Sunday, September 30, 2012
Video Reflection (Sep. 24, 2012/in class)
The video we watched in class was taken from a Japanese television program, I presume. It was showing us Newton's first law of inertia. The video was very good visually, but most of the video was a little bit pointless to people who do not know Japanese.The only parts that were relevant to all of us were the parts when they were showing us that both the baseball and the truck are movie at the same velocity. I liked the video, and it was cool for what it was. Very good visual aid for explaining the concept.
Video Reflections (Newton's 3rd law)
Newton's 3rd Law says that for every action there is an equal and opposite reaction. The first video i watched was talking about how in space, if you try to turn a screw, you end up turning instead of the screw. On earth, this doesn't happen because we have gravity to hold us down. In terms of Newton's third law, the reason why we turn in space when we try to turn a screw is because the opposite reaction is happening to us. We are trying to turn the screw, but instead of the screw turning, the opposite happens. The screw stays still while we spin around. The video also talks about ways to prevent this occurrence.
The second video was talking about how much of an impact we have on the position and spinning of the earth. It was mostly talking about whether or not we could actually move the earth if we had a bunch of people jump at once. The video said that we could, but our effect would be so little due to the mass of the earth being much, much greater than ours. In terms of Newton's third law, the reaction the earth would have to a bunch of people jumping at once is completely probable. If we jump up we are really pushing against the earth. As an opposite reaction, the earth would move downwards away from our feet, but only a tiny amount. The movement is so tiny it is not even detectable. However, it is still possible for the earth to move away from our feet due to Newton's 3rd Law.
The second video was talking about how much of an impact we have on the position and spinning of the earth. It was mostly talking about whether or not we could actually move the earth if we had a bunch of people jump at once. The video said that we could, but our effect would be so little due to the mass of the earth being much, much greater than ours. In terms of Newton's third law, the reaction the earth would have to a bunch of people jumping at once is completely probable. If we jump up we are really pushing against the earth. As an opposite reaction, the earth would move downwards away from our feet, but only a tiny amount. The movement is so tiny it is not even detectable. However, it is still possible for the earth to move away from our feet due to Newton's 3rd Law.
Newton’s Law of Motion Worksheet
1.A net force of 9.0 N east is used to push a 20.0 kg object. What is the acceleration of the object?
F=ma
9.0 = 20.0 (a)
a = 9.0/20.0
a= 0.45 m/s^2
2.A 16.0 kg object is accelerated at a rate of 2.0m/S2 by a net force. What is the magnitude of this force?
F=16.0*2.0
F=32 N
3. A 925 kg car accelerates uniformly from rest to a velocity of 25.0 mls south in 10.0 s. What is the net force acting on the car during this time?
F=(925*25)/10
F=2312.5 N to south
4. A net force of 6.6N east acts on a 9.0 kg object. lf this object accelerates uniformly from rest to a velocity of 3.0 m/s east. How far did the object travel and how long did it take?
6.6 = (9.0*3.0)/x
6.6x=9.0*3.0
x=(9.0*3.0)/6.6
x=4.09
5. A 12.0 kg object is pushed with a horizontal force of 6.0 N East across a table. If the force of friction is 2.0 N, what is the acceleration of the object?
6-2=4N
4/12=0.33
a=0.33
6. A 20.0 kg object is pulled horizontally along a level floor with a force of 27.0 N. lf the object is accelerating at a rate of 0.80 m/S2, what is the magnitude of the force of friction?
2*0.80=1.6
27.0-1.6=25.4
7. An object that has a mass of 36.0 kg is pushed along a horizontal surface with a force of 85.0 N. IF the force of friction is 72.0 N, what is the magnitude of the acceleration of the object?
85-72=13
13=36a
13/36=a
a=0.361m/s^2
8. A 7.0 kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal force that is required to accelerate it at the rate of 2.3 m/s2?
F=7.0*2.3
F=16.1N
9. You are traveling in your car at a velocity of 24.0 mls east when you slam on your brakes. The force of friction on your car tires is 1.80x104 N. IF the mass of you and your car is 1.50x103 kg, how far do you skid before stopping?
10. A 325 N box is sliding down a frictionless inclined plane. lf the inclined plane makes an angle of 30.0° with the horizontal, what is the acceleration along the incline?
F=ma
9.0 = 20.0 (a)
a = 9.0/20.0
a= 0.45 m/s^2
2.A 16.0 kg object is accelerated at a rate of 2.0m/S2 by a net force. What is the magnitude of this force?
F=16.0*2.0
F=32 N
3. A 925 kg car accelerates uniformly from rest to a velocity of 25.0 mls south in 10.0 s. What is the net force acting on the car during this time?
F=(925*25)/10
F=2312.5 N to south
4. A net force of 6.6N east acts on a 9.0 kg object. lf this object accelerates uniformly from rest to a velocity of 3.0 m/s east. How far did the object travel and how long did it take?
6.6 = (9.0*3.0)/x
6.6x=9.0*3.0
x=(9.0*3.0)/6.6
x=4.09
5. A 12.0 kg object is pushed with a horizontal force of 6.0 N East across a table. If the force of friction is 2.0 N, what is the acceleration of the object?
6-2=4N
4/12=0.33
a=0.33
6. A 20.0 kg object is pulled horizontally along a level floor with a force of 27.0 N. lf the object is accelerating at a rate of 0.80 m/S2, what is the magnitude of the force of friction?
2*0.80=1.6
27.0-1.6=25.4
7. An object that has a mass of 36.0 kg is pushed along a horizontal surface with a force of 85.0 N. IF the force of friction is 72.0 N, what is the magnitude of the acceleration of the object?
85-72=13
13=36a
13/36=a
a=0.361m/s^2
8. A 7.0 kg object rests on a horizontal frictionless surface. What is the magnitude of the horizontal force that is required to accelerate it at the rate of 2.3 m/s2?
F=7.0*2.3
F=16.1N
9. You are traveling in your car at a velocity of 24.0 mls east when you slam on your brakes. The force of friction on your car tires is 1.80x104 N. IF the mass of you and your car is 1.50x103 kg, how far do you skid before stopping?
10. A 325 N box is sliding down a frictionless inclined plane. lf the inclined plane makes an angle of 30.0° with the horizontal, what is the acceleration along the incline?
Tuesday, September 25, 2012
Tracker Analysis
Use Tracker to determine
the equations for the displacement (horizontal and vertical) of the
basketball. Take a screenshot showing your equations and graphs.
Determine the maximum
height of the basketball in the air (with respect to its initial
launching position). Take a screenshot showing your maximum height.
From Tracker, you can determine the time of the basketball flying in the air. Record this time.
t=1.75 seconds
With the horizontal displacement, the maximum height, and the time of the basketball traveling in the air that you determined previously, calculate the initial velocity, and the launching angle of the basketball.
Initial Velocity=7.06 m/s
Launch Angle=71.2 degrees
t=1.75 seconds
With the horizontal displacement, the maximum height, and the time of the basketball traveling in the air that you determined previously, calculate the initial velocity, and the launching angle of the basketball.
Initial Velocity=7.06 m/s
Launch Angle=71.2 degrees
Monday, September 24, 2012
Test Make-up
You can download the document if on click on File>Download as...
https://docs.google.com/spreadsheet/ccc?key=0AhZ8hA6fOtl9dE5fMU1MR2kyYkNJUi0tRE51bzhpT3c
https://docs.google.com/spreadsheet/ccc?key=0AhZ8hA6fOtl9dE5fMU1MR2kyYkNJUi0tRE51bzhpT3c
Sunday, September 23, 2012
Equilibrium Summary
According to the book when an object is in equilibrium it can be either moving or at rest, but it has to be at a constant velocity. In order for the object to be at equilibrium the net force acting upon the object must equal zero. This is all described by Newton in his first law. The example given in the book is one of a fishing bob, but I feel I can give my own example...which may or may not be correct. Imagine two children fighting over a stuffed toy. One child pulls the other arm while the other child pulls the other. The toy in the middle will be in a state of equilibrium if the two children pull with the same force. This is because it will be at a constant velocity and the forces exerted by the children will cancel out. Below is a diagram that represents my theoretical situation.
Sunday, September 16, 2012
Wednesday, September 5, 2012
Position Vs. Time Graph (Activity)
In the running activity, we recorded the distance we ran and the time it took to run it. The relationship between position and time allows us to get our velocity and our acceleration. The time it took me to reach each checkpoint was fairly constant at the end, but the begining part had some variations. The initial velocity was 0 m/s the ending velocity was 5m/s. However, from 0-2 seconds the velocity changed to 2.5 m/s. because of that change, acceleration is created. My accleration from 0-2 seconds/ 0-5 meters was 0.5m/s^2. My acceleration changed however, and remained constant for the rest of the data. From 2-6 seconds/ 5-25 meters my acceleration changed to 0.625m/s^2. After that Acceleration change, there is no more acceleration because the velocity remains constant. The velocity from 3 seconds onwards is 5m/s. It is constant therefore no acceleration. These are the conclusions I came to looking at my graph according to my running period.
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