Thursday, November 14, 2013
Tides
If you've ever been to the beach, I'm sure you've noticed how the water is high up on the shore at some parts of the day and it is very low at other points of the day. These are called high and low tides and they are caused by the moon and the sun and their position in relation to the Earth. Force is inversely proportional to distance squared, so the closer the moon is to a point on the Earth, the stronger the force between them will be. Likewise, with different points of water on the Earth, closer blobs of water to the moon will have a greater force on them than blobs of water on the other side of the Earth. This causes a bulge in the Earth's water; it has an ovular shape which allows for high tides to occur twice a day and low tides to occur twice a day across the globe, happening on opposite ends of the world at the same time. At different times of the year, the position of the moon in relation to the sun causes what are called spring tides (very high high-tides and very low low-tides) or neap tides (tides that are not as great in difference). Although I just gave a very brief summary of this relationship, the video below explains it quite well. Adam Hart-Davis uses food to demonstrate the moon's movement, the Earth's movement, and their effect on the Earth's tides. The visuals are very helpful. He could have included more detail about the process but I think this video gives a very good, simple explanation of tides.
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Friday, November 1, 2013
Unit 2 Blog Summary
Part A: What I Learned
In this unit, I learned about Newton's 2nd Law, skydiving, free fall, throwing objects straight up, throwing objects up at an angle, and falling at an angle.
Newton's 2nd law states that acceleration is directly proportional to force and inversely proportional to mass. We completed a lab to demonstrate this relationship. In conclusion, we found that as mass increases, acceleration will decrease, and if force increases, acceleration will increase in the same ratio.
You can write Newton's 2nd law using symbols...
a~F
a~ 1/m
or as an equation....
acceleration= net force/ mass
a= f(net)/m
An object's weight can be the force acting on the object. Weight is equal to the mass of an object times gravity (w=mg). If you know the mass of an object, you can find the weight or force on the object as gravity is always 9.8 m/s^2 (or 10m/s^2 for simple calculations).
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You can use the distance equation (d=1/2gt^2) to find how far an object fell in free fall, substituting gravity for acceleration. You can also use V=gt to find how fast the object was moving when it hit the ground.
In free fall, you can (obviously) fall straight down, But you can also throw things up.
The picture above illustrates the path a ball will take. In reality, it would fall straight up and straight down, but for the sake of the drawing, the ball is shown a little displaced from this path. As the ball travels upward, it is propelled by the initial velocity with which it is thrown while gravity is acting on the ball in the opposite direction (down), causing it to slow down as it reaches the top of its path. Since the ball is in free fall, its acceleration is 10 m/s^2 during the entire time in the air. In this particular drawing, the ball is thrown up with an initial velocity of 30 m/s and its speed decreases by 10m/s each second it falls up. When the ball is at the top of its path, it has a velocity of 0m/s, but the acceleration of the ball will still be 10m/s^2. The ball then falls back toward the earth with an increasing velocity because velocity and gravity will be in the same direction again.
Just from this picture you can tell a lot about the situation. You know the balls initial velocity as well as its velocity at each point along its journey. You know the hang time of the ball (total time in the air) is 6 seconds. To find how high the ball got, you can use the distance formula. However, this formula does not account for velocity in the upward direction, so you can only plug in values from the ball's downward fall. For example:
d=(1/2)gt^2
=(1/2) (10) (3)^2
= 1/2 (90)
d= 45 m
To find how far the ball is from the ground at any given second of the fall, you would use the value found for the total distance as shown just above. Then you would count the number of seconds from the top of the path down to the second you are finding the distance for and use that number of seconds as the t value in the distance formula. Once a distance is found, subtract it from the total distance and you will have the distance from the ground at that second.
After I learned about falling in free fall and throwing things straight up in free fall, I learned about things falling at an angle in free fall and things being thrown up at an angle in free fall.
When an object falls at an angle, the vertical acceleration is constant (9.8m/s^2) and the horizontal velocity is constant (the initial velocity the object is thrown with is the velocity that the object will have in the horizontal direction for the entire fall). The vertical velocity continues to increase as the object falls, so the path it takes to the ground will be curved, or parabolic. To find the vertical distance you can use the same distance equation we are familiar with (d=(1/2)gt^2) and you can use v=gt to find the vertical velocity at a given second. You could use either of these equations to find the time of the object's fall. The time the object falls is the same vertically and horizontally; t is one variable that will remain the same in any equations for vertical or horizontal values, besides g. The time of the object's fall is determined by the vertical distance, or the height from which the object falls, which is why the distance formula is so useful and is only used for vertical values. To find the horizontal distance, velocity, or the time of the fall, we can use the equation v=d/t. Problems like this can be applied to an object being dropped from a plane or something being thrown off a cliff.
It is also important to note that we use the vertical and horizontal velocities in our equations but we do not use the actual velocity of the object.
To find the actual velocity, we must know the vertical and horizontal velocities. When they are put together they make two sides of a box, through which a vector can be drawn diagonally to show the actual velocity of the object. In my physics class, we use either a 45 degree angle or a 3, 4, 5 triangle to find this value. The picture below is an example of how this is drawn.
We use this triangle system to find the actual velocity for objects thrown up at an angle as well. An example of this would be hitting a home-run in a baseball game. To find the horizontal velocity and distance, we again use the v=d/t equation. For vertical distance and velocity in this situation, we use d=1/2gt^2 and v=gt because these objects are still in free fall. We treat the vertical equations that use g in the same way we treat them in problems of objects being thrown straight up. They can only be used with values from the fall downward. The vertical height determines the hang time of the ball so these equations are really useful. If the vertical velocity of the object was 1m/s and the horizontal velocity was 1m/s, the actual velocity would be 1.41 (the square root of 2) at a 45 degree angle. If both velocities were 10m/s, the actual velocity would be 14.1 m/s, and if they were 100m/s, the actual velocity would be 141 m/s.
When an object is thrown up at an angle, vertical velocity decreases on the way up- at its highest point, the object has a vertical velocity of 0m/s. The horizontal velocity, however, is constant. If an object is thrown with a horizontal velocity of 10 m/s, it will still move horizontally at 10m/s at the top of its path when the vertical velocity is 0m/s. Since it keeps moving horizontally, it has a curved path. On the fall back to the ground, the vertical velocity will increase at a constant acceleration of 10m/s^2.
What I found difficult about this unit was the concept of skydiving with air resistance. At first, I just couldn't wrap my mind around the fact that you speed up so you have more air resistance and then once you open your parachute your air resistance will increase, but as speed decreases the air resistance decreases as well. After some practice and looking at tons of examples and diagrams, it makes a lot more sense now. The transition to falling with air resistance to falling in free fall was weird at first because I kept expecting the objects to reach a terminal velocity but I think I have gotten the hang of that concept now, too. Again, I made the lightbulb click with visual examples and just really breaking the concept down.
This unit, I think I had more effort in class but I could have put more effort into my work outside of class. I need to make more time for myself to not only get my work done but also to do it well. I do think, conversely, that the time I put into doing problems paid off. I always took the time to finish the problems and think through them which really helped me to be active and present in class. When we went over concepts and certain problems in class, I had already spent a lot of time on them and mostly understood them in my own way, so it was nice to be able to use class as a solidifier. I am getting to be very confident in physics- I feel like I really understand what is going on and I do not question everything I do. This comes from making sure I understand everything on my own time and then moving on in class, adding onto the information learned.
My goals for the next unit are to spend more time on my work outside of class and to finish my podcast early. I have found that the podcast gets really stressful when I have other work to do and my group starts finding less and less time to meet. Hopefully we can utilize the preceding weekend more efficiently int he next unit. I also want to have a very deep, holistic knowledge of the next unit. I think it would be a very bad time to start only half processing the information and I want to keep working hard all semester (all year, too, but I am thinking short-term until winter break).
Part B: Connections
I can connect this unit to soccer. I play right back, and one of my best assets in the game is my ability to clear the ball. This is similar to throwing things up at an angle; I use my foot to lift the ball and kick it across the field, in an arc through the air. I had usually just focused on my form for this, making sure I transferred my body weight over the ball and followed through my kick with my leg and foot pointed. I would love to find the actual velocity necessary to get the ball where I want it to be and then use this to find the horizontal velocity I need to swing my leg with to get this kind of power. I know that I will now start trying to kick the ball at a 45 degree angle to see if that gives it a smoother arc at all. It would also be really interesting to feel what different velocities feel like. For example, how hard will I need to swing my leg to hit a ball at 30m/s? And how will this feel on my leg? Is this an easy velocity to reach or would I need to really kick with power?
My group this unit made a podcast about things falling at an angle.
I can't wait to learn the next unit! Yay physics!
In this unit, I learned about Newton's 2nd Law, skydiving, free fall, throwing objects straight up, throwing objects up at an angle, and falling at an angle.
Newton's 2nd law states that acceleration is directly proportional to force and inversely proportional to mass. We completed a lab to demonstrate this relationship. In conclusion, we found that as mass increases, acceleration will decrease, and if force increases, acceleration will increase in the same ratio.
You can write Newton's 2nd law using symbols...
a~F
a~ 1/m
or as an equation....
acceleration= net force/ mass
a= f(net)/m
An object's weight can be the force acting on the object. Weight is equal to the mass of an object times gravity (w=mg). If you know the mass of an object, you can find the weight or force on the object as gravity is always 9.8 m/s^2 (or 10m/s^2 for simple calculations).
Above is an illustration of the experiment we performed in the lab, including labels of acceleration and force. If you add mass into the picture, depending on where it is placed (on the hanging weight or the cart), either the force and acceleration will increase or they will both decrease.
When you go skydiving, air resistance plays a large role in your fall through the air. As you fall, your speed increases which in turn causes your air resistance to increase. Conversely, as your speed decreases, your air resistance decreases as well. Air resistance is affected by two things: speed, and surface area.
To start at the beginning, when you first step out of a plane, your velocity is at 0m/s, which is its lowest point during the entire fall. In this same moment, acceleration and net force are at their highest points of the fall. This is because at t=0 seconds, you have not yet begun moving at speed, thus you have no velocity. However, your acceleration will decrease as you build up more air resistance. Keep in mind, you are still falling faster and faster (velocity increases), but you are gaining speed at a decreasing rate (acceleration decreases). Your net force is at its highest point at t=0 seconds as well, not only because acceleration is directly proportional to force, but also because your air resistance builds up to equal your weight (aka the force of gravity which is pulling you down). When you have no air resistance, your weight is the only force acting on your body. As the opposing force of air resistance builds, the net force decreases until it reaches 0 Newtons (the point when air resistance = your f weight). This point when the two forces are equal is also the time in your fall when your velocity has reached the fastest it can possibly be- this is called terminal velocity. When you are in terminal velocity, f net = 0, velocity is constant, and acceleration = 0 m/s^2. It is a total reversal from t=0s because velocity is at its highest point in terminal velocity while acceleration and net force are at their lowest points.
So that covers the falling part. But to skydive, there comes a point when you need a parachute. If you fell without a parachute, hitting the ground at terminal velocity would be very painful!
When you are falling and have reached terminal velocity, this is the point when you open your parachute. Immediately, your surface area is increased by the chute which increases the air resistance. The air resistance will be greater than the force of your weight, and since it is a force pulling you in the upward direction, your acceleration will be upward. This also causes the net force to be negative (it will no longer equal zero; at this point you are no longer in terminal velocity). As you slow down again and air resistance decreases, you reach a second terminal velocity. In this second terminal velocity, the acceleration and net force are both 0 again (equal to the first terminal velocity) and the air resistance will also be equal to the air resistance in the first terminal velocity as it has to equal the force of your weight for the net force to equal zero. The only thing that is different in the second terminal velocity is the velocity itself- it will still be constant but it will be much slower and safer in the second terminal velocity. From here you safely fall to the ground and the parachute has done its job.
When you fall straight down in free fall, another concept I learned in this unit, this means you are falling without any forces acting on you except for gravity. When you skydive, you are not in free fall. One cool thing about free fall is that mass does not matter. If you dropped a coin and a feather at the same time in a vacuum without air resistance, they would hit the bottom at the same time. This is because the only force acting on the objects is the force of gravity which is determined by an object's weight. You might be thinking, "I thought weight didn't matter?" You're wrong. Mass and weight are not the same thing. Weight is the mass of an object times gravity (w=mg). So, if you are finding the acceleration of an object, you use a=f net / mass; in free fall the f net = weight, so you can change the equation to say a= mg/m (acceleration is equal to mass times gravity divided by mass). Mass divided by itself equals one, so you can cancel out the m's. This leaves the equation as a=g (acceleration = gravity = 9.8m/s^2). So no matter what the mass of an object is, when an object is in free fall, it will always have the same acceleration. Here is a video to demonstrate:
You can use the distance equation (d=1/2gt^2) to find how far an object fell in free fall, substituting gravity for acceleration. You can also use V=gt to find how fast the object was moving when it hit the ground.
In free fall, you can (obviously) fall straight down, But you can also throw things up.
The picture above illustrates the path a ball will take. In reality, it would fall straight up and straight down, but for the sake of the drawing, the ball is shown a little displaced from this path. As the ball travels upward, it is propelled by the initial velocity with which it is thrown while gravity is acting on the ball in the opposite direction (down), causing it to slow down as it reaches the top of its path. Since the ball is in free fall, its acceleration is 10 m/s^2 during the entire time in the air. In this particular drawing, the ball is thrown up with an initial velocity of 30 m/s and its speed decreases by 10m/s each second it falls up. When the ball is at the top of its path, it has a velocity of 0m/s, but the acceleration of the ball will still be 10m/s^2. The ball then falls back toward the earth with an increasing velocity because velocity and gravity will be in the same direction again.
Just from this picture you can tell a lot about the situation. You know the balls initial velocity as well as its velocity at each point along its journey. You know the hang time of the ball (total time in the air) is 6 seconds. To find how high the ball got, you can use the distance formula. However, this formula does not account for velocity in the upward direction, so you can only plug in values from the ball's downward fall. For example:
d=(1/2)gt^2
=(1/2) (10) (3)^2
= 1/2 (90)
d= 45 m
To find how far the ball is from the ground at any given second of the fall, you would use the value found for the total distance as shown just above. Then you would count the number of seconds from the top of the path down to the second you are finding the distance for and use that number of seconds as the t value in the distance formula. Once a distance is found, subtract it from the total distance and you will have the distance from the ground at that second.
After I learned about falling in free fall and throwing things straight up in free fall, I learned about things falling at an angle in free fall and things being thrown up at an angle in free fall.
When an object falls at an angle, the vertical acceleration is constant (9.8m/s^2) and the horizontal velocity is constant (the initial velocity the object is thrown with is the velocity that the object will have in the horizontal direction for the entire fall). The vertical velocity continues to increase as the object falls, so the path it takes to the ground will be curved, or parabolic. To find the vertical distance you can use the same distance equation we are familiar with (d=(1/2)gt^2) and you can use v=gt to find the vertical velocity at a given second. You could use either of these equations to find the time of the object's fall. The time the object falls is the same vertically and horizontally; t is one variable that will remain the same in any equations for vertical or horizontal values, besides g. The time of the object's fall is determined by the vertical distance, or the height from which the object falls, which is why the distance formula is so useful and is only used for vertical values. To find the horizontal distance, velocity, or the time of the fall, we can use the equation v=d/t. Problems like this can be applied to an object being dropped from a plane or something being thrown off a cliff.
It is also important to note that we use the vertical and horizontal velocities in our equations but we do not use the actual velocity of the object.
To find the actual velocity, we must know the vertical and horizontal velocities. When they are put together they make two sides of a box, through which a vector can be drawn diagonally to show the actual velocity of the object. In my physics class, we use either a 45 degree angle or a 3, 4, 5 triangle to find this value. The picture below is an example of how this is drawn.
When an object is thrown up at an angle, vertical velocity decreases on the way up- at its highest point, the object has a vertical velocity of 0m/s. The horizontal velocity, however, is constant. If an object is thrown with a horizontal velocity of 10 m/s, it will still move horizontally at 10m/s at the top of its path when the vertical velocity is 0m/s. Since it keeps moving horizontally, it has a curved path. On the fall back to the ground, the vertical velocity will increase at a constant acceleration of 10m/s^2.
What I found difficult about this unit was the concept of skydiving with air resistance. At first, I just couldn't wrap my mind around the fact that you speed up so you have more air resistance and then once you open your parachute your air resistance will increase, but as speed decreases the air resistance decreases as well. After some practice and looking at tons of examples and diagrams, it makes a lot more sense now. The transition to falling with air resistance to falling in free fall was weird at first because I kept expecting the objects to reach a terminal velocity but I think I have gotten the hang of that concept now, too. Again, I made the lightbulb click with visual examples and just really breaking the concept down.
This unit, I think I had more effort in class but I could have put more effort into my work outside of class. I need to make more time for myself to not only get my work done but also to do it well. I do think, conversely, that the time I put into doing problems paid off. I always took the time to finish the problems and think through them which really helped me to be active and present in class. When we went over concepts and certain problems in class, I had already spent a lot of time on them and mostly understood them in my own way, so it was nice to be able to use class as a solidifier. I am getting to be very confident in physics- I feel like I really understand what is going on and I do not question everything I do. This comes from making sure I understand everything on my own time and then moving on in class, adding onto the information learned.
My goals for the next unit are to spend more time on my work outside of class and to finish my podcast early. I have found that the podcast gets really stressful when I have other work to do and my group starts finding less and less time to meet. Hopefully we can utilize the preceding weekend more efficiently int he next unit. I also want to have a very deep, holistic knowledge of the next unit. I think it would be a very bad time to start only half processing the information and I want to keep working hard all semester (all year, too, but I am thinking short-term until winter break).
Part B: Connections
I can connect this unit to soccer. I play right back, and one of my best assets in the game is my ability to clear the ball. This is similar to throwing things up at an angle; I use my foot to lift the ball and kick it across the field, in an arc through the air. I had usually just focused on my form for this, making sure I transferred my body weight over the ball and followed through my kick with my leg and foot pointed. I would love to find the actual velocity necessary to get the ball where I want it to be and then use this to find the horizontal velocity I need to swing my leg with to get this kind of power. I know that I will now start trying to kick the ball at a 45 degree angle to see if that gives it a smoother arc at all. It would also be really interesting to feel what different velocities feel like. For example, how hard will I need to swing my leg to hit a ball at 30m/s? And how will this feel on my leg? Is this an easy velocity to reach or would I need to really kick with power?
My group this unit made a podcast about things falling at an angle.
I can't wait to learn the next unit! Yay physics!
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