (D) C12. (D) C13. (E) C14. (C) Potassium dichromate is K2Cr2O7. C15. (C) C16. (E) C2H4 is not a valid empirical formula because the subscripts can be reduced to CH2 without changing their ratio. C17. (B) Ksp = 5.4 × 10–12 = [Ag+]2[C2H42–] Let [C2H42–] = x. [Ag+] = 2 × [C2H42–] = 2x. 5.4 × 10–12 = [2x]2[x] = 4x3. x = 1.105 × 10–4, so [Ag+] = 2.21 × 10–4 M C18. (A) The chemical formula is C8H10N4O2, so the molar mass is 194.22 g/mol. C19. (B) Each single bond is a sigma bond, and each double bond is made up of one sigma bond and one pi bond. C20. (E) PHYSICS SOLUTIONS – UIL INVITATIONAL A 2019 P01. (A) page 4: “The strangest thing was that about every two years it reversed its motion entirely for some ninety days and then switched back to its eastward journey. The Babylonians recorded this object and its peculiar backpedaling. We now understand this apparent motion of Mars to be a result of … ” P02. (C) page 52: “Cepheids could not be spotted easily and studied if they lay beyond about five million light- years, even with the unprecedented reach of the hundred-inch telescope. Hubble pushed out farther by using classes of some of the brightest stars – the O and B stars – as standard candles.” P03. (D) page 76: “He [Einstein] could not have anticipated our current reliance on his theory of general relativity to get our bearings. GPS [Global Positioning System] technology is built entirely on Einsteinian gravity…. This is no mean task, as it requires taking into account corrections predicted by both of Einstein’s theories – the special and the general theory of relativity.” P04. (E) The obliquity of a planet is the angle between the axis of the orbital plane of the planet and the axis of rotation of the planet. These angles are usually small: for example, the Earth’s obliquity is about 23° (this is responsible for the seasons we experience). Uranus, however, has an obliquity of 98° -- its axis of rotation is very nearly perpendicular to the axis of its orbital plane. It essentially rotates “on its side”.. P05. (C) This is a simple, though multi-step conversion: 𝑖𝑛𝑐ℎ𝑒𝑠 𝑐𝑚 1 𝑚𝑖𝑛𝑢𝑡𝑒 2 𝑐𝑚 285 𝑚𝑖𝑛𝑢𝑡𝑒 2 ∗ 2.54 𝑖𝑛𝑐ℎ ∗ (60 𝑠𝑒𝑐𝑜𝑛𝑑𝑠) = 0.201 𝑠2 P06. (E) Average acceleration is defined as the change in velocity divided by the change in time. In other words: ∆𝑣 𝑎̅ = ∆𝑡. Considering the direction East to be positive, then we have an initial velocity that is negative and a ∆𝑣 (30−(−42)) final velocity that is positive: 𝑣𝑖 = −42.0 m/s and 𝑣𝑓 = +30.0 m/s. Then 𝑎̅ = ∆𝑡 = 0.65 = 111 m/s2. P07. (C) Since the floor is frictionless, we only have three forces present: gravity (downward), the normal force (upward), and the applied force (angled at 20.0° below the horizontal). We must first break up the applied force into horizontal and vertical components: 𝐴𝑥 = 𝐴𝑐𝑜𝑠𝜃 = (35.0) cos(20.0) = 32.9 𝑁 and 𝐴𝑦 = 𝐴𝑠𝑖𝑛𝜃 = (35.0) sin(20.0) = 12.0 𝑁. The box only slides horizontally, so all of the vertical forces cancel out: ∑ 𝐹𝑦 = 𝐹𝑁 − 𝑚𝑔 − 𝐴𝑦 = 0. Since there is no vertical acceleration, we need look at that direction no more. Now, considering the horizontal 32.9 forces, we have only one: ∑ 𝐹𝑥 = 𝐴𝑥 = 32.9 = 𝑚𝑎 = 21.0𝑎. Solving for acceleration: 𝑎 = 21.0 = 1.57 m/s2. P08. (B) First, consider the work done by Gaston: 𝑊 = 𝐹𝑑 = (320𝑁)(5000𝑚) = 1.60 × 106 𝐽. Now we can 𝑊 1.60 × 106 𝐽 1.60 × 106 𝐽 calculate his power output, which is work divided by time: 𝑃 = = = = 889 𝑊. 𝑡 30 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 1800 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 P09. (D) Since it is balanced, we know that the torques must sum to zero. The torques in each case equal the force (mg) multiplied by the “torque arm”. Torque arms are measure from the pivot, which is at the 50.0cm mark. Thus, 𝑟1 = 50 − 30 = 20.0𝑐𝑚 = 0.200𝑚. The other torque arm is 𝑟2 = 90 − 50 = 40.0𝑐𝑚 = 0.400𝑚. Conveniently, the forces are perpendicular to the meter stick, so we don’t need to worry about any angles. Also, one torque is directed clockwise (negative) and the other counterclockwise (positive). Thus, the total torque due to the two hanging masses is: ∑ 𝜏 = 𝑚1 𝑔𝑟1 − 𝑚2 𝑔𝑟2 = 0. Plugging in values and solving: (0.150)(9.8)(0.200) (. 150𝑘𝑔)(9.8)(0.200𝑚) − (𝑚2 )(9.8)(0.400𝑚) = 0, or 𝑚2 = = 0.075𝑘𝑔 = 75.0𝑔. (9.8)(0.400) 𝑚 P10. (A) The period of oscillation of a mass-spring system is 𝑇 = 2𝜋√ 𝑘. Plugging in the given values: 0.500𝑘𝑔 0.500𝑘𝑔 𝑘 𝑇 = 0.390 = 2𝜋√ 𝑘 , This gives: 0.06207 = √ 𝑘 , and 259.6 = 0.500. So, 𝑘 = 129.8 ≈ 130N/m. P11. (B) Gauge pressure means that we get to ignore the pressure caused by the atmosphere – we need only consider the pressure created by the oil. We do need to convert the units on the density given. 𝑔 𝑘𝑔 𝜌 = 0.91 𝑐𝑚3 = 910 𝑚3. The pressure due to depth in a stationary fluid is given by: 𝑃 = 𝜌𝑔ℎ. So, 33890𝑃𝑎 𝑃 = (910)(9.8)(3.80) = 33890 𝑃𝑎. Converting to atmospheres: 𝑃 = 101300𝑃𝑎/𝑎𝑡𝑚 = 0.335 𝑎𝑡𝑚. P12. (E) Resistors in series combine by simply adding them, so the total resistance in the circuit is 𝑅0 = 𝑅1 + 𝑅2 = 27.0 + 47.0 = 74.0 Ω. Then using Ohm’s Law, we can get the total current: 𝑉 12.0 𝐼0 = 𝑅0 = 74.0 = 0.162𝐴 = 162 𝑚𝐴. For a series circuit, the current in each resistor is the same as the total 0 current, so 𝐼1 = 𝐼2 = 𝐼0 = 162𝑚𝐴. 𝑘𝑄1 𝑄2 (8.99 × 109 )(−12.0 × 10−9 )(90.0 × 10−6 ) P13. (C) Utilizing Coulomb’s Law: 𝐹 = 𝑟2 = (0.450)2 = −0.048 𝑁. The negative sign indicates the force is attractive, but the magnitude of the force is just: |𝐹| = 0.048 𝑁. P14. (B) The velocity and the magnetic field are perpendicular, so we can use the following equation for the radius 𝑚𝑣 of the circle traced by the beam: 𝑟 = |𝑞|𝐵. The beam is made of electrons, whose mass and charge are known. (9.11 × 10−31 )(4.50 × 105 ) 4.10 × 10−25 Thus, 𝑟 = 0.220 = (1.602 × 10−19 )𝐵 , which gives: 𝐵 = 3.52 × 10−20 = 1.16 × 10−5 𝑇 = 11.6 𝜇𝑇. P15. (C) This is a simple Snell’s Law problem, but the angles given are not the ones to use. We need the angles from the normal – the angles from the line perpendicular to the surface. It happens to be the same for the incident angle, but not for the refracted angle: 𝜃𝑖 = 90.0° − 45.0° = 45.0°, and 𝜃𝑟 = 90.0° − 56.0° = 34.0°. Now we can use Snell’s Law: 𝑛𝑖 𝑠𝑖𝑛𝜃𝑖 = 𝑛𝑟 𝑠𝑖𝑛𝜃𝑟 = (1.00) sin(45.0) = 𝑛𝑟 sin(34.0). This gives: 0.7071 𝑛𝑟 = 0.5592 = 1.26. 1 1 1 P16. (D) Since the lens is converging, the focal length is positive. Using the lens equation: 𝑝 + 𝑞 = 𝑓, we can get the 1 1 1 1 image location, 𝑞. Putting everything in meters: 2.20 + 𝑞 = 0.800. Solving gives 𝑞 = 0.795, or 𝑞 = 1.26𝑚. P17. (B) First, we should find the energy difference between the levels involved in this transition. There are various 1 1 formulas for this, but one of the easiest is: ∆𝐸 = (13.6 𝑒𝑉) (𝑛2 − 𝑛2 ). For 𝑛𝑖 = 5 and 𝑛𝑓 = 3 we get an 𝑓 𝑖 1 1 energy difference of ∆𝐸 = (13.6) (9 − 25) = 0.967 𝑒𝑉. Now the wavelength of the photon with this energy 1240 𝑒𝑉∙𝑛𝑚 1240 can be found by using: 𝜆 = ∆𝐸 = 0.967 = 1282 𝑛𝑚. P18. (C) Mesons are subatomic particles that consist of one quark bonded with one anti-quark. For example, a positive pion is composed of an up-quark and an anti-down-quark. A combination of three quarks is classified as a Baryon (for example, a proton). The other combinations (two quarks, a quark and a lepton, or an anti-quark and a lepton) are not color-neutral combinations, and thus would not be allowed to exist due to the nature of the strong force. P19. (A) We can analyze this experiment using conservation of energy. When the object is released, it has gravitational potential energy. When it reaches the ground, it has kinetic energy. Ignoring air resistance, we 1 would set these two energies equal: 𝐺𝑃𝐸 = 𝑚𝑔ℎ = 𝐾𝐸 = 2 𝑚𝑣 2. Solving for velocity gives us the formula: 𝑣 = √2𝑔ℎ. So, a graph of velocity as a function of height would give a graph resembling 𝑦 = √𝑥. The only choice resembling this kind of curve is choice A. 𝐿 P20. (D) The resistance depends on the length of the wire according to the equation 𝑅 = 𝜌 𝐴. The graph of resistance, R, as a function of length, L is linear as expected. Furthermore, we see from examining the 𝜌 equation that the line should have a slope equal to 𝑠𝑙𝑜𝑝𝑒 = 𝐴. We have the diameter, so we can calculate the radius, and the cross-sectional area of the wire: 𝐴 = 𝜋𝑟 2 = 𝜋(0.200 × 10−3 )2 = 1.26 × 10−7 m2. We (58Ω−12Ω) 46Ω Ω can also go to the graph and determine the slope: 𝑠𝑙𝑜𝑝𝑒 ≈ (50𝑚−10𝑚) = 40𝑚 = 1.15 𝑚. Then, solving for resistivity, we get: 𝜌 = (𝑠𝑙𝑜𝑝𝑒)𝐴 = (1.15)(1.26 × 10−7 ) = 1.4 × 10−7 Ω𝑚.  Science Contest Answer Sheet Conference ______ Grade Level ______ Contestant # ________ Biology Chemistry Physics B01 C01 P01 B02 C02 P02 B03 C03 P03 B04 C04 P04 B05 C05 P05 B06 C06 P06 B07 C07 P07 B08 C08 P08 B09 C09 P09 B10 C10 P10 B11 C11 P11 B12 C12 P12 B13 C13 P13 B14 C14 P14 B15 C15 P15 B16 C16 P16 B17 C17 P17 B18 C18 P18 B19 C19 P19 B20 C20 P20 B Score C Score P Score Grader Initials _____ _____ _____ OVERALL SCORE