Science Portfolio

Survey Graph Conclusion

1. Data Table

Sports Responses
Volleyball 1
Baseball 2
Football 6
Other
14

2. Graph

3. Write a short conclusion of 5 sentences

The overall purpose of doing this survey was to spell out which of the four choices is the most
popular sport to play. Out of all of the choices, volleyball had the lowest popularity with only one
vote, while other sports received fourteen votes. Baseball and football were in the middle with
baseball having two votes and six for football. Other sports were ahead of football by eight
votes, baseball by twelve votes, and volleyball by thirteen votes. In conclusion, sports other than
volleyball, baseball, and football, have more prevalence than they do currently.

Experiment Presentation

- Problem Statement: some balls weigh more than others so they will bounce higher
- Hypothesis: If the tennis ball, baseball, and the golf ball are dropped from the same height, then

golf ball will bounce the highest.
- Independent variable: Type of ball
- Dependent variable: height of bounce
- Control: height the ball is dropped from
- Constants: the constants would be how hard max pushes the ball off the , the floor material, and

the height

Data table:
Types of Balls Inches

Tennis ball 16

Golf ball 23

Baseball 11

1. Graph:

Quiz; Scientific Method

Mary Barto S7 9//13/16

Directions: R​ ead the following description of an experiment and complete the
components of the scientific method.

Experiment:​ Mr. Smithers believes that a special compound could help his workers
produce more “widgets” in one week. The chemical supply store sent him 3 different
compounds to try on his 100 workers. The following are the chemicals:

A. Sodium chloride
B. Magnesium hydroxide
C. Calcium sulfate
D. Water

*Help Mr. Smithers design an effective experiment and write a conclusion that analyzes
your results.

Problem Statement
How does the different compounds affect the amount of work done in one week?

Hypothesis

If the different compounds are tested, then the people that test the water will complete more
work than the others because it provides more nutrients to restore our body with energy.

Independent Variable (different compounds)

Sodium Chloride Magnesium Calcium Sulfate Water
Hydroxide

Dependent Variable Same job (Making widgets)
How many widgets are produced.

Constants​ (Pick 2)
Same amount of each compound (500 ml)
Same amount of time (One week)

Control
Water

Basic Procedures​:
(List 5-8 steps)

1. Start by splitting people into different but equal test groups.
2. Then, pour 500 ml of Sodium Chloride, Magnesium Hydroxide, Calcium Sulfate, and

Water for the people taking part in the experiment.
3. Give the workers the liquid to drink (make sure everyone drinks all of it.)
4. Have the people being experimented on work for one week.
5. Count up how many widgets each group made within the week.
6. Record the data from the experiment.
7. Share the results from the overall experiment.

Data Table:​ (Place data table here)

Type of Compound Amount of Widgets Made
Sodium Chloride 999
Magnesium Hydroxide 1,312
Calcium Sulfate 2,153
Water 1, 654

Graph:​ (Place graph here)

Conclusion:

After completing and recording the data from the experiment, I have come to a
conclusion that my hypothesis is inaccurate. I hypothesized that the water would have a greater
affect on how many widgets the workers made due to the nutrients provided, but my results
show that the Calcium Sulfate had a better effect. The calcium sulfate group made 2,153
widgets while the group with water made 1,654. There was a difference of 499 widgets made
between the calcium sulfate and water groups. Then the two lower groups were Magnesium
Hydroxide with 1,312 widgets made and Sodium Chloride with 999 widgets made. The overall
results make sense due to the fact that the calcium sulfate includes calcium which supports
growth and development with bones, joints, muscles and other parts of your body. For example,
a baby drinks milk which includes calcium because it develops growth and helps the muscular,
skeletal and respiratory systems. In conclusion, calcium sulfate had the greatest impact on the
amount of widgets made.

Density Lab Quiz

1. The scientist collected an object with a density of 6.4 g/cm3​ ​ and a
volume of 79 cm3​ ​. What is the mass of this object?

M=DxV
M=6.4g/cm3​ ​x79cm​3
M=505.6g/cm​3
2. An irregularly shaped stone was lowered into a graduated cylinder

holding a volume of water equal to 50.0mL. The height of the water rose
to 68 mL. If the mass of the stone was 125.0g, what was its density?
D=M/V
D=125g/18cm3​
D=6.94g/cm​3
3. A scientist had 350.0 grams of Gold (Au) and a 530.0 gram sample of
Silver on the lab table. Which metal would have a greater volume
(cm​3​)? Explain. ​*Show all work.

Gold=19.32g/cm​3 Silver=10.49g/cm3​

Gold:
V=M/D
V=350g/19.32g/cm​3
V=18.12cm​3
Silver:

V=M/D
V=530g/10.49g/cm3​
V=50.52cm​3
The silver would have the greatest volume in cm3​ b​ ecause the silver
has a greater mass than the gold but has a lower density than the
gold. In conclusion, the silver will have a higher mass than the gold
will.
4. Explain why the Titanic sank after hitting the iceberg. Use data to
explain your answer.
The Titanic was made of iron and steel. The density of iron and steel
is 7.87g/cm3 and 7.60g/cm3. The Titanic hit an iceberg which caused
water to rush into the ship. The average density of the iron, steel, and
water flooding the boat, was greater than the rest of the water’s
density therefore causing the ship to sink.

Phase Change of Water Lab

Mary Barto
S4
10/18/16
3. Activity: Phase Change of Water

Directions:
● Melt the ice water and record the temperatures every 30 seconds until you reach the
boiling point of water.
● Record the temperatures on the following data table:
​ C​ onstruct a graph of your results. *​Use Link on Classroom
● Respond to the Critical Thinking Questions

Graph:

Critical Thinking Questions:

1. When did the temperatures stay the same on the graph? Why did the
temperatures stay the same at 2 points during the lab?

The temperatures stayed the same at the beginning and end. This is because the
ice was a solid and not liquid yet, which was causing the temperature to stay
cold. The temperature stayed cold for a while because for water to be a solid, it
must be frozen. In the end it was ard for the temperature to increase because it

was starting to have a similar heat to the hot plate which was causing it to stay
the same for the while. Also, the molecules had already expanding a lot and it
was hard for them to keep expanding and let the water boil. The last reason for it
being harder was the water was giving of steam (gas) so the beaker was losing
little bits of water.

2. How would the graph be different if we tried this experiment with Gold?
Explain:

If we tried this experiment with gold, the experiment would be different because it
would take longer for the gold to melt because it has a greater mass and density
than ice does and it would take a longer time to melt, therefore causing the graph
to be in negative or low numbers for a while.

3. What is the role of energy during the phase changes?
The energy exchanges are playing the role of breaking the bond between the
molecules and the substance within each phase.

4. Describe the motion of the molecules throughout the experiment. Find
diagrams that show the motion.

Throughout the experiment, the molecules start to vibrate faster and faster
causing the molecules to separate farther from each other.

5. How does the Average Kinetic Energy change throughout the experiment?
The Average Kinetic Energy changes throughout the experiment because the
speed of the molecules to move faster when it is warmer, causing the
temperature to go up.

6. Suppose you had 200 mL of ice in one beaker and 400 mL of ice in another
beaker. Compare and explain the following in the beakers after they have
reached the boiling point:

A. Heat Energy​ The 400 mL of ice would have a greater Heat Energy because
there is a larger mass/quantity of the ice while the 400 mL would have a
smaller amount of Heat Energy.

B. Temperature ​Both the 400 and 200 mL of ice would have the same specific
heat because they have the same temperature.

C. Average Kinetic Energy B​ oth the 400 and 200 mL of ice would have the
same specific heat because they have the same temperature.

D. Specific Heat B​ oth the 400 and 200 mL of ice would have the same specific
heat because they have the same temperature.

E. Latent Heat L​ atent Heat is the heat required to convert substance- in this
case ice- throughout its different phases. They both would have the same
Latent Heat because for everything the freezing point is 0 degrees and the
boiling point is 100 degrees.

Quiz: Phase Change

Mary Barto
S7
Mr. Lopez
10/28/16

Directions: ​Analyze the following data table with data collected by a scientist that wanted to
study how Heat Energy affects the Phase Changes of 2 different metals. Respond to the
questions below and perform all necessary calculations.

Data Table:

Metal Mass Heat of Melting Boiling Heat of Specific Heat
Fusion Pt.​ (C) Pt. (​ C) Vaporization Heat Energy
(cal/g) (cal/gC) (cal)
(cal/g)

Aluminum 65 g 95 660 2467 2500 0.21
Gold 65 g
15 1063 2800 377 0.03

Scientific Method (___ out of 4)
Independent Variable:

Type of Metal

Dependent Variable:
Heat Energy

Constant:
The Mass of the Metal

Control:

Water

Calculate Heat Energy: * SH

Apply the following Equations: Boiling Heat of
Heat = Mass * Heat of Fusion Pt. (​ C) Vaporization
Heat = Mass * Change in Temperature
Heat = Mass * Heat of Vaporization (cal/g)
Data Table:

Metal Mass Heat of Melting Specific Heat
Fusion Pt.​ (C) Heat Energy
(cal/g) (cal/gC) (cal)

Aluminu 65 g 95 660 2467 2500 0.21 415289.
m 55 cal

Gold 65 g 15 1063 2800 377 0.03 28867.1
5 cal

*SHOW ALL MATH STEPS
Math Steps (____ out of 4)
A. Aluminum

Heat Energy = Mass * Heat of Fusion
Heat Energy = 65g * 95 cal/g
Heat Energy = 6175 cal

Heat Energy = Mass * Change in Temperature * Specific Heat
Heat Energy = 65g * 2467 C-660 C * 0.21 cal/gC
Heat Energy = 65g * 1807 C * 0.21 cal/gC
Heat Energy =​ 246614.55 cal

Heat Energy = Mass * Heat of Vaporization
Heat Energy = 65g * 2500 cal/g
Heat Energy = 162500 cal

Heat Energy= 6175 cal + 246614.55 cal + 162500 cal
Heat Energy= 415289.55 cal

B. Gold

Heat Energy = Mass * Heat of Fusion
Heat Energy = 65g * 15 cal/g
Heat Energy = 975 cal

Heat Energy = Mass * Change in Temperature * Specific Heat
Heat Energy = 65g * 2800 C- 1063 C * 0.03 cal/gC
Heat Energy = 65g * 1737 C * 0.21 cal/gC
Heat Energy = 3387.15 cal

Heat Energy = Mass * Heat of Vaporization
Heat Energy = 65g * 377 cal/g
Heat Energy = 24505 cal

Heat Energy= 975 cal + 3387.15 cal + 24505 cal
Heat Energy= 28867.15 cal

Graph your results (____ out of 4):

Write a Conclusion (____ out of 4):
Overall, the Aluminum requires more heat energy to melt and boil. To be exact, the

Aluminum needs 386422.4 more calories of heat energy to melt and boil. Although the
Gold has a higher melting point (1063 C) and boiling point (2800 C), Aluminum needs
more heat energy. Aluminum's melting point is 660 C and the boiling point is 2467 C. The
gold and the aluminum have the same mass of 65 grams but have different specific heats
(aluminum= 0.21cal/gC and gold= 0.03cal/gC) heat of vaporization (aluminum=2500cal/g
and gold= 377cal/g), heat of fusion (aluminum= 95cal/g and gold 15cal/g), melting point,
boiling point and the amount of heat energy needed (aluminum= 415289.55 cal and gold=
28867.15 cal). In conclusion, the aluminum requires more calories of heat energy than
gold does.

Questions:

1. How are Heat and Temperature different for the following pictures of​ ​boiling​ ​water?

Explain:​ (Hint: Use the Heat equation)

Heat and temperature are extremely different. For instance, two waters with
different masses such as the Atlantic Ocean and a small beaker of water can have the
same temperature but different amounts of heat. This is because the Atlantic Ocean has
a greater mass than the water in the beaker causing the mass to be different in the heat
energy equation. Then, the Atlantic Ocean will come out with a higher number because it
has a greater mass than the small beaker holding water does. If the water were boiling in
the beaker and in the Atlantic Ocean, it would be safer to pour the water from the beaker
of yourself instead of the Atlantic Ocean because there is a greater amount of boiling
water in the Atlantic Ocean than in the cup. In conclusion, mass plays a large role in the
difference of heat and temperature.

2. Water has a Specific Heat of 1.0 cal/gC and Gold has a Specific Heat of 0.03 cal/gC.
Use the data to explain the difference between their numbers.

There are various reasons for the difference in Specific Heat between water and
gold. For example, gold is a metal and water isn’t. Metals have a lower specific heat
because there is less heat required to raise the temperature of the metal than water. To
explain further, the gold requires 0.03cal/gC to raise the temperature while water needs
1cal/gC to raise its temperature. To conclude, water and gold have different specific
heats.

Boiling Point and Elevation Presentation

 

Mass % Practice

1. A scientist recorded the following data about a sample of rocks and sand:

37 grams of Large Rocks 75 grams of Fine Grained Sand
59 grams of Small Rocks 5 grams of Salt
125 grams of Coarse Grained Sand 25 grams of Copper (Cu)

2. D​ etermine the % of each component in this Heterogeneous Mixture and construct a pie
chart showing your results. 326

3. Data Table:

Components Mass (g) Percent (%)
Large Rocks 37 11.35
Small Rocks 59 18.1
Coarse Grained Sand 125 38.34
Fine Grained Sand 75 23.01
Salt
Copper 5 1.53
25 7.67

4. Pie Chart:
5. Math Examples

Large Rocks 37/326= 11.35
Small Rocks 59/326= 18.1
Coarse Grained Sand 125/326=
Fine Grained Sand 75/326= 38.34
Salt 23.01
Copper 5/326=
25/326= 1.53
7.67

1. A second scientist recorded the following data about a different sample of rocks and
sand:

48 grams of Large Rocks 175 grams of Fine Grained Sand
78 grams of Small Rocks 2 grams of Salt
56 grams of Coarse Grained Sand 17 grams of Copper (Cu)

2. Determine the % of each component in this Heterogeneous Mixture and construct a pie
chart showing your results.

3. Data Table: Mass (g) Percent (%)
48
Components 78 12.77
Large Rocks 56 20.74
Small Rocks 175
Coarse Grained Sand 14.9
Fine Grained Sand 2 46.54
Salt 17
Copper 0.53
4.52

4. Pie Chart:

5. Math Examples 48/376= 12.77
78/376= 20.74
Large Rocks
Small Rocks

Coarse Grained Sand 56/376= 14.9
Fine Grained Sand 175/376= 46.54
Salt
Copper 2/376= 0.53
17/376= 4.52

1. A third scientist received a 250 gram sample of Silver Nitrate - A​ gNO3​

Component Mass (g) Percent (%)
Silver 1 20
Nitrogen 1 20
Oxygen 3 60

2. Chart for Mass % of a Compound

Questions:
1. How are the samples from these scientists different?

The samples are different because in each mixture there are different amounts of each
component in each mixture.

2. How are Compounds different from Heterogeneous Mixtures? Provide evidence.
The mixtures are different than Heterogeneous Mixtures because Heterogeneous Mixtures can
be separated and compounds are all combined.

Quiz: Classifying Matter

I. Directions: I​ dentify the following as either a Heterogeneous Mixture, Homogeneous Mixture,

Element or Compound. Write the following letters in Column B for your choices:

A. Heterogeneous

B. Homogeneous

C. Element

D. Compound

Column A Column B

Salad A

Copper C

Lemonade B

Rocks, sand, gravel A

Salt Water B

Gold C

Sodium Chloride (NaCl) D

Air C

K2​ S​ O4​ D
Twix, snickers, pretzels, popcorn A

II. Directions:​ Determine the Mass % of each mixture and construct the appropriate graphs.

Mixture A Mass (g) %
Large Rocks 125 51.87
Small Rocks 75 31.12
Coarse Sand 32 13.28
Iron 9 3.73

Mixture B Mass (g) %
Large Rocks 205 52.70
Small Rocks 58 14.91
Coarse Sand 97 24.93
Iron 29 7.46

Calculation Examples (​ Provide 2 Examples showing how you determined the Mass %)

Mixture A Mass (g) %
Large Rocks 125/241✕100= 51.87
31.12
Small Rocks 75/241✕100= 13.28
3.73
Coarse Sand 32/241✕100=

Iron 9/241✕100=
51.87+31.12+13.28+3.73=100

Mixture B Mass (g) %
Large Rocks 205/389✕100= 52.70
14.91
Small Rocks 58/389✕100= 24.93
7.46
Coarse Sand 97/389✕100=

Iron 29/389✕100=
52.70+14.91+24.93+7.46=100

Graphs: ​Mixture A
Mixture B

Part III.​ Determine the Mass % of Elements in each Compound:

K2​ ​SO4​ ​ - Potassium Sulfate
(Show Math Here)

K (2) 39 = 78
S (1) 32 = 32
O (4) 16 = 64
78+32+64=174
78/174*100= 44.82
32/174*100= 18.40
64/174*100= 36.78
Chart and Graph

Element Mass (g) Percent (%)
K2​ 78 44.82
S 32 18.40

O​4 64 36.78

Na3​ P​ O​4​ - Sodium Phosphate
(Show Math Here)
Na3​ =​ (3) 23 = 69
P= (1) 31 = 31
O​4=​ (4) 16 = 64
Na​3=​ 69/164*100= 42.10
P= 31/164*100= 18.90
O4​ =​ 64/164*100= 39

Chart and Graph

Element Mass (g) Percent (%)
Na​3 69 42.10

P 31 18.90
O4​ 64 39

IV. Conclusion: ​Explain the difference between Mixtures and Compounds using data. Compare
the pie charts.

Mixtures and compounds have various differences. For example, a mixture is a
substance made by mixing other substances together. For instance, an example of a mixture
would be Large Rocks, Small Rocks, Coarse Sand and Iron. A compound is a thing that is
composed of two or more separate elements. For example, a compound would be considered
Na​3​PO​4.​ Mixtures are made from various mediums, while compounds are made out elements
from the periodic table. The mixtures tend to have different ranges than the compounds
because many of the compounds have to be multiplied to get the mass to find the percentage.
To show, the total mass of the mixture of (Mixture B) Large Rocks, Small Rocks, Coarse Sand
and Iron was 389 g and the total mass of the compound Na3​ P​ O4​ ​, was 174 g. The mixture B has
large rocks at 52.10%, small rocks at 14.91%, coarse sand at 24.93% and 7.46% of iron. This
all adds up to exactly 100%. In contrast, Na​3​PO​4 ​was 42.1% Na​3​, 18.9% P and 39% O​4​. As
shown in the data, the compounds percentages were numbered more closely as the mixtures
were more spread apart. The mass of each component or element in the mixture or compound
has a huge effect on the mass. In conclusion, mixtures and compounds have many differences.

Bonus:
Explain how you separated the Salt from the Sand. Use as much new vocabulary as you can.

I separated the salt from the sand by saturated the sand with water. The water grasped the
sand and came out of the funnel and into the beaker. Afterwards, I had the water meat its
boiling point. I kept it on the heater till all of the water evaporated and all that was left was the
salt. The salt was close to its melting point, so the salt was almost liquified.

Quiz: Solubility

Mary Barto
S4
12/6/16
Directions: Use the Solubility Graph to answer the following questions.
Graph

I. Solubility Graph

Questions:
1. What is the Solubility of KClO3​ ​ at 40 C?
15 g
2. What is the Solubility of NH4​ ​Cl at 70 C?
60 g
3. What Temperature would 80 grams of KNO3​ ​ completely dissolve and become saturated?
50 °​ C
4. Suppose you have 120 grams of NaNO3​ ​at 30 C. Is the solution Unsaturated, Saturated or
Supersaturated and how many grams can you add/or take away to make it Saturated?
Supersaturated - to make it saturated, it needs to have 25 g less. (95 g)

5. Suppose you have 120 grams of NaNO3​ ​ at 30 C. What could you do to the Beaker to make
the solution Saturated? (Use Data from graph here)

To make the 120 grams of NaNO3​ s​ aturated, you would need to increase the temperature of
the solvent from 30 °​ C ​to 55 °​ C​. (Increase temperature by 25 °​ C)​

6. Suppose you have 70 grams of KNO3​ ​ at 60 C. Is the solution Unsaturated, Saturated or
SuperSaturated and how many grams can you add/or take away to make it Saturated?

Unsaturated - to make it saturate​d, 70 g of solv​ ent must be added. (130 g)

7. Suppose you have 70 grams of KNO3​ ​ at 60 C. What could you do to the Beaker to make the
solution Saturated? (Use Data from graph here).

To make the solution in the beaker saturated at 70 °​ C,​ you can add 70 g of solvent into the
beaker for the g of solvent to increase from 60 g to 130 g.

II. Soluble vs. Insoluble
Directions: ​Use your Solubility Rules Chart to determine if the following compounds are Soluble
or Insoluble.

Compound Soluble or Insoluble Identify the Rule # Used

Sodium chloride Soluble 1 and 3

Silver nitrate Insoluble 4

Ammonium nitrate Soluble 1 and 2

Calcium carbonate Insoluble 6 and 8

Zinc sulfide Insoluble 7
AgCl Insoluble 3 and 4
Soluble
Na2​ S​ O4​ Insoluble 1
Calcium phosphate Insoluble 10
3
PbBr​2

III. Use your Solubility Rules to Determine how the beaker would look in the following chemical
reactions:

Reaction #1

Potassium Chloride + Silver Nitrate → Potassium Nitrate + Silver Chloride

Ions: K​+​Cl-​ + Ag+​ ​NO3​ ​- → K+​ N​ O3​ -​ + Ag​+​Cl-​

Reaction: KCl + AgNO3​ ​ → KNO​3 + AgCl
S S S I

Percent of Oxygen: Potassium Chloride vs. Potassium Nitrate

Potassium Chloride (​ 39+35= 74) Potassium Nitrate (​ 39+14+48= 101)

K (1) = 39(1)= 39/74 (100) = 52.7% K (1) = 39(1)= 39/101(100) = 38.61%
Cl (1) = 35(1)= 35/74 (100) = 47.3% N (1) = 14(1)= 14/101 (100)= 13.86%
O (3) = 16(3)= 48/101 (100) = 47.53%
---------
100% ---------
100%

Reaction #2

Lithium Phosphate + Calcium Sulfate → Lithium Sulfate + Calcium Phosphate

Ions: Li​+P​ O4​ 3​ - ​ + Ca​2+​SO​4​2- →​ Li+​ ​SO​4​2- + Ca2​ -P​ O​4​3-

Reaction: Li3​ P​ O4​ ​ ​ + CaSO4​ → Li2​ S​ O4​ + Ca​3(​ PO​4)​ 2​
I S S I

Percent of Oxygen: Lithium Phosphate vs. Lithium Sulfate

Lithium Phosphate Lithium Sulfate

Li (1) = 7(1)= 7/86 (100) = 8.14% Li (1) = 7(1)= 7/87 (100) = 8.05%
P (1) = 31(1)= 31/86 (100) = 36.05% S (1) = 32(1)= 32/8o7 (100)= 36.78%
O (3) = 16(3)= 48/86 (100) = 55.81 O (3) = 16(3)= 48/87 (100) = 55.17%

--------- ---------
100% 100%

IV. Conclusion:
Write a conclusion explaining the results of one of the reaction. You should focus on the
appearance of the final beaker. Your conclusion should also discuss the % of Oxygen between
2 of the compounds in the same reaction.

Overall, the reaction, Potassium Chloride, KCl, was soluble. The ionic charges of KCl
was K​+​Cl-​ ​. This became KCl because when the ionic charges crossed, they only had the ionic
charge of one. The reaction of Potassium Chloride was soluble based upon rule one. The rule
reads, “Salts containing Group 1 elements are soluble (Li+​ ,​ Na+​ ,​ K+​ ​, Cs​+​, Rb​+)​ . Exceptions to this
rule are rare. Salts containing the ammonium ion (NH4​ +​ )​ are also soluble.” The oxygens
percentages of Potassium Chloride are 52.7% Potassium and 47.3% Chloride. Moving on,
Potassium Nitrate, KNO3​ ,​ was soluble. The ionic charges of KNO3​ ​ was K+​ ​NO3​ -​ ​. This became
KNO3​ ​ because when the ionic charges crossed, they only had the ionic charge of one and the
three was left because there are three oxygen. The reaction of Potassium Nitrate was soluble
based upon rule one and two. The second rule explains, “Salts containing nitrate ion (NO3​ ​-)​ are
generally soluble.” The oxygens percentages of Potassium Nitrate are 38.61% Potassium,
13.86% of Nitrogen, and 43.57% Oxygen. In conclusion, both of the two chemical reactions
explained, KCl (Potassium Chloride) and KNO3​ ,​ (Potassium Nitrate) are soluble.

V. W​ hat is wrong with the following formula: ​ (PO4​ ​)​2N​ a

What make the formula, (PO​4​)​2​Na inaccurate is that the Monatomic Cation (Na) should
go before the Polyatomic Anion, in this case (PO​4​)2​ .​ The correct formula should be
Na(PO​4​)​2.​

Isotope Essay

Velocity Project

1. Define the following terms:

Motion- any movement or Speed- t​he rate at which Position- a place when
change in position or time. someone or something is someone or something is
located or has been put.
able to move or operate.

Distance- an amount of Acceleration- ​a vehicle's Terminal Velocity- the
space between two things capacity to gain speed internal velocity.
or people.
within a short time.

Time- the indefinite Initial Velocity- t​ he velocity Displacement- the moving
of something from its place
continued progress of at which motion begins. or position.

existence and events in the

past, present, and future

regarded as a whole.

Velocity- ​the speed of Final Velocity- ​the velocity Key Metric units- metric
something in a given
at which motion ends. units of length.
direction.

2. What is the difference between Speed and Velocity? Explain using an example
in your own words.

The difference between speed and velocity is that speed is the distance traveled by an
object where as, velocity is distance traveled by an object per unit time in a particular
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:

Cities: New York, New York, and Seattle, Washington

Map:
A. Fastest Runner
B. Model T Ford
C. Hindenberg
D. Tesla top speed
E. Fastest train
F. F35 Fighter Jet
G. Snail

*Provide a map showing your cities
*Show Detailed Math Steps

4. What would like to see in this city when you arrive?
What tourist attraction?
What would you like to eat in this city?
What is the basic history of this city?
When I arrive, I would like to see my cousins, aunt, and uncle. Also, I would enjoy

seeing the Space Needle and going inside. I would love to eat seafood in Seattle
because they have phenomenal sea cuisines (I know from experience). Seattle was
founded by members of the Denny Party on November 13, 1851.
5. Determine and graph an 18% increase in Velocity for each vehicle - Show how
the Times would be affected by the increase in speed.
*Include pictures and brief description of each mode of transportation

Fastest Runner- Usain Bolt (Four time Olympian).
Model T Ford- One of the First Cars.
Fastest Train- mag-lev Bullet Train (Japan).
Hindenburg- Caught Fire and was Destroyed.
Tesla Top Speed- Different modes of the Car (Fastest is Ludacris).
F35 Fighter Jet- Used for Air Defense and Ground Attack.
Snail- Thick, coiled shell makes them the slowest animal ever.
Graph (could not fit the snail speed on the graph- it was too slow)

6. Use a math calculation to show how long it would take the F35 Fighter Jet to
get to

A. Sun
B. Saturn
C. Neptune

(Use scientific notation)

GPE Project

Potential Energy Project
Due: Friday 3/17

Define and make note cards for the following words:

Energy-​ the property Joules-​ the SI unit of Chemical Potential Law of Conservation
that must be work or energy, equal Energy-​ is a form of of Energy-​ states that
transferred to an to the work done by a potential energy that the total energy of an
object in order to force of one newton can be absorbed or isolated system
perform work on – or when its point of released during a remains constant
to heat – the object, application moves chemical reaction or (it is said to be
and can be converted one meter in the phase transition. conserved over time.)
in form, but not direction of action of
created or destroyed. the force, equivalent
to one 3600th of a
watt-hour.

Kinetic Energy- Kilojoules-​ is the Elastic Potential Gravity-​ a natural
energy that a body Australian measure Energy-​ Potential phenomenon by
possesses by virtue of how much energy energy stored as a which all things with
of being in motion. people get from result of deformation mass are brought
consuming a food or of an elastic object, toward one another,
drink. such as the including planets,
stretching of a spring. stars and galaxies.

Potential Energy-​ the Gravitational Mechanical Energy-
energy possessed by Potential Energy-​ is the sum of kinetic
a body by virtue of its GPE = mgh, where m and potential energy
position relative to is the mass in in an object that is
others, stresses kilograms, g is the used to do work.
within itself, electric acceleration due to
charge, and other gravity (9.8 on Earth),
factors. and h is the height
above the ground in
meters.

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: You, gallon of milk, television (research the masses)
* 2.2 lbs = 1 kg

Pug - 20 lbs = 9.07185 kg
Armchair - 60 lbs = 27.2155 kg
Toilet - 100 lbs = 45.3592 kg

Data Table:
Your data table will need:
Object, mass, gravity, height, GPE

Videos: h​ ttp://www.youtube.com/watch?v=x5JeLiSBqQY
*Video shows you how to use the GPE equation.

Earth Mass ​(kg) Gravity​ (m/s2) Height​ (m) GPE (​ mgh) J
Object 9.07185 9.8 10 889.0413
Pug 27.2155 9.8 10 2667.119
Armchair 45.3592 9.8 10 4448.7296
Toilet

Determine the GPE of one of the masses on the following planets:
“Kotulskian” - 17% greater than Earth’s Gravity
“Danuzzitopia” - 39% less than Earth’s Gravity
“Carlucciville” - 82% greater than Earth’s Gravity
“Cheshire” - 63% less than Earth’s Gravity

Calculations:

A. Planet Kotulskian:
Gravity on Kotulskian: Earth’s Gravity x 1+Percent Growth or Decay
Gravity on Kotulskian: 9.8 m/s​2​ x 1+0.17
Gravity on Kotulskian: 9.8 m/s​2 ​x 1.17
Gravity on Kotulskian: 11.466 m/s​2

GPE on Kotulskian: Mass x Gravity x Height
GPE on Kotulskian: 9.07185 kg x 11.466 m/s2​ ​ x 10 m
GPE on Kotulskian: 1040.17832

B. Planet Danuzzitopia:

Gravity on Danuzzitopia: Earth’s Gravity x 1+Percent Growth or Decay
Gravity on Danuzzitopia: 9.8 m/s2​ ​ x 1-0.39
Gravity on Danuzzitopia: 9.8 m/s2​ x​ 0.61
Gravity on Danuzzitopia: 5.978 m/s​2

GPE on Danuzzitopia: Mass x Gravity x Height
GPE on Danuzzitopia: 9.07185 kg x 5.978 m/s2​ ​ x 10 m
GPE on Danuzzitopia: 542.315193

C. Planet Carlucciville:

Gravity on Carlucciville: Earth’s Gravity x 1+Percent Growth or Decay
Gravity on Carlucciville: 9.8 m/s​2​ x 1+0.82
Gravity on Carlucciville: 9.8 m/s2​ ​x 1.82
Gravity on Carlucciville: 17.836 m/s2​

GPE on Carlucciville: Mass x Gravity x Height
GPE on Carlucciville: 9.07185 kg x 17.836 m/s2​ ​ x 10 m
GPE on Carlucciville: 1618.05517

D. Planet Cheshire:

Gravity on Cheshire: Earth’s Gravity x 1+Percent Growth or Decay
Gravity on Cheshire: 9.8 m/s2​ ​ x 1-0.63
Gravity on Cheshire: 9.8 m/s​2 x​ 0.37
Gravity on Cheshire: 3.626 m/s​2

GPE on Cheshire: Mass x Gravity x Height
GPE on Cheshire: 9.07185 kg x 3.626 m/s​2​ x 10 m
GPE on Cheshire: 328.945281

Data Table: ​Pug

Planet Mass (kg) Gravity (m/s​2​) Height (m) GPE (mgh or J)
11.466 10 1040.17832
Kotulskian 9.07185 5.978 10 542.315193
17.836 10 1618.05517
Danuzzitopia 9.07185 3.626 10 328.945281

Carlucciville 9.07185

Cheshire 9.07185

Graph:

Critical Thinking Questions:
1. What factors affect Gravitational Potential Energy?

The factors that affect Gravitational Potential Energy are the gravity on the planet, the mass of
the object, and the height the object if falling from.

2. Why did the GPE change on the other planets?
The GPE changed on the other planets because they have different gravity.

3. Which planet would you be able to hit a golf ball further? Explain using data.
You would be able to hit a golf ball further on planet Cheshire because there is the least amount
of gravity on the planet (3.6 meters per second.)

4. How does GPE relate to Chemical Potential Energy?
GPE relates to Chemical Potential Energy because they both are based off of the amount of
energy something can have but in different ways.

5. How do Energy companies use GPE to generate Electrical Energy? Give an example
Energy companies use GPE to generate Electrical Energy by converting the energy of water
falling or moving at a fast pace. This is converted in the generator to energy. An example can be
a waterfall.

6. What happens to the GPE when the object falls to the ground? Describe the Energy
transformations along the way. Use a diagram.

As an object falls, its height above ground gets smaller and smaller therefore, the potential
energy stored in gravity gets less and less, too. The object speeds up as it falls and energy is
leaving the gravitational field and entering the falling object as kinetic energy.
Worksheet 1:
http://glencoe.mheducation.com/sites/0078600510/student_view0/unit1/chapter4/math_practice_2.html
Worksheet 2:​ ​http://go.hrw.com/resources/go_sc/ssp/HK1MSW65.PDF

KE Project

Kinetic Energy Project

Objective: 7​ .1.b. Energy can be stored in many forms and can be
transformed into the energy of motion.

Problem Statement:

1. Design an experiment to test how changing the angle of a ramp affects Kinetic
Energy?

2. What is the velocity of a roller coaster at the bottom of the hill?

Hypothesis:

If the angle of the ramp increases, the kinetic energy will increase.

Independent Variable:
The angle of the ramp

Dependent Variable:
Velocity

*Use the following angles: 8, 18, 28

Example:​ S​ uppose you want to make a ramp with a 23 degree angle. Look up 23

degrees on the Angle Chart and find the sine (decimal).
Sin 23 = 0.39
Substitute Sin 23 with 0.39

0.39 = opposite
hypotenuse

0.39 = opposite (height of ramp)
180 cm

solve for opposite (height of ramp)

*A​ ngle Chart

*Include diagrams of the 3 Angles
*Include math examples for determining the angles and KE.

Angle 8
Finding Opposite
Sin 8= 0.13197
Substitute Sin 8 with 0.13197

Opposite= 0.13197 x 120 cm
Opposite= 16.7 cm

Instantaneous Velocity
GPE=KE
mgh= 0.5mv2​
130 kg x 9.8 m/s2​ ​ x 1.67 m = 0.5 x 130v2​
21275.8 s​2/​ 65 = 65v2​ /​ 65
√327.32 = √2 v
1.81 m/s​2=​ v

Angle 18
Finding Opposite
Sin 18= 0.30902
Substitute Sin 18 with 0.30902

Opposite= 0.30902 x 120 cm
Opposite= 35.5 cm

Instantaneous Velocity
GPE=KE
mgh= 0.5mv2​
130 kg x 9.8 m/s​2 x​ 3.55 m = 0.5 x 130v2​
45227 s​2/​ 65 = 65v2​ ​/65
√695.8 = √2 v
2.64 m/s2​ ​=v

Angle 28
Finding Opposite
Sine 28= 0.46947
Substitute Sin 28 with 0.46947

Opposite= 0.46947 x 120 cm
Opposite= 56.3 m

Instantaneous Velocity

GPE=KE
mgh= 0.5mv​2
130 kg x 9.8 m/s2​ ​x 5.63 m = 0.5 x 130v​2
71726.2 m/s​2 ​/65 = 65v​2/​ 65
√1103.48 = √2 v

3.32 m/s2​ =​ v

Video Resources:
1. How to solve KE problems: ​https://www.youtube.com/watch?v=tryiwu4RhSM
2. PE/KE: h​ ttps://www.youtube.com/watch?v=Je8nT93dxGg
3. PE/KE: ​https://www.youtube.com/watch?v=BSWl_Zj-CZs
4. PE/KE: h​ ttps://www.youtube.com/watch?v=7K4V0NvUxRg

Data Analysis - W​ rite your data analysis paragraph here

According to the research gathered from our experiment, our hypothesis
of when the ramp is increased the instantaneous velocity will be faster, is
true. For example, when we were finding the GPE=KE, we found that for
angle 8’s instantaneous velocity we got 1.81 m/s. For the angle 18, we
found that the instantaneous velocity was 2.64 m/s2​ .​ The last angle, 28,
the instantaneous velocity ended up being 3.32 m/s​2​. For all of the
angles, the instantaneous velocity was much larger than the average
velocity (using the formula V=D/T). In summation, the data from our
experiment proves that our hypothesis is accurate.

Data Analysis Rubric (Self Evaluate) 34
Lab Rubric - Data Analysis Sections

12

Data/ ____Data is poorly ____Data is ____Data is ___Data is clearly
Observations organized or missing represented in a represented in the and accurately
altogether. table or graph, but table or graph with represented in a
No mention of it is incomplete or minor errors. More table or graph.
observations there are major complete discussion Observations
errors. Some of observations. include discussion
discussion of of both qualitative
observations and quantitative
observations.

Conclusion/ ____No conclusion is ____Somewhat ____Adequately ____Clearly explains

Analysis written in this report or it explains whether or explains whether or whether or not the
is very brief. No data is not the hypothesis not the hypothesis hypothesis was
cited. was supported. was supported. supported. Data
____No analysis is ____ Possible Data is cited to was cited to support
included or it is extremely sources of error are support hypothesis. the hypothesis.
brief no sources of error somewhat ____Possible ____Possible
are explained. explained. sources of error are sources of error are
____No discussion of ____ No discussion adequately clearly explained.
patterns or trends in the of patterns or explained.
data trends ____Some ____Trends and
discussion of Patterns in the data
patterns or trend in are clearly
the data discussed.

Self Evaluate Self Evaluate Score (1-4)
Category 4
4
Scientific Method 3.8
3.7
-Hypothesis
-Identify the variables 3.9

Math Examples

-Angle steps shown
-KE steps shown

Graph

-Accurate
-Informative

Data Analysis

-Hypothesis correct?
-Support for hypothesis
-Transition words

Reading
Comprehension

-Article questions

Article: E​ veryday Energy

Read the article and answer the questions.

1. A
2. B
3. D
4. C
5. D
6. A
7. C
8. The Hoover Dam provides power to California, Nevada and Arizona by harnessing the

converted potential energy of Lake Mead.
9. How much water is required downriver from the dam and the water levels of Lake Mead

determine the energy production of the Hoover Dam.
10. Students should communicate that if there is a drought, then Lake Mead will have less

water. If Lake Mead has less water, then there will be less potential energy stored in
Lake Mead. If there is less potential energy, then there will be less kinetic energy created
by water flowing through the Hoover Dam.

FINAL PART - Roller Coaster Physics

Objective:
1. When energy is transformed, the total amount of energy stays constant (is conserved).
2. Work is done to lift an object, giving it gravitational potential energy (weight x height). The
gravitational potential energy of an object moving down a hill is transformed into kinetic
energy as it moves, reaching maximum kinetic energy at the bottom of the hill.

Determine the velocity of a full roller coaster of riders at the bottom of the largest hill. You can
use the following roller coasters:
Watch these Videos for help:
http://www.youtube.com/watch?v=Je8nT93dxGg
http://www.youtube.com/watch?v=iYEWIuQBVyg

Use either:

GPEt​ op​ = KE​bottom

1. Cyclone - Coney Island

Height of largest hill - 2​ 5.908 m

Mass of Coaster/number of riders -​ 535 kg

GPE = KE

535kg(25.908m)(9.8m/s2​ ​) = 0.5(535kg)v2​
135835.6kg 267.5kgv2
267.5kg =​ 267.5

√507.8 = √v2

22.5 m/s = v

Velocity at Bottom of Hill in m/s - ​22.5 m/s

2. Pick your own coaster: ​Kingda Ka

Height of largest hill -​ ​139 m

Mass of Coaster/number of riders -​ 100 kg

GPE = KE

100kg(139m)(9.8m/s​2​) = 0.5(100kg)v2​
136220kg 50kgv2
50kg =​ 50

√2724.4 = √v2

52.2 m/s = v

Velocity at Bottom of Hill in m/s - 5​ 2.2 m/s

*Write about your results in a paragraph.

From the math, I discovered that the rollercoaster with the largest height of the largest hill had
the greatest velocity at the bottom of the hill. Cyclones height was 25.908 m while Kingda Ka’s
was a whopping 139 m tall. To show, the instantaneous velocity of Cyclone was 22.5 m/s2​ ,​ while
the instantaneous velocity of Kingda Ka was 52.2 m/s2​ .​ Also the rollercoaster with the smaller
mass to riders ratio had a greater instantaneous velocity. For example, the mass of Cyclone
was 535 kg and the mass for Kingda Ka was 100 kg. In conclusion, the rollercoaster with the
greater height and smaller mass (Kingda Ka) had a greater instantaneous velocity.

More resources:
http://www.youtube.com/watch?v=BSWl_Zj-CZs
Kinetic and Potential Energy

http://www.youtube.com/watch?v=7K4V0NvUxRg
Kinetic and Potential Energy

http://www.youtube.com/watch?v=btLU2lb3-xs
Bill Nye

http://www.youtube.com/watch?v=-dpBVtAbKJU
Roller Coasters

http://www.youtube.com/watch?v=iYEWIuQBVyg

Inclined Plane Project

Inclined Plane Project

Due: April 19, 2017
1. Define the following vocabulary: Use pgs. 124 - 153

Simple Machine Mechanical Work Input Force
Any of the basic
mechanical devices Advantage activity involving  The work input of a 
for applying force, mental or physical  machine is equal to 
such as an inclined The ratio of the force  effort done in order  the effort force 
plane, wedge or produced by a  to achieve a purpose  times the distance 
lever. machine to the force  or result. over which the 
applied to it, used in  effort force is 
assessing the  exerted.
performance of a 
machine.

Compound Machine Ideal Mechanical Power Output Distance

A machine Advantage supply (a device)  number of times 
consisting of two that a machine 
or more simple Mechanical  with mechanical or  increases an input 
machines force
operating advantage is a  electrical energy. 
together, as a
wheelbarrow measure of the   
consisting of a
lever, axle, and force amplification 
wheel.
achieved by using a 

tool,mechanical 

device or machine 

system.

Efficiency Actual Mechanical Input Distance Output Force

The ratio of the  Advantage Force exerted on a  Work output is 
energy delivered  machine. Distance  always less than 
(or work done) by a  Actual mechanical  the force acts  work input, so an 
machine to the  advantage takes  through actual machine 
energy needed (or  into account  cannot be 100% 
work required) in  energy loss due to  efficient.
operating the  deflection, friction, 
machine and wear

First Class Lever Second Class Lever Third Class Lever Energy

Class 1: Fulcrum in   Second-class lever  A lever where the  the strength and 
vitality required for 
the middle: the  share the load  effort is located  sustained physical or 
mental activity.
effort is applied on  between the effort  between the 

one side of the  and the fulcrum fulcrum and the 

fulcrum and the  load.

resistance (or load) 

on the other side, 

for example, a 

seesaw, a crowbar 
or a pair of 
scissors.

Block and Tackle Fixed Pulley Movable Pulley

Pulley A fixed pulley  A movable pulley is 
changes the  a pulley that is free 
Pulleys with a rope  direction of the  to move up and 
or cable threaded  force on a rope or  down, and is 
between them,  belt that moves  attached to a 
usually used to lift  along its  ceiling or other 
or pull heavy loads circumference object by two 
lengths of the 
same rope.

2. Experiment: How does the angle of an inclined plane affect:
A. Ideal Mechanical Advantage

B. Actual Mechanical Advantage
C. Efficiency
*Think about the scientific Method
DATA TABLE
***Why is the Actual Mechanical Advantage always less than the Ideal Mechanical

Trial Output Output Output Input Input Input IM AM Efficienc
Force Dist. Work Force Dist. Work AA y
Angle 1 - 10 N 6N
10 10 N 0.58 m 5.8 J 6N 1.2 m 7.2 J 5.7 1.67 29%
10 N 6N
Angle 1 - 10 N 0.58 m 5.8 J 7N 1.2 m 7.2 J 5.7 1.67 29%
10 10 N 7N
0.58 m 5.8 J 1.2 m 7.2 J 5.7 1.67 29%
Angle 1 -
10 0.58 m 5.8 J 0.87 m 6.09 J 4.1 1.4 34%

Angle 2 - 0.58 m 5.8 J 0.87 m 6.09 J 4.1 1.4 34%
14

Angle 2 -
14

Angle 2 - 10 N 0.58 m 5.8 J 7N 0.87 m 6.09 J 4.1 1.4 34%
14 10 N 0.58 m 5.8 J
10 N 0.58 m 5.8 J 9 0.5 m 4.5 J 2.4 1.1 46%
Angle 3 - 10 N 0.58 m 5.8 J
25 9 0.5 m 4.5 J 2.4 1.1 46%

Angle 3 - 9 0.5 m 4.5 J 2.4 1.1 46%
25

Angle 3 -
25

Conclusion:

*Write your OWN CONCLUSION HERE!

In the end, the evidence shows that the bigger the angle is the more force needed to be
able to lift the object. Looking at the data table, the IMA increase from our first trial (which was
our smallest angle) to our final trial (which was the largest angle). As you can see, once we
increased the angle the input force rose, ours rose about 1-2 N each trial. For this experiment
AMA calculate the amount of friction so this means that the AMA is always going to be smaller
than the IMA. In this experiment, our efficiency increases from our first trial to our final trial which
means that we did the experiment correct. Looking at our first trial we had an efficiency of 29%,
than for our second trial we had a efficiency of 34%, and finally for our last trial we had an
efficiency of 49%. Overall from this experiment we can conclude that when you increase the
angle the greater the ideal mechanical advantage, the less the actual mechanical advantage
and the greater the efficiency.

3. Critical Thinking (Include in Presentation):
A. How much WORK would be done to lift a 350 kg Piano to the top of the Empire State
Building using a ramp with an angle of 35 degrees?

350kg*9.8m/s2​ =​ 3430
2.5=3430 N/x
8575 N = x

W​in​= D​in​*Forcei​ n
Wi​ n=​ 834 ft(8575 N)
Win=7,151,550J

1. What is the length of the ramp to the top of the Empire State Building?

35 degrees = 0.57358
Hyp = Opp(0.57358)
Hyp = 1,454ft(0.57358)

Hyp = 834 feet

2. Suppose the Ideal Mechanical Advantage is 3.2

3. The Actual Mechanical Advantage (AMA) is 2.5.

4. What is the Efficiency of this Machine?

Worko​ ut ​= D​out*​ F​out
Worko​ ut =​ 1454 ft(3430 N)
Work​out​ = 4,987,220 J

E = Worko​ ut/​ Worki​ n
E = 4,987,220/7,151,550
E = .697*100
E = 69.7%

5. Provide a diagram of this example.

4. Explain how the Ideal Mechanical Advantage and Actual Mechanical Advantage is

determined for the following simple machines:

A. Inclined Plane

a. MA= D​in​/D​out
i. To find the IMA of an inclined plane, you must find the Distance input and the

Distance output first. Then, you do Distance input divided by the Distance output.

ii. AMA= Fo​ ut/​ F​in
1. To find the AMA of an inclined plane, you must find the Force output and

Force input first. Next, you divide the Force output by the Force input.

B. Lever

a. IMA = length of ef f ort arm
length of the resistance arm

i. The effort end of the lever is the lever is the side of the lever that effort is applied

to, while the resistance arm is the side of the lever which holds the load.

b. AMA = resistance f orce
ef f ort f orce

i. First you must find the resistance force and then divide it by the effort force.

C. Pulleys
a. n order to find the IMA and AMA for a pulley you have to count the amount of ropes you
have required to lift the object.

D. Wheel and Axle

a. IMA= radius of the wheel
radius of the axle

i. To find the IMA of a wheel and axis, you must divide the radius of the wheel by

the radius of the axle.

b. AMA= Resistance f orce
Actual Ef f ort f orce

i. To find the AMA of a wheel and axis, you must do the resistance force divided by

the actual effort force.

Quiz: Inclined Plane

QUIZ: Inclined Plane

Directions: A​ nalyze the Inclined Plane Data Table that is shared on
Classroom and determine which machine has the greatest Actual
Mechanical Advantage (AMA).
Problem Statement:
How does the angle of an inclined plane affect the Mechanical
Advantage?

Hypothesis: ​(Use proper form!)

If the angle of an inclined plane increases, then the Mechanical Advantage will
decrease.

Diagrams of Inclined Planes:​ (Label Diagrams)

Angle Chart: ​https://drive.google.com/open?id=0B4RmhXJlHvo1YXZhcDNMSDNSMXc

Angle 1:

SinL= opp
hyp

SinL= 30
150

SinL= 0.2

SinL= 12°

Angle 2:

SinL= opp
hyp

SinL= 30
90

SinL= 0.33

SinL= 19°

Angle 3:

SinL= opp
hyp

SinL= 30
50

SinL= 0.6

SinL= 37°

Calculations (​ Examples):

Angle IMA AMA Efficiency

Angle 1: 12° IMA= Din AMA= F out Efficiency
IMA= 1D5o0ut AMA= 7F in Wout= Fout × Dout
30 2 Wout= 7N × 30m

IMA= 5 AMA= 3.5

Wout= 210 N

Win= Fin × Din
Win= 2N × 150m

Win= 300 N

Efficiency= W out ×
W in
100
210
Efficiency= 300 ×

100

Angle 2: 19° IMA= Din AMA= F out Efficiency= 70%
Dout F in Efficiency

IMA= 90 AMA= 7
30 3
Wout= Fout × Dout
IMA= 3 AMA= 2.33 Wout= 7N × 30m