Day 1

Page 1

• 1000 mg = 1 g
• 160 cm = 1600 mm
• 109 g = 0.109 kg
• 1 l = 1000 ml
• 14 km = 14000 m
• 250 m = .250 km
• 56 cm < 6 m
• 7 g > 698 mg

Page 3

1. kg
2. m
3. g
4. ml
5. mm
6. l
7. km
8. cm
9. mg
1. 2 g
2. 104000 m
3. 4.8 m
4. 5600 g
5. 0.8 cm
6. 5000 ml
7. 0.198 kg
8. 0.075 l
9. 0.5 m
10. 560 cm
11. 160 mm
12. 2.5 km
13. 65000 mg
14. 63 mm
15. 0.12 g
16. <
17. >
18. =
19. =
20. <
21. >

Day 7

Conclusions/Questions

direct, straight line

1. 5.8cm, interpolated
2. 2.20N, interpolated
3. 1.58N, interpolated (note: 0.05m = 5.0cm)
4. 8.0cm, extrapolated

Day 8

1. Who Done It assignment on page 8

Nelson

straight line, directly height

Question for Thought: (answers will vary)

1. From our problem, we converted height of the tree from 2.1m to cm.

2.1m x 100cm = 210cm

1m

We could have changed the height of the dog from 55cm to m.

55cm x 1m = 0.55m

100cm

2. The direct proportion of the shadow versus height

(210-100)height,cm 100 height

*Answer is to 2 significant figures because the shadow measurement has the least at 2 significant figures

Day 11

Page 1

1. 27 degrees, 2cm
2. 106 degrees, 2cm
3. 338 degrees, 1.5cm
4. 218 degrees, 2cm
5. 165 degrees, 2cm
6. 247 degrees, 1.75cm
7. 1) 20 m/s, 2) 20m/s, 3) 15m/s, 4) 20m/s, 5) 20m/s, 6) 17.5m/s

Day 12

Page 2

1. a) 292 degrees, b) 202 degrees, c) 180 degrees, d) 156 degrees, e) 250 degrees, f) 353 degrees, g) 15 degrees, h) 164 degrees, i) 192 degrees
2. a) scale 2m=1cm

b) scale 1m/s=1cm

Distance

1. (use p.3 as an example)
1. Scale 2.5cm=50miles

14cm x 50miles = 280miles

2.5cm

1. 280miles x 1.6093km = 450km

1mile

Displacement

1. 41 degrees
2. 450km at 41 degrees
3. 495km
4. 495km – 450km = 45km/495km = 9% off
5. 495km at 131 degrees

Scalars and Vectors

distance = 18ft, displacement = 2ft

1. The head is at the state capital

Day 14

1. Scalars (distance, speed); Vectors (displacement, velocity, acceleration)
2. A
3. B
4. A
5. The object can move 5 seconds to the North and then 5 seconds to the South and end at the same place it began.
6. Path 1 & Path 3

Day 15

1. a) 6 miles, b) 0
2. 30
3. 3 km (b)
4. a) 360m, b) 120m, c) 120m, d) 80m

Day 22

1. mass, distance, time, volume, density, speed, etc.
2. velocity, displacement, force, weight, etc.
3. scalar, number, unit
4. magnitude, direction
5. arrows, length of the arrow
6. resultant
8. subtract
9. a) vector, b) vector, c) scalar, d) scalar
1. 79.8 units, 11.0 degrees N of W
2. 947 km/h, South (922 + 25 = 947)
3. West 235 units, South 769 units
4. 8.55 km/h, 35.8 degrees E of N
1. 310 km, 29 degrees E of N
2. 7 paces, 16 degrees S of W

Day 29

1. a = Δvt = (v  v0)/t, where it is assumed that t0 = 0 and the final velocity is represented as v. Solving for v yields v = v0 + at.
2. Its velocity is still 40 m/s because zero acceleration means “no change in the velocity.”
3. It also points to the right, because the velocity is increasing and the velocity vector points to the right.
4. v = v0 + at = +40 +7(4) = 40 + 28 = 68 m/s to the right
5.    Table:
6. Because time is a scalar quantity, and the product of a vector and any number of scalar quantities is still a vector.
7. With a = 0, the displacement becomes d = v0t = +4(4) = 16 meters to the right.
9. D = v0t + (1/2)at2 = +4(4) + (1/2)(1)(4)2 = 16 + 8 = 24 meters to the right.
10. Table:
13. If an object is not accelerating, it will maintain its original speed, v0. The at term in the equation represents the change in velocity that occurs because the object is accelerating. When at points in the same direction as v0, it tells you how much extra velocity is gained over its initial velocity, v0. When at points in the opposite direction of v0, the at term tells you by how much the final velocity vector is lower than v0.
14. The v0t term is the displacement of an object if it is not accelerating. However, if an object is accelerating, you would not expect it to have the same displacement. For example, if the object were speeding up, it would be expected to travel farther, and if it were slowing down, it would be expected to travel less far. The (1/2)at2 term indicates how much the displacement changes due to the acceleration of the object. If (1/2)at2 points in the same direction as v0, this term indicates how much greater the displacement becomes. If (1/2)at2points in the opposite direction, this term indicates how much less the displacement becomes.

Day 30

1. A — The speed is zero. Over time the position doesn’t change. It’s not moving.

B – Looks like it has gone about 6 meters in 10 seconds. .6 m/s

C – Looks like it has gone about 1 meter in 10 seconds. .1 m/s

Change in Velocity

1. A and B are the same (+10 m/s); C and D are the same (-10 m/s)

Average Acceleration

1. B, A, C (A acceleration is 0. B is 0 to 5 meters in 10 seconds = 5 m/s2. C is 7 meters to 2 in 10 seconds = -.5 m/s2)

4. A shows constant velocity. The others show variable.

5. 0, +, –

6. same as 5

7. Done in 3

8. [need answer – not provided]

Day 37

1. D

Some numerical values must be estimated in order to answer this question. Estimates should be made of the braking distance (the lighter of the two bars in Figure 1) for any two speeds that have a 2:1 ratio. Estimating the braking distance requires that one starts the measurement from where the reaction distance ends. For instance, the braking distance at 20.0 mi/hr is approximately 5 meters (from 6 meters to 11 meters). The braking distance at twice this speed – 40.0 mi/hr is 17 meters (from 13 meters to 30 meters). This doubling of the speed from 20.0 mi/hr to 40.0 mi/hr results in more than a doubling of the reaction distance.

1. B

Figure 1 data is based on a Toyota Prius driven by a person with a 0.75 second reaction time. The reaction distance at 60.0 mi/hr can be read off the bar chart. Reaction distance is the darker bar. Its length is approximately 20 meters at this speed.

1. B

Figure 1 represents the reaction distance by the darker bar and the stopping distance y the lighter bar. For most listed speeds, the two bar lengths are not equal. But at 30.0 mi/hr, the darker bar and the lighter bar have the same length – 10 meters each.

1. D

The total stopping distance is the sum of the reaction distance and the braking distance. These values are listed in Table 1 for various reaction times. The maximum reaction time listed in the table is 1.00 seconds. Since this question pertains to a reaction time of 1.5 seconds, one will have to extrapolate to determine the answer. Extrapolation involves using an observed trend to estimate the value of a quantity that falls outside the range of stated values. The trend in the dreaction values is that it increases by about 4.5 meters (estimated average) for every 0.20 second increase in trxn value. So at 1.5 seconds, the dreaction value would be approximately 13.5 m (three times 4.5 m) greater than the 20.3 m value at 0.90 seconds. This would yield an estimate of 33.8 m for dreaction. The trend in dbraking values is that it remains constant at 28.2 m for a 50.0 mi/hr Toyota Prius. Summing the dreaction and the dbraking values leads to an estimate of 62 m for the total stopping distance. The closest answer to the estimated 62 meters is the 61.7 m of choice D.

1. D

What is observed in Figure 1 is that braking distance increases with increasing speed. In fact, a doubling of the speed more than doubles the braking distance. One also observes that an increase in speed causes an increase in reaction distance. This effect is not as pronounced as the effect of car speed on braking distance for a doubling of car speed doubles the reaction distance. In Table 1, one observes that an increase in car speed and in reaction time each increase the reaction distance. A doubling of reaction time will double the reaction distance. Finally one learns that braking distances tend to be greater than reaction distances and thus the greater contributor to the total stopping distance. As such, one would reason that higher speeds would be the more dominant factor in answering this question. Choice D has the highest speed of all listed speeds and thus becomes an immediate first choice. Of course the question is do any of the lower speed vehicles compensate for its shortage in braking distance by a larger reaction distance. Choice C is the only one that needs to be considered since it has both the largest reaction time and the largest speed of the remaining choices. The 50 mi/hr speed of choice C makes its stopping distance findable in Table 1. The extra 0.20 seconds in reaction time would give the choice C car an approximately 5 meter advantage … and that is only if the choice C car had the same speed (and it doesn’t). This miniscule 5 meters is not nearly enough to compensate for the larger advantage that the choice D car has in terms of its braking distance. Choice D clearly has a larger stopping distance.

Day 45

1. a
2. a.  16
b. 32
c. 4
d. 4
e. 2
f. 4
g. 16
h. 96

Day 51

1. Ball hits bat. Car pushes man. Bug hits bus.

2. Athlete hand on ball pushing inward; Ball on athlete hands outward Foot pushes downwards on the floor; Floor pushes upwards on the foot Foot pushes rightwards on the ball; Ball pushes leftward on the ball.

3. false, your body pushes down on the earth

4. ↑ FN = 600N upward ↓ FG = 600N downward

5. a) equal to; greater than b) equal to; greater than c) equal to; less than d) equal to; equal to

To calculate acceleration: Vf 2=Vl 2+2a

Force of the bullet: F=ma

Day 53

1. B

Terminal velocity occurs when the velocity of the falling filters has become constant. When the filters achieve this terminal velocity, the line on the velocity graph will have leveled off to a constant value.

1. C

The first paragraph of the passage states that “when the upward air resistance force is equal to the downward gravity force, the object encounters a balance of forces and is said to have reached a terminal velocity.” The first paragraph also states “as objects move faster, they encounter more air resistance.” So one can reason that the gravity force is a constant value on the falling filters and it is the air resistance force that is increasing. It increases because of the gain in speed; the increasing continues until the upward force of air resistance equals the downward force of gravity.

1. A

The filters hit the ground before they achieved a constant, terminal velocity value. If they could have continued to fall, the air resistance would have increased as the speed increased and terminal velocity would eventually be reached. One way to fix this problem would be drop the filters from a higher initial height so that they have a longer time to fall. They would then reach a terminal velocity.

1. C

The statement made here assumes that one can interpolate the terminal velocity of a 2.5 gram penny by finding a value in Table 1 between the trial 2 (two filters) and trial 3 (three filters). This is based on the fact that the mass of a 2.5 gram penny is equal to the mass of 2.5 coffee filters. A penny may have the same mass as 2.5 filters but there are more variables that affect the terminal velocity than the mass. The person making the statement is assuming that mass is the only factor that matters. Another factor that plays a role is the profile area of the falling object.

1. D

As an object falls, air resistance will increase until it is equal to the force of gravity. The force of gravity is determined by the number of filters that are dropped. And so it reasons that “increasing the number of filters increases the upper limit on the amount of air resistance” that the filters can experience. 6. Answer: A Explanation: A more massive object is a heavier object. By inspecting either Table 1 or from Figure 3, one observes that objects of greater mass (heavier) have a greater terminal velocity. As such, heavier objects fall faster than lighter ones.

Day 61

Explanation: The maximum range occurs for a launch angle of 45°. At this angle, the range is 163 meters – read from the graph in Figure 1 and listed in the fourth row of Table 1.

Explanation: This answer can be determined by inspecting the trajectory plots in Figure 1 or the data in Table 1.There are five plots given in Figure 1 with the launch angle increasing by 15° increments beginning with a 15° launch angle. The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more. So the best answer is the launch angle of 85°.

Explanation: A careful inspection of either Figure 1 or Table 1 reveals that the range of a projectile has an identical value for more than one launch angle. For instance, the range is the same for a 15° and a 75° angle. The range is also the same for a 30° and a 60° launch angle. As a final example, the range is the same for a 40° and a 50° launch angle. What do all these angles have in common? The group of two angles with the same range always add up to 90°. This makes choice b an accurate answer. None of the other choices fit all the data provided in this passage.

Explanation: Given the description of the physical situation, one can conclude that the golf ball lies 35 meters from the nearest edge of the tree and 65 meters from the furthest edge of the tree. These values were determined by knowing the trunk is 50 meters from the golf ball and the branches stretch from the trunk a distance (radius) of 15 meters towards and away from the ball. Since the branches are as high as 30 meters, it is important that the trajectory of the ball carry it above a height of 30 meters by the time it is 35 meters horizontally from its launch position. When inspecting Figure 1, one can determine that neither the 30° nor the 45° trajectories meet this criteria. Knowing that 40° lies between the 30° and the 45° trajectory means that it can be ruled out as well. By the process of elimination, the only angle that works is the 60° launch angle.

Explanation: This question pertains to the development of an experimental design to isolate variables having a potential impact upon the flight time of a projectile. The two variables being considered are the original horizontal speed (vox) and the original vertical speed (voy). The best design to determine whether a variable affects the flight time would be a design in which only one variable changes and the other variable is held constant. Conducting an experiment with a changing angle will change both variables. This rules out the first two options. Conducting an experiment involving a constant launch angle but changing speed will also change both variables since vox and voy depend on angle and original speed. This rules out the third option. The best design (of the four options) involves a vertical launch; launching vertically will eliminate the horizontal speed. By changing the launch speed, one can observe the effect of the original vertical speed (voy) upon the time the projectile is in the air.

Day 66

1. The weight W of an object is the product of its mass and the acceleration due to the earth’s gravity  (9.81 m/s2 ). 2 W = (50 kg)(9.81 m/s ) = 490.5 N

2. According to Newton’s Third Law, for every force there is an opposing force, equal in magnitude and opposite in direction. In this case, the force is the object’s weight (which we know from Problem 1 was 490.5 N), and the opposing force is the normal force, FNorm. Therefore, FNorm = 490.5 N and is directed upward.

3. Newton’s Second Law states that the sum of the forces ( ) ∑F acting upon an object is the product of its mass m and its acceleration a. There are two forces acting on the crate, 1) its weight W, and 2) the upward force lifting it FL. We will define “up” as the positive direction. ∑F F W ma =-= L Solving for FL, we get: F ma W L =- = (50 kg)(1 m/s2 ) + 490.5 N = 50 N + 490.5 N = 540.5 N

4. Answer: Newton’s Second Law states that the sum of the forces acting upon an object is the product of its mass and acceleration. Since the crate has a constant velocity, its acceleration is zero. The sum of the forces is therefore zero. The lifting force must be equal to the weight of the object, because the the forces oppose each other. Therefore, the magnitude of the upward force on the crate is 490.5 Newtons.

5. The forces acting on the object are the horizontal force Fx and the frictional force Ff. Therefore the equation for the sum of the forces acting upon the object by Newton’s 2nd Law is: ∑F F F ma =-= x f
Solving for Ff, we get F F ma f x=-= 150 N – (50 kg)(1.20 m/s2 ) = 150 N – 60 N = 90 N

Day 80

1. *b. 8.9 m/s
2. *c. equal to zero because the force and the displacement of the object are perpendicular
3. *a. 61.7 N
4. *a. 29 m/s
5. *b. 2

Day 83

1. D. Heat energy in solids is transferred by the increased kinetic energy in the vibrations of the particles.
2. D. Rod B is a copper, which has a heat conductivity grade of 10.
3. B. The sequence in which the matches fall reflects the rate at which heat is conducted through the rods. Materials with higher heat conductivities will conduct heat faster.
4. B. The order in which the matches fall can be used to rank the heat conductivity and therefore identify the type of metal.
5. C. Glass and marble are not good conductors of heat but do have a measurable ability to conduct heat energy.
6. B. Marble has a heat conductivity grade of 4; glass has a heat conductivity grade of 3. These conductivity grades mean that heat from the boiling water will be transferred faster down the marble than the glass.
7. C.
8. D. The order in which the wax begins to melt on the rods is the same as the order of heat conductivities. Aluminum has a heat conductivity grade of 9 (rod A), steel has a heat conductivity index of 8 (rod B), and bronze has a heat conductivity grade of 7 (rod C).]

Day 146

1. Sound is a longitudinal pressure wave. When an object immersed in a fluid (such as air) vibrates, it sets up regions of high and low pressures that travel away from the object. When the wave strikes a membrane (such as an eardrum), the high pressure regions create a pressure difference across the membrane. This creates a force that points into the membrane, thus pushing the membrane in. When the low pressure wave strikes the membrane, just the opposite occurs, with the membrane being pulled out. The result is a membrane that vibrates at the same rate as the original object. The brain interprets this vibration as sound.
2. In a sample trial, four waves passed the first numbered tick mark in four seconds, indicating a frequency (f) = 1 cycle/sec.
3. The wavelength ( ) = 1 m, so the wave velocity (v) = 1 m/s.
4. On the right side (where the sound waves are moving toward the observer), the wavelength is now only ‘ = 0.50 meters. On the left side (where the sound waves are moving away from the observer), the wavelength has increased to ‘ = 1.50 meters. (The prime denotes the Doppler-shifted wavelength.)
5. The frequency on the right side of the source has increased. The frequency on the left side has decreased.
6. The original wavelength was = 1 meter. The constant obtained by multiplying wavelength ( ) by frequency (f) is, therefore, 1 m/s. On the right side of the source, the wavelength is 0.50 meters, resulting in a Doppler-shifted frequency of f’ = 2.0 cycles per second. On the left side of the source, the wavelength is 1.50 meters, producing a Doppler-shifted frequency of f’ = 2/3 cycle per second.
7. The ratio between vs and v is 0.50. Since the wave velocity, v, is 1.0 m/s (from Question 3), vs must be 0.50. When the source is approaching the observer (referring to the right side of the source), f’ = f v/(v -vs) = (1.0)(1.0)/(1.0 – 0.5) = 2.0 cycles per second, as found earlier. For the left side, use the equation for a source moving away from an observer: f’ = f v/(v + vs) = (1.0)(1.0)/(1.0 + 0.5) = 0.667 cycle per second,
which agrees with the earlier result of f’ = 2/3 cycle per second.
8. Both electromagnetic waves and sound waves have defined wavelengths and frequencies. Both propagate at well-defined velocities. Sound is a longitudinal pressure wave, whereas electromagnetic waves are transverse waves. Sound requires a medium in order to propagate, which is why there is no sound in outer space. Electromagnetic waves, however, do not require a medium and can travel through
outer space.
9. Celestial objects, like stars and planets, constantly emit electromagnetic waves. Because these bodies are moving with respect to the earth, the frequency of these electromagnetic waves will be shifted. This shift can be detected and used to ascertain whether the body is approaching the earth or moving away from it. When a body is approaching the earth, the frequencies will be higher than normal. This is called a red-shift. Conversely, when a body is moving away from the earth, the frequencies will be lower than normal, a blue-shift.
10. When a body reflects a wave, it acts like a wave emitter. Therefore, if the body is approaching the radar station, the electromagnetic wave will be higher in frequency than when it started (blue-shifted), and so on. By processing these shifts in frequency, the motion of the air molecules can be imaged to form a weather map of wind speed and direction. This process is called Doppler radar.
11. As the fire truck approaches the observer, sound frequency increases. The siren has a higher pitch than it does when it moves away from the observer, since the frequency will then decrease.

Day 150

Nature of Sound Waves

1. True
2. False
3. True
4. True
5. True
6. The particle of air vibrates back and forth about its fixed position. It does not translate; it vibrates. That is, the particle does not move to Kate. It simply wiggles back and forth about position A. The amplitude of its vibration is abnormally small, nowhere close in scale to that represented by the arrow on the diagram.
7. In longitudinal waves (like sound waves), particles of the medium (in this case, the coils) vibrate
back and forth parallel to the direction of wave motion. To cause this parallel motion, Tosh must
vibrate the first rightward and leftward.
8. Compressions:  A, C, E, G, I, and K
Rarefactions:  B, D, E, H and J
9. Sound waves are mechanical waves – they require a medium in order to move one location to another. The film disregards this property of sound waves by depicting their motion through space that is void of matter. Only electromagnetic waves could travel through the vacuum of outer space.

Properties of Sound Waves

1. C
2. D
3. A
4. B
5. E
6. 1.7
7. a. frequency
8. c. frequency, wavelength
9. d. decreases, decreases
10. a. frequency
11. properties of the medium through which it moves
12. c. air molecules to vibrate with greater amplitude
13. c. the loudness of the sound to be louder

The Doppler Effect

1.  Answer: FALSE. This shifting of the pitch is referred to as the Doppler effect.
2. As the car approaches and moves past Ken, he observes the Doppler effect. The Doppler effect is the shift in pitch perceived by an observer of the source of sound. The siren produces a sound wave of constant frequency – the same in approaching Ken as when moving away from him. But Ken perceives the sound to be shifted upward in pitch when it is approaching him and downward in pitch when it moves away from him.
3. Answer: FALSE. Electromagnetic waves and other waves exhibit this same behavior.
4. Answer: FALSE. The Doppler effect is associated with the perception of the pitch of sound, not the loudness. Pitch is related to the frequency at which waves reach the observer; loudness is related to the intensity of the sound wave’s vibrations.
5. c. both the same
6. a.  Jack
7. higher, lower