1

Answer KeysSkill Sheet 1.1: Solving EquationsPart 1 through 5:There are no questions to answer in parts 1 through 5.Part 6 answers:1. 6,797 points2. 18th floor3. 41 seconds

4. 148 psi5. Tim’s body temperature is 34.4°C. He will experience

hypothermia.6. 2.7 hours7. k = 10.68. horizontal distance = 8.6 feet

Skill Sheet 1.2: Galileo Galilei1. Galileo's telescope is the most likely student response,

because it so profoundly changed our understanding of thesolar system. However, students may choose anotherinvention as long as they provide valid reasons for theirdecision.

2. Galileo's many inventions include the thermometer, waterpump, military compass, microscope, telescope, andpendulum clock. Information and illustrations of theinventions can be found using the Internet or library.

Skill Sheet 2.1: International System of UnitsParts 1 through 6:There are no questions to answer in parts 1 through 6.Part 7 answers:1. 0.00345 grams2. 300.4 centimeters

3. 0.1123 kilograms4. 656.709 centimeters5. 5,200 milliliters6. 100,000 centimeters = 1 kilometer

Skill Sheet 2.2: Converting UnitsParts 1 to 3:There are no questions to answer in parts 1 to 3.Part 4 answers:1. Answers are:

2. 739.8 gallons (Since, building codes are listed in ranges,740 gallons is a precise enough answer.)

3. 365.8 m/sec4. 907.2 grams5. 1 quart = 0.946 liters, or 946 milliliters. An additional

54 milliliters of beverage must be added to a quart equal oneliter. By changing from quarts to liters, the beverage industryincreases the amount of product it sells per bottle and cancharge more money per bottle sold.

6. 7 ounces7. 93,000,000 miles = 150,000,000 kilometers; 150,000,000

kilometers = 150,000,000,000, meters; time = 437,317,784seconds; 437,317,784 seconds = 121,477 hours; 121,477hours = 13.9 years

Skill Sheet 2.3: Scientific NotationPart 1 answers:1. B2. APart 2 answers:There are no questions to answer in Part 2.Part 3 answers:1. 6.70×106

2. 3.00×108

3. 3.60×106

4. 6.45×10-4

5. 1.50×108

6. 1.00×10-3

7. 1.86×105

8. 1.34×10-3

Part 4 answers:1. 347,000,0002. 7,9403. 0.001964. 4,500,000,0005. 0.000009616. 202,0007. 0.0000000007018. 44.4Part 5 answers:There are no questions to answer in Part 5.Part 6 answers:1. 1.40×104

2. 2.57×107

Starting amount and unit Ending amount and unit

3.0 inches 0.076 meter3.7 gallons 14 liters47.0 pounds 21.3 kilograms3.0 pints 1.4 liters230 grams 0.23 kilogram42 millimeters 4.2 centimeters1,000 millimeters 1 liter24.3 meters 0.0243 kilometer

Answer Keys

2

3. 3.00×101

4. 1.22×1010

5. 2.20×10-4

6. 9.16×102

7. 2.40×109

8. 8.14×104

9. 4.70×10-4

10. 9.33×10-3

Skill Sheet 3.1: Speed ProblemsPart 1 answers:1. Sample problem: 28 m/sec2. 17 km/hr3. 55 mph4. 4.5 seconds5. 5.9 hours6. 4.0 kilometers7. 4.5 meters

Part 2 answers:1. Sample problem: 63 mph2. 490 mph3. 2.5 miles4. Answers are:

a. 2.54 cm/ inchb. 12 inches/min

Skill Sheet 3.2: Making Line GraphsParts 1 through 2:There are no questions to answer in parts 1 through 2.Part 3 answers:1. Graph of money earned vs. hours worked:

2. The amount earned is $4.50/hour.

Part 4 answers:

Part 5 answers:

Part 6 answers:

Data pair not necessarily

in order

Independent Dependent

Temp. Hours of heating

Hours of heating

Temp.

Reactiontime

Alcoholdrunk

Alcoholconsumed

Reactiontime

Number of people in family

Costper week for groceries

Number ofpeople in family

Costper week for groceries

Streamflow

Rainfall Amount of rainfall Rate of stream flow

Treeage

Averageheight

Tree age Averageheight

Lowest value Highest value Range

0 28 2810 87 770 4.2 4.2-5 23 280 113 113

100 1,250 1,150

Range Numberof lines

Range ÷ No.of lines

Calculated scale

(per line)

Adj. scale

(per line)

13 24 13 ÷ 24 0.54 183 43 83 ÷ 43 1.9 231 35 31 ÷ 35 0.88 14.2 33 4.2 ÷ 33 0.13 0.212 33 12 ÷ 33 0.36 0.5900 15 900 ÷ 15 60 60

Answer Keys

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Skill Sheet 3.3A: Analyzing Graphs of Motion Without NumbersParts 1 through 2:There are no questions to answer in parts 1 through 2.Part 3 answers:1. Little Red Riding Hood. Graph Little Red Riding Hood:

2. The Tortoise and the Hare. Use two lines to graph both thetortoise and the hare:

3. The Skyrocket. Graph the altitude of the rocket:

Part 4 answers:Each student story will include elements that are controlled bythe graphs and creative elements that facilitate the story. Only thegraph-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 towardCaroline's house slowed to a stop.

• Then the line rises steeply to the first peak. This indicates thatafter his stop, Tim continues toward Caroline's house fasterthan before.

• The first peak is sharp, indicating that Tim did not spend muchtime at Caroline's house on first arrival.

• The line then falls briefly, turns to the horizontal, and then risesto a second peak. This indicates that Tim left, paused, and thenreturned quickly to Caroline's house.

• The line then remains at the second peak for a long time, thendrops steeply to the baseline. This indicates that after spendinga long time at Caroline's house, Tim probably ran home.

Skill Sheet 3.3B: Analyzing Graphs of Motion With NumbersParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. The bicycle trip through hilly country.

2. A walk in the park.

3. Up and down the supermarket aisles.

Part 4 answers:1. The honey bee among the flowers.

Answer Keys

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2. Rover runs the street. 3. The amoeba.

Skill Sheet 4.1: Acceleration ProblemsPart 1 answers:1. 25 m/sec2 2. 2.7 m/sec2 3. -0.75 m/sec2 4. -8.9 m/sec2 5. 23 mph/sec6. 25 km/hr/sec

7. The cheetah. The cheetah’s acceleration in km/hr/sec is37 km/hr/sec, which is 12 km/hr/sec faster than the car.

Part 2 answers:1. 22 m/sec2. 2. 5.7 seconds3. 3. 7.5 seconds4. 4. 88 mph

Skill Sheet 4.2: Acceleration and Speed-Time GraphsParts 1 through 2:There are no questions to answer in parts 1 through 2.Part 3 answers:1. Acceleration = 5 miles/hour/hour or 5 miles/hour2

2. Acceleration = -2 meters/minute/minute or -2 meters/minute2

3. Acceleration = 0 feet/minute/minute or 0 feet/minute2 or noacceleration

4. Answers are:Segment 1: Acceleration = 2 feet/second/second, or 2 feet/second2

Segment 2: Acceleration = 0.67 feet/second/second, or 0.67 feet/second2

Part 4 answers:1. Distance = 1,400 meters2. Distance = 700 meters3. Distance = 75 kilometers

Skill Sheet 4.3: Acceleration Due to GravityParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. velocity = –14.7 meters/sec2. velocity = 11.3 meters/sec3. velocity = –76.4 meters/sec4. velocity = –15.8 meters/secPart 4:There are no questions to answer in Part 4.

Part 5 answers:1. depth = 86.4 meters2. height = 11.0 meters; yes3. height = 5.9 meters (The maximum height would have occur

in half the total time that the serve was in the air or 1.1seconds.)

Part 6 answers:1. time = 5.6 seconds2. time = 7.0 seconds3. Students use the equations in Part 6 to check their answers in

other parts of the skill sheet.

Skill Sheet 5.1: Isaac Newton1. The legend tells of Newton sitting in his garden in

Linconshire in 1666, watching an apple fall from a tree. Helater noted that “In the same year, I began to think of gravityextending to the orb of the moon.” However, he did not makepublic his musings about gravity until the 1680's, when heformulated his universal law of gravitation.

2. Newton claimed that 20 years earlier, he had invented thematerial that Leibnitz published. Newton accused Leibnitz ofplagiarism. Most historians today agree that the twodeveloped the material independently, and therefore they areknown as co-discoverers.

Answer Keys

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Skill Sheet 5.2: Newton's Second LawPart 1:There are no questions to answer in Part 1.Part 2 answers:

Part 3 answers:1. Sample problem: 6,000 N2. Sample problem: 10 kilograms3. 2.1 m/sec2

4. 83 m/sec2

5. 82 N6. 6 kilograms7. 9,800 N8. 900 kilograms9. 1.9 m/sec2

Skill Sheet 6.1: Mass and WeightPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Answers are:

a. On Earth: 758 Newtons = 170 pounds (force)b. On the asteroid: = 6.2 Newtons = 1.4 pounds (force)

2. 0.052 newtons = 0.012 pounds (force)

3. Answers are:

Skill Sheet 6.2: FrictionPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. More than 319 newtons2. More than 73.5 newtons3. More than 891 newtons4. The coefficient of static friction = 0.65. 815 newtons

Part 3 answers:1. The coefficient of sliding friction is 0.16.2. More than 24.5 newtons3. More than 324 newtons4. More than 536 newtons5. Half of the static coefficient of friction. 0.65/2 = 0.32 (Or 0.33

depending on rounding rule)

Skill Sheet 6.3: EquilibriumPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. The answer is:

2. 142 N3. A is 50 N; B is 50 N4. A is 20 N, B is 20 N, and C is 8 N.

What do youwant to know?

What do you know?

The formula you will use

acceleration (a) Force (F) and

mass (m)mass (m) acceleration (a)

andForce (F)

Force (F) acceleration (a) and

mass (m)

acceleration Forcemass-------------=

mass Forceacceleration-----------------------------=

Force mass acceleration×=

Force of gravity newtons pounds

Sun 274.4 N/kg 18,769 4,218Mercury 3.7 N/kg 253 57Venus 8.9 N/kg 609 137Mars 3.7 N/kg 253 57Saturn 10.4 N/kg 711 160Uranus 8.8 N/kg 602 135Neptune 10.7 N/kg 732 165Pluto 0.7 N/kg 48 11

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Part 3 answers: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 toavoid striking Earth. Any variation from a right angle forcewastes the available energy by simply making an insignificantchange in the asteroid's velocity.

2. From the outside of a balloon, two forces act inward. Theelastic membrane of the balloon and the pressure of Earth’satmosphere work together to balance the outward force of thehelium compressed inside. Together with the elastic force,atmospheric pressure near Earth’s surface applies enoughforce to maintain this equilibrium, but as the balloon rises,atmospheric pressure decreases. Although the inward force

supplied by the elastic membrane remains unchanged, thedecreasing atmospheric pressure force causes an imbalancewith the outward force of the contained helium and theballoon expands. At some point, the membrane of the balloonreaches its elastic limit and bursts.

Skill Sheet 7.1A: Adding Displacement VectorsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers: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.Part 3 answers:1. xR = (1, –5)m2. xR = (1, –5)m

3. xR = (3, 3)m

4. xR = (5, 5)m

Skill Sheet 7.1B: Vector ComponentsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. x = (6.1, 3.5)m

2. x = (5.0, 8.7)m;

3. Z =(0, 17)m/sec4. x = (100, 0) cm

Answer Keys

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5. The answer is: 6. The answer is:

7. Z =(–80,0)km/h8. x = (–3.5,–3.5)m/sec2

9. Z =(40,–69)km/h10. T = (–57,57)N11. Answers are:

a. x = (5, 5) cmb. x = (7.07, 45°)cm

Skill Sheet 7.1C: Pythagorean TheoremPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. 52. 53. 54. 13.42 centimeters5. Answers are:

a. (3.0, 3.0) mb. 4.2 m

6. 7 (rounded from 6.71)

7. 3 (rounded from 2.83)8. x = +/− 22 metersPart 3 answers:1. (4.2, 45°)m2. Z3 = (5,180°) = (−5, 0); adding all three vectors, the x-y

coordinates of the resultant vector are (0, −5); the magnitudeof the resultant vector is = (0)2 + (−5)2 = r2; r = 5

3. Z1 = (5,45°) = (3.5, 3.5); Z3 = (1,180°) = (-1, 0); adding allthree 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 = r2; r = 6.95 = 7.0.

Skill Sheet 7.2: Projectile MotionParts 1 through 3:There are no questions to answer in Parts 1 through 3.Part 4 answers:1. 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 = 1/2gt2; to find horizontal speed: vo = 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.

2. 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 = voy – 1/2gt2; to find the total horizontal distance: x = voxt.

d. 4.7 m/sec; 30 meters; the vertical distance is a negative number.-30 m = voy (3 sec) – 1/2 (9.8 m/sec2)(3 sec)2

voy = -30 m/3 sec + (44.1 m/sec)/3 sec = -10 m + 14.7 m = 4.7 m/sec

3. 1.4 sec4. 1.02 second (= 2 × 0.51 seconds; it takes 0.51 second to go up

and then 0.51 second to fall down)5. 58.7 m/sec This is very high speed (211 km/h). The ski

jumpers actually jump long distances by gliding in the air.6. vox = 14.1 m/sec; voy = 14.1 m/sec; t = 2.88 sec (2 × 1.44 sec);

x = 40.6 m; y =10.1 m

Answer Keys

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Skill Sheet 7.3A: Equilibrium in 2-DPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. (- 8.66, - 5.0)N2. T = (5,–10)kN

3. 10 kilograms × 9.8 m/sec2 = 98 N; 98 N × 0.5 = 49 N; 100 N -49 N = 51 N; 51 newtons is the pulling force required for thebox to be in equilibrium.

4. x component = –70.7 N; y component = 70.7 N5. x component: 51.96 N; y component: 111.4 N

Skill Sheet 7.3B: Inclined PlanesPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Component along the plane: 49 N; component perpendicular

to the plane: 85 N2. 4.9 m/sec2

3. Parallel force: 124 N; Perpendicular force: 266 N;Acceleration: 4.1 m/sec2

4. Friction has to be equal to the force along the plane.Therefore, mgsinθ = µmgcosθ; sinθ/cosθ = µ; tanθ = 0.30;θ = 16.7 degrees.

5. 28 N6. 94 N7. When the angle is 16.7°, the total force along the plane is

equal to 0 N because the force along the plane (28 N) cancelsthe frictional force (28 N); mgsinθ = µmgcosθ.

8. At 45°, acceleration = 4.85 m/sec2

Skill Sheet 8.1: Circular MotionPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Convert to radians:

a. 0 radiansb. 0.17 radiansc. 0.52 radiansd. 0.79 radianse. 1.6 radiansf. 3.14 radiansg. 4.71 radiansh. 6.28 radians

2. 60 degrees3. 143 RPM4. Front wheel: 17.8 rad/sec. 2.83 RPM; Back wheel: 12 rad/sec.

1.9 RPM 5. Angular speed is 8.38 radians/second. Linear speed is 8.38

meters/second.6. 0.88 m/sec7. 11 meters8. Answers are:

a. 200 radians/sec at 1 second; 600 radians/sec at 3 seconds.b. 32 revolutions/sec at 1 second; 96 revolutions/sec at 3

seconds.

Skill Sheet 8.3: Universal GravitationPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. F = 9.34 × 10–6 N. This is basically the force between you

and your car when you are at the door.2. 7.33 × 1022 kilograms

3. Answers are:a. 9.8 N/kg = 9.8 kg-m/sec2-kg = 9.8 m/sec2

b. Acceleration due to the force of gravity of Earth.c. Earth’s mass and radius.

4. 1.99 ×1020 N5. 4,848 N6. 3.52 × 1022 N

Skill Sheet 9.1: TorqueParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3:1. Example problem: −100 N2. 117.6 N-m3. Answers are:

a. m2 = 2 kilograms

b.

c. It should be placed to the right. The torque due to m1 is greater than the torque due to m2.

d. m3 = 8 kilograms4. Answers are:

a. 20 N force: 3.0 N-m clockwise or − 3.0 N-m15 N force: 3.75 N-m counterclockwise or + 3.75 N-m5 N force: 1.75 N-m clockwise or − 1.75 N-m

b. 1.0 N-m clockwise or –1.0 N-mc. The force creates a counterclockwise torque of 1.0 N-m to

balance the other torques. F = 3.33 N upward5. –118 N-m; clockwise, gravity is a downward force 6. Answers are:

a. 5 N-m clockwise or − 5 N-mb. 8.66 N-m clockwise or − 8.66 N-mc. 0 N-md. 13.66 clockwise or − 13.66 N-m

m1m2------- 3=

Answer Keys

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Skill Sheet 10.1: Mechanical AdvantagePart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Sample problem: mechanical advantage = 52. input force = 100 N3. output force = 26 N4. mechanical advantage = 3

5. mechanical advantage = 36. Ratio = 7.5; Lout = 0.23 metersPart 3 answers:1. Sample problem: mechanical advantage = 0.32. output force = 150 N3. mechanical advantage = 1.5

Skill Sheet 10.2: WorkPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Work is force acting upon an object to move it a certain

distance. In scientific terms, work occurs ONLY when theforce 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.Part 3 answers:1. Sample problem: 50 joules2. 100 joules3. 100,000 joules4. No work was done by the mouse. The force on the ant was

upward, but the distance was horizontal.5. 1.2 meters6. 31 pounds

7. 2,500 N or 562 pounds8. 54 joules9. 683 joules10. 0.5 meters11. 60,000 joules12. The people-mover accomplished 70,000 joules of work. The

person was 600 N plus 100 N for the bag. Therefore, the workdone by the people-mover was 700 N times 100 meters. Theperson did 100 joules of work. The total amount of work donein this situation was 70,100 joules.

Part 4 answers:1. 17.3 joules2. 3.47 joulesPart 5 answers:1. 1,768 joules2. 490 joules

Skill Sheet 10.3: Potential and Kinetic EnergyParts 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:

4. The object with more mass had more kinetic energy whilebeing lifted. The kinetic energy of the 2-kilogram object was4 joules. The kinetic energy of the 4-kilogram object was 18joules.

5. Although each object is lifted to the same height, the 4-kilogram object has more potential energy because it hasmore mass.

6. Answers are:a.

b. 1.7 metersc. 5.7 m/sec

7. Answers are:a. 450 joulesb. 450 joulesc. 46 meters

8. The potential energy is zero because the ball is not off theground. Height is zero.

9. 25,00 joules10. 4 m/sec11. 75 kilograms12. The energy of the ball at position B equals the potential

energy plus the kinetic energy at this position. Also, theenergy at position B equals the potential energy at position Aor the kinetic energy at position C.a. The equation that relates the energy at B to the potentialenergy at A:

where hA = 3 meters; hB = 1 meter.b. Solving for velocity at position B:

Shelf height (meters) Potential energy (joules)

1.0 51.5 7.52.0 10

588 kg m

sec2----------× 588 N=

EB EBPEBK

+ mghB12---mvB

2+ mghA= = =

v 2g hA hB–( ) 2 9.8× m

sec2---------- 3 1–( )m 6.26 m

sec2----------= = =

Answer Keys

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Skill Sheet 11.2: PowerPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Sample problem: 250 watts2. 50 watts3. 741 watts

4. 720 watts5. work = 500 joules; power = 33 watts6. work = 1,500 joules; time = 60 seconds7. force = 25 newtons; power = 250 watts8. distance = 100 meters; power = 1,000 watts9. force = 333 newtons; work = 5,000 joules

Skill Sheet 12.1: MomentumParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. Sample problem: 0.6 km-m/sec2. Sample problem: 2.5 kg-m/sec3. 350 kg-m/sec4. 375 kg-m/sec5. The fullback because the defensive back has more

momentum.

6. 6.7 kilograms7. 400 m/secPart 4 answers:1. Sample problem: - 3.33 m/sec or 3.33 m/sec to the left2. v3 = - 2 m/sec3. m1 = 60 kilograms4. v3 = 1.25 m/sec5. v3 = - 1 m/sec or 1 m/sec to the west6. v4 = 12.8 m/sec

Skill Sheet 12.2: Rate of Change of MomentumPart 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. Force = 200,000 N. At this level of force, after a couple of

hits with a wrecking ball, any impressive-looking wallcrumbles to pieces.

2. 3 seconds3. 250 N4. 750 N5. 75 million N6. Answers are:

a. The force created on the egg is about:

b. The force created on the egg by the person is:.

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:

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 MotionPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Answers are:

a. 2 secondsb. 0.5 Hzc. 24°

2. Amplitude decreases due to the friction between the swingchain and the frame to which it is connected. The gradual lossof amplitude of a pendulum (an oscillator) due to friction iscalled damping. Eventually, due to friction, the pendulum willstop moving and hang straight down.

3. Answers are:a. 100 centimetersb. 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.

0.05 kg 10 m/sec×0.001 sec

---------------------------------------------- 500 N=

50 kg 9.8 m/sec2× 491 N=

0.05 kg 10 m/sec×0.5 sec

---------------------------------------------- 1 N=

Answer Keys

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Skill Sheet 14.1 WavesPart 1 answers:1. The answer is:

2. Two wavelengths3. The amplitude of a wave is the distance that the wave moves

beyond the average point of its motion. In the graphic, theamplitude of the wave is 5 centimeters.

4. The frequency of a wave is the inverse of the period or1/period in seconds. The frequency is also the number ofwaves that pass a certain point per second.

5.

Part 2 answers: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 arethe same.

2. 2.5 meters

3. 50 m/sec4. The speed of wave A is 75 m/sec. The speed of wave B is

65 m/sec. Wave B is faster.

Part 3 answers:Answers are:

1. You can easily determine the harmonics of a vibrating stringby counting the number of “bumps” on the string. The firstharmonic (the fundamental) has one bump. The secondharmonic has two bumps and so on.

2. The wavelength of the fundamental harmonic of a 5-metervibrating string would be 10 meters. The “one bump” on thestring at this harmonic represents half of wavelength.Therefore, the entire wavelength is: 2 × 5 meters.

Skill Sheet 15.1: Decibel Scale ProblemsPart 1 answers: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.2. 50 dB3. Twice as loud.4. At 90 dB, the sound is doubled three times; 23 = 8 times as

loud.5. 10 dB6. A house in the city is 20.5 or 1.5 times as loud as a house in

the country.

Part 2 answers: 1. Example problem: Since every 20 dB has 10 times greater

amplitude, a jackhammer 10 feet away (90 dB) has anamplitude 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 afactor of 10 or 103. Therefore, the amplitude of the soundwaves from a jackhammer are 1,000 times greater than theamplitude of the sound waves from a country home.

4. The difference in decibels between the two levels of sound is20 dB. Therefore, the difference in amplitude of the soundwaves is 10 times. Normal conversation sound waves have a10 times greater amplitude than restaurant conversation soundwaves.

Skill Sheet 16.1: Light Intensity ProblemsPart 1 answers:1. Example problem: 4.8 W/m2

2. 0.0478 W/m2

3. 0.0119 W/m2

4. If distance from a light source doubles, then light intensitydecreases by a factor of 4. Example: 4 × 0.0119 W/m2

approximately equals 0.0478 W/m2 (see questions 2 and 3).5. Answers are:

a. 0.005 W/m2

b. 0.05 W/m2

c. 0.5 W/m2

d. 5 W/m2

6. The watts of a light source and light intensity are directlyrelated. This means that if you use a light source that has 10times 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 0.318

W/m2. At 1% efficiency, the light intensity is 0.003 W/m2.

frequency 2 waves0.05 second---------------------------- 40 waves

second---------------------- 40 Hz= = =

Table 1: Frequency, harmonic and wavelength dataHarmonic # Frequency

(Hz)Wavelength

(m)Speed of the Wave Frequency times

wavelength(m/sec)

1 3 6 18

2 6 3 18

3 9 2 18

4 12.0 1.5 18

5 15 1.2 18

6 18 1 18

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Skill Sheet 17.1A: The Law of ReflectionPart 1 answers:1a and 1b. The answers are:

2. The angle of reflection will be 30 degrees.3. Each angle will measure 45 degrees.

4. The answer is:

Part 2 answers: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. Theangle at point B is 41.5 degrees and the angles of incidenceand reflection at this point are each 20.8 or 21 degrees.

Skill Sheet 17.1B: RefractionPart 1 answers:1. We can solve for θr from the relation nisinθi = nrsinθr

and the angle θr is given by the inverse sine of 0.285:θr = sin–1(0.282) = 16.38° = 16°

2. 48.6 degrees3. 48.6 degrees4. 12 degrees5. The incident index of refraction is larger. The ratio of the

refraction indices is:

6. 41.8 degrees7. The critical “incident” angle corresponds to the case when the

angle of refraction becomes equal to 90 degrees. Therefore:

and θr = sin–1(0.43) = 24.6°8. 48.8 degrees

Skill Sheet 17.2: Ray DiagramsPart 1 answers:1. A is the correct answer. Light travels in straight lines and

reflects off objects in all directions. This is why you can seesomething from different angles.

2. C is the correct answer. When light goes from air to glass itbends about 13 degrees from the path of the light ray in air.The light bends toward the normal to the air-glass surfacebecause air has a lower index of refraction compared to glass.Then, when the light re-enters the air, it bends about 13degrees away for the light path in the glass and away from thenormal.

3. As a ray of light approaches glass at an angle, it bends(refracts) toward the normal. As it leaves the glass, it bendsaway from the normal. However, if a ray of light enters apiece of glass perpendicular to the glass surface, the light raywill slow, but not bend because it is already in line with thenormal. This happens because the index of refraction for air islower than the index of refraction for glass. The index ofrefraction is a ratio that tells you how much light is slowedwhen it passes through a certain material.

4. A is the correct answer. Light rays that approach the lens thatare in line with a normal to the surface pass right through,slowing but not bending. This is what happens at the principal

axis. however, due to the curvature of the lens, the parallellight rays above and below the principal axis, hit the lenssurface at an angle. These rays bend toward the normal (thisbending occurs toward the fat part of the lens) and are focusedat the focal point of the lens. Past the focal point, the rayscross.

5. The answer is:

Part 2 answers:There are no questions to answer in Part 2.

sinθrnrni----- sinθi

1.01.5------- sin25° 1.0

1.5------- 0.423 0.282====

nrni----- sin45

sin30------------- 1.4==

sinθr1.02.4------- sin90° 1.0

2.4------- 1.0( ) 0.43 ===

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Part 3 answers:1. The answer is:

2. The answer is:

3. A lens acts like a magnifying glass if an object is placed to theleft of a converging lens at a distance less than the focallength. The lens bends the rays so that they appear to becoming from a larger, more distant object than the real object.These rays you see form a virtual image. The image is virtualbecause the rays appear to come from an image, but don’tactually meet.

Skill Sheet 17.3: Thin Lens FormulaPart 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. 6.67 centimeters2. 6 centimeters3. 20 centimeters4. Answers are:

a. -30 centimetersb. A virtual image. It is on the same side as the object.

5. -4 centimeters6. It is a concave (diverging) lens. It is a virtual image.Part 4 answers:1. Using the thin lens formula to prove that di is 11.7

centimeters:

2. Answers are:a. Finding the object and image distances: . Given

this proportion, the image distance is x and the object distanceis 2x.

Therefore,

b. The answer is:

Skill Sheet 18.1: The Speed of LightPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Green2. Green3. 400 × 10–9 m4. 517 × 1012 Hz5. 652 × 10–9 m6. 566 × 1012 Hz

7. The answer is:

8. λ = 0.122 meter9. λ = 3.3 meters10. f = 6 × 1016 Hz11. 1.0 × 1015 Hz12. 1.0 × 10-4 Hz13. Answers are:

a. f = 3 × 10-19 mb. 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

111.7---------- 1

25------+ 1

8---=

0.085 0.04+ 0.125=

147------

dodi------=

12x------ 1

x---+ 1

10------=

32x------ 1

10------=

x 15 cm=2x 30 cm=

α tan-1 1430------⎝ ⎠

⎛ ⎞ tan-1 0.467( ) 25°= = =

λ1λ2------

f2f1----=

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Skill Sheet 18.3: Albert Einstein1. Scientists tested Einstein's theory of relativity by

photographing a solar eclipse on November 8, 1919 fromlocations in Brazil and the African island of Principe. IfEinstein's theory were correct, light from a cluster of starscalled the Hyades behind the dimmed sun should be bent intothe gravitational dimple created by the sun, making the starsappear slightly out of alignment. The photographs confirmedEinstein's predictions. While some scientists debated theaccuracy of this method and doubted Einstein's theory,conclusive evidence was provided by the European SpaceAgency's Hipparcos satellite, which in the years between1989 and 1993 charted the positions of the stars with greatprecision and confirmed Einstein's prediction that gravitybends light.

1. Brownian motion was first observed by British botanistRobert Brown in 1827. He observed through a microscopethat pollen grains suspended in water appeared to moveerratically. He thought maybe it was because they were alive,but then found that ground glass and other non-livingmaterials exhibited the same motion. He published his resultsin 1828. Eighty years later Einstein explained that the erraticmotion was due to water molecules bumping into the smallparticles. Brownian motion can be demonstrated bysuspending pollen grains (scraped from the anther of a lilyflower, available year-round from a florist shop), graphitescraped from a #2 pencil, or talcum powder in water. Placeseveral drops of the suspension in a cavity microscope slideand observe under a microscope.

Skill Sheet 19.2: Using an Electric MeterPart 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. First battery: 1.553 volts; second battery: 1.557 volts2. First bulb: 1.514 volts; second bulb: 1.586 volts3. 3.113 volts4. The answer is:5. post #1: 0.0980 amps;

post #2: 0.0981 amps;post #3: 0.0978 amps;post #4: 0.0980 amps;post #5: 0.0980 amps

6. Sample answer: Sometimes light bulbs are made with frostedglass, which makes it impossible to see the tungsten filamentinside. You can use an electrical meter to figure out if thetungsten 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 (wherethe bulb is threaded to fit into the socket). Place the other leadon the “bump” at the base of the light bulb.Check the meter reading. If the meter reports a small value forresistance, the tungsten filament is intact. If the meterdisplays “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 topass through air to complete the circuit. Because theresistance of air is too high for a small amount of current tocross, the meter reports very high resistance.

7. A fuse contains a wire that will melt or break if more than asafe amount of current (12 amps, for example) is applied.When the wire melts or breaks, the circuit is broken and thecurrent stops. If you measure the resistance of a fuse and themeter displays “OL” or the infinity symbol, the fuse musthave a melted or broken wire. The current would have to passthrough air to complete the circuit. As a result, the meterreports very high resistance.

8. Wires would normally have very low resistance, because theyare made of copper or other metals through which electricityeasily flows. If the measured resistance of a wire is very high,it must have a break in it.

Skill Sheet 19.3A: Ohm’s LawPart 1:There are no questions to answer in Part 1.Part 2 answers:1. 10 amps2. 3 ohms3. 120 volts4. Answers are:

a. Circuit A: 6 volts; circuit B: 12 voltsb. 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 ampse. Circuit A: 0.5 amp; circuit B: 1 ampf. The current would decrease in each circuit because the

resistance would increase. The current would be cut in half.

5. Answers are:a. Circuit A: 0.5 amp; circuit B: 1 amp

b. Circuit A: 0.25 amp; circuit B: 0.5 ampc. If voltage remains the same, adding bulbs in a series

circuit decreases the brightness of the bulbs.6. Current increases.7. Current decreases.

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Skill Sheet 19.3B: Ben Franklin1. If the kite had been struck by lightning, the amount of charge

coming down the hemp string would most likely haveelectrocuted Franklin.

2. A lightning rod is ametal rod attached tothe roof of a building. Athick cable stretchesfrom the rod to a metalstake buried in theground. When lightningstrikes the rod, itfollows the path of leastresistance-from the rod,through the cable, intothe ground, where thecharge can safelydissipate.

Skill Sheet 20.1: Parallel and Series CircuitsPart 1 answers:1. Answers are:

a. 12 voltsb. 4 ohmsc. 3 ampsd. 6 voltse. The answer is:

2. Answers are:a. 2 ohms per resistorb. The voltage drop per resistor is 1 volt.c. The answer is:

3. Answers are:a. 6 ohmsb. 1.5 ampsc. 2 ohm resistor: 3 volts

3 ohm resistor: 4.5 volts

1 ohm resistor: 1.5 voltsd. The answer is:

Part 2 answers:1. Answers are:

a. 12 voltsb. 6 ampsc. 12 volts

2. The current flow through each bulb in series is 3 amp.Through each bulb in parallel, current flow is 6 amps. Thebulbs in the parallel circuit are brighter.

3. Answers are:a. 9 voltsb. Branch with 2 ohm resistor: 4.5 amps

Branch with 3 ohm resistor: 3 ampsBranch 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.

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Skill Sheet 20.2: Network CircuitsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. IR2 = IR3 = 3 A; IR1 = 3 A2. RTotal = 2 Ω3. ITotal = 1 A4. Answers are:

a. No, current does notflow through R5,because it does nothave to. The currentwill go directly to R3and R4.b. RTotal = 9 Ω. c. ITotal = 2 Ad. IR3 = 1 Ae. The voltage across

R5 is zero because no current flows through this circuit.5. ITotal = 1.5 A;

VR4 = 4.5 V6. V5Ω = 5 V (ITotal = 1 A)7. For R4 = 10 Ω, VA = 8 V, VB = 4 V, VAB = 8 V - 4 V = 4 V

Skill Sheet 20.3: Electrical PowerParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. Sample problem: 270,000 joules2. 5,400 seconds, or 90 minutes3. 1,080 watts4. 240 volts5. 0.5 A6. 8.3 APart 4 answers:1. Answers are:

a. 1.2 kWb. 0.4 kWhc. 12 kWh/monthd. $1.80

2. 0.63 amps3. 960 W = 0.96 kW4. Answers are:

a. 3 Vb. 3 Ωc. 1 Ad. 3 W

5. Answers are:a. 24 Ωb. 600 W; 0.6 kW

6. Answers are:a. 20 Ab. 11 Ω

7. Answers are:a. 12 Ωb. 1,200 W = 1.2 kWc. 0.6 kWhd. 18 kWhe. $2.52

8. Your friend with the 300-watt stereo will pay more for theenergy used in an hour of play. This is because more twice asmuch energy is used per second for the 300-watt stereo thanfor the 150-watt stereo.

9. Since power is voltage times current, as current changes, thepower used per appliance will change. Each appliance can bedesigned to carry the amount of current needed to run well.Most electrical appliances in a dining room or living roomwould run on 10 amps of current or less. However, in akitchen you may have an electric stove and oven that requires40 amps of current. A special circuit with thicker wires isinstalled there so that the circuit does not overheat and pose afire hazard.

Skill Sheet 21.2: Coulomb’s LawPart 1 answers:There are no questions to answer in Part 1.Part 2 answers: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 = 9 × 109 N

9. The total charge equals 2.7 × 10-8 C. The square root of thisnumber is the value for each charge and is equal to1.7 × 10–4 C.

10. 4.5 N; a cat weighs 49 N, this is about eleven times moreforce than the force between the particles.

11. 5.5 × 10-3 C12. distance = 6.7 m13. distance = 0.03 m14. q = 2.96 × 10-12 C

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Skill Sheet 22.3: Magnetic EarthPart 1 answers:1. Scientists believe that the motion of molten metals in Earth’s

outer core create its magnetic field.2. Seven percent of 0.5 gauss is 0.035 gauss. In 100 years,

Earth’s strength could be 0.465 gauss.3. The poles could reverse within the next 2,000 years. During a

reversal, the field would not completely disappear. The mainmagnetic field that we use for navigation would be replacedby several smaller fields with poles in different locations.

4. Rock provides a good record because as the rock is made,atoms in the rock align with the magnetic field of Earth.(Actually, oceanic rock is made of a substance calledmagnetite!) Rock that was made 750,000 years ago wouldhave a north-south orientation that is exactly opposite thenorth-south orientation of rock that is made today. Therefore,we can use the north-south orientation of bands of rock on thesea floor to understand how many times the poles havereversed over geologic time.

Part 2 answers:NOTE: In many references, magnetic south pole is referred to as

‘magnetic north pole’ because it is located at the geographicnorth pole. This terminology can be confusing to studentswho know that opposite poles attract. The north pole of acompass needle is in fact the north end of a bar magnet. Thisis why we think it is best to use the term magnetic south poleas the point to which the north end of a compass needle isattracted. For more information about Earth’s magnetism seehttp://www.ngdc.noaa.gov/.

1. Answers are:a. Both the magnetic south pole and geographic north are

located at the Arctic.b. The magnetic south pole is about 1,000 kilometers from

the geographic north pole.c. Magnetic south pole is the point on Earth’s surface that

corresponds to Earth’s south pole if you think of Earth’s core as a bar magnet. Magnetic south pole is located at a distance from the geographic north pole or ‘true north.’ True north is a point on Earth’s surface that we call north. ‘True north’ and ‘true south’ follow Earth’s axis. If we want to go north, we need to head toward Earth’s geographic north. The tool we use to head north, however, points us toward the magnetic south pole. We use magnetic declination to correct for this.

2. A compass needle is a bar magnet. It’s north pole is attractedto Earth’s magnetic south pole (if you consider the interior ofEarth is like a bar magnet). Using magnetism, we can find ourway using north as a reference point. However, usingmagnetism actually points us a bit off course because themagnetic south pole is not located at the same position as truenorth.

3. Answers are:a. Example problem: northeast.b. southc. westd. easte. southeastf. northwest

Part 3 answers:1. If you didn’t correct your compass for magnetic declination,

you would be off course and possibly get lost.2. Yes, magnetic declination equals zero on Earth’s surface

along a line that goes from New Orleans through the easternedge of Minnesota up through Churchill, Canada. However,the location of the zero-degree line is always changing. Formore information about Earth’s magnetic declination andmagnetism, see http://www.ngdc.noaa.gov/.

Skill Sheet 23.1: Magnetic Fields and ForcesPart 1 and 2:There are no questions to answer in Parts 1 through 2.Part 2 answers:1. 4 × 10-6 tesla, 0.04 gauss2. 0.04 meter or 4 centimeters3. The answer is shown in the

graphic at right.

4. The force from each wire istoward the other wire. Thewires are attracting eachother.

5. The force would change by a factor of ½. Since the current inthe wires remains the same, the magnetic field at the wireswill be reduced by half as the distance increases from r to 2raccording to the formula:

6. 0.083 tesla7. The coil has 799 turns.

Skill Sheet 23.3 Michael Faraday1. The electric motor, invented by Faraday, is used in all sorts of

household appliances including electric fans, hair dryers, foodprocessors, and vacuum cleaners. Electromagnetic inductionis used by local power plants to generate the electricity that Iuse every day.

2. Iron filings are available from many science supply catalogs.Place some iron filings on a piece of clear plastic (such as anoverhead transparency). Place the plastic over a magnet toobserve the field lines. Investigation 23.3 providesinstructions on how to demonstrate electromagneticinduction.

B 2 10 7–× I

r--=

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Skill Sheet 24.3: Binary Number ProblemsPart 1 answers:1. Answers are:

2. 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 = 01100011b. 127 = 01111111

4. The number 127 cannot be represented by a 6-bit binarynumber. You need 8-bits to represent 128 (10 000 000).

Part 2 answers:1. Answers are:

a. Example problem: 204b. 170c. 255d. 9e. 6f. 2g. 3h. 2

Skill Sheet 25.2: Temperature ScalesParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers: 1. 7.2°C2. 177°C3. 107°CPart 4 answers:There are no questions to answer in Part 4.Part 5 answers:1. 375°F2. 450°F

3. The friend assumed you were speaking of degrees Celsius.Fifteen degrees Celsius equals 59°F, which is a much milderoutdoor temperature than 15°F (15°F = -9.4°C).

Part 6 answers:1. 98 K2. 275 K = 2°C

No, the mysterious, silver substance has a much highermelting point than mercury.

3. The thermometer is calibrated to the Fahrenheit scale. On theKelvin scale, 90 K is too cold (-298°F and -183°C), and 90°Cis too hot, just 10 degrees less than the boiling point of water.

Skill Sheet 26.3: Heat TransferPart 1 answers:1. Example answer: 10.7 watts2. 93.5°C3. 53°C4. 0.1 watt5. Sample answer: The heat conduction equation can be used to

evaluate roofing material. The surface area, thickness of theroof, and thermal conductivity of the material used can betaken into account using the heat conduction equation. It isimportant that roofing material conduct very little heat so thata house stays cool in the summer and warm in the winter.

Part 2 answers:1. Example answer: 210 watts2. 35 watts3. Sample answer: The area value used in the convection

equation is the area of the contacting fluid. My hypothesis forwhy the heat transfer coefficient values for water are so highis that gases have little contacting area—the molecules arevery spread out so not as much energy can be transferred, andoil is viscous so heating does not happen as quickly. Liquid

water has high molecular contact for heating efficiently andlow viscosity.

4. Free convection is the flow of fluids due to differences indensity or temperature. Forced convection is the flow offluids due to circulation by a fan or a pump. A heat transfercoefficient is the amount of watts per square meter for a giventemperature. Values are higher when convection is forcedbecause energy is used to heat a given area.

5. Sample answer: The convection equation can be used todesign tents for cold weather camping. You would need toknow the heat transfer coefficient for wind (forcedconvection), the area of the tent (or one side of the tent), andthe temperature differences between the outside environmentand a ‘safe’ temperature for inside the tent.

Part 3 answers:1. Example answer: 3,377 K2. 0.0365 watts3. 2.21 × 10-7 m2

4. 317 watts; 0°C

Binary DigitsNumber 8 4 2 1

0 0 0 0 01 0 0 0 12 0 0 1 03 0 0 1 14 0 1 0 05 0 1 0 16 0 1 1 07 0 1 1 18 1 0 0 09 1 0 0 110 1 0 1 011 1 0 1 112 1 1 0 013 1 1 0 114 1 1 1 015 1 1 1 1

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5. Heat transfer at 0°C is greater than heat transfer at absolutezero (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 usedto study stars. For example, the equation can be used to infer

the radius of a star (the surface area of a sphere is 4πr2). Also,knowing the amount of radiation from a star allows you toinfer its temperature and brightness and color. The higher itstemperature, the brighter a star will be.

Skill Sheet 27.1 Stress and StrainParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3:1. σ = 3.6 × 106 N/m2; e = 2.1 × 10-4

2. 2.5 centimeters3. 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

Skill Sheet 27.2 Archimedes1. How Archimedes solved the

problem of determining the goldcrown’s purity:

2. Inventions attributed toArchimedes include warmachines (such as a lever usedto turn enemy boats upsidedown), the Archimedes screw,compound pulley systems, and aplanetarium. There is somedebate about whether heinvented a water organ and asystem of mirrors and/or lensesto focus intense, burning lighton enemy ships. Students canuse the Internet or library to findmore information and diagramsof the inventions.

Skill Sheet 27.3: Gas LawsParts 1 through 2:There are no questions to answer in Parts 1 through 2.Part 3 answers:1. 22.9 L2. 563 kPa3. 570 L4. 40 L5. 62.5 L

Part 4:There are no questions to answer in Part 4.Part 5 answers:1. 2.45 kilograms2. The mass is 0.26 kilogram and the new volume is 0.4 liters.3. 4.2 atmospheres; the problem is solved using P1/P2 = V2/V1.

Because V2 = V1 + 0.20V1, V1 can be cancelled from theequation so that the only variable is P2.

Skill Sheet 28.1A: The Structure of the AtomPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Lithium, Li2. Carbon, C3. Hydrogen, H4. Hydrogen, H (a radioactive isotope, 3H, called tritium)5. Beryllium, BePart 3 answers:1. Atomic weight ≅ 72. Atomic weight ≅ 123. Atomic weight ≅ 14. Atomic weight ≅ 35. Atomic weight ≅ 96. The atomic mass of 1.00794 amu represents an average of the

masses and abundance of all to the hydrogen isotopes(hydrogen-1, hydrogen-2, and hydrogen-3). Because this

atomic mass is so close to “1,” we can assume that the mostabundant isotope is hydrogen-1.

Part 4 answers:1. 2H = 1 proton, 1 neutron2. 45Sc = 21 protons, 24 neutrons3. 27Al = 13 protons, 14 neutrons4. 235U = 92 protons, 143 neutrons5. 12C = 6 protons, 6 neutronsPart 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 areso light they contribute very little mass. The mass of theproton is 1,835 times the mass of the electron. Neutrons havea bit more mass than protons, but the two are so close in sizethat we usually assume their masses are the same.

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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 thischarge. Therefore, the overall charge of this atom (now, calledan ion) is +1.

3. Answers are: (a) one electron, (b) five electrons, and(c) 14 electrons.

4. This sodium atom has ten electrons, 11 protons, and 12neutrons.

Skill Sheet 28.1B: The Periodic TablePart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. Fluorine2. Argon3. Manganese4. Phosphorous5. TechnetiumPart 3 answers:1. Iron, 55.8 amu2. Cesium, 132.9 amu3. Silicon, 28.1 amu4. Sodium, 23.0 amu5. Bismuth, 209.0 amuPart 4 answers:1. Protons = 12; neutrons = 12

2. Protons = 7; neutrons = 73. Protons = 19; neutrons =20Part 5 answers:1. 211Po – α → 207Pb; 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 iscalled the transuranic series and takes millions years. Becausethis chain of events is so predictable, the proportion ofuranium to lead is the main “clock” used to date geologicevents in very deep time.

2. 226Ra– α → 222Rn; radon is a radioactive gas and can leakinto homes from underground sources.

3. 222Rn– α → 218Po; this isotope of polonium is possiblyradioactive.

Note: Work with radioisotopes is only done casually with theperiodic table of elements. Typically, a different table, thechart of radionuclides, is used because it includes data foreach isotope.

Skill Sheet 28.1 Lise Meitner1. The graphic at right

illustrates fission:2. Some topics students

may research anddescribe includenuclear powerplants, nuclearweapons, nuclear-powered submarinesor aircraft carriers.

Skill Sheet 28.2 Niels Bohr1. 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 DiagramsPart 1 answers:There are no questions to answer in Part 1.

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Part 2 answers:

Part 3 answers:

Skill Sheet 29.3 Chemical EquationsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers: Part 3 through 5 answers:

There are no questions to answer in Parts 3 through 5.Part 6 answers:1. 4Al + 3O2 → 2Al2O32. CO + 3H2 → H2O + CH43. 2HgO → 2Hg + O2 4. CaCO3 → CaO + CO2 (already balanced)5. 3C + 2Fe2O3 → 4Fe + 3CO2 6. N2 + 3H2 → 2NH37. 2K + 2H2O → 2KOH + H28. 4P + 5O2 → 2P2O5 9. Ba(OH)2 + H2SO4 → 2H2O + BaSO410. CaF2 + H2SO4 → CaSO4 + 2HF11. 4KClO3 → 3KClO4 + KCl

Skill Sheet 30.1A RadioactivityParts 1 through 2:1. There are no questions to answer in Parts 1 through 2.Part 3 answers:1. In the answers below, “a” is alpha decay and “b” is beta

decay.

2. 5.6 × 10-5 kilograms or 0.056 grams3. 1.4 × 10-9 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 thisradioactive isotope is 9 months.

Part 4 answers:1. 0.8 W/m2

2. 3.6 × 1013 reactions per second

Element Chemical Symbol

Total Electrons

# of Valence Electrons

Dot Diagram

Potassium K 19 1

Nitrogen N 7 5

Carbon C 6 4

Beryllium Be 4 2

Neon Ne 10 8

Sulfur S 16 6

Elements Dot Diagram for Each Element

Dot Diagram for Compound

Formed

Chemical Formula

Na and F NaF

Br and Br Br2

Mg and O MgO

Reactants Products Chemical EquationHydrochloric acid

HCland

Sodium hydroxideNaOH

WaterH2Oand

Sodium chlorideNaCl

HCl + NaOH → NaCl + H20

Calcium carbonateCaCO3

andPotassium iodide

KI

Potassium carbonateK2CO3

andCalcium iodide

CaI2

CaCo3 + KI → K2CO3 + CaI2

Aluminum fluoride

AlF3and

Magnesium nitrateMg(NO3)2

Aluminum nitrateAl(NO3)3

andMagnesium

fluorideMgF2

AlF3 + Mg(NO3)2 → Al(NO3)3 + MgF2

a. Answers are:

a→ b→ b→ a→ a→

a→ a→ a→ b→ b→

a→ b→ b→ a→

23892 U 234

90 Th 23491 Pa 234

92 U 23090 Th

22688 Ra 222

86 Rn 21884 Po 214

82 Pb 21483 Bi

21484 Po 210

82 Pb 21083 Bi 210

84 Po 20682 Pb

b. Answers are:

b→ a→ a→ b→ a→

a→ b→ a→ a→ a→

b→ a→ b→

24094 Pu 240

95 Am 23693 Np 232

91 Pa 23292 U

22890 Bi 224

88 Ra 22489 Ac 220

87 Fr 21685 At

21283 Bi 212

84 Po 20882 Pb 208

83 Bi

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Skill Sheet 30.1B: Marie and Pierre Curie1. Marie Curie proposed that uranium rays were an intrinsic part

of uranium atoms, which encouraged physicists to explore thepossibility that atoms might have an internal structure.

2. Marie and Pierre worked with uranium ores, separating theminto individual chemicals. They discovered two substances

that increased the conductivity of the air. They named the newsubstances polonium and radium.

3. Answers include nuclear physics, nuclear medicine, andradioactive dating.

Skill Sheet 30.3: Ernest Rutherford1. Rutherford first described two different kinds of particles

emitted from radioactive atoms, calling them alpha and betaparticles. He also proved that radioactive decay is possible.He developed the planetary model of the atom, and was thefirst to split an atom.

2. Rutherford's planetary model suggested that an atom consistsof a tiny nucleus surrounded by a lot of empty space in whichelectrons orbit in fixed paths. Subsequent research has shown

that electrons don't exist in fixed orbitals. The Heisenberguncertainty principle tells us that it is impossible to know bothan electron's position and its momentum at the same time.Scientists now discuss the probability that an electron willexist in a certain position. Computer models predict where anelectron is most likely to exist, and three-dimensional shapescan be drawn to show the most likely positions. The sum ofthese shapes produces the charge-cloud model of the electron.

Skill Builder: Internet Research SkillsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:1. “science museums” + “south carolina” not “columbia”2. “dog breeds” not “expensive”3. “producing electricity” not “coal” or “natural gas”Part 3 answers:There are no questions to answer in Part 3.Part 4 answers:1. Answers will vary. Sites that are authoritative may be non-

profit sites (recognizable by having “org” as the extension inthe web address). These sites often provide information tolarge, diverse, groups and are not supported by advertising.Sites that are supported by advertising can be authoritative,but may be biased in the information presented. Anothercharacteristic of authoritative sites are that they are activelybeing up dated on a regular basis.

2. Answers will vary. Reasons for why a source may not seem tobe authoritative include the author of the site is not affiliatedwith an organization and does not have obvious credentials,and the information seems to be biased.

3. Answers will vary. Intended audiences can be young children,pre-teens, teenagers, adults, or select groups of people(women, men, people who like dogs, etc.).

4. Answers will vary.

Answer Keys

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Skill Builder: Graphing SkillsPart 1 answers: There are no questions to answer in Part 1.

Answer to question: Twenty dollars is in the cash box at thebeginning of the car wash to make change.

Part 3 answers:

Part 4 answers:1. The plot on graph 1 is a straight line, whereas, the plot on

graph 2 is a line that curves upward. The y-values in graph 2steadily increase faster than the y-values for graph 1. Forexample, both data sets have x-values: 1, 2, 3, 4, and 5. Foreach x-value for graph 1, the y-value increases by $15 startingat $20. The increase in y-values for graph 2 is exponential.

2. The rate of earning money washing cars is $15/hour. For eachhour spent washing cars, the amount of money in the cash boxwill increase by $15.

3. As one person is added to the pool, a greater amount of wateris splashed out of the pool. The increase in the amount ofwater lost increases exponentially.

Part 5 answers:1. Graph 12. Graph 2

Table 1: Money in cash box vs. number of hours washing cars

Number of hours washing cars

(hours)x

Amount of money in cash box ($)

y

Coordinates(x,y)

0 20 (0,20)

1 35 (1,35)

2 50 (2,50)

3 65 (3,65)

4 80 (4,80)

5 95 (5,95)

Table 2: Number of people in a swimming pool vs. amount of water splashed out of the pool

Number of people in a swimming pool

x

Amount of water splashed out of

the pool(gallons)

y

Coordinates(x,y)

0 0 (0,0)

1 1 (1,1)

2 4 (2,4)

3 9 (3,9)

4 16 (4,16)

5 25 (5,25)

6 36 (6,36)

Answer Keys

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Skill Builder: Safety SkillsPart 1 through 3 answers:There are no questions to answer in Parts 1 through 3.Part 4 answers:1. Answers will vary.2. Answers will vary.3. Answers will vary. Example: (1) Be quite and listen, (2)

Locate your safety buddy, (3) In a safe and orderly manner,follow your teacher out of the lab to a designated safelocation.

4. In many cases, Investigations require the use of chemicalsthat may cause harm to your eyes or clothing if these are notprotected. Gloves are also important when working withchemicals. For Investigations that require heat, using a hotpad is very important.

5. Teamwork helps you to complete the lab efficiently, butkeeping safe also requires teamwork. Sometimes, you mayneed someone to help you pour a chemical or perform aprocedure that would be unsafe if you tried it by yourself. Ateam of people can also work together to each everyone onthe team safe. On your own, it is more difficult to be aware ofall the possible dangers in a laboratory setting.

6. Cleaning up after an Investigation prepares the work space forthe next day’s Investigation. A clean work space is more safebecause all chemicals and any sharp or dangerous objects areremoved. Clean up also involves turning off any appliancesthat could heat up and cause a fire.

7. The proper procedure is “a.” You should always pour anacidic solution into water. In this way, the acid is immediatelydiluted and made more safe. If you pour water into an acid,the interaction could be explosive and result in acid beingsplashed into your face or on your clothes.

8. (1) Immediately tell my teacher, (2) Listen carefully andfollow any safety instructions provided by the teacher, and (3)Follow the appropriate safety guidelines.

9. Answers are: a. First, I would make sure that my classmates know to stay

away from the broken beaker and I would tell my teacher what had happened. Then, I would clean up the glass with a dust pan and a brush. I would not use my hands to clean up the glass. I would place the broken glass in a cardboard box, seal the box, and label it “sharps.”

b. I would make sure my classmates know about the water so they don’t slip. I would tell my teacher as soon as possible. I would begin placing paper towels on the wet spot as soon as possible. Carefully, I would work with my classmates to clean up the spill. It would be best to use gloves during the clean up in case any chemicals are mixed in with the water.

c I would tell my teacher about the smell and follow any safety instructions given to me by the teacher. I would help to make sure that the lab is well ventilated (by helping to open windows and doors, for example). I would ask to leave the lab to get some fresh air if I needed to do so.

d. I would stop talking and listen to any safety instructions from my teacher. I would follow the classroom plan for exiting the lab as soon as possible. I would not worry about removing my lab apron. I may remove my goggles if it seems unsafe to keep them on.

e. I would take her hand and lead her to the eye wash station as soon as possible. I would tell a classmate to tell our teacher as soon as possible. When the teacher arrives, I would let her/him help my lab partner.

f. I would stop talking and listen to any safety instructions from my teacher. I would follow the classroom plan for exiting the lab as soon as possible. I may need to use the nearest classroom fire extinguisher if my teacher is unable to do so.

Skill Builder: Significant DigitsPart 1 answers:There are no questions to answer in Part 1.Part 2 answers:

Part 3 answers:There are no questions to answer in Part 3.Part 4 answers:1. 3.0 m2

2. 0.9 liters3. The average temperature is 24.2°C.4. 40 minutes, 32 seconds

5. Student answers will vary. Sample problem: The road race is10.0 kilometers long. Figure out how long it would take youto run the race if you were to keep up the pace of running eachmile in 7.5 minutes. (Hint: One mile is equal to 1.609 km.)The answer would be 47 minutes. Table 1: Number of Significant Digits

Value How many significant digits does each value have?

36.33 min 4100 miles 1120.2 mL 40.0074 km 20.010 kg 242 students infinite (students are counted, not

measured)