SCIENCE - Alvin ISD

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.

SCIENCEInvitational A • 2016

UNIVERSITY INTERSCHOLASTIC LEAGUEMaking a World of Difference

Periodic Table of the Elements

Some Standard Properties of Water

1A1

8A18

1H

1.0082A2

3A13

4A14

5A15

6A16

7A17

2He4.003

3Li

6.941

4Be9.012

5B

10.81

6C

12.01

7N

14.01

8O

16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

3B3

4B4

5B5

6B6

7B7

8B8

8B9

8B10

1B11

2B12

13Al26.98

14Si

28.09

15P

30.97

16S

32.07

17Cl35.45

18Ar39.95

19K

39.10

20Ca40.08

21Sc44.96

22Ti

47.87

23V

50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.41

31Ga69.72

32Ge72.64

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y

88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.07

45Rh102.91

46Pd106.42

47Ag107.87

48Cd112.41

49In

114.82

50Sn118.71

51Sb121.76

52Te

127.60

53I

126.90

54Xe

131.29

55Cs132.91

56Ba137.33

57La

138.91

72Hf

178.49

73Ta

180.95

74W

183.84

75Re186.21

76Os190.23

77Ir

192.22

78Pt

195.08

79Au196.97

80Hg200.59

81Tl

204.38

82Pb207.20

83Bi

208.98

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(261)

105Db(262)

106Sg(266)

107Bh(264)

108Hs(277)

109Mt(268)

110Ds(281)

111Rg(272)

112Cn(285)

58Ce 140.12

59Pr 140.91

60Nd 144.24

61Pm (145)

62Sm 150.36

63Eu 151.96

64Gd 157.25

65Tb 158.93

66Dy 162.50

67Ho 164.93

68Er 167.26

69Tm 168.93

70Yb 173.04

71Lu 174.97

90Th 232.04

91Pa 231.04

92U

238.03

93Np (237)

94Pu (244)

95Am (243)

96Cm (247)

97Bk (247)

98Cf (251)

99Es (252)

100Fm (257)

101Md (258)

102No (259)

103Lr (262)

property symbol valuedensity of water density of ice

ρwaterρice

1.000 g cm-3 0.9167 g cm-3

specific heats icewatersteam

cicecwatercsteam

2.09 J g-1 K-1 4.184 J g-1 K-1 2.03 J g-1 K-1

heat of fusion ∆Hfus or Lf 334 J g-1

heat of vaporization

∆Hvap or Lv2260 J g-1

index of refraction n 1.33

autoionization Kw 1.0 × 10-14

Pressure

1 atm = 760 torr= 101325 Pa= 14.7 psi

1 bar = 105 Pa= 100 kPa

Energy

1 cal = 4.184 J

1 L atm = 101.325 J

1 Cal = 1 kcal

1 hp = 746 W

1 eV = 1.602 × 10-19 J

Various Physical Constantsproperty symbol value

universal gas constant

R 8.314 J mol-1 K-1

62.36 L torr mol-1 K-1

0.08206 L atm mol-1 K-1

1.987 cal mol-1 K-1

Planck’s constant

h 6.626 × 10-34 J s

4.136 × 10-15 eV s

Planck’s reduced constant

h/2π 1.054 × 10-34 J s 6.582 × 10-16 eV s

Boltzmann constant kB 1.38 × 10-23 J K-1

Stefan-Boltzmann σ 5.67 × 10-8 W m-2 K-4

speed of light c 3.00 × 108 m s-1

speed of sound (at 20°C) vair 343 m s-1

acceleration of gravity g 9.80 m s-2

gravitational constant G 6.67 × 10-11 N m2 kg-2

Avogadro’s number

NA 6.022 × 1023 mol-1

elementary charge e 1.602 × 10-19 C

Faraday F 96485 C mol-1

Coulomb’s law constant k 8.988 × 109 N m2 C-2

Rydberg constant � 2.178 × 10-18 J

Some Other Conversion Factors

1 in = 2.54 cm

1 lb = 453.6 g

1 mi = 5280 ft = 1.609 km

1 gal = 4 quarts = 231 in3 = 3.785 L

property symbol value

electron rest mass

me 9.11 × 10-31 kg 0.000549 u 0.511 MeV c-2

proton mass mp 1.6726 × 10-27 kg 1.00728 u 938.3 MeV c-2

neutron mass mn 1.6749 × 10-27 kg 1.008665 u 939.6 MeV c-2

atomic mass unit

u 1.6605 × 10-27 kg

931.5 MeV c-2 earth'mass 5.972 × 1024 kg

earth'radius 6.371 × 106 m

moon'mass 7.348 × 1022 kg

sun'mass 1.989 × 1030 kg

distance'earth0moon

3.844 × 108 m

distance'earth0sun

1.496 × 1011 m

permittivity of free space ε0 8.85 × 10-12 F m-1

permeability of free space µ0 4π × 10-7 T m A-1

Some Average Bond Energies (kJ/mol)C–H 413 C–C 346 C–Cl 339 C–N 305

O–H 463 C=C 602 C–Br 285 N=N 418

N–H 391 C≡C 835 O=O 498 H–H 436

C–O 358 C=O 799 C≡O 1072 Br–Br 193

H–Cl 432 S–H 347 N≡N 945 Cl–Cl 242

H–Br 366 H–I 299 C≡N 887 I–I 151

HS Science • Invitational A • 2016

University Interscholastic League • page 1

Biology Questions (1 – 20) 1. Oxygen is more electronegative than hydrogen,

thus H2O is a A) non-reactive molecule. B) hydrophobic molecule. C) polar molecule. D) linear molecule.

2. RNA is to nucleotide as ___________ is to amino acid. A) polysaccharide B) polypeptide C) DNA D) nitrogen

3. Which of the following is not true regarding biological membranes? A) They facilitate communication. B) They are as fluid as salad oil. C) They allow diffusion of CO2. D) They rely solely on diffusion for passage of

materials.

4. A scientist is collecting wild sunflowers across their natural range to study the lipid composition of their seeds. She has discovered that as she travels from the equator northward, there is a graded change in the lipids. Given their different environmental conditions, you would expect plants from the more northern latitudes to make: A) increasing proportions of saturated fatty acids

to unsaturated fatty acids. B) increasing proportions of cholesterol to

saturated fatty acids. C) increasing proportions of unsaturated fatty

acids to saturated fatty acids. D) increasing proportions of both unsaturated

fatty acids and saturated fatty acids to cholesterol.

5. A catabolic process __________ the amount of entropy in the universe. A) is dependent on B) increases C) decreases D) maintains

6. A student is frustrated that the chemical reaction

they’ve set up is not going forward. Their lab partner reminds them to add an enzyme, which speeds the reaction considerably. What can be said of this reaction? A) It is not part of a natural, living system. B) It is exergonic. C) The reaction required energy. D) The products contain more energy than the

reactants.

7. A key aspect of an electron transport chain is that

A) hydrolysis of ATP fuels the chemiosmosis. B) it represents a increase in free energy, as organic

fuel gradually donates electrons. C) each electron carrier returns to its reduced form

by donating electrons. D) each subsequent electron carrier is more

electronegative than the last.

8. In a classic experiment, bacteriophages are first grown in a medium with radioactive sulfur (35S), and then allowed to infect bacteria. After this, the mixture is agitated such that the bacteriophages are shaken loose from the bacteria. The entire mixture is centrifuged such that the heavy bacteria form a pellet in the bottom, and the phages are suspended in the remaining liquid. The pellet and the liquid are then measured for radioactivity. Where should the radioactivity be found? A) In the liquid B) In both the liquid and the pellet C) In the pellet D) In neither the liquid nor the pellet

9. The mRNA for a protein destined to function in the nucleus at maturity will be translated A) at a free ribosome. B) at a ribosome within the nucleus. C) at the rough endoplasmic reticulum. D) at a ribosome bound to the nuclear pore.



10. An operon is A) an efficient way for eukaryotes to regulate

which proteins are made. B) a unit of mDNA that a virus inserts into its

host. C) a unit of multiple genes all controlled by the

same promoter. D) a unit of multiple genes all controlled by the

same start codon.

11. A dog that is heterozygous for coat color (Bb) has five puppies. Like their mother, two of the puppies are also heterozygous (Bb) for coat color. Is it likely they inherited both alleles from their mother?

A) Yes, the mother passed on her genotype to her puppies, so they are heterozygotes, like her.

B) Yes, the law of independent assortment states that the maternal chromosomes will appear together in the same daughter cell, independent of the paternal chromosomes.

C) No, the law of segregation states that maternal and paternal alleles separate into different cells during metaphase of meiosis I.

D) No, the law of independent assortment states that two alleles of the same gene sort into different chromosomes.

12. Consider a sexually-reproducing animal in which the sperm contain 14 chromosomes. In the cell cycle, just prior to cytokinesis, a cell from this animal’s epithelial tissue will contain A) 7 tetrads. B) 14 sister chromatids. C) 28 sister chromatids. D) 56 sister chromatids.

13. The statement that best reflects natural selection is that it depends on A) populations maintaining a certain degree of

gene flow with other populations. B) random mutations. C) systematic recruitment of fit individuals into a

population. D) mutations occurring as a response to the

environment.

14. One of the best lines of evidence that tetrapods evolved from lobed-fin fishes is seen in the discovery of a fossil fish with A) lungs as well as gills. B) scales that function as toes. C) a fin skeleton that includes an elbow and

primitive wrist. D) a “walking tail” that evolved into legs.

15. You have sampled some pond water and are examining many single-celled organisms in a powerful microscope. You recognize many as bacteria. What would convince you that you also are seeing members of domain Eukarya? A) Movement due to an outer capsule B) Movement by flagella C) Movement by pseudopodia D) Movement by cilia-like projections

16. Negative pressure within the xylem of a tall tree is greatest during A) warm, humid days. B) cold, foggy days. C) warm, windy days. D) a summer rainstorm.

17. The obligate relationship between a fungus and the ant that grows it for food is considered A) parasitism. B) mutualism. C) herbivory. D) predation.

18. An example of a density-dependent factor affecting population size in mice would be A) infection by mites. B) frequent flooding of the field in which they live. C) Deleterious mutations in their DNA. D) Severity of winter storms.



19. Considering two fruit-eating bird species inhabiting the same area, species 1 will fare best if species 2 A) eats the same berries as species 1. B) is less efficient at finding berries than species

1. C) eats larger berries than species 1. D) is more efficient at finding berries than species

1.

20. The burning of fossil fuels releases large quantities of sulfur dioxide and nitrogen oxide into the air. These chemicals mix with water, oxygen, and other chemicals in the atmosphere, and form acidic precipitation. One main reason this is harmful to forests is because A) the acid causes lesions in tree bark. B) excess sulfur makes leaves more attractive to

herbivores. C) The lower pH disrupts chemiosmosis in the

chloroplasts. D) H+ is attracted to negatively-charged soil

particles.

Chemistry Questions (21 – 40) 21. What is the name for the compound Mg(CN)2?

A) magnesium carbonitride B) manganese nitride C) manganese cyanide D) silver carbonitrate E) magnesium cyanide

22. What is the chemical formula for nitrous acid? A) HNO D) H2NO3 B) HNO3 E) HN2O3 C) HNO2

23. What is the molar mass of BaCl2·2H2O to the nearest integer? A) 244 g/mol D) 280 g/mol B) 218 g/mol E) 237 g/mol C) 262 g/mol

24. A compound has the formula C3HxOy. If the compound is 47.35% carbon by mass, what is the molar mass of this compound? A) 25.4 g/mol D) 76.1 g/mol B) 112.0 g/mol E) 54.0 g/mol C) 60.1 g/mol

25. A hydrocarbon is burned to produce water and carbon

dioxide. If 100.0 g of the hydrocarbon produces 302.9 g of carbon dioxide, what must be the empirical formula for this hydrocarbon? A) C2H5 C) C3H8 E) C3H7 B) CH2 D) C2H6

26. Balance the following reaction with whole numbers as coefficients.

K + KNO3 ⟶ K2O + N2

What are the respective coefficients? A) 12 : 2 ⟶ 3 : 1 B) 10 : 2 ⟶ 6 : 1 C) 1 : 1 ⟶ 3 : 1 D) 24 : 4 ⟶ 12 : 2 E) 8 : 2 ⟶ 5 : 1

27. Rxn: Na + KCl ⟶"" NaCl + K What is the maximum amount of NaCl that can be

produced when 20.0 grams of Na reacts with 60.0 grams of KCl? A) 80.0 g C) 50.8 g E) 58.4g B) 47.0 g D) 54.2 g

28. Which of the choices listed has the element with the

highest oxidation state ? A) BrO2

- C) Fe2O3 E) CO32-

B) MnO2 D) NO3

-



29. AgBr is made via the precipitation reaction AgNO3(aq) + NaBr(aq) ⟶ AgBr(s) +

NaNO3(aq) If 1.85 liters of 0.811 M AgNO3 are poured into a

solution of excess NaBr, and the resulting AgBr precipitate is dried and weighed to be 262 g, what is the percent yield of this reaction? A) 100% C) 93% E) 85% B) 96% D) 89%

30. What is the electron configuration of arsenic?

A) 1s2 2s2 2p6 3s2 3p6 4s2 3d12 4p1 B) 1s2 2s2 2p6 3s2 3p6 C) 1s2 2s2 2p6 3s2 3p6 4s2 3d9 4p3 D) 1s2 2s2 2p6 3s2 3p6 4s2 3d8 4p5 E) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3

31. Excited electrons in an atom are falling from an excited state of -2.1 eV down to -4.4 eV. What is the wavelength of this emission? A) 540 nm C) 470 nm E) 280 nm B) 620 nm D) 590 nm

32. What is the molecular geometry for the underlined carbon in acetone, CH3COCH3 ? A) trigonal pyramidal D) tetrahedral B) see-saw E) angular C) trigonal planar

33. Molecules X and Y have very similar sizes (radii).

X is non-polar and Y is polar. Which of the statements below is true about X and Y? A) The geometry of X must be tetrahedral. B) X will have a higher solubility in water than

that of Y. C) Y has no dispersion forces present while X has

many dispersion forces. D) The dipole moments of X and Y must be

identical. E) The boiling point of Y is higher than the

boiling point for X.

34. A rigid container at 25°C has a gas in it at a pressure of 513 torr. What is the pressure when the temperature is then raised to 75°C ?

A) 439 torr C) 575 torr E) 1539 torr B) 599 torr D) 652 torr

35. Water is added to 75 mL of 6.0 M hydrochloric acid

until the total volume is 3.0 L. What is the concentration of this solution?

A) 0.30 M C) 0.20 M E) 0.10 M B) 0.25 M D) 0.15 M

36. What is the pH of a 0.0016 M KOH solution? A) 10.75 C) 2.80 E) 1.79 B) 11.20 D) 12.21

37. Which of the following compounds has the highest normal boiling point? A) CH3CH2F D) CH3CH2CH3 B) CH3OH E) CH3F C) CH3CH2OH

38. An (generic) ionic compound is allowed to saturate an aqueous solution via the equilibria

MX2(s) ⇌ M2+(aq) + 2 X-(aq)

What is the value of Ksp for MX2 if the equilibrium concentration of M2+ is 0.0035 M ? A) 1.2×10-5 D) 8.6×10-8 B) 4.3×10-8 E) 1.7×10-7 C) 4.9×10-5

39. When 9.66 g of sodium is treated with excess oxygen, 87.0 kJ of heat is produced and sodium oxide is formed. 4 Na (s) + O2 (g) ⟶ 2 Na2O (s)

What is ∆Hrxn for the reaction show above?

A) -828 kJ/mol D) -414 kJ/mol B) -348 kJ/mol E) -87 kJ/mol C) -207 kJ/mol



40. How much heat is required to convert 1.50 moles of water at 28°C into steam at 175°C under 1 atm of pressure ?

A) 78.7 kJ C) 68.3 kJ E) 54.3 kJ B) 12.3 kJ D) 73.3 kJ

Physics Questions (41 – 60) 41. According to Feynman, if only one bit of scientific

information could be passed to the next generation, which statement contains the most information in the fewest words? A) Energy cannot be created nor destroyed. B) Particles behave like waves. C) Force equals mass time acceleration. D) All things are made of atoms. E) Experiment is the sole judge of science.

42. According to Feynman, as water evaporates from a glass of water, which statement is true? A) Evaporating molecules transfer momentum to

the liquid, heating it. B) Evaporating molecules carry away energy

from the liquid, cooling it. C) Returning molecules divide the energy and

cool the liquid. D) Returning molecules absorb energy from the

liquid and immediately evaporate, cooling the liquid.

E) The number of evaporating molecules always equals the number of returning molecules.

43. According to Feynman, if the electrical force between two grains of sand were entirely attractive, then the force between the grains separated by 30m would be approximately… A) thirty thousand tons. B) three million pounds. C) thirty million pounds. D) thirty million tons. E) three million tons.

44. According to Feynman, if an electron were confined in the nucleus of an atom, then the uncertainty principle would require that the electron have a… A) very large kinetic energy. B) very large potential energy. C) very small kinetic energy. D) very small potential energy. E) none of the above.

45. When a star exhausts its Hydrogen fuel, the inner core of the star collapses, heating the central portion of the star’s core and restarting fusion reactions. What element is consumed in this secondary fusion process? A) He B) Li C) C D) N E) O

46. A plane with an initial velocity of 200m/s accelerates at a constant rate for 3.5s, traversing a distance of 900m during the time it was accelerating. What is the final velocity of the plane? A) 110 m/s B) 260 m/s C) 310 m/s D) 510 m/s E) 600 m/s

47. A block of mass m1 sliding to the right at velocity v1 strikes a mass m2 which is initially at rest. After the collision, mass m2 is sliding to the right at velocity v2. Which formula below describes the final velocity of mass m1? A) (!!!! +!!!!)/!! B) (!!!! +!!!!)/!! C) (!!!! −!!!!)/!! D) (!!!! −!!!!)/!! E) None of the above.



48. Three resistors are connected to a 12.0 V battery as shown. What is the voltage across the 60.0 Ohm resistor in this circuit?

60 Ω 30 Ω 40 Ω 12.0 V

A) 8.0 V B) 4.0 V C) 12.0 V D) 5.5 V E) 36.0 V

49. A box of mass 40.0kg is being pushed to the left by a force of 160N across a horizontal floor. The coefficient of friction between the box and the floor is 0.22. What is the acceleration of the box? A) 6.2 m/s2 B) 5.8 m/s2 C) 4.0 m/s2 D) 2.2 m/s2 E) 1.8 m/s2

50. A 100.0g mass is attached to a spring hanging vertically. The mass is pulled by 15.0cm from the equilibrium point and released from rest. Harmonic motion ensues, and the period of the motion is 1.35s. What is the speed of the mass as it passes through the equilibrium point during each oscillation? A) 11.1 cm/s B) 32.4 cm/s C) 48.7 cm/s D) 69.8 cm/s E) 180 cm/s

51. A Carnot engine has an efficiency of 37%. If the Carnot engine does 60.0J of work each cycle, what amount of heat energy is lost in the exhaust each cycle? A) 162 J B) 102 J C) 60 J D) 38 J E) 22 J

52. A series AC-RLC circuit has a resonant frequency at ω0. What is true about the circuit when it is tuned to that resonant frequency? A) the circuit is capacitive at resonance B) the current leads the voltage at resonance C) the phase angle is zero at resonance D) the power output is a minimum at resonance E) none of these are true at resonance

53. Two charges, one positive (Q) and one negative (-2Q), are placed on the x-axis. The negative charge

is placed at the origin and the positive charge is placed at x = r. At what location on the x-axis is the electric field exactly zero? A) 2!/3 B) 2 − 2 ! C) 1 + 2 ! D) 2 + 2 ! E) −2 − 2 !

54. A singly-charged ion beam with a velocity of 2.5 x 105 m/s passes into a region of uniform magnetic field with a strength of 1200 Gauss. If the diameter of the circle formed by the ion beam is 65.4cm, what is the mass of the ions in atomic mass units (note: 1amu = 1.67 x 10-27kg). A) 63 amu B) 38 amu C) 30 amu D) 25 amu E) 15 amu



55. A concave mirror with a radius of curvature of 36.0cm is located 13.0cm to the right of a light bulb. If the light bulb is 15.0cm tall, how tall is the image of the light bulb? A) 54.0 cm B) 39.0 cm C) 23.5 cm D) 11.0 cm E) 8.71 cm

56. A spring with a spring constant of 50.0N/m is compressed 15.0cm by a block with a mass of 0.55kg. Once released, the block slides horizontally across a floor for a distance of 1.2m before coming to rest. What is the coefficient of friction between the floor and the block? A) 0.85 B) 0.58

C) 0.35 D) 0.17 E) 0.087

57. An isotope of Californium-251 decays by the following sequence of radioactive emissions: Alpha, Alpha, Beta-, Gamma, Beta-, Alpha, Gamma, Alpha. What is the isotope produced after this decay chain? A) Protactinium-234 B) Uranium-235 C) Protactinium-235 D) Thorium-232 E) Uranium-236

58. Which fundamental force is responsible for the

decay of the neutral pi-meson: π0 ! γ + γ ? The lifetime of the neutral pi-meson is 8.4 x 10-17 s A) Gravitational Force B) Weak Force C) Electromagnetic Force D) Strong Force E) Higgs Force

59. A 1000 W radio antenna produces 95.7MHz radio waves that expand uniformly in all directions. What is the maximum strength of the electric field of the radio waves at a location 400.0m from the antenna? (note: ε0 = 8.854 x 10-12 Nm2/C2) A) 0.053 N/C B) 0.38 N/C

C) 0.43 N/C

D) 0.61 N/C

E) 1.2 N/C

60. A certain atom has four energy states: a ground state at 0.0eV, and three excited states at energies of 1.77eV, 3.42eV, and 6.14eV. Which of the following photon wavelengths would you not expect to detect emitted from a collection of these atoms? (note: h = 6.626 x 10-34 Js) A) 202 nm B) 284 nm C) 456 nm D) 514 nm E) 752 nm

UIL HIGH SCHOOL SCIENCE CONTEST

Contestant Answer Sheet

Contestant # ________ 9 10 11 12 A 2A 3A 4A 5A CONTESTANT GRADE LEVEL CONFERENCE

BIOLOGY SCORE CHEMISTRY SCORE PHYSICS SCORE

OVERALL SCORE:

PLEASE WRITE ANSWERS WITH CAPITAL LETTERS

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UIL HIGH SCHOOL SCIENCE CONTEST ANSWER KEY

INVITATIONAL A • 2016

1. C

2. B

3. D

4. C

5. B

6. B

7. D

8. A

9. A

10. C

11. C

12. D

13. B

14. C

15. C

16. C

17. B

18. A

19. C

20. D

21. E

22. C

23. A

24. D

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26. B

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28. D

29. C

30. E

31. A

32. C

33. E

34. B

35. D

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37. C

38. E

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40. D

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42. B

43. E

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46. C

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49. E

50. D

51. B

52. C

53. D

54. E

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56. E

57. B

58. C

59. D

60. D

PHYSICS KEY for Science Contest • Invitational A Meet • 2016

41. (D) page 4: “…what statement would contain the most information in the fewest words? I believe it is the atomic hypothesis (or the atomic fact, or whatever you wish to call it) that all things are made of atoms…”

42. (B) page 12: “Therefore, since those that leave have more energy than the average, the ones that are left have less average motion than before. So the liquid gradually cools if it evaporates.”

43. (E) page 29: “…if everything attracted everything else instead of likes repelling, so that there were no cancellations, how much force would there be? There would be a force of three million tons between the two!”

44. (A) page 34: “…If they were in the nucleus, we would know their position precisely, and the uncertainty principle would then require that they have a very large (but uncertain) momentum, i.e., a very large kinetic energy.”

45. (A) When the hydrogen fuel is exhausted, the star’s core collapses until the core increases to a temperature of about 108 K. Then the triple-alpha process can begin fusing Helium into Carbon.

46. (C) ! = !! + !!! + 1 2 !!! 900 = 0 + (200)(3.5) + 1 2 !(3.5)! gives ! = 32.7!! !!

Then use ! = !! + !" ! = 200 + 32.7 3.5 = 314! ! 47. (C) Conservation of momentum: !!!! = !!!! +!!!!

so, (!!!! −!!!!) !! = !! 48. (A) Parallel Branches: V0 = V1 = V2 so the voltage across the upper branch V2 = 12.0 V

Resistance of the upper branch: resistors in series R = R1 + R2, so R = 60 + 30 = 90 Ohms Current in the upper branch: V2/R = 12/90 = 0.13333 A Voltage across the 60 Ohm: V = IR = (0.133)*(60) = 8.0 V

49. (E) Forces in the y-direction: N-mg = 0, so N = mg = (40)*(9.8) = 392 N Forces in the x-direction: F - ƒ = ma, and ƒ = µN = (0.22)(392) = 86.2 N So a = (160 – 86.2)/40 = 1.8 m/s2

50. (D) Period: ! = 2! ! ! = 1.35, so ! = ! (!)(2! !)! = (0.1!!")(2! 1.35)! = 2.17 N/m Energy: ! = !

! !!! = !

! 2.17 0.15!! ! = 0.0244!! = ! !!!!!"#!

So maximum velocity: !!"# = (2)(0.0244)/(0.1!!") = 0.698 m/s = 69.8 cm/s 51. (B) efficiency: = ! !!" = 0.37 , so : !!" = ! ! = 162!!!

!!"# = !!" −! = 162 − 60 = 102!! 52. (C) At resonance, the circuit is resistive (not capacitive, nor inductive); power output is maximum; and

the current and voltage are in phase, so the phase angle is zero. 53. (D) Vector electric field cannot be zero between two opposite charges; it will be zero at a point beyond

the smaller charge. Let the distance from the negative charge be X, then the distance from the positive charge is X – r. Then: !" (! − !)! − 2!" ! ! = 0, leading to 1 (! − !)! = 2 !! This produces the quadratic equation: !! − 4!" + 2!! = 0, the solution of which is ! = 2! + 2!

54. (E) For a charged particle in a magnetic field: ! = !" !", so ! = !"# ! m = (1.6 x 10-19 C)(0.12 T)(0.327 m)/(2.5 x 105 m/s) = 2.5 x 10-26 = 15 amu

55. (A) f = R/2 = 18.0 cm. 1 ! + 1 ! = 1 !, so 1 13 + 1 ! = 1 18, which gives an image location of q = -46.8 cm. Magnification M = -q/p = -(-46.8)/13 = 3.6. The image height is h’ = (3.6)(15) = 54 cm


56. (E) Use Energy: initially in the form of elastic potential energy: ! = !! !!

! = 0.5 50 (.15)! = ! .5625!! This energy is dissipated by the work done on the block by friction. ! = !" = .5625 = !(1.2) So the frictional force is ƒ = 0.469 N. From the force diagram, the normal force is given by

N = mg = (0.55)(9.8) = 5.4 N. Then the coefficient of friction µ = ƒ/N = (0.469)/(5.4) = 0.087

57. (B) Californium-251 has atomic number Z = 98 and atomic mass M = 251. Four alpha (Z = 2, M = 4), two Beta-minus (Z = -1, M = 0), and two gamma (Z = 0, M = 0) emissions lowers the atomic number and atomic mass as follows: Z = 98 – 4(2) + 2(1) = 92 → Uranium. M = 251 – 4(4) = 235

58. (C) The neutral pion consists of a quark and its corresponding anti-quark (uū, for example). The quark anti-quark pair annihilate and produce gamma rays (photons, the electromagnetic exchange bosons). This type of matter-antimatter decay is mediated entirely by the electromagnetic force.

59. (D) The intensity of the uniformly expanding radio waves is given by:

I = P/(4πr2) = (1000)/(4π*4002) = 4.97 x 10-4 W/m2. Intensity is related to electric field by: I = !!!!!!!, so ! = 2! !!! = 0.61 N/C

60. (D) Wavelengths in nanometers are given by ! = 1240 Δ! where ΔE (energy level difference) is given in eV. ΔE values for this atom are: 6.14 eV, 4.37 eV, 3.42 eV, 2.72 eV, 1.77 eV, and 1.65 eV. This gives wavelengths of 202 nm, 284 nm, 363 nm, 456 nm, 701 nm, and 752 nm.

GENERAL DIRECTIONS:













answers.








SCORING:



SCIENCEInvitational B • 2016




1A1

8A18

1H1.008

2A2

3A13

4A14

5A15

6A16

7A17

2He4.003

3Li6.941

4Be9.012

5B10.81

6C12.01

7N14.01

8O16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

3B3

4B4

5B5

6B6

7B7

8B8

8B9

8B10

1B11

2B12

13Al26.98

14Si28.09

15P30.97

16S32.07

17Cl35.45

18Ar39.95

19K39.10

20Ca40.08

21Sc44.96

22Ti47.87

23V50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.41

31Ga69.72

32Ge72.64

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.07

45Rh102.91

46Pd106.42

47Ag107.87

48Cd112.41

49In114.82

50Sn118.71

51Sb121.76

52Te127.60

53I

126.90

54Xe131.29

55Cs132.91

56Ba137.33

57La138.91

72Hf178.49

73Ta180.95

74W183.84

75Re186.21

76Os190.23

77Ir

192.22

78Pt195.08

79Au196.97

80Hg200.59

81Tl

204.38

82Pb207.20

83Bi208.98

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(261)

105Db(262)

106Sg(266)

107Bh(264)

108Hs(277)

109Mt(268)

110Ds(281)

111Rg(272)

112Cn(285)

58Ce 140.12

59Pr 140.91

60Nd 144.24

61Pm (145)

62Sm 150.36

63Eu 151.96

64Gd 157.25

65Tb 158.93

66Dy 162.50

67Ho 164.93

68Er 167.26

69Tm 168.93

70Yb 173.04

71Lu 174.97

90Th 232.04

91Pa 231.04

92U

238.03

93Np (237)

94Pu (244)

95Am (243)

96Cm (247)

97Bk (247)

98Cf (251)

99Es (252)

100Fm (257)

101Md (258)

102No (259)

103Lr (262)


density of water density of ice

ρwaterρice

1.000 g cm-3 0.9167 g cm-3


cicecwatercsteam

2.09 J g-1 K-1 4.184 J g-1 K-1 2.03 J g-1 K-1






Pressure

1 atm = 760 torr

= 101325 Pa

= 14.7 psi

1 bar = 105 Pa

= 100 kPa

Energy

1 cal = 4.184 J

1 L atm = 101.325 J

1 Cal = 1 kcal

1 hp = 746 W

1 eV = 1.602 × 10-19 J



R 8.314 J mol-1 K-1


0.08206 L atm mol-1 K-1

1.987 cal mol-1 K-1

Planck’s constant

h 6.626 × 10-34 J s

4.136 × 10-15 eV s


h/2π 1.054 × 10-34 J s 6.582 × 10-16 eV s







Avogadro’s number

NA 6.022 × 1023 mol-1






1 in = 2.54 cm

1 lb = 453.6 g

1 mi = 5280 ft = 1.609 km

1 gal = 4 quarts = 231 in3 = 3.785 L


electron rest mass

me 9.11 × 10-31 kg 0.000549 u 0.511 MeV c-2



atomic mass unit

u 1.6605 × 10-27 kg




sun'mass 1.989 × 1030 kg

distance'earth0moon

3.844 × 108 m

distance'earth0sun

1.496 × 1011 m




O–H 463 C=C 602 C–Br 285 N=N 418

N–H 391 C≡C 835 O=O 498 H–H 436

C–O 358 C=O 799 C≡O 1072 Br–Br 193

H–Cl 432 S–H 347 N≡N 945 Cl–Cl 242

H–Br 366 H–I 299 C≡N 887 I–I 151

HS Science • Invitational B • 2016


Biology Questions (1 – 20) 1. An example of evaporative cooling is

A) cellular respiration. B) transpiration in plants. C) blood vessels constricting on a hot day. D) a condensation reaction. E) hydrolysis during the summer months.

2. A student using a light microscope to find a prokaryotic cell among many unidentified cells could look for ______ to distinguish the prokaryotic cell. A) a cell wall B) a cell membrane C) a capsule D) DNA E) ribosomes

3. A membrane’s function is directly tied to the _________ structure of its lipids. A) amphipathic B) branched C) rigid D) pathetic E) tetrahedral

4. A chemical process that may take a year to occur without an input of energy is said to be A) anabolic. B) anemic. C) creative. D) catalyzed. E) spontaneous.

5. Given the atmosphere of Earth over 3.5 billion years ago, we can deduce that Earth’s earliest organisms used A) nitrogen in place of carbon. B) anaerobic respiration. C) photosynthesis. D) mitochondria. E) oxygen in place of CO2.

6. During the light reactions of photosynthesis, H2O is split due to A) the immediate presence of the electron carrier

NADP+. B) the electronegativity of chlorophyll “P680+”. C) a sufficient amount of light energy hitting the

water molecule. D) the need for ATP in the Calvin cycle. E) the need for oxygen as the ultimate electron

carrier. 7. If it were physically stretched out, the average human

chromosome would be about 4 cm long, yet the entire genome fits into a microscopic nucleus. This packing efficiency is largely due to A) histamines. B) histones. C) nucleosomes. D) chromatin. E) elastin.

8. Transcription of a eukaryotic gene that is 1,200 nucleotides long results in a functional protein that is just 200 amino acids long. The most likely explanation for the length of the protein is A) it takes more than one DNA nucleotide to code

for an amino acid. B) there was a mutation in the DNA that resulted in

a premature stop codon. C) the gene contained introns. D) the short polypeptide is just a sub-unit, and will

combine with other subunits to form the quaternary structure of the functional protein.

9. The gene p53 is a tumor-suppressor gene. Its protein product is a transcription factor that promotes the synthesis of proteins that inhibit the cell cycle. A serious but likely consequence of an insertion of 4 nucleotides into the coding region of p53 would be A) the p53 protein would not be transcribed. B) the p53 protein would be up-regulated. C) decreased cell division due to loss of p53 protein. D) increased cell division due to mutations in the

cell-cycle inhibitor proteins. E) increased cell division because cell-cycle

inhibitor proteins are not made.



10. You’ve landed a new job in a pharmaceutical laboratory, and your first task is to create bacteria that will make great quantities of a human-derived enzyme. You are familiar with gene cloning, and know that you need to A) include the proper eukaryotic transcription

factors in your gene so the protein can be expressed.

B) include an expression vector upstream of the eukaryotic gene so the protein can be expressed.

C) insert the gene into a bacterial operon so the gene will be expressed.

D) make sure the gene includes the proper introns so the gene will be expressed.

E) make sure the bacteria includes the spliceosome complex.

11. You have collected mice from a wild population in Nebraska, where the mice live within and feed in cornfields, and are preyed upon by hawks. Of these mice, 80 are black, 890 are tan, and 30 are white. You know that coat color expression in these mice is co-dominant and due to one gene with two alleles. How many total copies of the B allele are present in the mice you collected? A) 80 B) 160 C) 970 D) 1050 E) 1780

12. Considering this same population of mice you collected from cornfields of Nebraska, what kind of selection is the best explanation for the observed genotype frequencies shown above? A) Directional selection B) Frequency-dependent selection C) Sexual selection D) Disruptive selection E) Heterozygote advantage

13. Many students confuse “linked genes” with “sex-linked genes”. Which sentence best distinguishes the two? A) Linked genes are genes that code for proteins in

the same chemical pathway, whereas sex-linked genes can code for unrelated proteins.

B) Linked genes are two genes that are on the same chromosome, whereas sex-linked genes are two genes that are on the same X chromosome.

C) Linked genes are two genes that are on the same chromosome, whereas sex-linked genes refer to genes that are on a an X or a Y chromosome.

D) Linked genes are usually not inherited together, whereas sex-linked genes are always inherited by members one particular sex.

14. The human life cycle differs from that of plants in that A) humans somatic cells are diploid, whereas plant

leaf cells are haploid. B) humans gametes are haploid, whereas plant

gametes are diploid. C) the zygote is diploid in humans, whereas in

plants, the zygote is haploid. D) humans only create haploid cells by meiosis,

whereas plants can create haploid cells by mitosis.

E) in humans, only diploid cells undergo meiosis, whereas in plants, haploid cells undergo meiosis.

15. Community is to population as cell is to A) tissue. B) organ. C) organelle. D) organ system. E) organism.

16. The rate of exchange of O2 and CO2 for an organism is determined by A) the rate of diffusion. B) the surface area of the respiratory surface. C) the partial pressure of each gas on either side of

the membrane. D) ventilation of the respiratory medium. E) all of the above.



17. Vertebrate circulatory fluid consists of several types of cells suspended in A) plasma. B) cytosol. C) stroma. D) extracellular fluid. E) lymph.

18. Gene flow between two populations ________ genetic variation between them. A) increases B) decreases C) is enhanced by D) requires E) has no effect on

19. Population growth affected by carrying capacity is said to be A) exponential. B) unsustainable. C) logistic. D) deterministic. E) polarized.

20. The amount of energy available in a particular ecosystem is most closely linked to A) the total species richness. B) the total amount of photosynthetic production. C) the total amount of sunlight available per year. D) the number of trophic levels supported by that

ecosystem. E) the mean altitude of the ecosystem.

Chemistry Questions (21 – 40) 21. What is the name for the compound K3PO4?

A) sodium peroxide B) cobalt phosphite C) potassium phosphide D) potassium phosphate E) sodium phosphate

22. What is the chemical formula for cadmium sulfide? A) Ca2S D) CdS2 B) CdS E) CaSO3 C) CdSO4

23. What is the molar mass of La(NO3)3·6H2O to the

nearest integer? A) 433 g/mol D) 371 g/mol B) 219 g/mol E) 309 g/mol C) 343 g/mol

24. A specific iron oxide ore has the general formula

Fe3Ox. The iron oxide ore is known to be 72.36% iron by mass. What is the molar mass for this iron oxide? A) 154.4 g/mol D) 247.6 g/mol B) 215.6 g/mol E) 77.18 g/mol C) 231.6 g/mol

25. A hydrocarbon is burned to produce water and carbon

dioxide. If 120.0 g of the hydrocarbon produces 376.5 g of carbon dioxide, what must be the empirical formula for this hydrocarbon? A) C2H5 C) C3H8 E) C3H7 B) CH2 D) C2H6

26. Balance the following reaction with whole numbers

as coefficients.

Fe2O3 + CO ⟶ Fe + CO2

What are the respective coefficients? A) 1 : 2 ⟶ 2 : 2 B) 2 : 5 ⟶ 4 : 5 C) 3 : 4 ⟶ 6 : 9 D) 1 : 3 ⟶ 3 : 2 E) 1 : 3 ⟶ 2 : 3

27. Silicon carbide can be made via the reaction SiO2 + 3 C ⟶ SiC + 2 CO

What is the maximum amount of SiC that can be produced when 800 grams of SiO2 reacts with 600 grams of C? A) 534 g C) 668 g E) 1400 g B) 729 g D) 812 g



28. Which of the polyatomic ion choices listed has the element with the highest oxidation state ? A) ClO3

- C) Cr2O72- E) Al2O3

B) MnO4- D) SO4

2-

29. AgBr is made via the precipitation reaction AgNO3(aq) + NaBr(aq) ⟶ AgBr(s) +

NaNO3(aq) If 1.78 liters of 0.733 M AgNO3 are poured into a

solution of excess NaBr, and the resulting AgBr precipitate is dried and weighed to be 218 g, what is the percent yield of this reaction? A) 100% C) 93% E) 85% B) 96% D) 89%

30. What is the electron configuration of sulfur?

A) 1s2 2s2 2p6 3s2 3p2 B) 1s2 2s2 2p6 3s2 3p3 C) 1s2 2s2 2p6 3s2 3p4 D) 1s2 2s2 2p6 3s2 3p5 E) 1s2 2s2 2p6 3s1 3p5

31. A single photon is hits an atom and the energy absorbed promotes an electron from an energy level of -13.6 eV up to -1.5 eV. What is the wavelength of this photon? A) 830 nm C) 91.2 nm E) 103 nm B) 830 nm D) 535 nm

32. What is the molecular geometry for sulfur tetrachloride, SCl4 ? A) trigonal bipyramidal D) see-saw B) square planar E) tetrahedral C) trigonal planar

33. Consider the two alkanes, pentane (C5H12) and heptane (C7H16) and their properties. Which of the following statements must be true? A) The boiling point of heptane is higher than that of

pentane. B) Pentane has trigonal planar carbon geometries

while heptane has only tetrahedral. C) Heptane has more H-bonding than pentane. D) The dipole moment for pentane is more than the

dipole moment for heptane. E) Heptane is more polar than pentane.

34. A 2.5 L expandable container at 25°C has a gas in it at a pressure of 469 torr. What is the pressure when the container expands to 3.5 L?

A) 411 torr C) 335 torr E) 134 torr B) 377 torr D) 657 torr

35. A 50 mL sample of 0.50 M NaOH is added to 200 mL of 0.25 M NaOH. What is the concentration of NaOH in this new solution?

A) 0.28 M C) 0.36 M E) 0.30 M B) 0.32 M D) 0.40 M

36. What is the pH of a 0.00038 M HBr solution?

A) 7.00 C) 3.80 E) 4.38 B) 2.40 D) 3.42

37. Which of the following compounds has the highest

normal boiling point? A) CH3CH2I D) CH3CH2Cl B) CH3CH2CH3 E) CH3CH2F C) CH3CH2Br

38. An (generic) ionic compound is allowed to saturate

an aqueous solution via the equilibria MX3(s) ⇌ M2+(aq) + 3 X-(aq)

What is the value of Ksp for MX3 if the equilibrium concentration of M2+ is 0.0022 M ? A) 2.3×10-11 D) 4.3×10-8 B) 6.3×10-10 E) 2.1×10-10 C) 7.5×10-9



39. When 83.5 g of manganese is treated with excess oxygen, 738 kJ of heat is produced and an oxide is formed as shown. 4 Mn (s) + 3 O2 (g) ⟶ 2 Mn2O3 (s)

What is ∆Hrxn for the reaction show above?

A) -1568 kJ/mol D) -1122 kJ/mol B) -773 kJ/mol E) -1942 kJ/mol C) -971 kJ/mol

40. How much heat is required to convert 1.50 moles of water at 13°C into steam at 188°C under 1 atm of pressure ?

A) 13.4 kJ C) 75.7 kJ E) 92.0 kJ B) 58.9 kJ D) 81.2 kJ

Physics Questions (41 – 60) 41. According to Feynman, the principle of science is

that the test of all knowledge is _____________? A) imagination B) theory C) observation D) experiment E) philosophy

42. According to Feynman, which theory is the most useful for producing new ideas in the field of Biology? A) Energy flows from producers to consumers. B) The blueprint for every living thing is DNA. C) Living things are made of atoms. D) Living things are composed of cells. E) Biological processes are controlled by

proteins.

43. According to Feynman, no phenomenon directly involving a frequency has yet been detected above approximately __________ cycles per second. A) 1010

B) 1012

C) 1014

D) 1020

E) 1024

44. According to Feynman, which kind of particles do not interact strongly in nuclei and have nothing to do with a nuclear interaction? A) leptons B) mesons C) baryons D) hadrons E) bosons

45. Blue and White stars (O, B and A classes) are ______ when compared to Red and Orange stars (M and K classes). A) more luminous B) longer lived C) more abundant D) smaller in diameter E) closer to the Sun.

46. A balloon launches from rest from the ground with an upward acceleration of 4.20 m/s2. After 12.0 seconds, a sandbag is released from the balloon. From the moment it is released, how long does it take for the sandbag to hit the ground? Ignore air resistance. A) 4.25 s B) 7.86 s C) 12.0 s D) 14.5 s E) 29.0 s

47. A flywheel starting from rest spins up with an angular acceleration of 3.50 rad/s2 until it reaches a rotational velocity of 120.0 rad/s. Through how many complete revolutions did the flywheel spin while it was accelerating? A) 60.0 rev B) 327 rev C) 655 rev D) 2060 rev E) 4110 rev



48. Three resistors are connected to a 9.00 V battery as shown. What is the current through the 40.0 Ohm resistor in this circuit?

60 Ω 50 Ω 40 Ω 9.00 V

A) 270 mA B) 225 mA C) 122 mA D) 73 mA E) 49 mA

49. A box with a mass of 6.00 kg hangs from a string that passes over a pulley and is attached to a box of mass 2.50 kg that is sitting on a table (as shown). The coefficient of friction between the table and the 2.50 kg box is 0.35. Assuming that the pulley is massless and frictionless, what is the tension in the string?

2.50 kg µ = 0.35 6.00 kg A) 17.3 N B) 23.2 N C) 28.4 N D) 44.4 N E) 56.7 N

50. A sphere with a mass of 600.0 g is launched vertically from a spring loaded cannon. The spring in the cannon has a spring constant of 4500.0 N/m. To launch the mass, the spring was compressed by 15.0 cm and then released. Assuming that there is a constant drag force of 1.80 N during the flight, to what height above the cannon does the mass ascend? A) 6.59 m B) 8.61 m C) 12.4 m D) 13.2 m E) 43.9 m

51. A person is happily swimming in a pool when, suddenly, the water cools slightly and the person is ejected from the pool – the heat energy from the water having converted into kinetic energy of the person. Why don’t we see occurrences like this happen? A) This occurrence violates conservation of energy. B) Energy cannot transfer from one substance

(water) to another (person). C) This occurrence violates conservation of

momentum. D) This occurrence requires entropy to decrease. E) Heat energy cannot flow through water.

52. An AC-RLC series circuit has a resistance of 120.0 Ω, a capacitance of 400.0 pF and an inductance of 100.0 mH. What is the resonance frequency of this circuit? A) 726 Hz B) 25.2 kHz C) 158 kHz D) 3.31 MHz E) 3.98 GHz

53. Two charges, one positive (Q) and one negative (-2Q), are placed on the x-axis. The positive charge

is placed at the origin and the negative charge is placed at x = r. At what location on the positive y-axis is the electric potential exactly zero? A) ! = ! B) ! = ! 2 C) ! = ! 2 D) ! = ! 3 E) ! = ! 3



54. A railgun is constructed as shown. The railgun is powered by an 800.0 V battery, and the magnetic field is set to 3000 Gauss. The resistance of the rails is negligible, but the projectile resistance is 2.50 Ω. What is the velocity of the 5.00 g projectile after it accelerates down the 1.20 m long rails?

A) 30.4 m/s B) 87.6 m/s C) 96.0 m/s D) 152 m/s E) 304 m/s

55. You are trapped underwater in a spherical air bubble with a radius of 1.20 m. A fish swims at a distance of 50.0 cm from the edge of the bubble. From your perspective inside the bubble, how far from the bubble’s edge does the fish appear to be? Note: The index of refraction of water is 1.33. A) 34.1 cm B) 41.9 cm C) 44.0 cm D) 58.5 cm E) 77.1 cm

56. A human hair is placed in the beam of a laser of wavelength 577.0 nm. A diffraction pattern forms on a screen placed 2.00 m from the hair. The width of the central maximum (the separation of the innermost dark fringes) is 3.66 cm. What is the diameter of the hair? A) 31.5 µm B) 52.8 µm

C) 63.1 µm D) 85.5 µm E) 126 µm

57. You start with 22.5 g of a radioactive sample. After 36.0 hours, only 18.3 g of the radioactive sample remains. What is the half-life of the material? A) 44.3 hrs B) 63.9 hrs C) 87.1 hrs D) 121 hrs E) 174 hrs

58. A certain particle has a quark structure given as

“uus”. How would this particle be described? A) Meson, Strangeness = +1, Charge = 0. B) Meson, Strangeness = -1, Charge = +1. C) Baryon, Strangeness = +1, Charge = 0. D) Baryon, Strangeness = 0, Charge = +1. E) Baryon, Strangeness = -1, Charge = +1.

59. An astronaut is stranded at a distance of 4.20 m from her spacecraft. She decides to throw a 1.50 kg hammer to give her the velocity to get back to her ship. The astronaut in her spacesuit has a mass of 140.0 kg and she throws the hammer at a speed of 12.0 m/s directly away from the spacecraft. How long does it take before she gets within reach (0.5m) of the ship? A) 19.2 s B) 28.8 s

C) 32.7 s

D) 43.2 s

E) 49.1 s

60. A metal surface is illuminated by a light with a wavelength of 404 nm, resulting in the emission of photoelectrons. The stopping potential for these photoelectrons is measured to be 760 mV. What is the workfunction of the metal? A) 2.3 eV B) 3.1 eV C) 3.8 eV D) 4.0 eV E) 4.8 eV

20.0 cm B = 3000 G

1.20 m

m = 5.00 g

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24.!!________!

25.!!________!

26.!!________!

27.!!________!

28.!!________!

29.!!________!

30.!!________!

31.!!________!

32.!!________!

33.!!________!

34.!!________!

35.!!________!

36.!!________!

37.!!________!

38.!!________!

39.!!________!

40.!!________!

!

CHEMISTRY$SCORE$

!

41.!!________!

42.!!________!

43.!!________!

44.!!________!

45.!!________!

46.!!________!

47.!!________!

48.!!________!

49.!!________!

50.!!________!

51.!!________!

52.!!________!

53.!!________!

54.!!________!

55.!!________!

56.!!________!

57.!!________!

58.!!________!

59.!!________!

60.!!________!

$$PHYSICS$SCORE$$$


INVITATIONAL B MEET • 2016

1. B

2. C

3. A

4. E

5. B

6. B

7. B

8. C

9. E

10. B

11. D

12. E

13. C

14. D

15. C

16. E

17. A

18. B

19. C

20. B

21. D

22. B

23. A

24. C

25. B

26. E

27. A

28. B

29. D

30. C

31. E

32. D

33. A

34. C

35. E

36. D

37. A

38. B

39. E

40. C

41. D

42. C

43. B

44. A

45. A

46. D

47. B

48. D

49. B

50. A

51. D

52. B

53. E

54. C

55. A

56. C

57. D

58. E

59. B

60. A

PHYSICS KEY for Science Contest • Invitational B Meet • 2016

41. (D) page 2: “The principle of science, the definition, almost, is the following: The test of all knowledge is experiment.”

42. (C) page 20: “…there is nothing that living things do that cannot be understood from the point of view that they are made of atoms acting according to the laws of physics… it is the most useful theory for producing new ideas in the field of Biology.”

43. (B) page 36: “… no phenomenon directly involving a frequency has yet been detected above approximately 1012 cycles per second.”

44. (A) page 42: “… particles which do not interact strongly in nuclei, have nothing to do with the nuclear interaction… These are called leptons…”

45. (A) Blue and white stars are significantly larger, hotter and less common than red and orange stars. Because of their size and temperature they are much more luminous, but have very short life spans.

46. (D) Use ! = !! + !" for the balloon, giving the upward velocity after 12.0 seconds as ! = 0 + 4.20 ∗ 12 = 50.4!!/!. This is the initial velocity of the sandbag. Now use ! = !! + !!! + 1 2 !!! to get the height of the balloon when the sandbag is dropped. ! = 0 + 0 ∗ 12 + !1 2 ∗ 4.2 ∗ 12! = 302.4!! This is the initial height of the sandbag. Again, use = !! + !!! + 1 2 !!! , this time for the sandbag. ! = 302.4 + 50.4! + 1 2 −9.8 !! = 0 (the final position of the sandbag is back at the ground). This gives the quadratic: !! − 10.3! − 61.7 = 0, solutions are ! = 14.5,!-4.25 seconds. Only the

positive time solution is valid. 47. (B) Use the constant angular acceleration equation !! = !!! + 2!Δ! = 120! = 0 + 2(3.5)Δ!. This

gives an angle of rotation of Δ! = 14400 7 = 2057!"#. Convert to revolutions: Δ! = 2057 2! = 327!!"#$%&'($)* 48. (D) The resistance of the parallel group is !! = (1 60 + 1 40)!! = 24!!ℎ!". So the total resistance of

the circuit is !! = 50 + 24 = 74!!ℎ!". The total current is given by !! = 9.00 74 = 0.1216!!. This gives a voltage across the parallel group of !! = !!!! = . 1216 24 = 2.92!!. This is the

same voltage across each resistor in the parallel group. So the current through the 40 ohm resistor is given by: !!" = !! 40 = 2.92 40 = 0.073!! = 73!!".

49. (B) Using Newton’s second Law: ! = !" for each box. Forces on the hanging box are only vertical: ! = ! −!" = ! − 6 9.8 = ! − 58.8 = −!" = −6!. Acceleration is negative since it is downward. Vertical forces on the sliding box are: ! = ! −!" = ! − 2.5 9.8 = ! − 24.5 = 0. Vertical acceleration is zero since the sliding box moves only horizontally. This gives the Normal force N = 24.5 Newtons. Horizontal forces on the sliding box are:

! − ! = ! − !" = ! − 0.35 24.5 = ! − 8.575 = !" = 2.5!. Solving for T in this equation: ! = 2.5! + 8.575. Plug this into the equation from the hanging box:

! − 58.8 = 2.5! + 8.575 − 58.8 = −6!. −8.5! = −50.225!!"!! = 50.225 8.5 = 5.91!!/!!. Plug this back into the tension equation:

! = 2.5! + 8.575 = 2.5 5.91 + 8.575 = 23.2!!"#$%&'. 50. (A) Use conservation of energy. The initial energy stored in the spring is

! = 1 2 !!! = 1 2 4500 (0.15)! = 50.625!! This converts into gravitational potential energy and Work done by the drag force. The distance over

which work is done equals the height to which the sphere ascends. ! = 50.625 = !"ℎ +! = 0.6 9.8 ℎ + !ℎ = 5.88ℎ + 1.8ℎ = 7.68ℎ Solving for height gives: ℎ = 50.625 7.68 = 6.59!!.


51. (D) This scenario is the time reversal of a person jumping into a swimming pool; so all time-independent

laws, such as conservation of energy and momentum, will still be obeyed. One law that is time-dependent is the Second Law of Thermodynamics – the law that states that entropy will not decrease, and generally increases, as time passes. Our time reversed scenario could only happen if entropy decreased, which is why we do not see such occurrences in everyday life.

52. (B) ! = 1 2! 1 !" = (1/2!) 1 (400!x!10!!")(0.1)! = 25200 Hz = 25.2 kHz. 53. (E) The electric potential for one charge is given by ! = !" !. It is not a vector, so direction doesn’t

matter, just the absolute distance between charge and point. Assume that the point on the y-axis is a distance y up from the origin. Then the distance to charge Q is y, and the distance to charge -2Q is

!! + !!. Then ! = 0 = !" ! + !(−2!) !! + !!. So !" ! = 2!" !! + !!, which leads to 2! = !! + !! ! 4!! = !! + !! ! 3!! = !! ! ! = ! 3 54. (C) The current in the railgun is I = V/R = 800/2.5 = 320 A. Then the force on the projectile is F = IℓB = (320)(0.2)(0.3) = 19.2 N = ma, so the acceleration is a = F/m = 19.2/(.005) = 3840 m/s2. The velocity at the end of the rails is then !! = 0! + 2 3840 1.2 !→ !!! = 96.0!!/! 55. (A) Refracting surface, object (fish) is in water, image is in air: n1 = 1.33 and n2 = 1.00. A bubble is

convex as seen from the object (fish), so the sign of the radius is positive. !! ! + !! !′ = (!! − !!) ! = 1.33 0.5 + 1.00 !′ = (1.00 − 1.33) 1.2 !!→ !!2.66 + !1 !′ = −0.275!! → ! !!! = !−2.935 So !! = !−0.341 = !−34.1!!" or 34.1 cm from the bubble’s edge. 56. (C) For diffraction !"#$% = !" where !"#$ = ! !. The width of the central maximum equals 2!, so ! = 3.66 2 = 1.83!!". Then !"#$ = 1.83 200 = 0.00915! → !!! = 0.5242!!"#$""%. This is the angle for the first dark fringe, so ! = 1. So, ! ∗ !"# 0.5242 = (1)(577!x!10!!) ! = (577!x!10!!)/(0.00915) = 6.31!x!10!!!!! = 63.1!!" 57. (D) ! = !!!!!!" → !" = !!! (!! !) = !!! 22.5/18.3 = !0.2066 = !!(36). Then ! = 0.00574!ℎ!"#$!!. The half-life is then !! ! = !!!(2) ! = 0.693 0.00574 = 121!ℎ!". 58. (E) Three quarks → Baryon. Charge is given by u(+2/3) + u(+2/3) + s(-1/3) = +1. Strangeness for one s-quark is S = -1. 59. (B) Conservation of momentum: !!!""#$!!!""#$ = !!"#$%!!"#$% = 1.5 12 = 18 = (140)!!"#$% Then !!"#$% = 18 140 = 0.1286!!/! To get within reach of the spacecraft: ! = ! ! = (4.2 − 0.5)/0.1286 = 28.8!!"#$%&! 60. (A) !" = ℎ! − ! = ℎ! ! − ! = 1240!!"#$ 404!!" − ! = 3.07 − ! = 0.76 So ! = 3.07 − 0.76 = 2.31!!"

GENERAL DIRECTIONS:













answers.








SCORING:



SCIENCEDistrict 1 • 2016




1A1

8A18

1H1.008

2A2

3A13

4A14

5A15

6A16

7A17

2He4.003

3Li6.941

4Be9.012

5B10.81

6C12.01

7N14.01

8O16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

3B3

4B4

5B5

6B6

7B7

8B8

8B9

8B10

1B11

2B12

13Al26.98

14Si28.09

15P30.97

16S32.07

17Cl35.45

18Ar39.95

19K39.10

20Ca40.08

21Sc44.96

22Ti47.87

23V50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.41

31Ga69.72

32Ge72.64

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.07

45Rh102.91

46Pd106.42

47Ag107.87

48Cd112.41

49In114.82

50Sn118.71

51Sb121.76

52Te127.60

53I

126.90

54Xe131.29

55Cs132.91

56Ba137.33

57La138.91

72Hf178.49

73Ta180.95

74W183.84

75Re186.21

76Os190.23

77Ir

192.22

78Pt195.08

79Au196.97

80Hg200.59

81Tl

204.38

82Pb207.20

83Bi208.98

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(261)

105Db(262)

106Sg(266)

107Bh(264)

108Hs(277)

109Mt(268)

110Ds(281)

111Rg(272)

112Cn(285)

58Ce 140.12

59Pr 140.91

60Nd 144.24

61Pm (145)

62Sm 150.36

63Eu 151.96

64Gd 157.25

65Tb 158.93

66Dy 162.50

67Ho 164.93

68Er 167.26

69Tm 168.93

70Yb 173.04

71Lu 174.97

90Th 232.04

91Pa 231.04

92U

238.03

93Np (237)

94Pu (244)

95Am (243)

96Cm (247)

97Bk (247)

98Cf (251)

99Es (252)

100Fm (257)

101Md (258)

102No (259)

103Lr (262)



ρwaterρice

1.000 g cm-3 0.9167 g cm-3


cicecwatercsteam

2.09 J g-1 K-1 4.184 J g-1 K-1 2.03 J g-1 K-1






Pressure

1 atm = 760 torr

= 101325 Pa

= 14.7 psi

1 bar = 105 Pa

= 100 kPa

Energy

1 cal = 4.184 J

1 L atm = 101.325 J

1 Cal = 1 kcal

1 hp = 746 W

1 eV = 1.602 × 10-19 J



R 8.314 J mol-1 K-1


0.08206 L atm mol-1 K-1

1.987 cal mol-1 K-1

Planck’s constant

h 6.626 × 10-34 J s

4.136 × 10-15 eV s


h/2π 1.054 × 10-34 J s 6.582 × 10-16 eV s







Avogadro’s number

NA 6.022 × 1023 mol-1






1 in = 2.54 cm

1 lb = 453.6 g

1 mi = 5280 ft = 1.609 km

1 gal = 4 quarts = 231 in3 = 3.785 L


electron rest mass

me 9.11 × 10-31 kg 0.000549 u 0.511 MeV c-2



atomic mass unit

u 1.6605 × 10-27 kg




sun'mass 1.989 × 1030 kg

distance'earth0moon

3.844 × 108 m

distance'earth0sun

1.496 × 1011 m




O–H 463 C=C 602 C–Br 285 N=N 418

N–H 391 C≡C 835 O=O 498 H–H 436

C–O 358 C=O 799 C≡O 1072 Br–Br 193

H–Cl 432 S–H 347 N≡N 945 Cl–Cl 242

H–Br 366 H–I 299 C≡N 887 I–I 151

HS Science • District 1 • 2016


Biology Questions (1 – 20) 1. The attraction between water molecules allows

water to have A) high surface tension. B) low surface tension. C) low boiling point. D) low specific heat.

2. At 25ºC, if the concentration of H+ in a solution is 10-4 M, what is the concentration of OH– in this solution? A) 10-4 M B) 10-7 M C) 10-8 M D) 10-10 M

3. In a cell specialized for protein production, one would expect A) prominent lysosomes. B) frequent duplication of the genome. C) numerous prominent nucleoli. D) large vacuoles.

4. Epinephrine is produced and excreted by cells in the adrenal medulla, after which the hormone travels through the blood. Which cells will respond to the signal? A) Those cells with the proper receptors for that

hormone. B) Those cells close enough to the adrenal gland to receive the signal before it diffuses. C) All cells will respond to all hormones, but the responses will vary. D) Only cells that are close to blood vessels will be able to respond. E) Only cells with the proper transcription factors will respond.

5. An orange in a basket is best described as having what kind of energy? A) light energy B) chemical energy C) actual energy D) thermal energy

6. Most of the hydrolysis of ATP is used to _______ . A) create heat to keep animals warm B) facilitate diffusion C) drive endergonic reactions D) create the ADP required for cellular functions

7. Living organisms are carbon-based, and carbon is an

essential part of organic macromolecules. The carbon found in our proteins was originally non-organic carbon obtained from A) the atmosphere, via photosynthesis. B) the ground, via plant roots. C) the vegetables that we eat. D) the minerals we get from vitamins. E) using fire to cook meat.

8. In cellular respiration, oxygen becomes reduced as

___________ is oxidized. A) glucose B) water C) carbon dioxide D) ATP

9. The “Central Dogma of Biology” refers to

A) the fact that a hypothesis can be disproven but not proven.

B) the practice of having a control group as well as a test group.

C) the fact that all living organisms are made of one or more cells.

D) the idea that all experiments must be reproducible.

E) the concept that genetic information flows from DNA to RNA to protein.



10. The following distinguishes prokaryotic protein synthesis from eukaryotic protein synthesis. A) Prokaryotes translate the intron portion of a

gene, whereas eukaryotes splice introns out before translation.

B) Prokaryotes can begin translation of a gene while it is still being transcribed, but eukaryotes cannot.

C) Unlike eukaryotes, prokaryotes do not require the DNA to be unwound to single strands prior to transcription.

D) Prokaryotes can have many copies of RNA polymerase working on a gene simultaneously, but eukaryotes can only have one RNA. polymerase functioning on any particular gene.

E) Prokaryotes make proteins from just 20 amino acids, whereas eukaryotes use 40 different amino acids.

11. The trp operon in E. coli is “repressible”, and the

genes coding for the synthesis of tryptophan will be transcribed unless the presence of tryptophan A) deactivates the repressor B) activates the operator C) deactivates the transcription factors D) activates the repressor

12. The chromosomal DNA of the cloned sheep

“Dolly” originally came from A) the egg of the surrogate mother. B) the sperm of the donor father. C) a mammary cell of a female sheep. D) the fusion of sperm and egg from the donor

parents. E) a laboratory in which the DNA nucleotides

were created through chemical reactions.

13. A restriction enzyme is used to cut DNA from different sources so they can be combined to form recombinant DNA. A key aspect of a restriction enzyme’s usefulness is that it A) cuts the double-stranded DNA at an angle within

the restriction site, leaving “sticky ends”. B) originates from bacteria, so the bacterial immune

system does not see it as a foreign invader. C) is able to target the exact sequence of the genome

where the researcher wants the foreign gene inserted.

D) will sequence a length of DNA, using mRNA as a template.

E) will only make restriction fragments that contain functional operons.

14. You have a young friend whose maternal grandfather

is colorblind, which is an X-linked recessive trait. Everyone else in the family sees normally. The mother in this family is expecting a new baby girl. What is the chance that the new baby will also be colorblind? A) There is no chance that the daughter could be

colorblind. B) There is a 25% chance that the daughter will be

colorblind. C) There is a 50% chance that the daughter will be

colorblind. D) The daughter will definitely be colorblind (100%

chance). E) There is no way to determine this from the

information provided. 15. The term epigenetic inheritance refers to

A) a phenotype that is due to interactions between the genotype and the environment.

B) inheritance of traits due to something other than the sequence of DNA nucleotides.

C) inheritance of a single trait that is due to the interaction of two or more genes.

D) a situation in which one gene influences several different phenotypes.

E) a situation in which there are more than two alleles for a given gene.



16. A cell divides and produces non-identical daughter cells. This routinely happens at cytokenisis, ____ A) following meiosis, or following mitosis of

somatic cells in a growing embryo. B) following mitosis in the testes or ovaries. C) only following meiosis II. D) only following meiosis I. E) following meiosis I and following meiosis II.

17. Which of the following is an example of the

process of microevolution? A) A woman begins jogging as a way to get into

shape. At first she can barely go around the block, but a year later she can easily jog five miles.

B) A family from the coast goes on a week-long winter ski vacation. During the first two days they are all tired because of the high altitude, but on the third day they find their bodies are able to get enough oxygen, and they no longer feel the effects of the altitude.

C) Our side yard is full of red and white poppy flowers in the Spring. Our daughter loves the red ones and picks them the minute they open. There seem to be fewer red ones each year.

D) A man with blond hair and blue eyes marries a woman with green eyes and black hair. Their three children all have dark hair, two have blue eyes, and one has green eyes.

E) A man got chicken pox as a child, and when his own child gets chicken pox, the father is exposed to the virus but does not fall ill.

18. Although extensive, the fossil record is an incomplete record of Earth’s biological history, because it is skewed toward A) species with internal or external skeletons or

shells. B) species that lived (and died) in the proper sort

of environment. C) species that existed for a long time before

going extinct. D) species that had large or widespread. E) All of the above.

19. Prey animals have many adaptations to prevent being caught, killed, and eaten. In contrast, plants have evolved a mechanism of sustaining attack by herbivores while still remaining healthy and contributing to the next generation (reproducing). This ability depends on their unique _______. A) asexual reproduction. B) ability to use insect pollinators. C) ability to grow tall. D) meristematic growth. E) relationship to mycorrhizal fungi.

20. Vertebrate circulatory fluid consists of several types of cells suspended in A) plasma. B) cytosol. C) stroma. D) extracellular fluid. E) lymph.

Chemistry Questions (21 – 40) 21. A compound containing only C, H, and O is analyzed

through combustion analysis. Combustion of 2.553 g of this substance gives 5.501 g of CO2, and 2.252 g of H2O. What is the empirical formula for this substance?

A) C3H5O C) CHO E) C7H12O2 B) C5H10O2 D) C3H6O2 22. Some old instructions say to dissolve one half pound

of Na3PO4 into some water. After it all dissolves the volume is then brought to exactly one gallon by adding more water. What is the molarity of this solution of Na3PO4?

A) 0.425 M D) 2.04 M B) 1.38 M E) 0.525 M C) 0.366 M



23. Which of the following formulas corresponds to the one for sulfurous acid?

A) H2SO4 C) H2S2O3 E) H2SO2 B) H2S D) H2SO3 24. A 458 g sample of CoCl2·6H2O is put in a drying

oven and is converted to anhydrous CoCl2. What is the weight of the anhydrous CoCl2?

A) 196 g D) 308 g B) 232 g E) 250 g C) 215 g 25. What are the allowed values for the quantum

number ℓ, when ! = 4, !ℓ = −2, and !s = +½ ? A) 2 and 3 B) 3 only C) 0, 1, 2, and 3 D) 1 and 2 E) 2, 3, and 4 26. A central atom in a molecule is surrounded by

three single bonds and one lone pair. The molecular geometry for this atom is .

A) trigonal pyramidal B) angular C) tetrahedral D) trigonal planar E) T-shaped 27. A sample of neon gas at 300 K and a pressure of !

is compressed to one third its original volume and the temperature is increased to 400 K. What is the new pressure?

A) 3! B) 4!/3 C) !/3 D) 4! E) 3!/4

28. What volume of 0.18 M NaOH will it take to completely neutralize 36.0 mL of 0.54 M H2SO4 ?

A) 288 mL B) 216 mL C) 120 mL D) 60 mL E) 108 mL 29. Estimate ∆H for the following gas phase reaction:

CH3CH2Cl + CH3OH ⟶CH3CH2OCH3 + HCl A) –12 kJ D) –5 kJ B) +5 kJ E) +19 kJ C) +12 kJ 30. An amount of 16.4 g of HBr is dissolved into 1250 L

of pure water. What is the pH of this solution? A) 3.79 C) 2.77 E) 4.74 B) 1.95 D) 3.87 31. A 0.038 M solution of a weak base has a pH of 10.25

at room temperature. What is the Kb of this weak base?

A) 8.4×10-7 D) 5.6×10-10 B) 1.8×10-5 E) 8.3×10-9 C) 1.2×10-8 32. Which reaction below corresponds to Ka3 for arsenic

acid? A) H3AsO4 ⇌ 3H+ + AsO4

3– B) H3AsO4 ⇌ H+ + H2AsO4

– C) H2AsO4

– ⇌ 2H+ + AsO43-

D) H2AsO4– ⇌ H+ + HAsO4

2– E) HAsO4

2– ⇌ H+ + AsO43–

33. pentafluorophosphate (PF5) forms a molecular solid

at low temperatures. What type of forces hold it in a solid configuration?

A) only dipole-dipole forces B) dispersion, dipole-dipole,

and H-bonding forces C) only H-bonding forces D) only dispersion forces E) dispersion and dipole-dipole forces



34. Which of the following statements about intermolecular forces (IMFs) is true?

A) IMFs tend to increase as the symmetry of a molecule increases.

B) IMFs tend to decrease as the symmetry of a molecule increases.

C) IMFs are independent of the symmetry of a molecule.

35. Consider the following reaction.

CO(g) + 2H2(g) ⇌ CH3OH(g) At a given temperature, an equilibrium mixture of this reaction contains 0.12 atm of CO, 0.15 atm of H2, and 0.21 atm of CH3OH. What is the value of Kp for this reaction?

A) 97 B) 12 C) 35 D) 52 E) 78 36. Ag2SO3 has a molar solubility of 1.55×10-5 M.

What is the value of Ksp for this compound? A) 3.7×10-15 B) 2.4×10-10 C) 1.5×10-14 D) 4.8×10-10 E) 7.4×10-15 37. A first order reaction (unimolecular) is 88%

complete in 1.5 hours. What is the rate constant, k, for this reaction?

A) 4.7×10-4 s-1 D) 2.4×10-5 s-1 B) 2.4×10-2 s-1 E) 1.8×10-3 s-1 C) 3.9×10-4 s-1

38. Consider the following reaction: 2 N2O5(g) ⟶ 4NO2(g) + O2(g)

At a certain time period during this reaction, 0.125 moles of N2O5 are being consumed in a 5.00 L container during each second. What is the reaction rate for NO2 during this time period?

A) +.0125 M/s D) –0.0250 M/s B) +0.0250 M/s E) +0.0500 M/s C) +0.100 M/s 39. An aluminum anode in a voltaic cell is oxidized to

Al3+ ions as the cell is allowed to run. What amount of aluminum is converted (consumed) when the cell runs for 3.5 hours at a constant current of 1.8 amps?

A) 4.56 g C) 3.17 g E) 8.74 g B) 6.34 g D) 2.11 g 40. When ½ mol of a specific fuel is burned at constant

pressure, it produces 1820 kJ of heat and does 15 kJ of work. What is the value of ΔU for this process (combustion)?

A) +1805 kJ D) –1820 kJ B) –1835 kJ E) –1805 kJ C) +1835 kJ

Physics Questions (41 – 60)

41. According to Feynman, when a piston moves inward and slowly compresses a gas, what happens when an atom of the gas hits the moving piston? A) the atom’s momentum slows the piston’s motion. B) the atom loses energy during the collision. C) the atom picks up speed from the collision. D) the atom adsorbs onto the piston’s surface. E) the atom emits energy in the form of light.

42. According to Feynman, when we try and tell if we “understand” the world, one way to determine whether our ideas are right is by… A) rough approximation. B) mathematical symmetry. C) checking self consistency. D) observation and reason. E) reducing the number of variables.



43. According to Feynman, nature behaves in such a way that it is fundamentally impossible to… A) unify all of the fundamental forces of nature.

B) explain complex chemistry and biology in terms of the laws of physics.

C) predict statistical averages for a repeated set of experiments.

D) make a precise prediction of exactly what will happen in a given experiment.

E) describe atomic behavior in an analytic way.

44. According to Feynman, Yukawa predicted the existence of a particle associated with the force that holds protons and neutrons together in a nucleus. What is the name of that particle? A) the muon (µ-meson) B) the pion (π-meson) C) the gluon D) the photon E) the K-meson

45. Once a star dies, if the stellar remnant has a mass

below the Chandrasekhar limit of 1.44 solar masses, then the stellar remnant is most likely a… A) black hole B) neutron star C) pulsar D) white dwarf E) quasar

46. A two-stage toy rocket launches from rest with a

net upward acceleration of 5.20 m/s2. Exactly 8.00 seconds after launch, the first stage separates from the rocket and free-falls back to Earth. How fast is the first stage moving when it hits the ground? Ignore air resistance. A) 15.5 m/s B) 29.1 m/s C) 39.1 m/s D) 57.1 m/s E) 70.6 m/s

47. An 80.0 kg skier sits atop a snow-covered hill that is 150.0 m above the valley floor. The skier starts from rest and descends along the frictionless snowy slope. After reaching the bottom, the skier slides horizontally across some rough ice, and the coefficient of friction between the skis and the rough ice is 0.2. How far across the rough ice does the skier slide before coming to rest? A) 750 m B) 400 m C) 375 m D) 30.0 m E) 13.8 m

48. Three capacitors are connected in the following circuit to a 12.0V battery. What is the charge stored on the 40.0 µF capacitor?

40.0 µF 80.0 µF 60.0 µF 12.00 V

A) 0.0692 µC B) 111 µC C) 213 µC D) 267 µC E) 2080 µC



49. A box with mass M1 hangs from a string that is attached to a box of mass M2 sitting on a table. The coefficients of friction between the table and M2 are µs (static) and µk (kinetic). Assuming that the pulley is massless and frictionless, and the boxes start from rest, what condition is necessary for the boxes to accelerate?

M2

µ M1

A) M1 > µsM2 B) M1 > µkM2 C) M1 > M2

D) M1 – M2 > µsM2 E) M1 – M2 > µkM2

50. A small moon is observed orbiting an exoplanet of unknown mass. It is noted that the moon completes a circular orbit once every 16.5 days. The moon is measured to be 195,000 km from the planet. What is the approximate mass of the planet? A) 1.24 x 1027 kg B) 2.05 x 1025 kg C) 2.16 x 1024 kg D) 5.47 x 1022 kg E) 2.51 x 1021 kg

51. A small mirror needs to be able to magnify an object by a factor of 5.00 when located at a distance of 1.50cm from the object. Furthermore, the image needs to be upright. Which of the following mirror specifications will work? A) convex mirror, f = -1.25cm. B) convex mirror, f = -0.375cm. C) concave mirror, f = 0.25cm. D) concave mirror, f = 1.25cm. E) concave mirror, f = 1.88cm.

52. An AC circuit consists of an AC source (set to 24.0 Vrms at a frequency of 200 Hz) connected to a 2.70 µF capacitor. What is the rms-current flowing through the capacitor and what is the relationship of the phase of the current to that of the voltage? A) 13.0 mA, current leads voltage B) 13.0 mA, current lags voltage C) 57.6 mA, current leads voltage D) 81.4 mA, current leads voltage E) 81.4 mA, current lags voltage

53. Two charges are placed on the x-axis. One charge, Q1 = +3.50 µC, is located at the origin (x = 0.00 cm). Another charge, Q2 = +1.50 µC, is located at x = 9.00 cm. What is the Electric Field magnitude at x = 6.00 cm? A) 75.0 kV/m B) 975 kV/m C) 6.25 MV/m D) 15.0 MV/m E) 23.8 MV/m

54. A bundle of wire carrying a DC current is oriented in a North-South direction. A compass held directly above the bundle of wire deflects to an angle of 45o when the compass is exactly 5.50cm above the center of the bundle. If there are 15 strands in the bundle, how much current, on average, is flowing in a single strand? Note: the magnetic field of the Earth is 0.5 Gauss. A) 0.292 A B) 0.648 A C) 0.917 A D) 4.38 A E) 13.8 A

55. The average cat has a body temperature of 38.6oC and a surface area of approximately 0.30m2. If a black cat (with an emissivity of 0.95) is located in a room in which the temperature is 20oC, what is the net energy loss of the cat due to radiation emission? A) 1.93 cal/day B) 33.2 kcal/day C) 686 kcal/day D) 2400 kcal/day E) 3144 kcal/day



56. When stationary, an ambulance has a siren that produces a sound with a frequency of 540 Hz. While you are driving at 20.0 m/s, you hear an ambulance approaching from behind you. The frequency that you hear is 660 Hz. How fast is the ambulance driving? A) 101 m/s B) 79 m/s

C) 62 m/s D) 52 m/s E) 46 m/s

57. Consider the simple harmonic oscillations of a mass attached to a spring. Which of the following combinations of spring constant and mass will result in the greatest frequency of oscillation? A) spring constant = k, mass = m B) spring constant = k, mass = 2m C) spring constant = ½k, mass = m D) spring constant = 2k, mass = 2m E) spring constant = k, mass = ½m

58. Which discovery proved that a neutrino must have

a rest mass greater than zero? Note: this discovery resulted in the awarding of the Nobel Prize in Physics in 2015 to Takaaki Kajita and Arthur B. McDonald. A) the discovery that neutrino collisions with

nuclei can produce electrons. B) the discovery that neutrinos oscillate between

different identities. C) the discovery of the deflection of neutrinos by

a strong magnetic field. D) the discovery that there exist three different

identities of neutrinos (e, µ, τ) E) the discovery that neutrinos are

supersymmetric partners to quarks.

59. A geologic sample is found to contain a large amount of lead-208. Measurements on the sample find that it emits two distinct energies of negative beta particles, two distinct energies of alpha particles, and three distinct energies of gamma radiation. If all of these emissions are part of a single long decay chain, what is the parent radioactive element that eventually decays into lead-208? A) polonium-216 B) bismuth-216

C) polonium-218

D) radon-218

E) radon-219

60. A sample of four-level atoms is found to produce the following spectral lines when energized: 295nm, 500nm, 544nm, 642nm, 717nm, 2254nm. If the ground state is at E = 0, which of these could be the energy of the second excited state? A) 0.55 eV B) 1.73 eV C) 1.93 eV D) 2.28 eV E) 4.21 eV


DISTRICT 1 • 2016

1. A

2. D

3. C

4. A

5. B

6. C

7. A

8. A

9. E

10. B

11. D

12. C

13. A

14. A

15. B

16. E

17. C

18. E

19. D

20. A

21. B

22. C

23. D

24. E

25. A

26. A

27. D

28. B

29. C

30. A

31. A

32. E

33. D

34. B

35. E

36. C

37. C

38. E

39. D

40. B

41. C

42. A

43. D

44. B

45. D

46. E

47. A

48. C

49. A

50. C

51. E

52. D

53. C

54. C

55. C

56. B

57. E

58. B

59. A

60. D

Science Contest • District Meet I • 2016 41. (C) page 8: “What happens when an atom hits the moving piston? Evidently it picks up speed from the

collision. You can try… and you will find that it comes off with more speed than that with which it struck.”

42. (A) page 25: “If we know the rules, we consider that we “understand” the world….The third way to tell our ideas are right is relatively crude but probably the most powerful of them all. That is, by rough approximation.”

43. (D) page 35: “…nature, as we understand it today, behaves in such a way that it is fundamentally impossible to make a precise prediction of exactly what will happen in a given experiment.”

44. (B) pages 38-39: “Yukawa suggested that the forces between neutrons and protons also have a field of some kind, and that when this field jiggles it behaves like a particle. … a little while later, in 1947 or 1948, another particle was found, the π-meson, or pion, which satisfied Yukawa’s criterion.”

45. (D) When small stars such as our Sun die, they will leave a white dwarf as a stellar remnant. The Chandrasekhar limit of 1.44 Solar masses is the maximum mass that a white dwarf can have. Larger mass remnants will collapse into neutron stars or black holes.

46. (E) First, we need to know the height at which the first stage separates from the rocket: ! = !! + !!! + 1 2 !!! = 0 + 0 + 1 2 5.2 8 ! = 166.4! We also need to know the velocity of the first stage at the moment of separation: ! = !! + !" = 0 + 5.2 8 = 41.6 !/! Using these as the initial position and velocity, we can calculate the final velocity using free-fall

equations: !! = !!! + 2!∆! = (41.6)! + 2 −9.8 −166.4 = 1731 + 3261 = 4992 So ! = 4992 = 70.6 !/! 47. (A) Use conservation of energy. Gravitational potential energy à Kinetic energy à work done by

friction. You don’t really need to find the velocity at the bottom of the slope, but I’ll do it anyway: !"ℎ = 1 2!!! = 80 9.8 150 = 117,600 ! = 1 2 (80)!!. This gives ! = 54.2 !/! The work done by friction removes all of the energy from the skier. So ! = !" = 117,600 ! On a horizontal surface: ! = !" = !"# = 0.2 80 9.8 = 156.8 !"#$%&' Thus ! = (117,600)/(156.8) = 750! 48. (C) First, determine the total capacitance of the circuit: Parallel branches add: !! = 40 + 60 = 100 !" Series combination add inversely: 1 !! = 1 80 + 1 100 Thus, !! = 44.4 !" is the total

capacitance.

Now find the total charge stored in the circuit: !! = !!! = 44.4 12 = 533 !" Charge is the same for series, so !! = !! = 533 !"; Thus the voltage across the parallel branch is given by !! = !! !! = !""

!"" = 5.33 !"#$%. Because they are in parallel, this is the voltage across both the 40µF and the 60µF capacitors. !!" = !!" = !! Then the charge stored on the 40µF is given by: !!" = !!"!!" = 40 5.33 = 213 !". 49. (A) Since the boxes start from rest, it is static friction that must be overcome to begin acceleration. Static

friction is generally larger than kinetic friction, so if the forces are large enough to overcome static friction, then those forces will certainly be large enough to overcome kinetic friction.

The forces on M1 are tension and gravity, while the forces on M2 are tension and friction. In order to accelerate, the tension must be greater than friction: ! > !. Also recall for static friction: ! = !!!. For a horizontal surface ! = !!!. In other words, to accelerate we need ! > !!!!!

In order to accelerate M1, we need gravity to be greater than tension. In other words: !!! > !. Combining these two inequalities: !!! > ! > !!!!!. Or, reducing: !! > !!!!.

Science Contest • District Meet I • 2016 50. (C) Since the moon orbits in a circle, then it is subjected to a centripetal force equal to !! = !!! !. The

force providing the centripetal acceleration is gravity, which for such a long range is given by !! = !"# !!. Equating these gives: !!! ! = !"# !!, which reduces to ! = !!! !. To find velocity, we consider the orbital path length (circumference) and the orbital period. We need both in better units: R = 195,000 km = 1.95 x 108 m. T = 16.5 days = 1.43 x 106 seconds. Thus, ! = 2!" ! = 2!(1.95 ×10!) (1.43 ×10!) = 859 !/! So, ! = (859)!(1.95 ×10!)/(6.67 ×10!!!) = 2.16 ×10!" !". 51. (E) Magnification = +5.00 and p = 1.50 cm. ! = −! ! = 5, so ! = −5! = −5 1.5 = −7.50!". Also 1 ! + 1 ! = 1 !, so 1 (1.5) + 1 (−7.5) = 1 !. This gives ! = +1.88!". Since f is

positive, then the mirror is concave. 52. (D) Frequency is 200Hz, so the angular frequency is given by ! = 2!" = 2! 200 = 1257 !"#/!"# Capacitive reactance is !! = 1 !" = 1 (1256)(2.70 ×10!!) = 295 Ω = Z Current is ! = ! ! = !"

!"# = 81.4!". In capacitive circuits, the current leads the voltage. 53. (C) !! = !!! !!! = (8.998 ×10!)(3.50 ×10!!)/ . 06 ! = 8.75 !"/!. E1 is directed to the right. !! = !!! !!! = (8.998 ×10!)(1.50 ×10!!)/ . 03 ! = 15.0 !"/!. E2 is directed to the left. So, the electric field magnitude is !! − !! = 6.25 !"/!. 54. (C) Deflecting the compass by exactly 45o means that the East-West B-field produced by the current in

the wire exactly equals the North-South B-field of the Earth. The B-field produced by a straight segment of wire is ! = !!! 2!". This equals the Earth’s

magnetic field (converted to Teslas). So, !!! 2!" = 5×10!! ! = (4!×10!!)! 2!(0.055) Thus, I = 13.75A for the bundle. There are 15 strands, so each strand get ! = 13.75 15 = 0.917! 55. (C) Energy transfer by radiation is given by the Stefan-Boltzmann Law: ! = !"#(!!! − !!!), where

temperatures are in Kelvin. Thus, ! = 5.67×10!! 0.95 0.3 38.6 + 273 ! − 20 + 273 ! = 33.2 ! Converting Watts to kcal/day: 33.2 !/! (86400 !/!"#)/(4186 !/!"#$) = 686 !"#$/!"# 56. (B) Doppler shift is given by ! = !! (! + !!) (! + !!) . Since the observer is moving away from the

ambulance, !! is negative. Since the ambulance is approaching the observer, !! is also negative. Using 343 m/s as the speed of sound, we get: 660 = (540) (343 − 20) (343 − !!)

This gives 343 − !! = 264, or !! = 78.7 !/!. 57. (E) Frequency of oscillation of a mass-spring system is ! = 2! ! !. So maximum frequency is

achieved with the largest ratio of k/m. The ratios are : A.1k/m, B.½k/m, C.½k/m, D.1k/m, E.2k/m. 58. (B) The Nobel Prize in Physics 2015 recognises Takaaki Kajita and Arthur B. McDonald for their key

contributions to the experiments which demonstrated that neutrinos change identities. This metamorphosis requires that neutrinos have mass. (Press Release, Royal Swedish Academy of Sciences, October 2015)

59. (A) Lead-208 has Z = 82 and M = 208. If these emission are all part of one long decay chain, then the parent element is two negative betas, two alphas, and two gammas removed from the final daughter element. Negative betas have Z = -1, M = 0; Alphas have Z = 2, M = 4; Gammas have Z = 0, M = 0. Working backwards to the parent: Z = 82 + (-1) + (-1) + (2) + (2) + (0) + (0) + (0) = 84.

And M = 208 + (0) + (0) + (4) + (4) + (0) + (0) + (0) = 216. So, the parent element has Z = 84 and M = 216, which is Polonium-216.

Science Contest • District Meet I • 2016 60. (D) There are four energy states, with one at !! = 0 (ground state). The highest state can be found from

the most energetic spectral line: !! − !! = !! = (1240!"#$) (295!") = 4.21 !". Here we are using ℎ! = 1240 !"#$. The other spectral lines are given by differences between energy levels, but it is not known which spectral line goes with which energy level difference. So you have !! − !!, !! − !!, !! − !!, !! − !!, and !! − !! for the other spectral lines. The energies of these spectral lines are: 2.48 eV, 2.28 eV, 1.93 eV, 1.73 eV, 0.55 eV. A little trial and error give two possible choices for the energy of the second excited state: !! = 2.28 !" !" 2.48 !". Only one of these is given as an answer option.


$ OVERALL$SCORE:!$!



!! !! !

!1.!!!!________!

2.!!!!________!

3.!!!!________!

4.!!!!________!

5.!!!!________!

6.!!!!________!

7.!!!!________!

8.!!!!________!

9.!!!!________!

10.!!________!

11.!!________!

12.!!________!

13.!!________!

14.!!________!

15.!!________!

16.!!________!

17.!!________!

18.!!________!

19.!!________!

20.!!________!

!

BIOLOGY$SCORE!

!

21.!!________!

22.!!________!

23.!!________!

24.!!________!

25.!!________!

26.!!________!

27.!!________!

28.!!________!

29.!!________!

30.!!________!

31.!!________!

32.!!________!

33.!!________!

34.!!________!

35.!!________!

36.!!________!

37.!!________!

38.!!________!

39.!!________!

40.!!________!

!

CHEMISTRY$SCORE$

!

41.!!________!

42.!!________!

43.!!________!

44.!!________!

45.!!________!

46.!!________!

47.!!________!

48.!!________!

49.!!________!

50.!!________!

51.!!________!

52.!!________!

53.!!________!

54.!!________!

55.!!________!

56.!!________!

57.!!________!

58.!!________!

59.!!________!

60.!!________!

$$PHYSICS$SCORE$$$

GENERAL DIRECTIONS:













answers.








SCORING:



SCIENCEDistrict 2 • 2016




1A1

8A18

1H1.008

2A2

3A13

4A14

5A15

6A16

7A17

2He4.003

3Li6.941

4Be9.012

5B10.81

6C12.01

7N14.01

8O16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

3B3

4B4

5B5

6B6

7B7

8B8

8B9

8B10

1B11

2B12

13Al26.98

14Si28.09

15P30.97

16S32.07

17Cl35.45

18Ar39.95

19K39.10

20Ca40.08

21Sc44.96

22Ti47.87

23V50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.41

31Ga69.72

32Ge72.64

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.07

45Rh102.91

46Pd106.42

47Ag107.87

48Cd112.41

49In114.82

50Sn118.71

51Sb121.76

52Te127.60

53I

126.90

54Xe131.29

55Cs132.91

56Ba137.33

57La138.91

72Hf178.49

73Ta180.95

74W183.84

75Re186.21

76Os190.23

77Ir

192.22

78Pt195.08

79Au196.97

80Hg200.59

81Tl

204.38

82Pb207.20

83Bi208.98

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(261)

105Db(262)

106Sg(266)

107Bh(264)

108Hs(277)

109Mt(268)

110Ds(281)

111Rg(272)

112Cn(285)

58Ce 140.12

59Pr 140.91

60Nd 144.24

61Pm (145)

62Sm 150.36

63Eu 151.96

64Gd 157.25

65Tb 158.93

66Dy 162.50

67Ho 164.93

68Er 167.26

69Tm 168.93

70Yb 173.04

71Lu 174.97

90Th 232.04

91Pa 231.04

92U

238.03

93Np (237)

94Pu (244)

95Am (243)

96Cm (247)

97Bk (247)

98Cf (251)

99Es (252)

100Fm (257)

101Md (258)

102No (259)

103Lr (262)



ρwaterρice

1.000 g cm-3 0.9167 g cm-3


cicecwatercsteam

2.09 J g-1 K-1 4.184 J g-1 K-1 2.03 J g-1 K-1






Pressure

1 atm = 760 torr

= 101325 Pa

= 14.7 psi

1 bar = 105 Pa

= 100 kPa

Energy

1 cal = 4.184 J

1 L atm = 101.325 J

1 Cal = 1 kcal

1 hp = 746 W

1 eV = 1.602 × 10-19 J



R 8.314 J mol-1 K-1


0.08206 L atm mol-1 K-1

1.987 cal mol-1 K-1

Planck’s constant

h 6.626 × 10-34 J s

4.136 × 10-15 eV s


h/2π 1.054 × 10-34 J s 6.582 × 10-16 eV s







Avogadro’s number

NA 6.022 × 1023 mol-1






1 in = 2.54 cm

1 lb = 453.6 g

1 mi = 5280 ft = 1.609 km

1 gal = 4 quarts = 231 in3 = 3.785 L


electron rest mass

me 9.11 × 10-31 kg 0.000549 u 0.511 MeV c-2



atomic mass unit

u 1.6605 × 10-27 kg




sun'mass 1.989 × 1030 kg

distance'earth0moon

3.844 × 108 m

distance'earth0sun

1.496 × 1011 m




O–H 463 C=C 602 C–Br 285 N=N 418

N–H 391 C≡C 835 O=O 498 H–H 436

C–O 358 C=O 799 C≡O 1072 Br–Br 193

H–Cl 432 S–H 347 N≡N 945 Cl–Cl 242

H–Br 366 H–I 299 C≡N 887 I–I 151



Biology Questions (1 – 20) 1. Raw fish left in lime juice for 30 minutes is as safe

to eat as cooked fish because A) Fish oil is rich in unsaturated fatty acids,

which react with the lime juice, essentially “frying” the fish.

B) Bacteria are repelled by lime juice. C) The lime juice changes the primary structure

of proteins. D) The lime juice interrupts the weak bonds

between R-groups. 2. According to the pH scale, a solution with a pH of

4 is ___ as acidic as a solution with a pH of 8. A) Twice B) Half C) 1,000 times D) 10,000 times

3. An example of the connection between structure

and function can be seen in the cristae of mitochondria, which serve to __________ in the organelle. A) increase surface area B) increase CO2 uptake C) remove more glucose D) increase circulation

4. A friend of yours is happy because they just got a new treatment for their asthma. When you look at the drug pamphlet, it shows the chemical diagram for the drug as:

Without knowing anything else about the drug or asthma, (but given your understanding of cellular communication), where in a cell do you think the receptor protein for this drug has its active site? A) Within a peroxisome. B) Within the centrosome. C) Within the cytoplasm. D) On the outside of the cell membrane.

5. A transfer of potential to kinetic energy will always

produce _________. A) light B) heat C) metabolism D) more energy

6. What can be said of a “phosphorylated intermediate”?

A) It is highly stable. B) It is a source of multiple phosphate groups. C) It has received a phosphate group. D) It has given up a phosphate group.

7. Plants are able to harvest light because they have

pigment protein complexes containing ______. A) chloroplasts B) mitochondria C) thylakoids D) chromatids E) chlorophyll

8. Imagine an individual for which the electron transport chain of ATP production is inefficient, such that extra food must be consumed to produce the same number of ATPs. One consequence of this would be that this individual would ____ A) consume less O2 but produce a normal amount of

CO2. B) consume a normal amount of O2 but produce

more CO2 than normal. C) consume less than normal O2 and produce less

than normal CO2. D) create more body heat. E) gain weight.



9. A major difference between eukaryotic and bacterial DNA replication is that A) eukaryotes, being multicellular, take much

longer to replicate their genomes. B) the eukaryotic DNA polymerase works at a

much faster rate than bacterial DNA polymerase.

C) eukaryotes must move their mRNA out of the cell for translation, whereas bacteria do not.

D) eukaryotes replicate their genomes using DNA polymerase, whereas bacteria use RNA polymerase instead of DNA polymerase.

E) the efficiency of eukaryotic DNA replication is enhanced by having many origins of replication per chromosome, whereas bacteria have just one origin of replication.

10. Which of the following is true for eukaryotic gene expression? A) Proteins called transcription factors are

enzymes required for post-transcriptional modification.

B) The functional mRNA transcript of a gene may contain many fewer nucleotides than the actual gene.

C) Unlike in prokaryotes, after transcription eukaryotic RNA is exported out of the nucleus without modification.

D) If a protein is to function in the nucleus, then the mRNA is translated by nuclear ribosomes.

E) Do to enzymatic proofreading capabilities, eukaryotic mRNA does not have to be in reading frame during translation.

11. The lac operon in E. coli is “inducible”, and the

gene coding for the synthesis of beta-galactosidase will not be transcribed unless _____ . A) the lacI repressor gene is up-regulated B) the bacterium requires lactose C) the bacterium is in the presence of glucose D) allolactose, the inducer, inactivates the

repressor

12. One novel aspect of the experiments that led to the cloned sheep “Dolly” was that A) researchers had never before cloned an animal. B) both the sperm and the egg came from the same

parent. C) they started with a nucleus from a fully-

differentiated adult cell. D) the entire organism was created in vitro, without

the need for a surrogate uterus. E) researchers had never before performed nuclear

transplantation. 13. Mature cells with the potential to dedifferentiate and

give rise to all the specialized cell types of an organism are called A) totipotent cells. B) growth factors. C) flexor cells. D) omnipotent cells. E) facultative cells.

14. A population of gophers lives in a field. Of these gophers, 910 are brown and 90 are white. You know that color expression in gophers is due to one gene with two alleles. Assuming the population is in Hardy Weinberg equilibrium, what is q for this population? A) 0.60 B) 0.90 C) 0.91 D) 0.30 E) 0.09

15. Chromatin is made from A) DNA associated with various lengths of fatty acid

tails. B) DNA associated with protein. C) double-stranded RNA wrapped around histones. D) multiple nucleoli bundled together. E) four strands of nucleotides, twisted into a double-

helix.



16. A cell actively moving through the cell cycle will pause at the M checkpoint, resuming only after ___. A) the cell has completed mitosis B) all chromosomes are attached to spindle fibers

from both poles C) a hormone signals that cell division is required

in that tissue D) all sister chromatids have successfully

separated from one another E) the metaphase plate has been constructed

17. Which of the following scenarios is an example of genetic drift? A) During migration, several gazelles from

population A happen to become disoriented and end up in population B, where they stay and reproduce.

B) Tube-shaped flowers with red petals are especially attractive to hummingbirds.

C) A population of frogs has tan individuals and green individuals. A flood carries a few of the tan frogs down river into a side stream, where they start a new population.

D) Many plants are able to reproduce clonally by sending out underground shoots which give rise to new offspring.

E) Infant mortality is extremely high for some fish. This is offset by the great number of eggs produced, such that the population is sustained even though the proportion of survivors is low.

18. Scientists are able to estimate the absolute age of

fossils by A) measuring radioisotopes in rocks surrounding

the fossil. B) evaluating the fossil’s order in the rock strata. C) measuring the ratio of certain isotopes

contained within the fossil. D) determining how deeply the fossils were

buried. E) both A and C are correct.

19. We animals depend, directly and indirectly, on plants for our food. For their nutrition, plants rely greatly on relationships with non-plant organisms, including A) bee and bat pollinators. B) humans to provide fertilizer. C) predatory wasps that kill plant-eating caterpillars. D) soil bacteria and mycorrhizal fungi. E) fruit-eating birds to disperse seeds.

20. In the human cardiovascular system, the rate of blood flow greatly decreases as blood travels from the heart away to the body tissues. Although we need a steady delivery of oxygen to our cells, the slower rate has the advantage of A) allowing the oxygen time to diffuse from the

capillary into cells that need it. B) preventing the thin-walled capillaries from

breaking. C) maintaining sufficient blood pressure to get the

blood back to the heart. D) allowing interstitial fluid to leak out from the

capillary and bath the surrounding cells. E) there is no real advantage to a decrease in blood

flow rate.

Chemistry Questions (21 – 40) 21. A organic amine containing only C, H, and N is

analyzed through combustion analysis. Combustion of 2.757 g of this substance gives 5.048 g of CO2, and 2.755 g of H2O. What is the empirical formula for this substance?

A) C3H9N2 C) C2H4N E) C5H12N3 B) C4H11N2 D) C3H8N2 22. Some old instructions say to dissolve 1.50 pounds of

Na2HPO4 into some water. After it all dissolves the volume is then brought to exactly 5 gallons by adding more water. What is the molarity of this solution of Na2HPO4?

A) 0.591 M D) 1.29 M B) 0.468 M E) 0.253 M C) 0.366 M



23. Which of the following formulas corresponds to the one for bromous acid?

A) HBrO4 C) HBrO E) HBrO2 B) HBr D) HBrO3 24. A 500 g sample of CuSO4·5H2O is put in a drying

oven and is converted to anhydrous CuSO4. What is the weight of the anhydrous CuSO4?

A) 284 g D) 344 g B) 449 g E) 392 g C) 320 g 25. What are the allowed values for the quantum

number ℓ, when ! = 5, !ℓ = −2, and !s = −½ ? A) 2, 3, and 4 B) 4 only C) 0, 1, 2, 3, and 4 D) 1, 2, and 3 E) 3, 4, and 5 26. A central atom in a molecule is surrounded by two

single bonds and two lone pairs. The molecular geometry for this atom is .

A) trigonal pyramidal B) tetrahedral C) angular D) trigonal planar E) T-shaped 27. A sample of hydrogen gas at 250 K and a pressure

of ! is expanded to twice its original volume and the temperature is increased to 500 K. What is the pressure of the gas after these changes?

A) 2! B) ! C) !/4 D) 4! E) !/2

28. What volume of 0.24 M KOH will it take to completely neutralize 32.0 mL of 0.60 M H2SO4 ?

A) 160 mL B) 64 mL C) 13 mL D) 80 mL E) 125 mL 29. Estimate ∆H for the following gas phase reaction:

C2H4 + 3 O2 ⟶ 2 CO2 + 2H2O A) +298 kJ D) –837 kJ B) –1556 kJ E) –1763 kJ C) –1300 kJ 30. An amount of 13.6 g of HBr is dissolved into 1250 L

of pure water. What is the pH of this solution? A) 3.96 C) 5.12 E) 4.04 B) 2.93 D) 3.87 31. A 0.024 M solution of a weak base has a pH of 11.12

at room temperature. What is the Kb of this weak base?

A) 7.7×10-5 D) 8.6×10-5 B) 8.9×10-6 E) 7.2×10-5 C) 1.3×10-10 32. Which reaction below corresponds to Ka2 for

phosphoric acid? A) HPO4

2– ⇌ H+ + PO43–

B) H3PO4 ⇌ H+ + H2PO4–

C) H2PO4– ⇌ 2H+ + PO4

3- D) H2PO4

– ⇌ H+ + HPO42–

E) H3PO4 ⇌ 2H+ + HPO42–

33. Chloromethane (CH3Cl) forms a molecular solid at

low temperatures. What type of forces hold it in a solid configuration?

A) only dispersion forces B) only dipole-dipole forces C) only H-bonding forces D) dispersion, dipole-dipole,

and H-bonding forces E) dispersion and dipole-dipole forces



34. Compare the physical properties of NH3 and PH3 under the same set of conditions. Which is true?

A) NH3 has a higher vapor pressure than PH3. B) PH3 has a higher vapor pressure than NH3. C) NH3 and PH3 have the same vapor pressures. 35. Consider the following reaction.

2H2S(g) ⇌ 2H2(g) + S2(g) At a given temperature, an equilibrium mixture of this reaction contains 0.45 atm of H2S, 0.018 atm of H2, and 0.034 atm of S2. What is the value of Kp for this reaction?

A) 1.4×10-3 B) 5.4×10-5 C) 7.6×10-4 D) 2.4×10-5 E) 3.2×10-6 36. La(IO3)3 has a molar solubility of 7.26×10-4 M.

What is the value of Ksp for this compound? A) 2.8×10-13 B) 2.5×10-12 C) 1.0×10-8 D) 7.5×10-12 E) 4.1×10-13 37. A first order reaction (unimolecular) is 70%

complete in one hour. What is the rate constant, k, for this reaction?

A) 3.3×10-4 s-1 D) 9.9×10-5 s-1 B) 2.0×10-2 s-1 E) 5.2×10-3 s-1 C) 6.5×10-4 s-1 38. Consider the following reaction:

4 PH3(g) ⟶ P4(g) + 6H2(g) At a certain time period during this reaction, 0.0660 moles of PH3 are being consumed in a 3.00 L container during each second. What is the reaction rate for H2 during this time period?

A) +.0220 M/s D) +0.132 M/s B) +0.0330 M/s E) –0.0220 M/s C) +0.0147 M/s

39. An iron anode in a voltaic cell is oxidized to iron(II) ions as the cell is allowed to run. What amount of iron is converted (consumed) when the cell runs for 2.5 hours at a constant current of 1.5 amps?

A) 10.4 g C) 3.91 g E) 5.18 g B) 7.81 g D) 2.83 g 40. When ¾ mol of a specific fuel is burned at constant

pressure, it produces 2850 kJ of heat and does 21 kJ of work. What is the value of ΔU for this process (combustion)?

A) +2871 kJ D) –2850 kJ B) –2829 kJ E) –2871 kJ C) +2829 kJ

Physics Questions (41 – 60) 41. According to Feynman, when carbon burns in the

presence of oxygen, the reaction can produce flames because… A) carbon dioxide molecules from the reaction move

so fast as to heat up from air friction and emit light.

B) unreacted carbon atoms that escape from the solid carbon have enough potential energy to emit light.

C) carbon and oxygen atoms combine into an excited molecular state, which emits light when it decays.

D) carbon and oxygen atoms snap together with tremendous energy, generating heat and light.

E) only carbon heated to the point at which it glows can react with oxygen.

42. According to Feynman, the jiggling of a particle in water – the Brownian motion – is caused by… A) fluctuations of the bond lengths of the molecules

in the particle. B) bombardment of the particle by the atoms of

water. C) chemical reactions taking place on the surface of

the particle. D) changes in the kinetic energy of the particle as the

temperature of the water fluctuates. E) dissolution of the particle as molecules are

dissolved from it and go into the water.



43. According to Feynman, the aim of fundamental physics and the primary problem in basic theoretical physics today is… A) to amalgamate the different classes of physics.

B) to unify Quantum theory and Relativity.

C) to explain heat in terms of the laws of mechanics.

D) to finish the laws of electromagnetic theory

E) to find phenomena that do not follow our current understanding of physics.

44. According to Feynman, the ultimate basis of an

interaction between atoms is… A) magnetic B) electrical C) gravitational D) the strong force E) the weak force

45. The Milky Way galaxy, the galaxy in which we reside, is classified as what type of galaxy? A) irregular B) elliptical C) spiral D) barred spiral E) lenticular

46. Two cars are racing along a straight stretch of

road. Car A is initially at rest, but accelerates at a rate of 10.5 m/s2 beginning exactly when Car B is passing it. Car B is moving at a constant speed of 42.0 m/s. The finish line is 400.0 m from the point at which Car B passes Car A. Which car wins, and what is their separation at the finish? A) Car A wins by 141 m B) Car A wins by 76.2 m C) Car A wins by 33.4 m D) Car B wins by 76.2 m E) Car B wins by 33.4 m

47. A 2.00m aluminum rod is balanced about a pivot point located 40.0cm from the left end of the rod. Two masses are attached to the rod: a 100.0g mass located 20.0cm to the right of the pivot point, and a 750.0g mass located 30.0cm to the left of the pivot point. If the rod is of uniform size and density, what is the mass of the rod? A) 342g B) 375g C) 408g D) 512g E) 613g

48. Four resistors are connected to a 24.00 V battery as shown. What is the power dissipated in the 120.0 Ohm resistor in this circuit?

120.0 Ω 60.0 Ω 80.0 Ω 100.0 Ω 24.00 V

A) 10.8 W B) 4.80 W C) 3.32 W D) 0.953 W E) 0.423 W



49. A box with a mass of 10.00 kg hangs from a string that is attached to a box of mass 5.00 kg sitting on a table. Another box of mass 4.00 kg hangs from a string that is also attached to the 5.00 kg box, but on the opposite side (as shown). The coefficient of friction between the table and the 5.00 kg box is 0.250. Assuming that the pulleys are massless and frictionless, what is the acceleration of the 5.00 kg box?

5.00 kg

µ = 0.25 4.00 kg 10.00 kg A) 2.45 m/s2 B) 3.07 m/s2 C) 3.72 m/s2 D) 6.60 m/s2 E) 9.23 m/s2

50. A disk of steel (linear coefficient of thermal expansion 12.0 x 10-6 K-1), is heated from room temperature (20oC) until its area has increased by 1.00%. What is the final temperature of the steel? A) 437oC B) 853 oC C) 710 oC D) 1126 oC E) 1687 oC

51. A 90.0g ball is thrown downward from a height of 5.30m above an uncompressed spring. When the ball strikes the spring, it compresses the spring by 48.0cm. Knowing that the spring constant is 60.0 N/m, what was the initial downward velocity of the ball? A) 6.35 m/s B) 7.19 m/s C) 7.82 m/s D) 14.4 m/s E) 16.3 m/s

52. An AC circuit consists of a constant voltage AC source in series with a resistor and a capacitor. Which of the following is true about this circuit? A) the voltage and current are in-phase at low

frequencies B) the voltage leads the current at low frequencies. C) the voltage across the capacitor is greatest at high

frequencies D) the voltage across the resistor is greatest at high

frequencies. E) the voltage across the resistor is independent of

frequency.

53. A pendulum is used in an elevator to measure the vertical acceleration of the elevator car. When the car is stationary, the period of the pendulum is measured to be 1.22 seconds. When the car is accelerating, the period of the pendulum is measured to be 1.02 seconds. What is the acceleration of the elevator car? A) 1.61 m/s2 B) 1.92 m/s2

C) 2.95 m/s2 D) 4.22 m/s2 E) 9.44 m/s2

54. A 20.0 cm length of wire carrying a current of 30.0A levitates above another wire carrying a current of 75.0A. The second wire is secured to a lab table, and the two wires are separated by 3.30mm. What is the mass of the levitating wire? A) 0.0928 g B) 2.78 g C) 13.9 g D) 17.5 g E) 27.3 g

55. An object is placed in front of a concave mirror at exactly the center of the radius of curvature of the mirror. What kind of image will be formed? A) virtual, upright, same size B) virtual, inverted, reduced C) real, upright, magnified D) real, inverted, magnified E) real, inverted, same size



56. A laser of unknown wavelength is directed through a diffraction grating which has 800 lines/mm. The resulting pattern is shown onto a screen located 1.20m from the grating. The first maximum is located 88.0cm from the central (zeroth) maximum on the screen. What is the frequency of the laser? A) 240 THz B) 327 THz

C) 406 THz D) 654 THz E) 812 THz

57. A certain capsule of radioactive waste has a half-life of 140 years. If the activity of the sample today is 5.00 Ci, how many years are required for it to decay down to a safe level of 10 µCi? A) 800 years B) 1150 years C) 1840 years D) 2650 years E) 2930 years

58. Predicted by Steven Weinberg, and discovered in

1983, which neutral (uncharged) particle mediates the weak force? A) the photon, γ B) the pion, π C) the gluon, g D) the W boson E) the Z boson

59. A 150.0g arrow is shot horizontally at a speed of 35.0 m/s through an apple that is initially sitting at rest on a table. The arrow passes cleanly through the apple, emerging with a speed of 28.0 m/s. The apple flies off the table horizontally at a speed of 11.0 m/s. What is the mass of the apple? A) 37.5 g B) 72.9 g

C) 95.5 g

D) 129 g

E) 327 g

60. Two charges are placed on a coordinate plane as described: Q1 = +12.0 µC located at the origin (0, 0), and Q2 = +8.0 µC located at (3.0m, 0). How much energy would be required to bring a third charge Q3 = +5.0 µC from very far away and place it at location (0, 4.0m)? A) 0.29 J B) 0.21 J C) 0.096 J D) 0.063 J E) 0.048 J


DISTRICT MEET 2 • 2016

1. D

2. D

3. A

4. C

5. B

6. C

7. E

8. D

9. E

10. B

11. D

12. C

13. A

14. D

15. B

16. B

17. C

18. E

19. D

20. A

21. D

22. E

23. E

24. C

25. A

26. C

27. B

28. A

29. C

30. D

31. A

32. D

33. E

34. B

35. B

36. D

37. A

38. B

39. C

40. E

41. D

42. B

43. A

44. B

45. D

46. C

47. A

48. E

49. A

50. A

51. A

52. D

53. D

54. B

55. E

56. C

57. D

58. E

59. C

60. B

PHYSICS KEY for Science Contest • District Meet 2 • 2016

41. (D) page 16: “…the oxygen may arrive with only a little energy, but the oxygen and carbon will snap together with a tremendous vengeance and commotion, and everything near them will pick up energy….we are getting heat from the combination of oxygen and carbon….in certain circumstances it can be so enormous that it generates light. This is how one gets flames.”

42. (B) page 20: “Therefore, if we look at very tiny particles (colloids) in water through an excellent microscope, we see a perpetual jiggling of the particles, which is a result of the bombardment of the atoms. This is called the Brownian motion.”

43. (A) page 26: “At first the phenomena of nature were roughly divided into classes, like heat, electricity, mechanics... …the aim is to see complete nature as different aspects of one set of phenomena. That is the problem in basic theoretical physics today – to find the laws behind experiment; to amalgamate these classes.”

44. (B) pages 29: “The ultimate basis of an interaction between the atoms is electrical.” 45. (D) Our Galaxy has been known for a long time to be a sprial-type galaxy; however, investigations in

recent decades have produced significant evidence that the Milky Way has a bar, or barlike structure, in its central region. Thus the Milky Way now has a classification of barred spiral of type SB.

46. (C) First we need to determine which car wins. To do this, calculate the time for each car to traverse the distance of 400.0m:

Car A: = !! + !!! + 1 2 !!! = 0 + 0 + 1 2 10.5 !! = 400 , this gives !! = 8.73 !"# Car B: = !! + !!! + 1 2 !!! = 0 + 42 ! + 0 = 400 , this gives !! = 9.52 !"# Since Car A has the faster time, we know that Car A wins. So how far does Car B travel in the time

to take Car A to win? ! = !! + !!! + 1 2 !!! = 0 + 42 (8.73) + 0 = 366.6! Thus, Car B is 33.4m from the finish line when Car A wins. 47. (A) This is a matter of calculating torques. Since everything is in equilibrium, we know that the torques

must sum to zero. We will take the pivot point to be the origin (x = 0). Three forces are involved: !! = 0.1!" 9.8 = 0.98 ! !" 0.2! !! !ℎ! !"#ℎ!. (clockwise rotation)

!! = 0.75!" 9.8 = 7.35 ! !" 0.3! !" !ℎ! !"#$. (counterclockwise rotation) !! = (!) 9.8 !" 0.6! !" !ℎ! !"#ℎ!. (clockwise rotation). M is the unknown mass of the rod, with

its action located at the center of the rod – which lies 60cm to the right of the pivot. Summing torques: − 0.98 0.2 + 7.35 0.3 − ! 9.8 0.6 = 0 Solving gives ! = 0.342 !" = 342 !

48. (E) First, find the total resistance of the circuit. The first series group: !! = 120 + 60 = 180 Ω This group is in parallel with another resistance: 1 !! = 1 180 + 1 100, which gives !! = 64.3 Ω Finally, this is in series with the last resistor: !! = 64.3 + 80 = 144.3 Ω Then !! = !! !! = 24/144.3 = 0.166 ! = !!. Note that currents in series are equal to the total

current through the series combination. Now, !! = !!!! = 0.166 64.3 = 10.7 !"#$% = !!. Note that voltages in parallel are equal to the

total voltage across the parallel combination. So, !! = !! !! = 10.7/180 = 0.0594 ! = !!"# Which gives a power dissipated by the 120 Ω resistor as: ! = !!! = (0.0594)! 120 = 0.423 !.


49. (A) Sum the forces on each box: For the 10.00 kg box: !" − !! = !" = 98 − !! = 10! For the 5.00 kg box: !! − !! − ! = !" = 5!. Note ! = !" = !"# for a horizontal surface. This

gives ! = !"# = 0.25 5 9.8 = 12.25 !"#$%&'. For the 4.00 kg box: !! −!" = !" = !! − 39.2 = 4!

This gives three equations with three unknows. Solving for the Tensions and substituting into the middle equation gives: !! = 98 − 10! and !! = 4! + 39.2

Then !! − !! − ! = 5! = 98 − 10! − 4! − 39.2 − 12.25

Solving for acceleration gives: 19! = 46.55, or ! = 2.45 !/!!.

50. (A) The coefficient for area expansion is twice that for linear expansion, so Δ! = 2!!!Δ!

We know that Δ! !! = 1% = 0.01. So, 0.01 = 2(12.0 ×10!!)Δ!

This gives a ∆! = 417 ℃. Therefore, the final temperature is !! = 20 + 417 = 437℃ 51. (A) Use conservation of energy: Initial energies are Kinetic and Gravitational Potential, final energy type

is Elastic Potential. !" + !"# = !"!. Be careful to include the extra height of compression in the GPE calculation.

!!!!! +!" ℎ + ! = !

!!!! = !! 0.09 !! + 0.09 9.8 5.3 + 0.48 = !

!(60)(0.48)! Solving for velocity gives: ! = 6.35 !/!

52. (D) Capacitive reactance is !! = 1 !", which decreases as frequency increases. Therefore, the voltage across the capacitor decreases as frequency increases, so the voltage across the resistor must increase as frequency increases. Also, in AC-RC circuits, the current leads the voltage at all frequencies.

53. (D) The period of a stationary pendulum is given by ! = 2! ! ! = 1.22 = 2! ! 9.8.

This gives ! = 0.369!. Now, for an accelerating pendulum, we have ! = 2! ! (! + !). So, 1.02 = 2! 0.369 (! + 9.8), which gives ! = 4.22 !/!!.

54. (B) The force between two parallel conductors is given by ! = !!!!!!! 2!". In order to levitate the upper wire, then this force must equal the gravitational force on the upper wire.

Thus, !!!!!!! 2!" = !" = (4!×10!!)(30)(75)(0.2) 2! 0.0033 = 0.0273

So, ! = 0.00278!" = 2.78!.

55. (E) The focal length of a mirror is: ! = ! 2. For this case, we also have ! = !. Then, 1 ! + 1 ! = 2 !, leading to ! = !. Also, ! = −! ! = −1.

Since q is positive, the image is real. Since the magnitude of magnification is 1, the image is the same size as the object. Since magnification is negative, the image is inverted.

56. (C) First, find the angle of the first maximum: !"#$ = ! ! = 0.88/1.2 = 0.733. So, ! = 36.25° The grating spacing is given by ! = 1 800 = 0.00125!! = 1.25×10!!!.

The wavelength is ! = !"#$% = 7.39×10!!!. And the frequency is then ! = ! ! = (3.00×10!)/(7.39×10!!) = 4.06×10!" !" = 406 !"#.


57. (D) Activity ! = !!!!!". Here ! = ℓ!(2)/!!

!= 0.693/140 = 0.00495 !

!"#$%. So, 10×10!! = (5)!!(!.!!"#$)!, which gives ℓ! 2×10!! = − 0.00495 ! Solving for time: ! = 2650 !"#$%. 58. (E) The Weak force is mediated by two charged particles, the W+ and the W-, and by one neutral particle,

the Z. Gluons mediate the strong force, pions mediate the nuclear force, and photons medaite the electromagnetic force.

59. (C) Conservation of momentum: !!""!!!"" = !!""!!"## +!!""!!"". Plugging in: (0.15)(35) = (0.15)(28) +!!""(11), and solving gives !!"" = 0.0955!" = 95.5!. 60. (B) When the charge is located “very far away” it has an electric potential energy of zero. Once it is

brought near to the other charges, it has an electrical potential energy given by: ! = !!!!! !!" + !!!!! !!". Note !!" = 4.0! and !!" = 3.0! + 4.0! = 5.0!

! = (8.988 ×10!)(12 ×10!!)(5×10!!)/(4) + (8.988 ×10!)(8 ×10!!)(5×10!!)/(5) ! = 0.1348 + 0.0719 = 0.21 ! which is the energy needed to bring the charge near to the other

charges.


$ OVERALL$SCORE:!$!



!! !! !

!1.!!!!________!

2.!!!!________!

3.!!!!________!

4.!!!!________!

5.!!!!________!

6.!!!!________!

7.!!!!________!

8.!!!!________!

9.!!!!________!

10.!!________!

11.!!________!

12.!!________!

13.!!________!

14.!!________!

15.!!________!

16.!!________!

17.!!________!

18.!!________!

19.!!________!

20.!!________!

!

BIOLOGY$SCORE!

!

21.!!________!

22.!!________!

23.!!________!

24.!!________!

25.!!________!

26.!!________!

27.!!________!

28.!!________!

29.!!________!

30.!!________!

31.!!________!

32.!!________!

33.!!________!

34.!!________!

35.!!________!

36.!!________!

37.!!________!

38.!!________!

39.!!________!

40.!!________!

!

CHEMISTRY$SCORE$

!

41.!!________!

42.!!________!

43.!!________!

44.!!________!

45.!!________!

46.!!________!

47.!!________!

48.!!________!

49.!!________!

50.!!________!

51.!!________!

52.!!________!

53.!!________!

54.!!________!

55.!!________!

56.!!________!

57.!!________!

58.!!________!

59.!!________!

60.!!________!

$$PHYSICS$SCORE$$$

GENERAL DIRECTIONS:













answers.








SCORING:



SCIENCERegion • 2016




1A1

8A18

1H1.008

2A2

3A13

4A14

5A15

6A16

7A17

2He4.003

3Li6.941

4Be9.012

5B10.81

6C12.01

7N14.01

8O16.00

9F

19.00

10Ne20.18

11Na22.99

12Mg24.31

3B3

4B4

5B5

6B6

7B7

8B8

8B9

8B10

1B11

2B12

13Al26.98

14Si28.09

15P30.97

16S32.07

17Cl35.45

18Ar39.95

19K39.10

20Ca40.08

21Sc44.96

22Ti47.87

23V50.94

24Cr52.00

25Mn54.94

26Fe55.85

27Co58.93

28Ni58.69

29Cu63.55

30Zn65.41

31Ga69.72

32Ge72.64

33As74.92

34Se78.96

35Br79.90

36Kr83.80

37Rb85.47

38Sr87.62

39Y88.91

40Zr91.22

41Nb92.91

42Mo95.94

43Tc(98)

44Ru101.07

45Rh102.91

46Pd106.42

47Ag107.87

48Cd112.41

49In114.82

50Sn118.71

51Sb121.76

52Te127.60

53I

126.90

54Xe131.29

55Cs132.91

56Ba137.33

57La138.91

72Hf178.49

73Ta180.95

74W183.84

75Re186.21

76Os190.23

77Ir

192.22

78Pt195.08

79Au196.97

80Hg200.59

81Tl

204.38

82Pb207.20

83Bi208.98

84Po(209)

85At(210)

86Rn(222)

87Fr(223)

88Ra(226)

89Ac(227)

104Rf(261)

105Db(262)

106Sg(266)

107Bh(264)

108Hs(277)

109Mt(268)

110Ds(281)

111Rg(272)

112Cn(285)

58Ce 140.12

59Pr 140.91

60Nd 144.24

61Pm (145)

62Sm 150.36

63Eu 151.96

64Gd 157.25

65Tb 158.93

66Dy 162.50

67Ho 164.93

68Er 167.26

69Tm 168.93

70Yb 173.04

71Lu 174.97

90Th 232.04

91Pa 231.04

92U

238.03

93Np (237)

94Pu (244)

95Am (243)

96Cm (247)

97Bk (247)

98Cf (251)

99Es (252)

100Fm (257)

101Md (258)

102No (259)

103Lr (262)

property symbol valuedensity of water density of ice

ρwaterρice

1.000 g cm-3 0.9167 g cm-3


cicecwatercsteam

2.09 J g-1 K-1 4.184 J g-1 K-1 2.03 J g-1 K-1






Pressure

1 atm = 760 torr= 101325 Pa= 14.7 psi

1 bar = 105 Pa= 100 kPa

Energy

1 cal = 4.184 J

1 L atm = 101.325 J

1 Cal = 1 kcal

1 hp = 746 W

1 eV = 1.602 × 10-19 J



R 8.314 J mol-1 K-1


0.08206 L atm mol-1 K-1

1.987 cal mol-1 K-1

Planck’s constant

h 6.626 × 10-34 J s

4.136 × 10-15 eV s


h/2π 1.054 × 10-34 J s 6.582 × 10-16 eV s







Avogadro’s number

NA 6.022 × 1023 mol-1






1 in = 2.54 cm

1 lb = 453.6 g

1 mi = 5280 ft = 1.609 km

1 gal = 4 quarts = 231 in3 = 3.785 L


electron rest mass

me 9.11 × 10-31 kg 0.000549 u 0.511 MeV c-2



atomic mass unit

u 1.6605 × 10-27 kg




sun'mass 1.989 × 1030 kg

distance'earth0moon

3.844 × 108 m

distance'earth0sun

1.496 × 1011 m




O–H 463 C=C 602 C–Br 285 N=N 418

N–H 391 C≡C 835 O=O 498 H–H 436

C–O 358 C=O 799 C≡O 1072 Br–Br 193

H–Cl 432 S–H 347 N≡N 945 Cl–Cl 242

H–Br 366 H–I 299 C≡N 887 I–I 151

HS Science • Regional • 2016


Biology Questions (1 – 20) 1. In pure water at 25ºC, the ratio of H+ ions to OH–

ions is A) equal B) 0.0 M C) 7.0 M D) 107 M

2. Humans have exploited yeast in making bread for thousands of years. What trait do yeast cells have that has proven most useful to bakers? A) Yeast releases O2, which makes the bread rise. B) Bread mold is very common and easy to use. C) The baker’s yeast organisms are facultative

anaerobes. D) Yeast release CO2 during alcohol

fermentation. E) Yeast are single-celled eukaryotes that can

make spores.

3. Human cells are surrounded by A) interstitial fluid B) circulatory fluid C) lymph fluid D) hemolymph E) plasma

4. The function of this feature necessitates a gas exchange system, with intake of O2 and release of CO2. A) the lungs B) the chloroplasts C) the gills D) the mitochondria E) the trachea

5. If there has been no recent rain and the ground is dry, a plant might close its stomata to preserve water. The greatest drawback of this strategy is that _____. A) the plant could not take up CO2 B) water would build up in the plant, creating too

much pressure in the xylem C) CO2 could reach toxic concentrations in the leaf D) turgor pressure would decrease E) the roots could not take up water

6. Gene expression is A) the process through which an allele is passed on

to the next generation through the gamete. B) the process through which the presence of a

particular allele is expressed in the phenotype. C) the process in which a sequence of DNA directs

the synthesis of a protein. D) the process of creating RNA nucleotides from a

DNA template using base-pairing rules.

7. “Gene sequencing” refers to A) plotting the location of particular genes on a

particular chromosome. B) determining the exact sequence of the four

nucleotides that make up a particular gene. C) determining the sequence of amino acids that

make up the primary structure of a protein. D) determining the sequence of when adaptive traits

evolved in a particular lineage.

8. A difference between a benign tumor cells and malignant (cancerous) tumor cells is that A) a benign tumor cell is specialized and constrained

to its tissue type, whereas a malignant tumor cell has dedifferentiated and can grow in new tissues.

B) benign tumor cells exhibit density-dependent inhibition and thus grow in a layer, whereas malignant tumor cells grow out in all directions.

C) benign tumor cells can divide indefinitely, whereas malignant tumor cells will slow and stop after about 50 divisions.

D) benign tumors grow in a location that is not threatening to health, whereas a malignant tumor is one that creates a harmful mass of tissue due to its particular location on the body.



9. In a cross between an individual that is heterozygous at three loci (AaBbCc) with an individual that is homozygous recessive at the same loci (aabbcc), how many different genotypes would you expect to find in the offspring due to independent assortment? A) Two B) Four C) Six D) Eight E) Nine

10. A potato farmer grows red potatoes that are either

round or fingerling (long), with no intermediate shapes. The red phenotype is due to a homozygous genotype (ww). The shape is controlled by another gene, where the R_ genotype is round, and rr genotype is fingerling. With regard to color and shape, how many genotypes are represented in this farmer’s crop? A) Two B) Three C) Four D) Six

11. The farmer from the previous question is going to

add some variety to his crop. He breeds red fingerling potatoes with white (WW) fingerling potatoes. All of the F1 offspring are pink. What percent of the F2 generation can we predict will have the pink fingerling phenotype? A) None will be pink fingerling. B) 25% will be pink fingerling. C) 50% will be pink fingerling. D) 75% will be pink fingerling. E) All the offspring will be pink fingerling.

12. What can you infer from the graph below? A) If you remove an individual penstemon growing

at 7000 ft. and replant it at a 2000-f.t location, it will bloom earlier.

B) If you remove an individual penstemon growing at 7000 ft. and replant it at a 2000-ft. location, it will change allele frequency.

C) The population of P. cochinea is in Hardy-Weinberg equilibrium.

D) The FTa allele is more prevalent at lower altitudes.

E) For P. cochinea there is a graded change in flowering time along a geographic axis.

FloweringtimeinPenstemoncochinea

13. A massive die-off of trout happened in a lake. The

biologists studying the dead fish believe that the die-off was caused by infection from a species of bacterium that is new to science. Which species concept would they not be using, as they gather evidence to support or deny their hypothesis that this is a new species of bacteria? A) The morphological species concept. B) The biological species concept. C) The ecological species concept. D) The phylogenetic species concept. E) None of the above species concepts would be

appropriate.



14. Which of the following scenarios is an example of gene flow? A) During the spring, some male robins happen to

find a mate, and others do not. B) A species of desert bee collects pollen only

from one species of desert daisy. The bee often travels three miles from daisy population A, on the east side of a canyon, to daisy population B, on the west side of the canyon.

C) In a population of penguins, half have tufts over their eyes, and the other half do not. An avalanche moves a group of penguins 50 miles away, where they start a new colony. The new colony members comprise 13 tufted penguins and 4 non-tufted penguins.

D) A hydra is able to create offspring by mitosis, in a process called “budding”. The offspring grows as an extension of the parent, and then is released into the water as an independent individual.

E) For the gene “A”, an individual with a genotype of “aa” has a selective advantage over those with a genotype of “AA”, and thus more of the next generation carries the “a” allele.

15. Two species of beetle used to be separated by a

river, but the river has gone dry. The beetles look very similar to one another (to the scientists, at least!). However, the males of species 1 conduct an elaborate courtship ritual, whereas the males of species 2 do not. After close observation, it is clear that females of species 1 ignore the advances of males of species 2. Should gene flow eventually dissolve the differences between these two populations? A) Yes, because the species look the same and are

sympatric, so they are clearly the same species.

B) Yes, because there are no post-zygotic barriers, so fertile offspring will be created.

C) No, because they exhibit behavioral isolation. D) No, because they exhibit gametic isolation.

16. You have crossed an individual that is heterozygous at three loci (AaBbCc) with an individual that is homozygous recessive at the same loci (aabbcc). Most of the 1600 offspring had the following genotypes, with numbers of offspring for each also shown:

Genotype # of offspring AaBbCc 344 aaBbCc 354 Aabbcc 353 aabbcc 349 Which of the following statements is the best explanation

for your results?

A) Genes “A”, “B”, and “C” must be on the same chromosome.

B) The genotypes shown have a higher reproductive fitness.

C) All three genes are inherited in a simple dominance pattern.

D) Two hundred of the offspring died. E) Genes “B” and “C” must be on the same

chromosome.

17. Which of the following illustrates the principal of maintaining homeostasis? A) After a large meal of pasta, your liver converts

glucose into glycogen. B) If your body temperature is reduced due to cold

surroundings, blood vessels in the skin dilate, allowing more blood to flow to the extremities.

C) On a very warm day, blood traveling through a rabbit’s ears may pick up heat from the environment and dissipate it to the internal organs.

D) A fish living in the ocean constantly loses body water due to osmosis.

E) the suckling motion of a baby mammal causes milk to be released, thus causing the baby mammal to suckle more.



18. Which taxon is now considered a non-monophyletic group? A) Animalia B) Archaea C) Protista D) Eukarya E) Bacteria

19. A growth factor is released from a cell and diffuses to a neighboring cell, which has the proper receptors to recognize and respond to the signal. This is an example of A) endocrine signaling B) autocrine signaling C) hormone signaling D) ubiquitin signaling E) paracrine signaling

20. In most animal cells, a newly fertilized egg will

undergo a rapid succession of cell divisions without growth, this is called _________. A) Meiosis B) Cleavage C) Organogenesis D) Morphogenesis E) Embryogenesis

Chemistry Questions (21 – 40) 21. What is the name of LiNO2·H2O ? A) lithium nitroxide aqueous B) nitrolithiate hydronium C) lithium nitrite monohydrate D) lithium nitride dioxide aqueous E) lithium nitrate hydrate 22. What is the mass of 4.55×1022 molecules

of CCl2F2 ? A) 13.6 g D) 9.14 g B) 1600 g E) 19.2 g C) 121 g

23. An ore has the formula FexCryO4. Analysis showed that the mass percent of oxygen is 28.59. Determine the values for x and y. What is the correct formula for this ore?

A) FeCrO4 D) Fe2Cr3O4 B) Fe2CrO4 E) FeCr2O4 C) Fe3Cr2O4 24. Balance the following chemical equation.

Co(NO3)3 + (NH4)2S ⟶ Co2S3 + NH4NO3 What are the coefficients for this reaction (in order as shown)?

A) 2 : 3 : 1 : 6 B) 2 : 1 : 1 : 4 C) 3 : 2 : 1 : 6 D) 1 : 2 : 1 : 3 E) 1 : 1 : 1 : 1 25. Balance the following precipitation reaction.

Na3PO4(aq) + CuCl2(aq) ⟶ Cu3(PO4)2(s) + NaCl(aq) What is the correct stoichiometric volume of 0.125 M CuCl2 needed to completely react with 120 mL of 0.150 M Na3PO4 ?

A) 144 mL D) 80 mL B) 216 mL E) 240 mL C) 184 mL 26. A gas mixture of oxygen, nitrogen, and helium is

made and the overall pressure is 1150 torr. There are twice as many moles of oxygen as there are nitrogen. The partial pressure of oxygen is 540 torr. What is the partial pressure of the helium?

A) 70 torr D) 340 torr B) 220 torr E) 180 torr C) 610 torr



27. H2 gas will partially decompose to monatomic hydrogen (H) at high temperatures. If exactly 0.12 moles of H2 gas is put into a 15 L container and the temperature is brought to 2800 K, the pressure is found to be 2.14 atm. What is the partial pressure of the H(g) in this container at this high temperature?

A) 1.07 atm D) 1.89 atm B) 2.14 atm E) 0.306 atm C) 0.526 atm 28. Rank the following elements according to their

electronegativities. Rank from highest down to lowest. sodium, phosphorus, nitrogen, aluminum.

A) P > Al > Na > N B) N > Na > Al > P C) N > Al > P > Na D) N > P > Al > Na E) P > Al > N > Na 29. Consider the following phase diagram for a

substance. What is the normal boiling point of this substance?

A) 180 K C) 370 K E) 220 K B) 300 K D) 170 K 30. Which statement best describes the relative

solubilities of CCl4 and CH2Cl2 in water? A) CH2Cl2 is more soluble in water than CCl4. B) CCl4 and CH2Cl2 virtually the same

solubilities in water. C) CCl4 is more soluble in water than CH2Cl2 D) No conclusion can be made about their

solubilities without more information.

31. Which of the following is the best (true) statement about endothermic reactions?

A) Each of the bonds in the reaction lengthen during the reaction.

B) Double bonds tend to convert to single bonds. C) The bonds that are broken tend to be stronger

bonds than those that are made. D) The bond energies of the products are

approximately equal to the bond energies of the reactants.

32. A sample of 0.25 moles of a soluble metal salt, MyXz

is dissolved into 400 g of water. The freezing point of this salt solution is found to be 268.50 K. What is the formula of the salt? Kf for water = 1.86°C/m.

A) MX2 D) MX B) MX3 E) MX5 C) M2X3 33. The following schematic represents a gas phase

reaction reaching equilibrium. What is the value for K for this reaction? A) 5 C) 3 E) 9 B) 2 D) 18 34. The following reaction has an equilibrium constant,

K, equal to 2.5×105 at 350 K. CO(g) + 2 H2 (g) ⇌ CH3OH (g)

What is the value for the equilibrium constant for the following reaction.

½ CH3OH (g) ⇌ ½ CO(g) + H2 (g) A) 1.25×105 D) 5.0×102 B) 2.0×10-6 E) 8.0×10-6 C) 2.0×10-3

100

10

1

0.1

0.01150

temperature (K)

pres

sure

(atm

)

250 350



35. The weak acid, HX, has a Ka value of 2.5×10-5. What is the pH of a 0.036 M NaX (sodium salt of HX) solution.

A) 9.40 D) 8.58 B) 10.98 E) 7.92 C) 12.56 36. A 0.005 M solution is made from each of the

following substances. Which solution will have the highest pH?

A) Sr(OH)2 D) NaOH B) NH3 E) N2H4 C) CH3COOH 37. The reaction AB(g) ⟶ A(g) + B(g) has a rate

constant of 0.0050 atm-1·min-1 at 50°C. AB is put in a reaction chamber at a pressure of 4.00 atm. How long will it take for the reaction to reach 75% completion?

A) 167 min D) 187 min B) 58 min E) 277 min C) 150 min 38. How many grams of nickel are deposited from a

Ni2+ solution at the cathode when 2.5 A of current is passed for 3.5 hours?

A) 9.58 g D) 19.2 g B) 6.41 g E) 5.84 g C) 12.7 g 39. The following reaction at 25°C

2 Ca(s) + O2(g) 2 CaO(s) has ∆Hrxn°= –1270 kJ, and ∆Srxn°= –365 J/K.

What is the value of ∆Grxn° for this reaction? A) +108 kJ D) –1379 kJ B) –1161 kJ E) +1075 kJ C) –1260 kJ

40. 100 mL of 0.050 M HNO3 was added to 500 mL of 0.040 M methylamine solution. What is the equilibrium pH of this solution? Kb for methylamine is 4.4×10-4.

A) 10.64 D) 10.04 B) 11.24 E) 11.12 C) 10.17

Physics Questions (41 – 60)

41. According to Feynman, which physical principle was

first demonstrated by Mayer in connection with heat flow in and out of a living creature? A) The second law of thermodynamics. B) The law of conservation of energy. C) The law of conservation of momentum. D) The principle of statistical mechanics. E) The principle of thermal equilibrium.

42. According to Feynman, an old physical problem that is common to many of the sciences, and that has not been solved, is the problem of… A) finding the most fundamental particles. B) predicting how things change with time. C) analyzing circulating or turbulent fluids. D) dealing mathematically with large numbers of

atoms. E) synthesizing complex molecules in the

laboratory.

43. According to Feynman, we can use the principle of virtual work, the argument that we imagine a structure moves a little even though it is not really moving or moveable, to deduce which laws? A) the laws of balance.

B) the laws of thermodynamics.

C) the laws of momentum conservation.

D) the laws of friction.

E) the laws of potential energy.



44. According to Feynman, in quantum mechanics, the proposition that experimental results are the same no matter where you do the experiment is associated with which conservation law? A) conservation of energy. B) conservation of leptons. C) conservation of charge. D) conservation of angular momentum. E) conservation of linear momentum.

45. A planetary nebula, a small emission nebula

consisting of an expanding glowing shell of ionized gas, is formed when…

A) a dying star explodes in a supernova. B) a protostar, a star being born, initiates nuclear

fusion in its core. C) layers of gas are ejected from an old red giant

star late in its life. D) a pulsar ionizes a nearby gas cloud. E) a white dwarf in a binary system absorbs too

much gas from its partner star and explodes. 46. A wooden block with a mass of 450.0 g is hanging

at the end of a string that is 1.50m long. A 60.0 g dart, flying horizontally, strikes the wooden block and sticks into it. The block, with the dart stuck in it, swings up like a pendulum until the string makes an angle of 26.0o with respect to the vertical. What was the original velocity of the dart before it struck the block? A) 12.9 m/s B) 14.7 m/s C) 25.3 m/s D) 30.5 m/s E) 43.7 m/s

47. A box with a mass of 8.00 kg hangs from a string that is attached to a box of mass 4.00 kg sitting on an inclined plane. The inclined plane is angled at 22o relative to horizontal, and the coefficient of friction between the incline and the 4.00 kg mass is 0.17. Assuming that the pulley is massless and frictionless, what is the acceleration of the 4.00 kg mass? A) 6.53 m/s2 B) 5.98 m/s2 C) 5.31 m/s2 D) 4.79 m/s2 E) 3.30 m/s2 4.00 kg µ = 0.17 8.00 kg 22o

48. A resistor network consists of four resistors and two batteries as shown. What is the power dissipated in the 40.0 Ω resistor?

50.0 Ω 12.6 V

60.0 Ω 40.0 Ω 7.80 V 30.0 Ω

A) 0.0233 W B) 0.0858 W C) 0.188 W D) 0.294 W E) 0.483 W



49. A horizontal solid circular disk has a mass of 300.0g and a radius of 25.0 cm. It is accelerated from rest by a torque of 0.17 Nm, resulting in it spinning about its center. A small beetle sits on the disk at a location 15.0cm from the center. If the coefficient of friction between the beetle and the disk is 0.50, how much time will pass from when the disk begins to spin until the beetle slides off of the disk? A) 0.24 sec. B) 0.32 sec. C) 0.39 sec. D) 0.48 sec. E) 0.69 sec.

50. A golf ball is driven from ground level with an initial velocity directed at an angle of 34.0º above the horizontal. The golf ball strikes a tree located 120.0m horizontally from the tee location. Examination of the tree shows that the ball hit the tree at a point 23.7m above the ground. What was the initial velocity of the ball? A) 35.6 m/s B) 38.3 m/s C) 42.4 m/s D) 65.8 m/s E) 102 m/s

51. What is the magnitude of the net electric field at

the origin produced by these two charges: +16.0 µC charge placed at (-2.00m, 3.00m) and +12.0 µC charge placed at (0.00m, -3.00m). A) 6750 N/C B) 8930 N/C C) 16,300 N/C D) 23,100 N/C E) 30,700 N/C

52. You have 2.00 moles of a diatomic ideal gas at a temperature of 80.0 oC occupying a volume of 15.0 L. This sample of gas is adiabatically compressed until it reaches a pressure of 6.00 x 105 Pa. What is the final temperature of the gas? (Note γ = 1.4 for a diatomic ideal gas). A) 80 oC B) 126 oC C) 268 oC D) 399 oC E) 541 oC

53. A mass attached to a spring is oscillating horizontally on a frictionless surface with an amplitude of 12.0 cm. You observe that the mass is moving with a speed of 34.0 cm/s when it is 8.00 cm from the equilibrium position. What is the period of oscillation of the mass? A) 4.82 sec B) 2.66 sec C) 2.15 sec D) 1.65 sec E) 0.605 sec

54. The mass of an atom of Nickel-62 is 61.928345 amu. What is the binding energy per nucleon of Nickel-62? A) 8.441 MeV B) 8.565 MeV C) 8.671 MeV D) 8.721 MeV E) 8.796 MeV

55. The lifetime of a negative pion (π-) at rest is 2.6 x 10-8 seconds. If a pion is travelling at a speed of 0.98c (98% the speed of light), how far does it appear to travel -- from the perspective of a stationary observer -- before it decays? A) 7.6 m B) 38 m C) 54 m D) 190 m E) 380 m



56. An AC circuit consists of a 100.0 Ω resistor in series with a 300.0 mH inductor. These are connected to an AC voltage source producing 24.0 Vrms at a frequency of 60.0 Hz. What is the phase angle between the voltage and the current in this circuit? A) the current and voltage are in phase, 0.0o B) the current leads the voltage by 90.0o

C) the voltage leads the current by 18.4o D) the current leads the voltage by 41.5o E) the voltage leads the current by 48.5o

57. A magnetic field of 10.0 Gauss is directed around a rectangular area as shown. The top region of the field is directed horizontally to the left and the bottom region of the field is directed horizontally to the right. The rectangular area is 5.00 cm by 2.00 cm in size. Given this magnetic field, what must be the current flowing through the rectangular area?

B = 10.0 Gauss 2.00 cm 5.00 cm

A) 15.9 A directed out of the page. B) 31.8 A directed into the page. C) 39.8 A directed into the page. D) 79.6 A directed out of the page. E) 112 A directed out of the page.

58. A proton and an electron collide at high energy in a particle accelerator. The collision produces a muon, a pion, two neutrinos, and an unknown neutral particle, X0. The unknown particle then decays into two pions, a positron and a neutrino. What conservation law appears to be violated by this interaction/decay?

P+ + e- µ- + π+ + !! + !! + X0

X0 π0 + π- + e+ + !!

A) conservation of baryon number. B) conservation of lepton number. C) conservation of charge. D) conservation of linear momentum. E) conservation of energy.

59. A point charge of +15.0 nC is placed at the exact center of a conductive spherical shell. The shell has an inner radius of 4.00 cm and an outer radius of 7.00 cm, and the shell has a net charge of +5.00 nC. What is the magnitude of the electric field at a distance of 6.00 cm from the point charge? A) E = 0.00 N/C B) E = 12,500 N/C

C) E = 37,500 N/C

D) E = 50,000 N/C

E) E = 84,400 N/C

60. A lens with a focal length of +12.0 cm is located 8.00 cm to the left of a convex mirror that has a radius of curvature of 30.0 cm. An object is placed 20.0 cm to the left of the lens. What is the distance from the lens to the final image? A) 55.1 cm B) 47.1 cm C) 41.3 cm D) 36.0 cm E) 15.3 cm


REGIONAL • 2016

1. A

2. D

3. A

4. D

5. A

6. C

7. B

8. A

9. D

10. B

11. C

12. E

13. B

14. B

15. C

16. E

17. A

18. C

19. E

20. B

21. C

22. D

23. E

24. A

25. B

26. D

27. E

28. D

29. B

30. A

31. C

32. B

33. E

34. C

35. D

36. A

37. C

38. A

39. B

40. E

41. B

42. C

43. A

44. E

45. C

46. B

47. D

48. D

49. B

50. C

51. A

52. B

53. D

54. E

55. B

56. E

57. D

58. A

59. A

60. E

PHYSICS KEY for Science Contest • Meet • 2015

41. (B) page 50: “…in which biology helped physics in the discovery of the conservation of energy, which was first demonstrated by Mayer in connection with the amount of heat taken in and given out by a living creature.”

42. (C) pages 65-66: “Finally, there is a physical problem that is common to many fields, that is very old, and that has not been solved…. Nobody in physics has really been able to analyze it mathematically satisfactorily in spite of its importance to the sister sciences. It is the analysis of circulating or turbulent fluids.”

43. (A) pages 79-80: “In this way we can work out the laws of ‘balance’…. This approach is called the principle of virtual work, because in order to apply this argument we had to imagine that the structure moves a little – even though it is not really moving or even moveable.”

44. (E) page 84: “The conservation of momentum is associated in quantum mechanics with the proposition that it makes no difference where you do the experiment, the results will always be the same.”

45. (C) Planetary nebulae are formed when a red giant star ejects layers of gas as it dies. The expanding layers of gas are ionized by the ultraviolet radiation produced by the dying star’s core remnant. Planetary nebulae glow with a variety of colors, and appear in a myriad of different shapes.

46. (B) This is essentially a ballistic pendulum. First, we need to find how to high the block and dart swing: ℎ = ! 1 − !"#$ = 1.50 1 − cos 26 = 0.152 ! Now use energy conservation to find the velocity of the block and dart right after the dart stuck into

the block: !"ℎ = !!!!! or solving for velocity: ! = 2!ℎ = 2 9.8 (0.152) = 1.72 !/!

Now we can use conservation of momentum to find the original velocity of the dart: !!!! = !!!!!!!! = (0.45 !" + 0.06 !")(1.72 !/!) = (0.06 !")!! Which gives !! = 14.7 !/! 47. (D) The forces acting on the 8.00kg box are tension and gravity, giving the equation: (8)! − ! = (8)! The forces acting on the 4.00kg box are tension, gravity, friction, and the normal force. If we take

“up the incline” to be the positive x-axis, and if we break gravity into components, then the forces in the x-direction give: ! − ! − 4 !"#$% = 4 !. The forces on the 4.00kg box that are in the y-direction, perpendicular to the incline, sum up to zero since there is no motion in that direction:

! − 4 !"#$% = 0. This gives the normal force: ! = 4 9.8 cos 22 = 36.35 !"#$%&'. Then we can get the frictional force: ! = !" = 0.17 36.35 = 6.179 !"#$%&'.

Then we get ! − 6.179 − 4 9.8 sin 22 = 4 ! = ! − 20.86 = 4!, or ! = 4! + 20.86 Substituting this into the equation for the 8.00kg box: (8)(9.8) − (4! + 20.86) = (8)! Then 78.4 − 20.86 = 12!; or ! = 4.79 !/!!. 48. (D) This network requires the use of Kirchhoff’s rules. I do the two smaller loops and the upper node.

Define currents as I1 flowing up through the middle branch, I2 flowing to the left out of the upper node, and I3 flowing to the right out of the upper node. Then the node rule gives: !! = !! + !!

Going around the left loop couterclockwise gives: −50!! + 7.8 − 60!! = 0 Going around the right loop clockwise gives: 12.6 − 40!! − 30!! − 60!! = 0 Eliminating I1 in these equations and rearranging gives these two equations: 110!! + 60!! = 7.8 and 60!! + 130!! = 12.6 Multiplying the second equation by -110/60, gives: −110!! − 238.33!! = −23.1. Adding this

equation to the first equation to do variable elimination gives: −178.33!! = −15.3. Giving: I3 = 0.08579 A. Then the power dissipated in the 40 Ω resistor is ! = !!!! = 0.08579 ! 40 = 0.294 !


49. (B) First we need to find the angular acceleration of the disk. This requires that we have the rotational inertia of the disk. For a solid disk: ! = !

!!!! = 0.5 0.3!" 0.25! ! = 0.009375 !"!! Now, ! = !

! = (0.17)/(0.009375) = 18.133 rad/s2 Switching to the beetle, the forces acting on the beetle are: x-direction: friction; and y-direction:

gravity and normal. Using the y-direction forces: ! −!" = 0, so ! = !" Then we can get the maximum frictional force: ! = !" = !"# Since friction is the force keeping the beetle moving in a circle, then it must provide the centripetal

force: !"# = !"!!. This gives a maximum angular velocity of: ! = !" ! = (0.5)(9.8) 0.15 = 5.715 !"#/!

Finally, ! = !" = 5.715 = 18.133!. Solving for time gives: t = 0.32 seconds. 50. (C) First, we have expressions for the components of the initial velocity: !!! = !!cosθ and !!! = !!sinθ. Using the horizontal distance travelled: ! = !!!! = 120 = !! cos 34 !, giving !!! = 144.75 Now consider the vertical distance travelled: ! = !!!! − !

!!!! = 23.7. This gives: !! sin 34 ! − !

! 9.8 !! = 23.7 = 0.5592 !!! − 4.9!!. Combining this with the previously derived expression for !!!: 23.7 = 0.5592 144.75 − 4.9!!. Giving: 57.24 = 4.9!!, or ! = 3.42 !"#$%&!. Then we get !! = !"".!"

!.!" = 42.4!/!. 51. (A) The electric field magnitude from the first charge is: !! = !!! !!! = (9.00×10!)(16.0×10!!) (2! + 3!) = 11077 !/!.

This has components based on its location relative to the origin. Keep in mind that the vector direction is directed from the charge towards the origin, so the signs are flipped.

!!! = !!(2)/ 13 = 6144.4 !/! !!! = !!(−3)/ 13 = −9216.6 !/!

Now, the electric field magnitude for the second charge is: !! = !!! !!! = (9.00×10!)(12.0×10!!) ( 3!) = 12000 !/!

Since this charge is located on the y-axis, its vector direction is entirely in the y-direction. In other words:

!!! = 0 !/! !!! = !! = 12000 !/!

Summing the components gives the total electric field at the origin: !! = 6144.4 !/!

!! = −9216.6 + 12000 = 2783.4 !/! And then the magnitude is found using the Pythagorean theorem: ! = (6144.4)! + (2783.4)! = 6750 !/!.


52. (B) First, let’s get temperature into Kelvin and use the ideal gas law to get the initial pressure: ! = 80 + 273 = 353 ! and !" = !"# = ! 0.015 !! = (2)(8.314)(353) Which gives !! = 391,000 !" For an adiabatic compression, we know !!!!

! = !!!!!

So, 391,000 0.015 !.! = (600,000)!!!.!, which gives a final volume of !! = 0.011 !! Using the ideal gas law again to get the final temperature: !! = !!!! !" = (600,000)(0.011)/ 2 8.314 = 399 ! = 126 ℃ 53. (D) First, find the total energy of the system in terms of k: ! = !

!!!! = 0.5 ! 0.12 ! = 0.0072! Now, we know that the total energy must be the same at the other position: ! = 0.0072! = !

!!!! + !!!!! = 0.5 ! 0.34 ! + 0.5 ! 0.08 ! = 0.0578! + 0.0032!

This gives a relationship between k and m: ! = 0.0072 − 0.0032 !/0.0578 = 0.0692! The period of oscialltion of a mass-spring system is ! = 2! !/! = 2! 0.0692 = 1.65 !"#. 54. (E) It is important to carry a lot of significant figures on this problem. The atomic number of Nickel is

28, so Ni-62 has 28 protons, 34 neutrons and 28 electrons. The mass defect is given by: ∆! = 28!! + 34!! + 28!! −!!" =

28 1.00728 + 34 1.008665 + 28 0.000549 − 61.928345 = 0.585477 !"# Converting to energy and dividing by the number of nucleons (62) gives the binding energy per

nucleon: !! = (0.585477!"#)(931.5 !"#/!"#)/(62 !"#$%&!') = 8.796 !"# 55. (B) A particle travelling so close to the speed of light will be subject to time dilation as seen by an outside

observer. Therefore, it will have a lifetime that is longer than its rest lifetime. ! = !!/ 1 − !! !! = (2.6×10!!)/ 1 − 0.98! ! !! = 1.3×10!!!"# Using that time and the particle’s speed gives the distance it will travel: ∆! = !" = 0.98 3×10! 1.3×10!! = 38 ! 56. (E) In an AC circuit, the inductive reactance is given by: !! = !" = 2!"# = 2! 60 0.3 = 113 Ω Since there is only an inductor and a resistor, the total reactance equals the inductive reactance. Then

the phase angle is given by: ! = !"#!!(!/!) = !"#!!(113/100) = 48.5° In inductive circuits, the current always lags behind the voltage, so the voltage leads the current.


57. (D) This requires the use of Ampere’s Law and the right hand rule. First, to get the magnitude of the

current, use Ampere’s Law. Normally an integral, we can reduce Ampere’s Law to a sum of terms, one for each side of the rectangle. This is possible since the magnetic field is constant and uniform.

! ∙ ! = !!! = ! ∙ !!"# + ! ∙ !!"#$ + ! ∙ !!"##"$ + ! ∙ !!"#!! These are dot products, so anytime B and s are perpendicular, the result of the dot product is zero.

Notice that B is perpendicular to the rectangle on both the left and right sides. So those two terms will be zero. On the top and bottom, B and s are parallel, so the dot product can be reduced to a scalar product.

! ∙ ! = !!! = !!!"# + !!!"##"$ = 10 !"#$$ 5.0 !" + (10 !"#$$)(5.0 !") !!! = 2 . 001 !"#$% 0.05 ! = 1.0×10!! !" Which gives: ! = 79.6 ! As for direction, using the right hand rule (curl fingers in the direction of the magnetic field, then the

thumb points in the direction of the current), it can be seen that the current must be directed out of the page to generate a magnetic field oriented this way.

58. (A) Counting the charges, the electrons and electron neutrinos, the muons and the muon neutrinos, and the baryons in the first reaction leads to the conclusion that the X particle must be a baryon (on the left side, the proton is a baryon). However, in its decay process, the X particle does not conserve baryon number, as it decays into only leptons and mesons. Leptons and charges are conserved throughout both the reaction and decay, and neither energy nor momentum can be examined without velocity information. Therefore, the only law clearly violeted is the conservation of baryon number.

59. (A) The point charge is located at the center of a conductive spherical shell that extends from 4.0cm to 7.0cm from the charge. So, 6.0cm from the point charge places us within the spherical shell. Since the shell is conductive, the electric field inside it must be zero (the static electric field inside any conductor is zero). Therefore, the electric field at a distance of 6.0cm from the point charge is zero.

60. (E) This requires three steps: passing through the lens, reflecting from the mirror and passing back through the lens after the reflection. Each part can be worked separately: First, through the lens:

1 20 + 1 !! = 1 12, so the first image is at !! = 30.0!" to the right of the lens. The first image becomes the object for part two: the mirror. First, the object distance: !! = ! − !! = 8.0 − 30.0 = −22.0!" Now to find the image formed by the mirror. Note that convex mirrors have negative focal lengths. 1 (−22) + 1 !! = 2 (−30) So the second image is located at !! = −47.1 !". That is 47.1 cm to the right of the mirror. The second image becomes the object for part three: back through the lens. The object distance is: !! = ! − !! = 8.0 − (−47.1) = 55.1!" So, the image formed after going back through the lens is given by: 1 (55.1) + 1 !! = 1 12, which gives a final image location of !! = 15.3!" from the lens.


$ OVERALL$SCORE:!$!



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SCIENCE - Alvin ISD

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