# 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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1 Skill Sheet 1.1: Solving Equations Part 1 through 5: There are no questions to answer in parts 1 through 5. Part 6 answers: 1. 6,797 points 2. 18th floor seconds Skill Sheet 1.2: Galileo Galilei 1. Galileo's telescope is the most likely student response, because it so profoundly changed our understanding of the solar system. However, students may choose another invention as long as they provide valid reasons for their decision. Skill Sheet 2.1: International System of Units Parts 1 through 6: There are no questions to answer in parts 1 through 6. Part 7 answers: grams centimeters psi 5. Tim s body temperature is 34.4 C. He will experience hypothermia hours 7. k = horizontal distance = 8.6 feet 2. Galileo's many inventions include the thermometer, water pump, military compass, microscope, telescope, and pendulum clock. Information and illustrations of the inventions can be found using the Internet or library kilograms centimeters 5. 5,200 milliliters ,000 centimeters = 1 kilometer Skill Sheet 2.2: Converting Units Parts 1 to 3: There are no questions to answer in parts 1 to Answers are: Starting amount and unit Ending amount and unit 3.0 inches meter 3.7 gallons 14 liters 47.0 pounds 21.3 kilograms 3.0 pints 1.4 liters 230 grams 0.23 kilogram 42 millimeters 4.2 centimeters 1,000 millimeters 1 liter 24.3 meters kilometer gallons (Since, building codes are listed in ranges, 740 gallons is a precise enough answer.) m/sec grams 5. 1 quart = liters, or 946 milliliters. An additional 54 milliliters of beverage must be added to a quart equal one liter. By changing from quarts to liters, the beverage industry increases the amount of product it sells per bottle and can charge more money per bottle sold ounces 7. 93,000,000 miles = 150,000,000 kilometers; 150,000,000 kilometers = 150,000,000,000, meters; time = 437,317,784 seconds; 437,317,784 seconds = 121,477 hours; 121,477 hours = 13.9 years Skill Sheet 2.3: Scientific Notation 1. B 2. A There are no questions to answer in Part ,000, , ,500,000, , Part 5 answers: There are no questions to answer in Part 5. Part 6 answers:

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

9 Skill Sheet 10.1: Mechanical Advantage 1. Sample problem: mechanical advantage = 5 2. input force = 100 N 3. output force = 26 N 4. mechanical advantage = 3 5. mechanical advantage = 3 6. Ratio = 7.5; L out = 0.23 meters 1. Sample problem: mechanical advantage = output force = 150 N 3. mechanical advantage = 1.5 Skill Sheet 10.2: Work 1. Work is force acting upon an object to move it a certain distance. In scientific terms, work occurs ONLY when the force is applied in the same direction as the movement. 2. Work is equal to force multiplied by distance. 3. Work can be represented in joules or newton-meters. 1. Sample problem: 50 joules joules ,000 joules 4. No work was done by the mouse. The force on the ant was upward, but the distance was horizontal meters pounds 7. 2,500 N or 562 pounds joules joules meters ,000 joules 12. The people-mover accomplished 70,000 joules of work. The person was 600 N plus 100 N for the bag. Therefore, the work done by the people-mover was 700 N times 100 meters. The person did 100 joules of work. The total amount of work done in this situation was 70,100 joules joules joules Part 5 answers: 1. 1,768 joules joules Skill Sheet 10.3: Potential and Kinetic Energy Parts 1 through 3: There are no questions to answer in Parts 1 through 3. Part 4: 1. Sample problem: The kinetic energy of the boy was 625 joules. The kinetic energy of the father was 1,250 joules. 2. Sample problem: The potential energy of the book is 25 joules. 3. Answers are: Shelf height (meters) Potential energy (joules) The object with more mass had more kinetic energy while being lifted. The kinetic energy of the 2-kilogram object was 4 joules. The kinetic energy of the 4-kilogram object was 18 joules. 5. Although each object is lifted to the same height, the 4- kilogram object has more potential energy because it has more mass. 6. Answers are: a. m 588 kg sec 2 = 588 N b. 1.7 meters c. 5.7 m/sec 7. Answers are: a. 450 joules b. 450 joules c. 46 meters 8. The potential energy is zero because the ball is not off the ground. Height is zero ,00 joules m/sec kilograms 12. The energy of the ball at position B equals the potential energy plus the kinetic energy at this position. Also, the energy at position B equals the potential energy at position A or the kinetic energy at position C. a. The equation that relates the energy at B to the potential energy at A: E B E BP + E BK mgh 1 2 = = B + --mv 2 B = mgh A where h A = 3 meters; h B = 1 meter. b. Solving for velocity at position B: v 2g( h A h B ) m = = sec 2 ( 3 1 )m = m sec 2 9

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

14 Skill Sheet 18.3: Albert Einstein 1. Scientists tested Einstein's theory of relativity by photographing a solar eclipse on November 8, 1919 from locations in Brazil and the African island of Principe. If Einstein's theory were correct, light from a cluster of stars called the Hyades behind the dimmed sun should be bent into the gravitational dimple created by the sun, making the stars appear slightly out of alignment. The photographs confirmed Einstein's predictions. While some scientists debated the accuracy of this method and doubted Einstein's theory, conclusive evidence was provided by the European Space Agency's Hipparcos satellite, which in the years between 1989 and 1993 charted the positions of the stars with great precision and confirmed Einstein's prediction that gravity bends light. 1. Brownian motion was first observed by British botanist Robert Brown in He observed through a microscope that pollen grains suspended in water appeared to move erratically. He thought maybe it was because they were alive, but then found that ground glass and other non-living materials exhibited the same motion. He published his results in Eighty years later Einstein explained that the erratic motion was due to water molecules bumping into the small particles. Brownian motion can be demonstrated by suspending pollen grains (scraped from the anther of a lily flower, available year-round from a florist shop), graphite scraped from a #2 pencil, or talcum powder in water. Place several drops of the suspension in a cavity microscope slide and observe under a microscope. Skill Sheet 19.2: Using an Electric Meter Part 1 through 2: There are no questions to answer in Parts 1 through First battery: volts; second battery: volts 2. First bulb: volts; second bulb: volts volts 4. The answer is: 5. post #1: amps; post #2: amps; post #3: amps; post #4: amps; post #5: amps 6. Sample answer: Sometimes light bulbs are made with frosted glass, which makes it impossible to see the tungsten filament inside. You can use an electrical meter to figure out if the tungsten filament is burned out. First, set the meter dial to measure resistance. Then place one lead on the side of the metal portion of the light bulb (where the bulb is threaded to fit into the socket). Place the other lead on the bump at the base of the light bulb. Check the meter reading. If the meter reports a small value for resistance, the tungsten filament is intact. If the meter displays OL or the infinity symbol, the filament must be burned out. A burned out filament has a break in it, creating an open circuit. The electrical current would have to pass through air to complete the circuit. Because the resistance of air is too high for a small amount of current to cross, the meter reports very high resistance. 7. A fuse contains a wire that will melt or break if more than a safe amount of current (12 amps, for example) is applied. When the wire melts or breaks, the circuit is broken and the current stops. If you measure the resistance of a fuse and the meter displays OL or the infinity symbol, the fuse must have a melted or broken wire. The current would have to pass through air to complete the circuit. As a result, the meter reports very high resistance. 8. Wires would normally have very low resistance, because they are made of copper or other metals through which electricity easily flows. If the measured resistance of a wire is very high, it must have a break in it. Skill Sheet 19.3A: Ohm s Law Part 1: amps 2. 3 ohms volts 4. Answers are: a. Circuit A: 6 volts; circuit B: 12 volts b. Circuit B has a greater current because it has more voltage but the same resistance as circuit A. c. The brightness is greater in circuit B because there is more current. d. Circuit A: 1 amp; circuit B: 2 amps e. Circuit A: 0.5 amp; circuit B: 1 amp f. The current would decrease in each circuit because the resistance would increase. The current would be cut in half. b. Circuit A: 0.25 amp; circuit B: 0.5 amp c. If voltage remains the same, adding bulbs in a series circuit decreases the brightness of the bulbs. 6. Current increases. 7. Current decreases. 5. Answers are: a. Circuit A: 0.5 amp; circuit B: 1 amp 14

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

16 Skill Sheet 20.2: Network Circuits 1. I R2 = I R3 = 3 A; I R1 = 3 A 2. R Total = 2 Ω 3. I Total = 1 A 4. Answers are: a. No, current does not flow through R 5, because it does not have to. The current will go directly to R 3 and R 4. b. R Total = 9 Ω. c. I Total = 2 A d. I R3 = 1 A e. The voltage across R 5 is zero because no current flows through this circuit. 5. I Total = 1.5 A; V R4 = 4.5 V 6. V 5Ω = 5 V (I Total = 1 A) 7. For R 4 = 10 Ω, V A = 8 V, V B = 4 V, V AB = 8 V - 4 V = 4 V Skill Sheet 20.3: Electrical Power Parts 1 through 2: There are no questions to answer in Parts 1 through Sample problem: 270,000 joules 2. 5,400 seconds, or 90 minutes 3. 1,080 watts volts A A 1. Answers are: a. 1.2 kw b. 0.4 kwh c. 12 kwh/month d. \$ amps W = 0.96 kw 4. Answers are: a. 3 V b. 3 Ω c. 1 A d. 3 W 5. Answers are: a. 24 Ω b. 600 W; 0.6 kw 6. Answers are: a. 20 A b. 11 Ω 7. Answers are: a. 12 Ω b. 1,200 W = 1.2 kw c. 0.6 kwh d. 18 kwh e. \$ Your friend with the 300-watt stereo will pay more for the energy used in an hour of play. This is because more twice as much energy is used per second for the 300-watt stereo than for the 150-watt stereo. 9. Since power is voltage times current, as current changes, the power used per appliance will change. Each appliance can be designed to carry the amount of current needed to run well. Most electrical appliances in a dining room or living room would run on 10 amps of current or less. However, in a kitchen you may have an electric stove and oven that requires 40 amps of current. A special circuit with thicker wires is installed there so that the circuit does not overheat and pose a fire hazard. Skill Sheet 21.2: Coulomb s Law 1. It becomes 1/9 the original amount. 2. It becomes 1/16 the original amount. 3. It quadruples. 4. It doubles. 5. It quadruples. 6. It remains the same. 7. It becomes 16 times as great. 8. F = N 9. The total charge equals C. The square root of this number is the value for each charge and is equal to C N; a cat weighs 49 N, this is about eleven times more force than the force between the particles C 12. distance = 6.7 m 13. distance = 0.03 m 14. q = C 16

18 Skill Sheet 24.3: Binary Number Problems 1. Answers are: Binary Digits Number It takes 5 bits (16, 8, 4, 2, 1) to convert the decimal number 16 to a binary number (10 000). 3. Answers are: a. 99 = b. 127 = The number 127 cannot be represented by a 6-bit binary number. You need 8-bits to represent 128 ( ). 1. Answers are: a. Example problem: 204 b. 170 c. 255 d. 9 e. 6 f. 2 g. 3 h. 2 Skill Sheet 25.2: Temperature Scales Parts 1 through 2: There are no questions to answer in Parts 1 through C C C There are no questions to answer in Part 4. Part 5 answers: F F 3. The friend assumed you were speaking of degrees Celsius. Fifteen degrees Celsius equals 59 F, which is a much milder outdoor temperature than 15 F (15 F = -9.4 C). Part 6 answers: K K = 2 C No, the mysterious, silver substance has a much higher melting point than mercury. 3. The thermometer is calibrated to the Fahrenheit scale. On the Kelvin scale, 90 K is too cold (-298 F and -183 C), and 90 C is too hot, just 10 degrees less than the boiling point of water. Skill Sheet 26.3: Heat Transfer 1. Example answer: 10.7 watts C C watt 5. Sample answer: The heat conduction equation can be used to evaluate roofing material. The surface area, thickness of the roof, and thermal conductivity of the material used can be taken into account using the heat conduction equation. It is important that roofing material conduct very little heat so that a house stays cool in the summer and warm in the winter. 1. Example answer: 210 watts watts 3. Sample answer: The area value used in the convection equation is the area of the contacting fluid. My hypothesis for why the heat transfer coefficient values for water are so high is that gases have little contacting area the molecules are very spread out so not as much energy can be transferred, and oil is viscous so heating does not happen as quickly. Liquid water has high molecular contact for heating efficiently and low viscosity. 4. Free convection is the flow of fluids due to differences in density or temperature. Forced convection is the flow of fluids due to circulation by a fan or a pump. A heat transfer coefficient is the amount of watts per square meter for a given temperature. Values are higher when convection is forced because energy is used to heat a given area. 5. Sample answer: The convection equation can be used to design tents for cold weather camping. You would need to know the heat transfer coefficient for wind (forced convection), the area of the tent (or one side of the tent), and the temperature differences between the outside environment and a safe temperature for inside the tent. 1. Example answer: 3,377 K watts m watts; 0 C 18

19 5. Heat transfer at 0 C is greater than heat transfer at absolute zero (0 K). At zero Kelvin, there would be no heat transfer (heat transfer would equal zero). 6. Sample answer: The Stefan-Boltzmann equation can be used to study stars. For example, the equation can be used to infer the radius of a star (the surface area of a sphere is 4πr 2 ). Also, knowing the amount of radiation from a star allows you to infer its temperature and brightness and color. The higher its temperature, the brighter a star will be. Skill Sheet 27.1 Stress and Strain Parts 1 through 2: There are no questions to answer in Parts 1 through 2. Part 3: 1. σ = N/m 2 ; e = Skill Sheet 27.2 Archimedes 1. How Archimedes solved the problem of determining the gold crown s purity: 2. Inventions attributed to Archimedes include war machines (such as a lever used to turn enemy boats upside down), the Archimedes screw, compound pulley systems, and a planetarium. There is some debate about whether he invented a water organ and a system of mirrors and/or lenses to focus intense, burning light on enemy ships. Students can use the Internet or library to find more information and diagrams of the inventions. Skill Sheet 27.3: Gas Laws Parts 1 through 2: There are no questions to answer in Parts 1 through L kpa L L L Skill Sheet 28.1A: The Structure of the Atom 1. Lithium, Li 2. Carbon, C 3. Hydrogen, H 4. Hydrogen, H (a radioactive isotope, 3 H, called tritium) 5. Beryllium, Be 1. Atomic weight 7 2. Atomic weight Atomic weight 1 4. Atomic weight 3 5. Atomic weight 9 6. The atomic mass of amu represents an average of the masses and abundance of all to the hydrogen isotopes (hydrogen-1, hydrogen-2, and hydrogen-3). Because this centimeters 3. Answers are: (a) radius = 3.42 mm; diameter = 6.84 mm, and (b) radius = 10.8 mm; diameter = 21.6 mm. 4. ε = 0.002; 0.01 meter or 1 centimeter Part 4: There are no questions to answer in Part 4. Part 5 answers: kilograms 2. The mass is 0.26 kilogram and the new volume is 0.4 liters atmospheres; the problem is solved using P 1 /P 2 = V 2 /V 1. Because V 2 = V V 1, V 1 can be cancelled from the equation so that the only variable is P 2. atomic mass is so close to 1, we can assume that the most abundant isotope is hydrogen H = 1 proton, 1 neutron Sc = 21 protons, 24 neutrons Al = 13 protons, 14 neutrons U = 92 protons, 143 neutrons C = 6 protons, 6 neutrons Part 5 answers: 1. Most of an atom s mass is concentrated in the nucleus. The number of electrons and protons is the same but electrons are so light they contribute very little mass. The mass of the proton is 1,835 times the mass of the electron. Neutrons have a bit more mass than protons, but the two are so close in size that we usually assume their masses are the same. 19

20 2. Answers are: a. Yes, it is hydrogen-3. It has one proton and two neutrons. b. No, it has a proton (+1) and no electrons to balance this charge. Therefore, the overall charge of this atom (now, called an ion) is Answers are: (a) one electron, (b) five electrons, and (c) 14 electrons. 4. This sodium atom has ten electrons, 11 protons, and 12 neutrons. Skill Sheet 28.1B: The Periodic Table 1. Fluorine 2. Argon 3. Manganese 4. Phosphorous 5. Technetium 1. Iron, 55.8 amu 2. Cesium, amu 3. Silicon, 28.1 amu 4. Sodium, 23.0 amu 5. Bismuth, amu 1. Protons = 12; neutrons = Protons = 7; neutrons = 7 3. Protons = 19; neutrons =20 Part 5 answers: Po α 207 Pb; this alpha decay to stable lead is the last of a succession of radioactive decays that begin with uranium. The radioactive decay succession from uranium to lead is called the transuranic series and takes millions years. Because this chain of events is so predictable, the proportion of uranium to lead is the main clock used to date geologic events in very deep time Ra α 222 Rn; radon is a radioactive gas and can leak 3. into homes from underground sources. 222 Rn α 218 Po; this isotope of polonium is possibly radioactive. Note: Work with radioisotopes is only done casually with the periodic table of elements. Typically, a different table, the chart of radionuclides, is used because it includes data for each isotope. Skill Sheet 28.1 Lise Meitner 1. The graphic at right illustrates fission: 2. Some topics students may research and describe include nuclear power plants, nuclear weapons, nuclearpowered submarines or aircraft carriers. Skill Sheet 28.2 Niels Bohr 1. Niels Bohr described atoms as existing in specific orbital pathways, and explained how atoms emit light. 2. Bohr s model of the atom: Skill Sheet 29.2 Dot Diagrams 20

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

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