miércoles, 10 de junio de 2015

Hypothesis

As said the above formula of the weight of an object, the weight equals it mass multiplied by acceleration (which is always the same quantity, 9.8). What I am going to do in this work is how changing the surface area of an object, can affect the velocity of the same. The object is going to be a square of clay which is going to have the same weight, but I am going to change the surface area of the square by using the formula of this object (A = l2), also, the height from which I’m going to throw it is not going to change, always is 1.75 m of altitude.
What I think is going to happen is that if there is a change on the surface area, if there is an increase on its surface area, the square would reach more slowly the floor.
On this experiment, the formula more indicated to see the change is the one of velocity = distance/time.

LAB REPORT CHANGE IN SURFACE AREA OF AN SQUARE OF CLAY

Height from where we are going to throw the ball of clay: 1,75 cm
Area of the square:


Mass with 25m2: 115,66 g
5 cm is the length of one side of the square
A = 5 * 5 = 25 m2
Number of the experiment
Surface area
Time (s)
1
25 m2
0.50 s
2
25 m2
0.62 s
3
25 m2
0.59 s
4
25 m2
0.63 s
5
25 m2
0.59 s

Mass with 36m2 = 113, 15 g
6 cm is the length of one side of the square
A = 6 * 6 = 36 m2

Number of the experiment
Surface area
Time (s)
1
36 m2
0.66 s
2
36 m2
0.62 s
3
36 m2
0.62 s
4
36 m2
0.59 s
5
36 m2
0.60 s

Mass with 49m2: 112,13 g
7 cm is the length of one side of the square
A = 7 * 7 = 49 m2
Number of the experiment
Surface area
Time (s)
1
49 m2
0.63 s
2
49 m2
0.62 s
3
49 m2
0.50 s
4
49 m2
0.56 s
5
49 m2
0.60 s

Mass with 64m2: 113,13 g
8 cm is the length of one side of the square
A = 8 * 8 = 64 m2
Number of the experiment
Surface area
Time (s)
1
64 m2
 73 s
2
64 m2
 70 s
3
64 m2
 72 s
4
64 m2
60 s
5
64 m2
62 s

Average of the time
Length of the side of the square
Average of the time
5 cm
0.59 s
6 cm
0.62 s
7 cm
0.58 s
8 cm
0.67 s



Conclusion:
As we can observe in our graph and table, our results weren’t conclusive. Our result weren’t the same as we expected them to be in the hypothesis, and this may be because an experimental change or fault in our experiment such as not the exact precision.
The time had to decrease as the length of the side of the square increase, and it didn’t: first, it increased, then it decreased and finally the time increased again.
“What I think is going to happen is that if there is a change on the surface area, if there is an increase on its surface area, the square would reach more slowly the floor”.
This is what we supposed in the hypothesis but what finally did happened was different:
What finally happened in our experiment is that as there is an increase on the surface area, the square reaches the floor first, quicker, then slower, and finally quicker.
The right result should have been equal to the hypothesis.
Equations related with the motion in the air of an object:
V= D/T

Velocity (m/seconds or km/hours) = distance (m)/time (seconds)
If we are throwing an object at a high altitude, obviously, it would reach the floor slowly than if the altitude is smaller, and also, depending on how is the surface area or the shape of the object, the velocity upon the object would be quicker or not.
 W= m*g
Weight of an object = mass of the object x acceleration of the gravity
If the mass of an object is bigger, the acceleration of the object is going to be slower and the weight of the object would be as well as its mass bigger.
 Ec= 1/2 m*v^2
Kinetic energy = ½ x mass x volume square
If there is in an object which has a volume square and mass is bigger and has a higher quantity, the kinetic energy of the same, would have an increase.
F= m*a
Force of an object = mass of the object x acceleration
Frictional force is any force opposite to a movement, which manifests itself in the contact surface of two bodies that always one moves or tends to move over another.
If there is an object that has a big mass and acceleration, its force would increase also, as it depends on this two factors to show its force.
 Fair= -cv^2 
The air resistance = air constant x object’s velocity (square)
Air Resistance Formula is used to find the air resistance, air constant and velocity of a body if some of these quantities are known. This is the formula more appropriate for my experiment.
These are some formulas that helped us in our project.
Evaluation:
Problem: Precision. Maybe we should have been more cautious at the time we were doing the experiment because there are several factors that could have affect the results:
                Not measuring the time exactly. To be more precise with time, we might have film the experiment so that way we were more precise.
                Not throwing the clay with the same force: Maybe we added some force at a specific time when we were throwing the clay to the floor.
I know that we aren't adding force because we are just dropping the plasticene but perhaps we are pushing the clay a little when we let it drop, this could be prevented just being precise and standing still without moving.
Solution: The solution to these problems is to be very careful at the time we are doing an experiment which can be easily affected for small faults
Problem: The clay had not the same weight always. We used the same piece of clay every time. So, maybe, at the time it reached the floor, it was losing some its weight leaving some clay in the floor. And that could have affected the result.

Solution: Making one piece of clay for each experiment and making sure they al  have the same weight.
  
REFERENCES:
En el texto: (Hyperphysics.phy-astr.gsu.edu, 2015)

Bibliografía: Hyperphysics.phy-astr.gsu.edu,. (2015). Mass, Weight, Density. Retrieved 13 June 2015, from http://hyperphysics.phy-astr.gsu.edu/hbase/mass.html

En el texto: (Infoplease.com, 2015)
Bibliografía: Infoplease.com,. (2015). formula weight. Retrieved 13 June 2015, from http://www.infoplease.com/encyclopedia/science/formula-weight.html

En el texto: ('Weight', 2015)
Bibliografía: Weight. (2015). Boundless. Retrieved from https://www.boundless.com/physics/textbooks/boundless-physics-textbook/the-laws-of-motion-4/other-examples-of-forces-47/weight-239-6231/

2 comentarios:

  1. I have just noticed that all the times are in seconds! I don´t think it took 60 seconds for a piece of plasticene to fall 1.75m!!! I suppose they should be 0.60 seconds?

    Conclusion: It is a shame that you only have 4 data points as it is difficult to draw a justified conclusion from them. To me, one of the data points seems to be anomalous compared to the others? If this one is ignored then the others seem to be in a clear pattern. Please include a relevant equation (with in-text reference) to support what you have said about increasing surface area and increasing time.

    Evaluation: What do you mean by "be careful with the recording of the time"? How much effect do you think the use of the stopwatch would have? For example, if human reaction time is 0.25 seconds then that is 50% of some of your recordings! Could you maybe film the experiment to record the time more accurately?

    How would you have added force if you were dropping the plasticene? How could you prevent this from happening?

    What would you use to measure the MASS of the plasticene?

    References: You need 3 references.

    3/8 --> 3.8

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  2. You have included equations but not used them to explain your results. Also you have not referenced them correctly. The equation about kinetic energy contains v which is velocity not volume!

    The evaluation is slightly more detailed but you have not read my comments carefully!

    5/8 --> 6.3

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