# Answer Keys. Skill Sheet 1.1: Solving Equations. Skill Sheet 1.2: Galileo Galilei. Skill Sheet 2.1: International System of Units

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2 Skill Sheet 3.1: Speed Problems 1. Sample problem: 28 m/sec km/hr mph seconds hours kilometers meters Sample problem: 63 mph mph miles 4. Answers are: a cm/ inch b. 12 inches/min Skill Sheet 3.2: Making Line Graphs Parts 1 through 2: There are no questions to answer in parts 1 through Graph of money earned vs. hours worked: 2. The amount earned is \$4.50/hour. Data pair not necessarily in order Temp. Hours of heating Reaction time Number of people in family Stream flow Tree age Part 5 answers: Alcohol drunk Part 6 answers: Range Number of lines Cost per week for groceries Independent Hours of heating Alcohol consumed Number of people in family Dependent Temp. Reaction time Cost per week for groceries Rainfall Amount of rainfall Rate of stream flow Average height Tree age Average height Lowest value Highest value Range ,250 1,150 Range No. of lines Calculated scale (per line) Adj. scale (per line)

3 Skill Sheet 3.3A: Analyzing Graphs of Motion Without Numbers Parts 1 through 2: There are no questions to answer in parts 1 through Little Red Riding Hood. Graph Little Red Riding Hood: 3. The Skyrocket. Graph the altitude of the rocket: 2. The Tortoise and the Hare. Use two lines to graph both the tortoise and the hare: Each student story will include elements that are controlled by the graphs and creative elements that facilitate the story. Only the graph-controlled elements are described here. The line begins and ends on the baseline, therefore Tim must start from and return to his house. The line rises toward the first peak as a downward curved line that becomes horizontal. This indicates that Tim's pace toward Caroline's house slowed to a stop. Then the line rises steeply to the first peak. This indicates that after his stop, Tim continues toward Caroline's house faster than before. The first peak is sharp, indicating that Tim did not spend much time at Caroline's house on first arrival. The line then falls briefly, turns to the horizontal, and then rises to a second peak. This indicates that Tim left, paused, and then returned quickly to Caroline's house. The line then remains at the second peak for a long time, then drops steeply to the baseline. This indicates that after spending a long time at Caroline's house, Tim probably ran home. Skill Sheet 3.3B: Analyzing Graphs of Motion With Numbers Parts 1 through 2: There are no questions to answer in Parts 1 through The bicycle trip through hilly country. 3. Up and down the supermarket aisles. 1. The honey bee among the flowers. 2. A walk in the park. 3

4 2. Rover runs the street. 3. The amoeba. Skill Sheet 4.1: Acceleration Problems m/sec m/sec m/sec m/sec mph/sec km/hr/sec 7. The cheetah. The cheetah s acceleration in km/hr/sec is 37 km/hr/sec, which is 12 km/hr/sec faster than the car m/sec seconds seconds mph Skill Sheet 4.2: Acceleration and Speed-Time Graphs Parts 1 through 2: There are no questions to answer in parts 1 through Acceleration = 5 miles/hour/hour or 5 miles/hour 2 2. Acceleration = -2 meters/minute/minute or -2 meters/minute 2 3. Acceleration = 0 feet/minute/minute or 0 feet/minute 2 or no acceleration Skill Sheet 4.3: Acceleration Due to Gravity Parts 1 through 2: There are no questions to answer in Parts 1 through velocity = 14.7 meters/sec 2. velocity = 11.3 meters/sec 3. velocity = 76.4 meters/sec 4. velocity = 15.8 meters/sec Part 4: There are no questions to answer in Part Answers are: Segment 1: Acceleration = 2 feet/second/second, or 2 feet/second 2 Segment 2: Acceleration = 0.67 feet/second/second, or 0.67 feet/second 2 1. Distance = 1,400 meters 2. Distance = 700 meters 3. Distance = 75 kilometers Part 5 answers: 1. depth = 86.4 meters 2. height = 11.0 meters; yes 3. height = 5.9 meters (The maximum height would have occur in half the total time that the serve was in the air or 1.1 seconds.) Part 6 answers: 1. time = 5.6 seconds 2. time = 7.0 seconds 3. Students use the equations in Part 6 to check their answers in other parts of the skill sheet. Skill Sheet 5.1: Isaac Newton 1. The legend tells of Newton sitting in his garden in Linconshire in 1666, watching an apple fall from a tree. He later noted that In the same year, I began to think of gravity extending to the orb of the moon. However, he did not make public his musings about gravity until the 1680's, when he formulated his universal law of gravitation. 2. Newton claimed that 20 years earlier, he had invented the material that Leibnitz published. Newton accused Leibnitz of plagiarism. Most historians today agree that the two developed the material independently, and therefore they are known as co-discoverers. 4

5 Skill Sheet 5.2: Newton's Second Law Part 1: What do you want to know? acceleration (a) mass (m) Force (F) What do you know? Force (F) and mass (m) acceleration (a) and Force (F) acceleration (a) and mass (m) The formula you will use acceleration = mass = Force mass Force acceleration Force = mass acceleration 1. Sample problem: 6,000 N 2. Sample problem: 10 kilograms m/sec m/sec N 6. 6 kilograms 7. 9,800 N kilograms m/sec 2 Skill Sheet 6.1: Mass and Weight 1. Answers are: a. On Earth: 758 Newtons = 170 pounds (force) b. On the asteroid: = 6.2 Newtons = 1.4 pounds (force) newtons = pounds (force) 3. Answers are: Force of gravity newtons pounds Sun N/kg 18,769 4,218 Mercury 3.7 N/kg Venus 8.9 N/kg Mars 3.7 N/kg Saturn 10.4 N/kg Uranus 8.8 N/kg Neptune 10.7 N/kg Pluto 0.7 N/kg Skill Sheet 6.2: Friction 1. More than 319 newtons 2. More than 73.5 newtons 3. More than 891 newtons 4. The coefficient of static friction = newtons 1. The coefficient of sliding friction is More than 24.5 newtons 3. More than 324 newtons 4. More than 536 newtons 5. Half of the static coefficient of friction. 0.65/2 = 0.32 (Or 0.33 depending on rounding rule) Skill Sheet 6.3: Equilibrium 1. The answer is: N 3. A is 50 N; B is 50 N 4. A is 20 N, B is 20 N, and C is 8 N. 5

6 1. A force applied at right angles to the path of the asteroid will result in a new acceleration that may turn its path enough to avoid striking Earth. Any variation from a right angle force wastes the available energy by simply making an insignificant change in the asteroid's velocity. supplied by the elastic membrane remains unchanged, the decreasing atmospheric pressure force causes an imbalance with the outward force of the contained helium and the balloon expands. At some point, the membrane of the balloon reaches its elastic limit and bursts. 2. From the outside of a balloon, two forces act inward. The elastic membrane of the balloon and the pressure of Earth s atmosphere work together to balance the outward force of the helium compressed inside. Together with the elastic force, atmospheric pressure near Earth s surface applies enough force to maintain this equilibrium, but as the balloon rises, atmospheric pressure decreases. Although the inward force Skill Sheet 7.1A: Adding Displacement Vectors 1. The total displacement is 5 meters east and 5 meters north. 2. The total displacement is 2 meters east and 2 meters south. 3. The total displacement is zero. Total distance traveled is 40 meters. 1. x R = (1, 5)m 2. x R = (1, 5)m 3. x R = (3, 3)m 4. x R = (5, 5)m Skill Sheet 7.1B: Vector Components 1. x = (6.1, 3.5)m 2. x = (5.0, 8.7)m; 3. Z =(0, 17)m/sec 4. x = (100, 0) cm 6

7 5. The answer is: 6. The answer is: Skill Sheet 7.1C: Pythagorean Theorem 7. Z =( 80,0)km/h 8. x = ( 3.5, 3.5)m/sec 2 9. Z =(40, 69)km/h 10. T = ( 57,57)N 11. Answers are: a. x = (5, 5) cm b. x = (7.07, 45 )cm centimeters 5. Answers are: a. (3.0, 3.0) m b. 4.2 m 6. 7 (rounded from 6.71) Skill Sheet 7.2: Projectile Motion Parts 1 through 3: There are no questions to answer in Parts 1 through Answers are: a. The time it takes for the cat to go 5 m horizontally is the same as the time it will take for the cat to drop 3 m under the effect of gravity. The initial vertical velocity is zero. b. The length of time for the cat to fall 3 meters and the cat s horizontal speed. c. To find the time: y = 1 / 2 gt 2 ; to find horizontal speed: v o = d/t. d. The time it takes for the cat to fall 3 meters is 0.78 seconds; the horizontal speed is 6.39 m/sec (rounded from 2.83) 8. x = +/ 22 meters 1. (4.2, 45 )m 2. Z 3 = (5,180 ) = ( 5, 0); adding all three vectors, the x-y coordinates of the resultant vector are (0, 5); the magnitude of the resultant vector is = (0) 2 + ( 5) 2 = r 2 ; r = 5 3. Z 1 = (5,45 ) = (3.5, 3.5); Z 3 = (1,180 ) = (-1, 0); adding all three vectors, the x-y coordinates of the resultant vector are (2.5, - 6.5); magnitude of the resultant vector is = (2.5) 2 +(- 6.5) 2 = r 2 ; r = 6.95 = Answers are: a. The horizontal speed is 10 m/sec; the time before the object hits the ground is 3 seconds; and the vertical distance is 30 meters down (use - 30 meters in the calculations). b. The initial vertical velocity and the total horizontal distance. c. To find the initial vertical velocity: y = v oy 1 / 2 gt 2 ; to find the total horizontal distance: x = v ox t. d. 4.7 m/sec; 30 meters; the vertical distance is a negative number. -30 m = v oy (3 sec) 1 / 2 (9.8 m/sec 2 )(3 sec) 2 v oy = -30 m/3 sec + (44.1 m/sec)/3 sec = - 10 m m = 4.7 m/sec sec second (= seconds; it takes 0.51 second to go up and then 0.51 second to fall down) m/sec This is very high speed (211 km/h). The ski jumpers actually jump long distances by gliding in the air. 6. v ox = 14.1 m/sec; v oy = 14.1 m/sec; t = 2.88 sec ( sec); x = 40.6 m; y =10.1 m

10 Skill Sheet 11.2: Power 1. Sample problem: 250 watts watts watts Skill Sheet 12.1: Momentum Parts 1 through 2: There are no questions to answer in Parts 1 through Sample problem: 0.6 km-m/sec 2. Sample problem: 2.5 kg-m/sec kg-m/sec kg-m/sec 5. The fullback because the defensive back has more momentum. Skill Sheet 12.2: Rate of Change of Momentum Part 1 through 2: There are no questions to answer in Parts 1 through Force = 200,000 N. At this level of force, after a couple of hits with a wrecking ball, any impressive-looking wall crumbles to pieces seconds N N million N 6. Answers are: a. The force created on the egg is about: kg 10 m/sec = 500 N sec watts 5. work = 500 joules; power = 33 watts 6. work = 1,500 joules; time = 60 seconds 7. force = 25 newtons; power = 250 watts 8. distance = 100 meters; power = 1,000 watts 9. force = 333 newtons; work = 5,000 joules kilograms m/sec 1. Sample problem: m/sec or 3.33 m/sec to the left 2. v 3 = - 2 m/sec 3. m 1 = 60 kilograms 4. v 3 = 1.25 m/sec 5. v 3 = - 1 m/sec or 1 m/sec to the west 6. v 4 = 12.8 m/sec b. The force created on the egg by the person is:. 50 kg 9.8 m/sec 2 = 491 N c. The force created by the person is close to the amount of force that broke the egg. Therefore, if the person fell on the egg, it would probably break. d. As a result the force will be 500 times smaller: kg 10 m/sec = 1 N 0.5 sec e. The egg would probably not break if it fell on the pillow because the force is 500 times smaller than if it fell on the hard floor. Skill Sheet 13.1: Harmonic Motion 1. Answers are: a. 2 seconds b. 0.5 Hz c Amplitude decreases due to the friction between the swing chain and the frame to which it is connected. The gradual loss of amplitude of a pendulum (an oscillator) due to friction is called damping. Eventually, due to friction, the pendulum will stop moving and hang straight down. 3. Answers are: a. 100 centimeters b. The graphic shows 3.5 wavelengths. A wavelength is defined as the harmonic motion from peak-to-peak or from trough-to-trough of a wave. 4. Answers are: a. The length of the string. b. The two trials in which the string length is 30 centimeters. c. The two trials in which the string length is 10 centimeters. d. Due to the acceleration of gravity, the force that pulls on the pendulum increases as mass increases. However, increased mass means greater inertia and less acceleration. The two factors, force and mass, offset each other. This is Newton s second law of motion. For this reason, adding mass to the pendulum does not alter its period. 10

11 Skill Sheet 14.1 Waves 1. The answer is: 2. Two wavelengths 3. The amplitude of a wave is the distance that the wave moves beyond the average point of its motion. In the graphic, the amplitude of the wave is 5 centimeters. 4. The frequency of a wave is the inverse of the period or 1/period in seconds. The frequency is also the number of waves that pass a certain point per second. 5. frequency = waves = waves = 40 Hz 0.05 second second 1. The wavelength divided by the period is the same a multiplying the wavelength by the inverse of the period (1/period). The frequency of a wave is equal to 1/period, therefore, these two ways of calculating the wave speed are the same meters m/sec 4. The speed of wave A is 75 m/sec. The speed of wave B is 65 m/sec. Wave B is faster. Answers are: Table 1: Frequency, harmonic and wavelength data Harmonic # Frequency (Hz) Wavelength (m) Speed of the Wave Frequency times wavelength (m/sec) You can easily determine the harmonics of a vibrating string by counting the number of bumps on the string. The first harmonic (the fundamental) has one bump. The second harmonic has two bumps and so on. 2. The wavelength of the fundamental harmonic of a 5-meter vibrating string would be 10 meters. The one bump on the string at this harmonic represents half of wavelength. Therefore, the entire wavelength is: 2 5 meters. Skill Sheet 15.1: Decibel Scale Problems 1. Example problem. Answers are: a. According to the table, city traffic has a decibel reading of 70 db. b. Since every 20 db increase sounds about twice as loud, the sound relating to 90 db (70 db + 20 db) would sound twice as loud. c. A jackhammer 10 feet away corresponds to 90 db db 3. Twice as loud. 4. At 90 db, the sound is doubled three times; 2 3 = 8 times as loud db 6. A house in the city is or 1.5 times as loud as a house in the country. Skill Sheet 16.1: Light Intensity Problems 1. Example problem: 4.8 W/m W/m W/m 2 4. If distance from a light source doubles, then light intensity decreases by a factor of 4. Example: W/m 2 approximately equals W/m 2 (see questions 2 and 3). 5. Answers are: a W/m 2 b W/m 2 1. Example problem: Since every 20 db has 10 times greater amplitude, a jackhammer 10 feet away (90 db) has an amplitude 10 times greater than city traffic (70 db). 2. The amplitude would be ten times greater. 3. Because the difference between the loudness of the sounds is 60 db, the difference in amplitude should be three times a factor of 10 or Therefore, the amplitude of the sound waves from a jackhammer are 1,000 times greater than the amplitude of the sound waves from a country home. 4. The difference in decibels between the two levels of sound is 20 db. Therefore, the difference in amplitude of the sound waves is 10 times. Normal conversation sound waves have a 10 times greater amplitude than restaurant conversation sound waves. c. 0.5 W/m 2 d. 5 W/m 2 6. The watts of a light source and light intensity are directly related. This means that if you use a light source that has 10 times the wattage, then light intensity will increase 10 times. 7. The light is more intense by five times. 8. The light is less intense by five times. 9. If the light was 100% efficient, light intensity would be W/m 2. At 1% efficiency, the light intensity is W/m 2. 11

12 Skill Sheet 17.1A: The Law of Reflection 1a and 1b. The answers are: 4. The answer is: 2. The angle of reflection will be 30 degrees. 3. Each angle will measure 45 degrees. 1. The angle is 96 degrees. There for the angles of incidence and reflection will each be 48 degrees. 2. The angle at point A is 137 degrees. Therefore the angles of incidence and reflection at point A are each 68.5 degrees. The angle at point B is 41.5 degrees and the angles of incidence and reflection at this point are each 20.8 or 21 degrees. Skill Sheet 17.1B: Refraction 1. We can solve for θ r from the relation n i sinθ i = n r sinθ r n sinθ r = ---- r sinθ n i = sin25 = = i and the angle θ r is given by the inverse sine of 0.285: θ r =sin 1 (0.282) = = degrees degrees degrees 5. The incident index of refraction is larger. The ratio of the refraction indices is: Skill Sheet 17.2: Ray Diagrams 1. A is the correct answer. Light travels in straight lines and reflects off objects in all directions. This is why you can see something from different angles. 2. C is the correct answer. When light goes from air to glass it bends about 13 degrees from the path of the light ray in air. The light bends toward the normal to the air-glass surface because air has a lower index of refraction compared to glass. Then, when the light re-enters the air, it bends about 13 degrees away for the light path in the glass and away from the normal. 3. As a ray of light approaches glass at an angle, it bends (refracts) toward the normal. As it leaves the glass, it bends away from the normal. However, if a ray of light enters a piece of glass perpendicular to the glass surface, the light ray will slow, but not bend because it is already in line with the normal. This happens because the index of refraction for air is lower than the index of refraction for glass. The index of refraction is a ratio that tells you how much light is slowed when it passes through a certain material. 4. A is the correct answer. Light rays that approach the lens that are in line with a normal to the surface pass right through, slowing but not bending. This is what happens at the principal n ---- r = sin = 1.4 n i sin degrees 7. The critical incident angle corresponds to the case when the angle of refraction becomes equal to 90 degrees. Therefore: sinθ 1.0 r = sin 9 0 = 2.4 and θ r =sin 1 (0.43) = degrees ( 1.0) = axis. however, due to the curvature of the lens, the parallel light rays above and below the principal axis, hit the lens surface at an angle. These rays bend toward the normal (this bending occurs toward the fat part of the lens) and are focused at the focal point of the lens. Past the focal point, the rays cross. 5. The answer is: There are no questions to answer in Part 2. 12

13 1. The answer is: 2. The answer is: Skill Sheet 17.3: Thin Lens Formula Part 1 through 2: There are no questions to answer in Parts 1 through centimeters 2. 6 centimeters centimeters 4. Answers are: a. -30 centimeters b. A virtual image. It is on the same side as the object centimeters 6. It is a concave (diverging) lens. It is a virtual image. 1. Using the thin lens formula to prove that d i is 11.7 centimeters: = = Answers are: 14 d a. Finding the object and image distances: o = Given 7 d i 3. A lens acts like a magnifying glass if an object is placed to the left of a converging lens at a distance less than the focal length. The lens bends the rays so that they appear to be coming from a larger, more distant object than the real object. These rays you see form a virtual image. The image is virtual because the rays appear to come from an image, but don t actually meet. this proportion, the image distance is x and the object distance is 2x = x x 10 Therefore, x x = 2x = b. The answer is: = cm 30 cm α = tan = tan -1 ( 0.467) = Skill Sheet 18.1: The Speed of Light 1. Green 2. Green m Hz m Hz λ 1 f 2 7. The answer is: = ---- λ 2 8. λ = meter 9. λ = 3.3 meters 10. f = Hz Hz Hz 13. Answers are: f 1 13 a. f = m b. It is the maximum frequency. 14. radio waves, microwaves, infrared, visible, ultraviolet, x rays, and gamma rays 15. gamma rays, x rays, ultraviolet, visible, infrared, microwaves, and radio waves

15 Skill Sheet 19.3B: Ben Franklin 1. If the kite had been struck by lightning, the amount of charge coming down the hemp string would most likely have electrocuted Franklin. 2. A lightning rod is a metal rod attached to the roof of a building. A thick cable stretches from the rod to a metal stake buried in the ground. When lightning strikes the rod, it follows the path of least resistance-from the rod, through the cable, into the ground, where the charge can safely dissipate. Skill Sheet 20.1: Parallel and Series Circuits 1. Answers are: a. 12 volts b. 4 ohms c. 3 amps d. 6 volts e. The answer is: 1 ohm resistor: 1.5 volts d. The answer is: 2. Answers are: a. 2 ohms per resistor b. The voltage drop per resistor is 1 volt. c. The answer is: 1. Answers are: a. 12 volts b. 6 amps c. 12 volts 2. The current flow through each bulb in series is 3 amp. Through each bulb in parallel, current flow is 6 amps. The bulbs in the parallel circuit are brighter. 3. Answers are: a. 9 volts b. Branch with 2 ohm resistor: 4.5 amps Branch with 3 ohm resistor: 3 amps Branch with 1 ohm resistor: 9 amps c. The 2 ohm resistor uses 40.5 watts. The 3 ohm resistor uses 27 watts. The 1 ohm resistor uses 81 watts. d. The more current drawn by a resistor, the more power it uses. As current increases, power increases. 3. Answers are: a. 6 ohms b. 1.5 amps c. 2 ohm resistor: 3 volts 3 ohm resistor: 4.5 volts 15

21 Element Chemical Total # of Valence Symbol Electrons Electrons Potassium K 19 1 Nitrogen N 7 5 Carbon C 6 4 Beryllium Be 4 2 Neon Ne 10 8 Dot Diagram Elements Dot Diagram for Each Element Na and F Dot Diagram for Compound Formed Chemical Formula NaF Br and Br Br 2 Mg and O MgO Sulfur S 16 6 Skill Sheet 29.3 Chemical Equations Reactants Products Chemical Equation Hydrochloric acid Water HCl H 2 O and and HCl + NaOH NaCl + Sodium hydroxide Sodium chloride H NaOH 2 0 NaCl Calcium carbonate CaCO 3 and Potassium iodide KI Aluminum fluoride AlF 3 and Magnesium nitrate Mg(NO 3 ) 2 Potassium carbonate K 2 CO 3 and Calcium iodide CaI 2 Aluminum nitrate Al(NO 3 ) 3 and Magnesium fluoride MgF 2 CaCo 3 + KI K 2 CO 3 + CaI 2 AlF 3 + Mg(NO 3 ) 2 Al(NO 3 ) 3 + MgF 2 Part 3 through 5 answers: There are no questions to answer in Parts 3 through 5. Part 6 answers: 1. 4Al + 3O 2 2Al 2 O 3 2. CO + 3H 2 H 2 O + CH HgO 2Hg + O 2 4. CaCO 3 CaO + CO 2 (already balanced) 5. 3C + 2Fe 2 O 3 4Fe + 3CO 2 6. N 2 + 3H 2 2NH K + 2H 2 O 2KOH + H P + 5O 2 2P 2 O 5 9. Ba(OH) 2 + H 2 SO 4 2H 2 O + BaSO CaF 2 + H 2 SO 4 CaSO 4 + 2HF 11. 4KClO 3 3KClO 4 + KCl Skill Sheet 30.1A Radioactivity Parts 1 through 2: 1. There are no questions to answer in Parts 1 through In the answers below, a is alpha decay and b is beta decay. a. Answers are: U 234 a 90 Th 234 b 91 Pa 234 b 92 U 230 a 90 Th a a a a b b Ra Rn Po Pb Bi Po 210 a 82 Pb 210 b 83 Bi 210 b 84 Po 206 a 82 Pb b. Answers are: Pu Am Np Pa U b a a b a a b a a a Bi Ra Ac Fr At Bi 212 b 84 Po 208 a 82 Pb 208 b 83 Bi kilograms or grams gram; This is a problem, so they would have to make it as they use it. Hospitals have laboratories to do this. 4. The amount after 30,000 years would be 0.026m where m is the mass of the sample. 5. For one-fourth of the original mass to be left, there must have been time for two half-lives. Therefore, the half-life for this radioactive isotope is 9 months W/m reactions per second 21

### Practice final for Basic Physics spring 2005 answers on the last page Name: Date:

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