Thursday, November 14, 2013
Tides are cool
Friday, November 1, 2013
Unit 2 Reflection
In this unit, we learned about...
Newtons 2nd Law:
During our 2nd unit we learned first about Newton's Second Law which states...
a = fnet or acceleration = total force
m mass
This equation in mathematical terms is that acceleration is directly proportional to force and inversely proportional to mass. Meaning, that if acceleration increases, the force will also increase and the mass will decrease.
a ~ f and a ~ 1
m
I the example to the right, the 10kg box is being pushed with 20N of force to the right. to find the acceleration, we simply plug in 20N for force and 10kg for mass and solve to get 2 m/s^2. In the example to the left, the 10kg box is being pushed downward with a force of 10N. We know that the force is 10N because whenever an object is falling down, the only force acting upon it will be gravity which is always 10. So we plug the 10kg and 10N of gravity into the weight = mass x gravity and find that the weight (the force) is 100N. We then take this and plug it into the acceleration equation to find that the acceleration will be 10 m/s^2.
Newtons 2nd Law Lab:
We did a lab involving Newtons 2nd law to find how the acceleration of a system is related to the mass and the force on a system...
In experiment A, we set up the lab so that the force was being kept constant (the hanging weight), buy we added mass to the system each round. On the first round, we didn't add anything to the cart, and the acceleration was fast. For the 2nd round, we added 10 grams of weight to the cart, and continued to after each round. As we added more weight, we noticed a trend: the acceleration was decreasing. We related this back to Newtons 2nd law where the acceleration is inversely proportional to mass. In this case the acceleration was decreasing because the mass was increasing, thus, proving Newtons 2nd law to be true.
In experiment B, we did something different in the sense that the mass was kept constant throughout this entire experiment, but we increased the force. We started out with all the weight in the cart, and for each round we removed a 10kg weight from the cart and moved it over to the force (the hanging weight). In this experiment, as the force of the cart increased, the acceleration increased, thus, proving again that Newtons 2nd law saying that acceleration is proportional to force, is true because they both increased in this example.
Newtons 2nd law can also be written where a = fnet x 1/m. So for experiment A, we kept the force constant, and since we know that whatever is being kept constant is the slope on a graph, we can compare the two:
y = m (slope) x X
Falling Through the Air:
We continued to learn about Newtons 2nd law, but now we were learning about how it was involved with skydiving.
While skydiving, we learned that when you first jump out of the plane, your velocity will be 0 m/s. Lets say that your fweight is 100N, the total air resistance, or fair, is going to be 100-0 which is 0N. The acceleration when you first jump out let say is 10 m/s^2. Once you start moving, lets say that your fair is now 80N. This will mean that the total force will be 100-80 which is 20N. Once you finally reach terminal velocity which is just traveling at a constant speed, your fair will equal your fweight, therefore, in this case, it will be 100-100 = 0N (terminal velocity A). This is the time where you open your parachute so that you can fall slower. Once the parachute is opened your fnet will now be in the opposite direction, meaning that the acceleration is upward, but the velocity is downward. Once the fair goes back down and reaches the same as the fweight, the fnet will be 0N, therefore you are at your second terminal velocity. The only difference between terminal velocity A and B is that B is slower, there fnet, and fair are the same. Two things that increase the force of air resistance are speed and surface area.
Connecting it back to Newtons 2nd law, we learned that the acceleration is going to decrease as an object falls thought the air until it reaches terminal velocity adn stays at that speed. Because fo Newtons 2nd law we know that acceleration dn net force are directly proportional, therefore since acceleration is decreasing, we know that fnet is also decreaseing.
We can change the situation to a lead ball and a ping pong ball and find out which one hits the ground first as well. When the lead ball and the ping pong ball are dropped from a short hight, they will land at the same time because they didn't have much time to really start speeding up. But when the two balls are dropped from a higher hight, the lead ball will land first because it needs a greater fair to reach its terminal velocity.
Free Fall- Falling Straight Down:
When objects are falling in free fall, it means that the only force acting on it is gravity, there is no air force. When an object is dropped straight down, at 0 seconds, it will have a velocity of 0m/s. It is important to remember that even when the object is first dropped, the acceleration will always be 10m/s^2. At 1 second, the velocity will have increased by 10 m/s, therefore at 1 sec, the velocity is 10m/s. At 2 seconds, the velocity will be 30m/s and so on. To find the distance that this ball has been dropped, the equation d = 1/2gt^2. So to find how far the ball has dropped at 2 seconds, we plug in
d = 1/2 (10)(2)^2, which when solved comes out to be 20 m. To find the velocity the ball was going at 2 seconds, you use the equation v = gt, which is v = (10)((2), and it equals 20 m/s.
When an object falls straight down, the equations used are:
Free Fall- Falling at an Angle:
100 + 100 = c^2
200 +c^2
c = 14.1m/s
Free Fall- Throwing things up at an Angle:
The last thing we learned in our unit about Newtons 2nd law was throwing things up at an angle like a baseball player. To the right, there is a picture of a baseball player throwing a ball will an initial vertical velocity of 30m/s, and a horizontal velocity of 5m/s with 1 second intervals between them. We know that the ball will continue to travel 5m/s in the horizontal direction, but in the vertical direction, the ball will start slowing down, therefore the vertical speed will go from 30 to 20 to 10 to 0, and then start speeding up again as it starts falling down. To find how far downfield the ball travels, you will use the horizontal equation v = d/t:
v = d/t
5 = d/6
d = 30m
We can also find how high the ball got at the top of its path using the vertical equation. Be careful because the vertical equation is meant for things falling down, so we will use it in terms of it falling down:
d = 1/2gt^2
d = 1/2 (10)(3)^2
d = 5 x 9
d = 45m
Newtons 2nd Law:
During our 2nd unit we learned first about Newton's Second Law which states...
a = fnet or acceleration = total force
m mass
This equation in mathematical terms is that acceleration is directly proportional to force and inversely proportional to mass. Meaning, that if acceleration increases, the force will also increase and the mass will decrease.
a ~ f and a ~ 1
m
I the example to the right, the 10kg box is being pushed with 20N of force to the right. to find the acceleration, we simply plug in 20N for force and 10kg for mass and solve to get 2 m/s^2. In the example to the left, the 10kg box is being pushed downward with a force of 10N. We know that the force is 10N because whenever an object is falling down, the only force acting upon it will be gravity which is always 10. So we plug the 10kg and 10N of gravity into the weight = mass x gravity and find that the weight (the force) is 100N. We then take this and plug it into the acceleration equation to find that the acceleration will be 10 m/s^2.Newtons 2nd Law Lab:
We did a lab involving Newtons 2nd law to find how the acceleration of a system is related to the mass and the force on a system...
In experiment A, we set up the lab so that the force was being kept constant (the hanging weight), buy we added mass to the system each round. On the first round, we didn't add anything to the cart, and the acceleration was fast. For the 2nd round, we added 10 grams of weight to the cart, and continued to after each round. As we added more weight, we noticed a trend: the acceleration was decreasing. We related this back to Newtons 2nd law where the acceleration is inversely proportional to mass. In this case the acceleration was decreasing because the mass was increasing, thus, proving Newtons 2nd law to be true.
In experiment B, we did something different in the sense that the mass was kept constant throughout this entire experiment, but we increased the force. We started out with all the weight in the cart, and for each round we removed a 10kg weight from the cart and moved it over to the force (the hanging weight). In this experiment, as the force of the cart increased, the acceleration increased, thus, proving again that Newtons 2nd law saying that acceleration is proportional to force, is true because they both increased in this example.
Newtons 2nd law can also be written where a = fnet x 1/m. So for experiment A, we kept the force constant, and since we know that whatever is being kept constant is the slope on a graph, we can compare the two:
a = fnet x 1/m
y = m (slope) x X
But in experiment B, we kept the mass of the system constant, so in this case, the mass was being compared tot he slope:
a = 1/m x fnety = m (slope) x X
Falling Through the Air:
We continued to learn about Newtons 2nd law, but now we were learning about how it was involved with skydiving.
While skydiving, we learned that when you first jump out of the plane, your velocity will be 0 m/s. Lets say that your fweight is 100N, the total air resistance, or fair, is going to be 100-0 which is 0N. The acceleration when you first jump out let say is 10 m/s^2. Once you start moving, lets say that your fair is now 80N. This will mean that the total force will be 100-80 which is 20N. Once you finally reach terminal velocity which is just traveling at a constant speed, your fair will equal your fweight, therefore, in this case, it will be 100-100 = 0N (terminal velocity A). This is the time where you open your parachute so that you can fall slower. Once the parachute is opened your fnet will now be in the opposite direction, meaning that the acceleration is upward, but the velocity is downward. Once the fair goes back down and reaches the same as the fweight, the fnet will be 0N, therefore you are at your second terminal velocity. The only difference between terminal velocity A and B is that B is slower, there fnet, and fair are the same. Two things that increase the force of air resistance are speed and surface area.Connecting it back to Newtons 2nd law, we learned that the acceleration is going to decrease as an object falls thought the air until it reaches terminal velocity adn stays at that speed. Because fo Newtons 2nd law we know that acceleration dn net force are directly proportional, therefore since acceleration is decreasing, we know that fnet is also decreaseing.
We can change the situation to a lead ball and a ping pong ball and find out which one hits the ground first as well. When the lead ball and the ping pong ball are dropped from a short hight, they will land at the same time because they didn't have much time to really start speeding up. But when the two balls are dropped from a higher hight, the lead ball will land first because it needs a greater fair to reach its terminal velocity.
Free Fall- Falling Straight Down:
When objects are falling in free fall, it means that the only force acting on it is gravity, there is no air force. When an object is dropped straight down, at 0 seconds, it will have a velocity of 0m/s. It is important to remember that even when the object is first dropped, the acceleration will always be 10m/s^2. At 1 second, the velocity will have increased by 10 m/s, therefore at 1 sec, the velocity is 10m/s. At 2 seconds, the velocity will be 30m/s and so on. To find the distance that this ball has been dropped, the equation d = 1/2gt^2. So to find how far the ball has dropped at 2 seconds, we plug in
d = 1/2 (10)(2)^2, which when solved comes out to be 20 m. To find the velocity the ball was going at 2 seconds, you use the equation v = gt, which is v = (10)((2), and it equals 20 m/s.
When an object falls straight down, the equations used are:
how far --> d = 1/2gt^2 and how fast --> v = gt
Free Fall- Throwing things straight up:
When working with free falling objects that are thrown straight up into the air, the most helpful thing to do is draw a picture. When an object is thrown straight up, it will always start at the initial speed and as it gets higher, it will lose its speed.
Lets say a ball is thrown up with an initial velocity of 40m/s. At 0 seconds, the ball will be traveling at 40m/s. The acceleration will also always be 10m/s^2. In the example picture to the right, the ball is thrown up with initial velocity 40m/s, and you can also see at which point the ball is at what time.
Looking at the example picture, it takes 4 seconds for the ball to get to the top of its path. It is in the air for 8 seconds. The velocity at the top of the path is 0m/s. The acceleration at the top of the path is 10m/s^2. The velocity after 6 seconds is going to be 20m/s. I was just able to find all those answers from drawing a picture, thus: draw a picture, it makes life easier.
In order to find how high the ball is at the top of its path, you use the d = 1/2gt^2:
d = 1/2gt^2
d = 1/2 (10)(4)^2
d = 80m
After 6 seconds, how high is the ball:
total it fell d = 1/2(10)(2)^2 80
- distance it fell to reach the point --> d = 20 --> - 20
hight at 6 sec 60m
In the example picture to the right, the person is moving 10m/s in the horizontal direction and 0m/s in the vertical direction at 0 sec. As the person jumps, the horizontal direction doesn't change, but the vertical direction does, speeding up 10m/s^2. As the person jumps they are going to move in a diagonal direction. To find how long the person is in the air before hitting the ground, you use the vertical equation for distance:
d = 1/2gt^2
45 = 1/2 (10)t^2
9 = t^2
3 = t
So the person is in the air for 3 sec. To find how far away the person is from the cliff when they hit the ground, you use the horizontal equation v = d/t. Since you are trying to find how far it is away from the cliff, you ignore the vertical distance and just find the horizontal distance. You know that the person is traveling at a constant velocity of 10m/s, and you know what the time, so just plug it in:
v =d/t
10 = d/3
d = 30m
You now know that the distance the person traveled is 30m in 3sec. Now the hardest part I believe is finding the velocity. For this, you use an equation from geometry. You know the vertical and the horizontal velocity, so to find the diagonal velocity, the velocity you are actually moving at a certain point, you use a^2 + b^2 = c^2. So lets find the velocity the person was moving at 1 seconds:
a^2 + b^2 = c^2
10^2 + 10^2 = c^2d = 1/2gt^2
45 = 1/2 (10)t^2
9 = t^2
3 = t
So the person is in the air for 3 sec. To find how far away the person is from the cliff when they hit the ground, you use the horizontal equation v = d/t. Since you are trying to find how far it is away from the cliff, you ignore the vertical distance and just find the horizontal distance. You know that the person is traveling at a constant velocity of 10m/s, and you know what the time, so just plug it in:
v =d/t
10 = d/3
d = 30m
You now know that the distance the person traveled is 30m in 3sec. Now the hardest part I believe is finding the velocity. For this, you use an equation from geometry. You know the vertical and the horizontal velocity, so to find the diagonal velocity, the velocity you are actually moving at a certain point, you use a^2 + b^2 = c^2. So lets find the velocity the person was moving at 1 seconds:
a^2 + b^2 = c^2
100 + 100 = c^2
200 +c^2
c = 14.1m/s
Free Fall- Throwing things up at an Angle:
v = d/t
5 = d/6
d = 30m
We can also find how high the ball got at the top of its path using the vertical equation. Be careful because the vertical equation is meant for things falling down, so we will use it in terms of it falling down:
d = 1/2gt^2
d = 1/2 (10)(3)^2
d = 5 x 9
d = 45m
Just by looking at the picture we know that the ball is in the air for a total of 6 seconds. And we also know by drawing a picture that at the top of its path, the ball is moving 5m/s because it is always moving 5m/s horizontally.
To summarize the unit, I put together all of the equations that are needed involving vertical direction at constant acceleration, and horizontally at constant velocity:
What I have found difficult about what I've studied:
During this unit, I found it difficult to grasp all of the concepts. I would do fine during class, but then go and try to do the homework and find it very difficult. I missed quite a few classes as well, which made it even more difficult. I overcame these difficulties by trying to come in during me free, and I also used my peers around me well. I put in quite a lot of effort towards labs and class. And after a while, as I started to really use the resources around me, I found that things were starting to make more sense. Something I would like to start doing in the upcoming units, is start my blog reflection during the unit. I think this will help me because I will be able to summarize what I know, but at the same time, it will help me to realize what I don't know. I would use this to help me study for quizzes, and it will also help me to ask better questions.
A connection I made to the real world is diving in swimming. I know understand why when swimmers dive, they need a lot of force in the horizontal direction in order to be able to get out faster. I guess now I should be able to have the perfect dive off the block!
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