28,650 3.125
34,380 1.06
40,110 .5
45,840 .25
51,570 .125
57,300 0
Hint: Remember to add gridlines
Graph: (place graph here)
Questions: (Use your graph above to answer the questions below)
1. How old is the following fossil?
Fossil A - 73% of Isotope A remaining
The fossil is about 3,500 years old.
2. How old is the following fossil?
Fossil B - 15% of Isotope A remaining
The fossil is about 16,000 years old.
3. What percentage of Isotope A is remaining if the fossil is 1200 years old?
(Use your graph)
92% of Isotope A remaining
Average Atomic Mass Calculations
1. Naturally occurring chlorine that is put in pools is 75.53 percent 35Cl (mass = 34.969
amu) and 24.47 percent 37Cl (mass = 36.966 amu). Calculate the average atomic
mass of chlorine.
34.969 * 0.7553 = 26.412
36.966 * 0.2447 = 9.0455
26.412 + 9.0455 = 35.45 => AVERAGE ATOMIC MASS
2. Calculate the atomic mass of silicon. The three silicon isotopes have atomic masses
and relative abundances of 27.9769 amu (92.2297%), 28.9765 amu (4.6832%) and
29.9738 amu (3.0872%).
27.9769 * 0.922297 = 25.803
28.9765 * 0.046832 = 1.357
29.9738 * 0.030872 = 0.925
25.803 + 1.357 + 0.925 = 28.085 => AVERAGE ATOMIC MASS
Writing:
Use one of the examples above to discuss how you determine the number of neutrons for
each isotope. You also need to discuss how the %abundance contributed to the Average
Atomic Mass of the element. (HINT: Think of the M&M Lab!)
Key Terms to use: Isotope, nucleus, neutrons, average atomic mass, Mass%, M&Ms, protons,
atomic number, element, however, therefore, additionally, for instance, in conclusion, data,
% abundance
For the isotopes above, I calculated three silicon isotopes. The first isotope has an
atomic mass of 27.9769 and a relative abundance of 92.2297%. I divided 92.2297 by
100 to put it into decimal form. After that I multiplied 27.9769 by 0.922297 to find the
number of neutrons in the first isotope, 25.803. I continued to do the same process
for each isotope. The second isotope has an atomic mass of 28.9765 and a relative
abundance of 4.6832, then I multiplied 0.046832 by 28.9765 to get 1.357 neutrons.
For the last isotope, there was an atomic mass of 29.9738 and a relative abundance
of 3.0872. I then multiplied 29.9738 by 0.030872 to get 0.925. The number of
neutrons goes inside the nucleus of the atom. To find the average atomic mass, I
added all of the number of neutrons that I got together to come out with an
equation of, 25.803 + 1.357 + 0.925. My final answer for the average atomic mass of
the silicon atom was 28.085 which is the actual mass of a silicon atom. Although I
was absent for the M&M lab, I can see how this problem relates to it. The percent
abundance contributed to the average atomic mass of the element because that is
the number you needed to multiply by.
Velocity
Velocity Story
Name: Avery DePodesta D ate: January 12, 2018
Directions: Work in a group to tell a story of a classmate in motion. You must include 3
turns (change in direction) and 3 different velocities. Your story must also have an amount
of time where the classmate does not move. What did the person do when they stopped?
Where were they going?
Data Table:
Example: Velocity = Distance/Time
V = 12 m/3 sec
V = 4 m/sec.
Description Distance (m) Time (sec.) Velocity
(m/s)
Walking 4 meters 4 seconds V= 1 m/sec
Cartwheel 2 meters 4 seconds V= 0.5 m/sec
Skipping 12 meters 6 seconds V= 2 m/sec
Jogging 15 meters 5 seconds V= 3 m/sec
Stopping (Unlocking Lock 0 meters 16 seconds V= 0 m/sec
19 seconds
and Grabbing Notebook from V= 1.42
m/sec
Locker) V= 0.56
m/sec
Hopping 27 meters V= 0 m/sec
Lunging 5 meters 9 seconds V= 5.25
m/sec
Stopping (Drinking from the 0 meters 3 seconds V= 0 m/sec
Water Fountain) 42 meters 8 seconds
Sprinting
Stopping (Leaving School) 0 meters 3 seconds
Graph: (X-axis is Time; y axis is Distance)
Story:
-
- Story:
- Julie needed to get her science notebook from her locker, so she
began her adventure outside of Mr. Lopez’s room. She walked 4 meters and
then did a cartwheel over the span of 2 meters because she is a crazy cat.
The velocity of this action was 1 m/sec. At this time, she realized that she
needed to change her ways of life and get her notebook as quickly as
possible. She had to get back to class like the goody-goody she is. At the
bottom of the ramp she decided to skip 12 meters to the top, which took her 6
seconds, meaning the velocity of this action was 2 m/sec. From the top of the
ramp, she took 5 seconds to jog 15 meters to her locker. The velocity of this
action was 3 m/sec. It took her 16 seconds to unlock her lock and grab her
science notebook to go back to class. Julie then hopped 27 meters to the
bottom of the ramp again to travel back to science class in 19 seconds, giving
this action a velocity of 1.42 m/sec. When she got to the end of the ramp, she
noticed A-Dog and Emmy in math class. Julie forgot to go to pilates class, so
she lunged 5 meters to the water fountain in only 9 seconds. The velocity of
this action was 0.56 m/sec. Emily was a little parched, so she rested and took
a nice drink for 3 seconds. Emily and Avery were feeling stressed and
overwhelmed and they just couldn’t take it anymore. They made a drastic
move, sprinting 42 meters down the hall in only 8 seconds, which made the
velocity of this action 5.25 m/sec, and threw open the door, only taking 3
seconds to leave Dodd Middle School forever, to never come back again.
Emily and Avery left Mrs. Montano in the dust!!!!!
Velocity Project 2018
Due: Wednesday Night February 23,2018
1. Define the following terms and include pictures if possible:
Motion: t he action or Speed: The distance an Position: the location
object travels per unit of relative to a reference
process of moving or being time point
moved
Distance: The motion of Acceleration: the rate of Terminal Velocity: t he
an object is to describe change of velocity
how far it has moved constant speed that a freely
falling object eventually
reaches when the
resistance of the medium
through which it is falling
prevents further
acceleration
Time: the indefinite Initial Velocity: t he rate Displacement: The
that the position of an distance and direction of
continued progress of
object changes relative to an object’s change in
existence and events in the
time position from the starting
past, present, and future
point
regarded as a whole
Velocity: The speed of an Final Velocity: the Key Metric units:
object and the direction velocity at the final point ● Meter (to measure
length)
of its motion of time ● Seconds (to
measure time)
● Kilogram (for mass
or weight)
● Liter (for volume)
● Degree Celsius (for
temperature)
2. What is the difference between Speed and Velocity? Explain using an example
in your own words.
a. There are many differences between speed and velocity. The definition of
speed is how fast an object moves and velocity is speed in a given direction.
3. Pick 2 cities (minimum 500 miles apart) in the United States or world and
construct a data table and graph showing the amount of hours that it would take
to travel between the 2 cities with the following modes of transportation:
A. Fastest Runner
B. Model T Ford
C. Hindenburg
D. Tesla top speed
E. Fastest train
F. F35 Fighter Jet
G. Vehicle of your choice
Lake Placid to Pittsburg
*Provide a map showing your cities
*Show Detailed Math Steps
Modes of Transportation Amount of Hours (kilometers per hour)
Top Speed: 44.7
Fastest Runner (Usain Bolt) 889.967/44.7 = 19.9
Model T Ford
Hindenburg T = D/V
Tesla (Model S) T = 889.967km/44.7h
T = 19.9km/h
Top Speed: 72
889.967/72 = 12.36
T = D/V
T = 889.967km/72h
T = 12.36km/h
Top Speed: 135
889.967/135 = 6.59
T = D/V
T = 889.967km/135h
T = 6.59km/h
Top Speed: 250
889.967/250 = 3.56
T = D/V
T = 889.967km/250h
T = 3.56km/h
Fastest Trains (Bullet Train)
Top Speed: 320
889.967/320 = 2.78
T = D/V
T = 889.967km/320h
T = 2.78km/h
F35 Fighter Jet
School Bus
Top Speed: 1,930
889.967/1,930 = 0.46
T = D/V
T = 889.967km/1930h
T = 0.46km/h
Top Speed: 110
889.967/110 = 8.09
T = D/V
T = 889.967km/5h
T = 8.09km/h
Lake Placid
1) What would like to see in this city when you arrive?
a) Whiteface Mountain
b) Mirror Lake
2) What tourist attraction?
a) The 1980 Olympic Buildings and the Ski Jumping Area
b) The Olympic Torch
c) The Olympic Oval
3) What restaurant would you like to visit in this city?
a) Emma’s Lake Placid Creamery
4) What is the basic history of this city?
a) Founded in the early 19th century
b) Lake Placid is America’s first winter resort
c) Lake Placid hosted two Olympic games, one in 1932 and one in 1980
Pittsburgh
1) What would you like to see in this city when you arrive?
a) Abby Lee Dance Company
b) PPG Paints Area
2) What tourist attraction?
a) Kennywood- Amusement Park
3) What restaurant would you like to visit in this city?
a) Altius
4) What is the basic history of this city?
a) Founded on November 27, 1758
b) Named after British secretary William Pitt
5. Determine and graph an 18% increase in Velocity for each vehicle - Show how
the Times would be affected by the increase in speed. Show a double bar graph
with the 2 different times for each vehicle.
*Include pictures and brief description of each mode of transportation
Modes of Transportation Amount of Hours (kilometers per hour)
Fastest Runner (Usain Bolt) Top Speed: 44.7
889.967/44.7 = 19.9
44.72 * 0.18 = 8.0496
8.0496 + 44.72 = 52.7696
889.967/52.7696 = 16.87
Model T Ford
Hindenburg
Tesla (Model S) Top Speed: 72
889.967/72 = 12.36
72 * 0.18 = 12.96
12.96 + 72 = 84.96
889.967/84.96 = 10.48
Top Speed: 135
889.967/135 = 6.59
135 * 0.18 = 24.3
24.3 + 135 = 159.3
889.967/159.3 = 5.59
Top Speed: 250
889.967/250 = 3.56
3.56 * 0.18 = 0.6408
0.6408 + 250 = 250.6408
889.967/250.6408= 3.55
Fastest Trains (Bullet Train)
Top Speed: 320
889.967/320 = 2.78
2.78 * 0.18 = 0.5004
0.5004 + 320 = 320.5004
889.967/320.5004 = 2.78
F35 Fighter Jet
School Bus
Top Speed: 1,930
889.967/1,930 = 0.46
0.46 * 0.18 = 0.0828
0.0828 + 1,930= 1,930.0828
889.967/1,930.0828 = 0.46
Top Speed: 110
889.967/110 = 8.09
8.09 * 0.18 = 1.4562
1.4562 + 110 = 111.4562
889.967/111.4562 = 7.98
6. Use a math calculation to show how long it would take the F35 Fighter Jet to
get to
A. Sun
Time = distance/velocity
Time = 1.46 x 10 8 kilometers
1.93 x 10 3km/hour
Time = 0.7565 x 105 hours = 75,650 hrs/24 = 3152.08333 days/365 = 8.63 years
It would take the F35 Fighter Jet 8.63 years to get to the Sun.
B. Saturn
Time = distance/velocity
Time = 1.4 x 10 9kilometers
1.93 x 10 3 km/hour
Time = 0.7254 x 106 hours = 725,400 hrs/24 = 30,225 days/365 = 82.81 years
It would take the F35 Fighter Jet 82.81 years to get to Saturn.
C. Neptune
Time = distance/velocity
Time = 4.3 x 10 9 kilometers
1.93 x 10 3 km/hour
Time = 2.228 x 106 hours = 2,288,000 hrs/24 = 95333.33 days/365 = 254.34 years
It would take the F35 Fighter Jet 254.34 years to get to Neptune.
(Use scientific notation)
Name: Avery Date:
DePodesta Februa
ry 23,
2018
Hypothesis: if the angle of the ramp is higher then, the acceleration will be greater
Independent
Variable: Angle
Dependent
Variable:
Distance
Meters
Write Units --> Second meters/s Second meters/se
Acceleratio
s econds Meters s conds
Velocity Time Velocity
Trial Dist. 1 Time 1 1 Dist. 2 2 2 n
angle 1 = 23
degrees 0.61 0.5 0.61 0.22
angle 1 = 23
degrees 0.61 0.46 0.61 0.14
angle 1 = 23
degrees 0.61 0.53 0.61 0.15
avg. 0.61 0.49 1.24 0.61 0.17 3.59 13.82
angle 2 = 11 0.61 0.75 0.61 0.44
degrees 0.61 0.73 0.61 0.57
0.61 0.73 0.61 0.53
angle 2 = 11
degrees
angle 2 = 11
degrees
avg. 0.61 0.74 0.81 0.61 0.51 1.2 0.76
Acceleration Conclusion
Problem Statement: How does the angle of the ramp affect the acceleration of the car?
Conclusion: (use data)
The purpose of the experiment was to find if the angle of the ramp affects the
acceleration of the car. Our hypothesis was, if the angle of the ramp is higher then, the
acceleration will be greater. We found that our hypothesis was correct, the higher the angle,
the greater the acceleration is. In our experiment, when the ramp was at 23° the acceleration
was 13.82 meters per second. When the ramp was at 11°, the acceleration was 0.76 meters
per second. This means that the acceleration had a decrease of 13.06 from the first angle to
the second angle. The velocity in the first angle (of the top half of the ramp) had an average
of 1.24 meters per second and an average time of 0.49 seconds. The first angle (at the
bottom half of the ramp) had an average velocity of 3.59 meters per second and an average
time of 0.17 seconds. The second angle (at the top half of the ramp) had an average velocity
of 0.81 meters per second and an average time of 0.74 seconds. The second angle (at the
bottom half of the ramp) had an average velocity of 1.2 meters per second and an average
time of 0.51 seconds.
Key words: Purpose of experiment, Hypothesis, data to prove your hypothesis
Acceleration Worksheet. Name: Avery DePodesta
14.2 Acceleration Date: February 26, 2018
Acceleration is the rate of change in the speed of an object. To determine the rate of
acceleration, you use the formula below. The units for acceleration are meters per
second per second or m/s2.
A positive value for acceleration shows speeding up, and negative value for
acceleration shows slowing down. Slowing down is also called deceleration.
The acceleration formula can be rearranged to solve for other variables such as final
speed (v2) and time (t) .
EXAMPLES
1. A skater increases her velocity from 2.0 m/s to 10.0 m/s in 3.0 seconds. What is the
skater’s acceleration?
Looking for Solution
Acceleration of the skater
Given
Beginning speed = 2.0 m/s The acceleration of the skater is 2.7 meters per
second per second.
Final speed = 10.0 m/s
Change in time = 3 seconds
Relationship
2. A car accelerates at a rate of 3.0 m/s2. If its original speed is 8.0 m/s, how many
seconds will it take the car to reach a final speed of 25.0 m/s?
Looking for Solution
The time to reach the final speed.
Given `
Beginning speed = 8.0 m/s; Final speed = 25.0 The time for the car to reach its final speed is
m/s 5.7 seconds.
Acceleration = 3.0 m/s2
Relationship
1. While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds.
What is the automobile’s acceleration? (Remember that a negative value indicates a
slowing down or deceleration.)
A = (V2 - V1)/T2
A = (15 m/s - 24 m/s)/12 Sec.
A = -9 m/s/12 sec.
A = -0.75 meters
2. A parachute on a racing dragster opens and changes the speed of the car from 85
m/s to 45 m/s in a period of 4.5 seconds. What is the acceleration of the dragster?
a = v2 -v1 / t2
a = 45m/s - 85m/s / 4.5s
a = -40m/s / 4.5s
a = -8.89m/s
3. The table below includes data for a ball rolling down a hill. Fill in the missing data
values in the table and determine the acceleration of the rolling ball.
Time (seconds) Speed (km/h)
0 (start) 0 (start)
2 3
6
9
8
10 15
Acceleration = ___________________________
4. A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a
complete stop (0 m/s). How much time will it take for the car to stop if it decelerates
at -4.0 m/s2?
t = v2 - v1 / a
t = 0m/s - 30m/s / -4m/s2
t = -30m/s / -4m/s2
t = 7.5s
5. If a car can go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after
5.0 seconds if its starting speed were 50 mi/hr?
a = v2 - v1 / t
a = 50mi/hr - 60mi/hr
6. A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the
cart has a beginning speed of 2.0 m/s, what is its final speed?
V2 = V1 + (a*t)
V2 = 2 + (4 * 5)
V2 = 2+ 20
V2 = 22 m/s
7. A helicopter’s speed increases from 25 m/s to 60 m/s in 5 seconds. What is the
acceleration of this helicopter?
a = v2 - v1 / t
a = 60m/s - 25m/s / 5s
a = 35m/s / 5s
a = 7m/s2
8. As she climbs a hill, a cyclist slows down from 25 mi/hr to 6 mi/hr in 10 seconds.
What is her deceleration?
a = v2 - v1 / t
a = 6mi/hr - 25mi/hr / 10s
a = -19mi/hr / 10
a = -1.9 mi/hr2
9. A motorcycle traveling at 25 m/s accelerates at a rate of 7.0 m/s2 for 6.0 seconds.
What is the final speed of the motorcycle?
V2 = V1 + (a*t)
V2 = 25m/s + (7m/s2 * 6)
V2 = 25m/s+ 42
V2 = 67 m/s
10. A car starting from rest accelerates at a rate of 8.0 m/s/s. What is its final speed at
the end of 4.0 seconds?
V2 = V1 + (a*t)
V2 = 0 + (8 * 4)
V2 = 0+ 32
V2 = 32 m/s
11. After traveling for 6.0 seconds, a runner reaches a speed of 10 m/s. What is the
runner’s acceleration?
a = V2 - V1 / t
a = 10 -0 / 6
a = 10/6
a = 1.67m/s2
12. A cyclist accelerates at a rate of 7.0 m/s2. How long will it take the cyclist to reach a
speed of 18 m/s?
T = V2 - V1 / a
t = 18 - 0 / 7
6 = 18/7
t = 2.57s
13. A skateboarder traveling at 7.0 meters per second rolls to a stop at the top of a ramp
in 3.0 seconds. What is the skateboarder’s acceleration?
a = v2 - v1 / t
a = 0m/s - 7m/s / 3s
a = -7m/s / 3s
a = -2.33m/s
QUIZ: Motion
Name: A very DePodesta Date: March 1, 2018
Formulas:
A= v2 −v1 V2 = V1 + (a * T) T= V2 − V1
T2 a
1. After traveling for 14.0 seconds, a bicyclist reaches a speed of 89 m/s. What is
the runner’s acceleration?
A= v2−v1
t
A= 89m/s−0m/s
14sec
A= 89m/s
14s
A = 6.357m/s2 is the runner’s acceleration
2. A car starting from rest accelerates at a rate of 18.0 m/s2. What is its final
speed at the end of 5.0 seconds?
V2 = V1 + (a x t)
V2 = 0m/s + (18.0m/s2 x 5sec)
V2 = 90m/s is the final speed of the car
3. A cyclist accelerates at a rate of 16.0 m/s2. How long will it take the cyclist to
reach a speed of 49 m/s?
T= V 2−V 1
a
T= 49m/s − 0m/s
16.0m/s2
T= 49m/s
16.0m/s2
T = 3.0625 seconds
3. During an Apollo moon landing, reflecting panels were placed on the moon.
This allowed earth-based astronomers to shoot laser beams at the moon's
surface to determine its distance. The reflected laser beam was observed 4.6
seconds after the laser pulse was sent. The speed of light is 3.0 × 108 m/s. What
was the distance between the astronomers and the moon?
D = v x t
D = (3.0 x 108 m /s) (4.6 seconds)
D = 13.8 x 108m /s is the distance between the astronomers and the moon
Directions: Choose 4 or 5
4. It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be late for
French class for the third time this week. She must get from one side of the
school to the other by hurrying down three different hallways. She runs down
the first hallway, a distance of 65.0 m, at a s peed of 5.2 m/s. The s econd
hallway is filled with students, and she covers its 32.0 m length at an average
speed of 1.46 m/s. The f inal hallway is empty, and Suzette sprints its 6 0.0 m
length at a speed of 7.3 m/s.
a. Does Suzette make it to class on time or does she get detention for
being late again?
FIRST HALLWAY
T = d/v
T = 65m / 5.2m/s
T = 12.5s
SECOND HALLWAY
T = d/v
T = 32m / 1.46m/s
T = 21.92s
THIRD HALLWAY
T = d/v
T = 60m / 7.3m/s
T = 8.22s
12.5 + 21.92 + 8.22 = 42.64
* Suzette makes it on time to class with 17.36 seconds left to spare. *
5. The tortoise and the hare are in a road race to defend the honor of their breed.
The tortoise crawls the entire 1000. m distance at a speed of 0.35 m/s while the
rabbit runs the first 200.0 m at 1.85 m/s The rabbit then stops to take a nap for
1.200 hr and awakens to finish the last 800.0 m with an average speed of 4.2 m/s.
Who wins the race and by how much time?
6. What is the Acceleration of the Cart on the Ramp? Determine the Angle of the
Ramp (A).
Angle Chart:
https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc
Which Angle had the greatest Acceleration? Write a Conclusion based on your
findings. Create a Graph if you have time.
HYPOTHESIS: If the height of the ramp (opposite) increases then the acceleration
will increase because the car will have more time to get down the ramp and will be
going at a faster speed in the end.
Height of Velocity Time 2 Velocity
Ramp Dist. 1 Time 1 1 Dist. 2 2 Acceleration
(Opposite)
50 m 100 m 10 sec. 10m/s 100 m 5 sec. 20m/s 5m/s2
100 m 100 m 5 sec. 20m/s 100 m 2 sec. 50m/s 15m/s2
The angle of the first ramp is 15°.
V = d/t
V = 100m / 10s
V = 10m/s
V = d/t
V = 100m / 5s
V = 20m/s
A = V2 - V1 / time
A = 20m/s - 10m/s / 5s
A = 10m/s / 5s
A = 5m/s2
The angle of the second ramp is 30°.
V = d/t
V = 100m / 5s
V = 20m/s
V = d/t
V = 100m / 2s
V = 50m/s
A = V2 - V1 / time
A = 50m/s - 20m/s / 2s
A = 30m/s / 2s
A = 15m/s2
Graph:
Conclusion:
In this experiment, I needed to determine the acceleration of the car on the
ramp and the ramps angle. There were two ramps that I needed to test. The first
ramp had an angle of 15° and the second ramp had an angle of 30°. Before I
completed this experiment I made a hypothesis that states, if the height of the ramp
(opposite) increases then the acceleration will increase because the car will have
more time to get down the ramp and will be going at a faster speed in the end. After
the experiment, I found that my hypothesis was correct. When the height of the
ramp increases the acceleration increased. For example the height of the ramp for
the first trial was 50 meters tall and had an angle of 15°. With these measurements,
the acceleration was 5m/s2. The height of the second ramp was 100 meters tall and
had an angle of 30°. This data concluded to an acceleration of 15m/s2. To find these
accelerations, I found the velocity of the first half of the first ramp which was 10m/s
and the velocity of the second half of the first ramp which was 20m/s. I used the
formula, velocity = distance / time to find the two velocities. Then, to find the
acceleration I used the formula, acceleration = velocity 2 - velocity 1 / time 2. I found
out that the acceleration of the first ramp was 5m/s2. To find the second
acceleration, for the second ramp, I found the velocity of the first half of the second
ramp which was 20m/s and the velocity of the second half of the second ramp
which was 50m/s. I used the formula, velocity = distance / time to find the two
velocities. Then, to find the acceleration I used the formula, acceleration = velocity 2
- velocity 1 / time 2. I found out that the acceleration of the second ramp was
15m/s2.
EXTRA CREDIT:
Light from another star in the galaxy reaches the earth in 46 minutes. The speed of
light is 3.0 × 108 m/s. In kilometers, how far is the earth from the star?
Answer must be in scientific notation
46 minutes = 2760 seconds
D = v x t
D = (3.0 x 108 m/s) x 2760 sec
D = 8280 x 108 m/s
D = 828,000,000 km/s
The earth is 828,000,000 kilometers per second away from the sun.
SCIENTIFIC NOTATION: 8.288 km/s
Potential Energy Project
Due: Friday 3/17
Define and make note cards or QUIZLET for the following words:
Energy: Joules: Chemical Potential Law of
The capacity or The SI unit of work Energy: Conservation of
power to do work, or energy Energy stored in Energy:
such as the capacity chemical bonds States that energy
to move an object cannot be created or
(of a given mass) by destroyed
the application of
force
Kinetic Energy: Kilojoules: Elastic Potential Gravity:
The energy a A unit of measure of Energy: AN attractive force
moving object has energy, in the same Energy stored by between any two
because of its way that kilometres something that can objects that
motion measure distance stretch or compress, depends on the
such as a rubber masses of the
band or spring objects and the
distance between
them
Potential Energy: Gravitational Mechanical Energy:
Stored energy due Potential Energy: The total amount of
to its position Energy stored by potential and kinetic
objects due to their energy in a system
position above the and can be
Earth’s surface expressed by the
equation-
mechanical energy =
potential energy +
kinetic energy
Resource: h ttp://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy
Gravitational Potential Energy
Determine the Gravitational Potential Energy (GPE) of 3 different masses (g) at 3
different heights.
3 objects: A verage Person, Basketball, Chevy Camaro (research the masses)
*2.2 lbs = 1 kg
Data Table:
Your data table will need: Object, mass, gravity, height, GPE
Object mass (kg) gravity (9.8 m/s2) H1 = 5 m GPE
810.91
1: Candace Flynn (15) 51.71 9.8 1.6002 693.43
496.79
2: Ferb Fletcher (13) 45.3 9.8 1.562
3: Phineas Flynn (11) 36.29 9.8 1.397
Videos: h ttp://www.youtube.com/watch?v=x5JeLiSBqQY
*Video shows you how to use the GPE equation.
Determine the GPE of one of the masses on the following planets:
Planet Meap - 17% greater than Earth’s Gravity
9.8 m/s2 x 1.17 = 11.47m/s2 = gravity
Whalemingo Planet - 39% less than Earth’s Gravity
9.8 m/s2 x .39 = 3 .82m/s2
9.8 - 3.8 = 6m/s2 = gravity
Planet Drusselstein - 82% greater than Earth’s Gravity
9.8m/s2 x .82 = 8 .04m/s2
9.8 + 8.04 = 17.84m/s2 = gravity
*Use the height of your favorite Roller Coaster. You will use this to figure out the
Velocity at the bottom of the hill on the Star Wars Planets.
HEIGHT OF MY FAVORITE ROLLER COASTER (COOLEST COASTER
EVER): 324 meters
Calculations:
Choose 3 planets from the Star Wars Universe and use 3 different
Examples:
A. Planet Meat:
GPE = KE
Mass x gravity x height = 0.5mv2
36.29kg x 11.47m/s2 x 324m = 0.5 (36.29kg) (v2)
134863.8 = 1 8.15v2
18.15 18.15
7430.51 = v2
√7430.51 = √v 2
86.2 = velocity
B. Fish Planet:
GPE = KE
Mass x gravity x height = 0.5mv2
36.29kg x 6m/s2 x 324m = 0.5 (36.29kg) (v2)
70547.76 = 18.15v2
18.15 18.15
3886.93 = v2
√3886.93 = √v 2
62.35 = velocity
C. Planet Rice:
GPE = KE
Mass x gravity x height = 0.5mv2
36.29kg x 17.84m/s2 x 324m = 0.5 (36.29kg) (v2 )
209762.01 = 18.15v2
18.15 18.15
11557.14 = v2
√11557.14 = √v 2
107.5 = velocity
Data Table:
Planet #1
Object mass (kg) gravity (m/s2) H1 = your coaster (m) GPE
1 36.29 11.47 324 134,863.8
Planet #2
Object mass (kg) gravity (m/s2) H1 = your coaster (m) GPE
1 36.29 6 324 70,547.76
Planet #3
Object mass (kg) gravity (m/s2) H1 = your coaster (m) GPE
1 36.29 17.84 324 209,762.01
Use the formula: GPE = mass * acceleration due to gravity (Earth is 9.8 m/s2) * height of object
Graph:
X - axis: Planet
Y -axis: Potential Energy
Critical Thinking Questions:
1. What factors affect Gravitational Potential Energy?
a. The mass of an object and the gravitational field that it is in affect the
Gravitational Potential Energy.
2. Why did the GPE change on the other planets?
a. The GPE changed on the other planets because each planet has a different
gravity.
3. Which planet would you be able to hit a golf ball further? Explain using data.
a. On Whalemingo Planet, you would be able to hit the golf ball the furthest
because its gravity is less than the other two planets. Whalemingo Planet has
a gravity of 6m/s2, while Planet Meap has a gravity of 11.47m/s2 and Planet
Drusselstein has a gravity of 17.84m/s2.
4. How does GPE relate to Chemical Potential Energy?
a. CPE is the chemical bond in gasoline and in food. GPE is the potential energy
that an object contains. They relate to each other because they both are
types of potential energy and are the amount of potential energy in an object.
5. How do Energy companies use GPE to generate Electrical Energy? Give an
example
a. The water flows down the pipes (potential to kinetic energy) to turn the
turbine. The tubine is connected to a generator to produce electricity (kinetic
to potential energy).
6. What happens to the GPE when the object falls to the ground? Describe the
Energy transformations along the way.
a. When an object falls to the ground, the GPE transfers into KE. This means that
the potential velocity is now turning into the actual velocity.
Worksheet 1:
http://glencoe.mheducation.com/sites/0078600510/student_view0/unit1/chapter4/math_practic
e_2.html
Worksheet 2: http://go.hrw.com/resources/go_sc/ssp/HK1MSW65.PDF
QUIZ: GPE/KE
Scenario: You are an engineer for a major engineering firm that will design the lift motor
and safety restraints for the next roller coaster on the planet Hoth in Star Wars. Hoth has a
gravity equal to 37% greater than Earth’s. The Star Wars Theme Park needs to provide you
with the velocity of the roller coaster on this planet to help you with your design. Your roller
coaster will be called the Millenium Falcon and will have a height of 125 m. Your roller
coaster will “The Falcon” will have a mass of 7000 kg. You will need to compare the needs
for safety on Earth to the needs on Hoth. Explain your reasoning for the changes on Hoth.
Hoth:
Directions: Provide a data table showing the comparisons between the Millenium Falcon
Roller Coaster on Earth and Hoth. Describe the types of restraints that you would need on
the faster coaster.
Calculations:
Earth Hoth
9.8 * 0.37 = 3.626
GPE = KE
M * G * H = 0.5mv2 9.8 + 3.626 = 13.426
7,000kg * 9.8m/s2 * 125m = 0.5(7000kg)v2
8,575,000 = 3 ,500kg (v2) GPE = KE
3,500 3,500 M * G * H = 0.5mv2
√2450 = √v2 7,000kg * 13.426m/s2 * 125m = 0.5(7000kg)v2
49.5m/s2 = v 11,747,750 = 3,500kg (v2)
3,500 3,500
√3356.5 = √v2
57.94m/s2 = v
Mass (kg) Height (m) Gravity GPE (joules) Velocity
(m/s2) (m/s)
9.8 8,575,000 49.5
Data Table: 13.426 11,747,750 57.94
Planet
Earth 7000 125
7000 125
Hoth
Graph:
Conclusion:
In conclusion, the purpose of the experiment was to design the lift motor and safety
restraints for the next roller coaster on the planet Hoth in Star Wars. I needed to find the
velocity on each planet to compare the needs for safety on Earth to the needs on Hoth. The
velocity on Earth is 49.5m/s and the velocity on Hoth is 57.94m/s. To find these I used the
equation GPE = KE. I then plugged in all of the information I knew to get the equation,
7,000kg * 9.8m/s2 * 125m = 0.5(7000kg)v2 for Earth and 7,000kg * 13.426m/s2 * 125m =
0.5(7000kg)v2 for Hoth. After that, I simplified the equations and isolated v2 since that was
what I was looking for. Next, I found the square root of v2 for each planet and the square
root of 2450 for Earth and 3356.5 for Hoth. Finally the results came out to be that the velocity
on Earth is 49.5m/s and the velocity on Hoth is 57.94. Since Hoth’s velocity is larger, that
planet will need more safety restraints on the roller coaster than Earth. Hoth’s velocity is
larger so the roller coaster will be going faster. On a faster roller coaster you would need
more safety restraints like helmets, goggles, and extra seat belts and lap bars.
Extra Problems:
1. The Millenium Falcon Roller Coaster has a mass of 3200 kg on Planet Tatooine.
The height of the roller coaster is 15 m which results in a Potential Energy of
800,000 J. What is the gravity on Planet Tatooine?
G = GPE / M*H
G = 800,000J / 3200kg * 15m
G = 800,000J / 48000
G = 16.67m/s2
The gravity on Planet Tatooine is 16.67m/s2 .
2. The Tie Fighter Roller Coaster has a height of 150 m. on Planet Hoth. Hoth has a
gravity of 5.2 m/s2. This roller coaster has a Potential Energy of 600,000 J. What is the
mass of the Tie Fighter?
M = GPE / G*H
M = 600,000J / 5.2m/s2 * 150m
M = 600,000J / 780
M = 769.23 kg
The mass of the Tie Fighter is 769.23kg.
Scenario: Suppose you would like to bring a 175 N box up to a height of 29 m. You decide
to use an inclined plane because you just learned about them in science class. The ramp
you design has a distance of 48 m. You also measure the Force (N) needed to push the box
up the ramp which is 85 N. What is the Work Output, Work Input, Ideal Mechanical
Advantage, Actual Mechanical Advantage, and Efficiency of the machine?
A. Use “Drawing” to label a triangle (Inclined Plane)
B. Calculate the angle of the ramp.
OPPOSITE
HYPOTENUSE
29m
48m
0.6m = 37 degrees
C. Calculate the Ideal Mechanical Advantage (IMA)
Input distance / output distance
48m / 29m
1.66m = IMA
D. Calculate the Actual Mechanical Advantage (AMA)
Output force / input force
175N / 85N
2.06N = AMA
E. Calculate the Efficiency (%)
Work output / work input X 100
(175 * 29) / (48 * 85)
5075 / 4080 X 100
1.24 X 100
124 = efficiency
Questions:
1. Is this machine possible? Explain using evidence from the problem.
a. This machine is not possible because the efficiency is 124% which is larger
than 100%. This means that it can not work.
2. How could you change the Input Force or Distance or to make it possible?
a. To make this machine possible, we can change the height of the opposite,
causing the inclined plane to be less steep. This change would require less
force , reducing the efficiency.
3. How would this problem be different on another planet?
a. The gravity would be different on other planets so it would take more or less
force to push the box up the ramp.
QUIZ: Inclined Plane
QUIZ: W ednesday and Thursday
The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.
Avery DePodesta (Class of 2022) - Blue Science Portfolio (1)
Discover the best professional documents and content resources in AnyFlip Document Base.
Search