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Ajwrighter
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1. Which of the following is an example of an object possessing Kinetic energy
A. a ball rolling downhill
B. a bird perched in a tree
C. a stick of dynamite
D. a boulder resting on a cliff
E. Water at the top of a dam
3. Without hesitation I answered E. Water at the top of a dam
The reason I answered E. It is to my belief that flowing water (/large bodies of water) always has a constant (ever changing) kinetic energy (no matter how small it may be.) If the Water is at the top of a dam, (regardless if the water is over flowing over the dam or the turbines are on and the water is flowing through the spillways.) The water that is moving against the dam will always have kinetic energy.
4. Proof: KE = (1/2)M * V^2 and for reference (V = [tex]\sqrt{(4/3)gh}[/tex] ) for the velocity of the ball rolling down hill. In order for an object to have Kinetic energy it must have mass and velocity. Water has mass and if you were to measure the waters velocity I bet it would be at least a tenth of the objects Ke if given the same mass (assuming we neglect the area of the water all together.) Here are my thoughts on why. Regardless if the dam was on a Lake or a river we must assume that water is flowing from point A to point B (the top of the Dam). Now let’s create two scenarios and freeze frame them.
For simplicity the two scenarios will involve Scene 1 which will include the Ball and measure its KE in three different Quadrants (1/4, ½ and ¾ the distance traveled down the hill) I want to make the ball rolling down the hill the exact time in which (Scene 2) the body of water is considered at the “…top of a dam.”
Scene 1. We release a ball with Mass = 1kg from rest on top of a hill of 20 meters. It rolls down our X distance and we freeze frame it (we assume the angle is a perfect 90 degree angle and we are neglecting any air resistance.) T = D/V
(V1/4 = [tex]\sqrt{(4/3)(9.8) (5)}[/tex] = 8.08m/s
Ke = (1/2)(1kg)(8.08m/s)^2 = 32.64 J
T = 5/8.08 = .62s
(V1/2 = [tex]\sqrt{(4/3)(9.8) (10)}[/tex] = 11.43m/s
Ke = (1/2)(1kg)(11.43m/s)^2 = 65.32 J
T= 10/ 11.43 = .875s
(V3/4 = [tex]\sqrt{(4/3)(9.8) (15)}[/tex] = 14m/s
Ke3/4 = (1/2)(1kg)(14m/s)^2 = 98 J
T = 15/14 = 1.07s
Scene 2. Remember the velocity of any object (for our equation particle of water) is always tangent to the path taken by the object. So the particle of water’s (which is a fluid element) velocity is always tangent to the streamline (the direction headed to the dam.) Also I would like to note that no two streamlines will ever intersect giving off multiple velocities at the same time. The weight of the water is exactly the same as the ball M2 = 1kg. As for its velocity we will obtain the mean from previously recorded velocity of water that exists near a dam and use this data to determine the average velocity if the data seems a bit skewed I will select a sample of the data using a random generator provided by excel. After putting in the numbers I get an average water speed of
3.2mph = 1.43m/s
and a speed of
2.9mph = 1.29m/s
Without putting up the formulas the KE for both (in order) is as follow (both in a freeze frame at the top of a dam so that it is in contrast to the ball rolling down hill
KE1 = 1.022 J
KE2 = .832 J
If you notice they are exactly as I predicted (given already the common knowledge of about how fast water flows down a stream) the KE is about 1/10th of the ball rolling down hill.
But Wait A Minute. The question never asked in regards if rather or not the water was the same weight, nor did it say it was the same area or mass. All it asked is “Which of the following is an example of an object possessing kinetic energy” Logically a person would assume that a ball rolling downhill would be the first and correct answer and that all other answers are considered objects possessing potential energy and not kinetic energy. But how big is this ball what is the individual’s initial thoughts on the answers? Initially my thoughts where a bowling ball rolling down the streets of San Francisco, and my initial thoughts of the water was about a billion gallons of water hitting the side of the Hoover dam instantaneously at a undetermined velocity.
Alright now we have a new format what’s the weight of the water surrounding the top of the Hoover dam. Well we know the density of water to be 1m per 1v. Well we can’t get any easier than a depth of 1 meter. So for simplicity we will use a depth of 1 meter and since we are talking three dimensions and volume we will use the cubic meter. Since we are already talking about the Hoover dam let's use it as the example. The Hoover dam is 379 meters long. So if water was at the top of the Hoover dam joining it adjacently with a volume of 379 cubic meters then by conversion to liters we would get roughly 379,000 Liters, and if we were to convert to Gallons we would get 100121.2gal, and finally if were to convert it from gallons to Kilograms we would get 379,000kg of water. (ok now where’s are little bouncing ball…
Ok San Francisco right? Well let’s pick an adequate ball? When I was a kid I always wanted to roll a bowling ball down one of the streets of San Francisco from about 9 blocks away to the bay that’s about a half a mile or more. Since we don’t know the exact distance or dimension, let’s use our old 90 degree angle perfect slope. And let’s make our bowling ball 20lbs = 9.07 kilograms. Since we don’t actually know the height of which the ball is initially released let’s do a crazy experiment and see what the Kinetic energy would be from 1 half mile (804.672 meters) up at a 90 degree angle. This time we are going to measure the Ke just before it hits bottom (to avoid any other formulas we are going to assume this means 99.9% of the distance.) Here is our formula V = [tex]\sqrt{(4/3)gh}[/tex] ) and our velocity is…
V = 102.5 m/s
And our Ke ? = (1/2)(9.07kg)(102.5)^2 = 47,677 J .
What about our water?
The average speed of the Colorado River that flows through the Hoover Dam is between 1.6 mi/hr and 4 mi/hr (based on multiple resources). For this experiment we are going to give it the later of the two speeds.
1.6mi/hr = .715264m/s
Now let’s fill in the equation for KE = (1/2)( 379,000kg)( .715264)^2 = 96,948.69 J
So our answer is Water at the top of the Hoover dam is roughly equal to 97kJ .versus a bowling ball a half a mile up rolling down a 90 degree angle slope producing roughly 48KJ
Are these numbers completely accurate? (of course not I never used collisions in two dimensions for the water to dam nor did I use Ideal fluids in Motion). But I do believe they give a good representation of rather or not Water on top of a dam could be used as an alternative answer.
Given this information wouldn’t it seem reasonable that Water at the top of a dam possesses kinetic energy? If I’m incorrect on this information by all means please show me how I am wrong.
Sincerely
AJ
A. a ball rolling downhill
B. a bird perched in a tree
C. a stick of dynamite
D. a boulder resting on a cliff
E. Water at the top of a dam
3. Without hesitation I answered E. Water at the top of a dam
The reason I answered E. It is to my belief that flowing water (/large bodies of water) always has a constant (ever changing) kinetic energy (no matter how small it may be.) If the Water is at the top of a dam, (regardless if the water is over flowing over the dam or the turbines are on and the water is flowing through the spillways.) The water that is moving against the dam will always have kinetic energy.
4. Proof: KE = (1/2)M * V^2 and for reference (V = [tex]\sqrt{(4/3)gh}[/tex] ) for the velocity of the ball rolling down hill. In order for an object to have Kinetic energy it must have mass and velocity. Water has mass and if you were to measure the waters velocity I bet it would be at least a tenth of the objects Ke if given the same mass (assuming we neglect the area of the water all together.) Here are my thoughts on why. Regardless if the dam was on a Lake or a river we must assume that water is flowing from point A to point B (the top of the Dam). Now let’s create two scenarios and freeze frame them.
For simplicity the two scenarios will involve Scene 1 which will include the Ball and measure its KE in three different Quadrants (1/4, ½ and ¾ the distance traveled down the hill) I want to make the ball rolling down the hill the exact time in which (Scene 2) the body of water is considered at the “…top of a dam.”
Scene 1. We release a ball with Mass = 1kg from rest on top of a hill of 20 meters. It rolls down our X distance and we freeze frame it (we assume the angle is a perfect 90 degree angle and we are neglecting any air resistance.) T = D/V
(V1/4 = [tex]\sqrt{(4/3)(9.8) (5)}[/tex] = 8.08m/s
Ke = (1/2)(1kg)(8.08m/s)^2 = 32.64 J
T = 5/8.08 = .62s
(V1/2 = [tex]\sqrt{(4/3)(9.8) (10)}[/tex] = 11.43m/s
Ke = (1/2)(1kg)(11.43m/s)^2 = 65.32 J
T= 10/ 11.43 = .875s
(V3/4 = [tex]\sqrt{(4/3)(9.8) (15)}[/tex] = 14m/s
Ke3/4 = (1/2)(1kg)(14m/s)^2 = 98 J
T = 15/14 = 1.07s
Scene 2. Remember the velocity of any object (for our equation particle of water) is always tangent to the path taken by the object. So the particle of water’s (which is a fluid element) velocity is always tangent to the streamline (the direction headed to the dam.) Also I would like to note that no two streamlines will ever intersect giving off multiple velocities at the same time. The weight of the water is exactly the same as the ball M2 = 1kg. As for its velocity we will obtain the mean from previously recorded velocity of water that exists near a dam and use this data to determine the average velocity if the data seems a bit skewed I will select a sample of the data using a random generator provided by excel. After putting in the numbers I get an average water speed of
3.2mph = 1.43m/s
and a speed of
2.9mph = 1.29m/s
Without putting up the formulas the KE for both (in order) is as follow (both in a freeze frame at the top of a dam so that it is in contrast to the ball rolling down hill
KE1 = 1.022 J
KE2 = .832 J
If you notice they are exactly as I predicted (given already the common knowledge of about how fast water flows down a stream) the KE is about 1/10th of the ball rolling down hill.
But Wait A Minute. The question never asked in regards if rather or not the water was the same weight, nor did it say it was the same area or mass. All it asked is “Which of the following is an example of an object possessing kinetic energy” Logically a person would assume that a ball rolling downhill would be the first and correct answer and that all other answers are considered objects possessing potential energy and not kinetic energy. But how big is this ball what is the individual’s initial thoughts on the answers? Initially my thoughts where a bowling ball rolling down the streets of San Francisco, and my initial thoughts of the water was about a billion gallons of water hitting the side of the Hoover dam instantaneously at a undetermined velocity.
Alright now we have a new format what’s the weight of the water surrounding the top of the Hoover dam. Well we know the density of water to be 1m per 1v. Well we can’t get any easier than a depth of 1 meter. So for simplicity we will use a depth of 1 meter and since we are talking three dimensions and volume we will use the cubic meter. Since we are already talking about the Hoover dam let's use it as the example. The Hoover dam is 379 meters long. So if water was at the top of the Hoover dam joining it adjacently with a volume of 379 cubic meters then by conversion to liters we would get roughly 379,000 Liters, and if we were to convert to Gallons we would get 100121.2gal, and finally if were to convert it from gallons to Kilograms we would get 379,000kg of water. (ok now where’s are little bouncing ball…
Ok San Francisco right? Well let’s pick an adequate ball? When I was a kid I always wanted to roll a bowling ball down one of the streets of San Francisco from about 9 blocks away to the bay that’s about a half a mile or more. Since we don’t know the exact distance or dimension, let’s use our old 90 degree angle perfect slope. And let’s make our bowling ball 20lbs = 9.07 kilograms. Since we don’t actually know the height of which the ball is initially released let’s do a crazy experiment and see what the Kinetic energy would be from 1 half mile (804.672 meters) up at a 90 degree angle. This time we are going to measure the Ke just before it hits bottom (to avoid any other formulas we are going to assume this means 99.9% of the distance.) Here is our formula V = [tex]\sqrt{(4/3)gh}[/tex] ) and our velocity is…
V = 102.5 m/s
And our Ke ? = (1/2)(9.07kg)(102.5)^2 = 47,677 J .
What about our water?
The average speed of the Colorado River that flows through the Hoover Dam is between 1.6 mi/hr and 4 mi/hr (based on multiple resources). For this experiment we are going to give it the later of the two speeds.
1.6mi/hr = .715264m/s
Now let’s fill in the equation for KE = (1/2)( 379,000kg)( .715264)^2 = 96,948.69 J
So our answer is Water at the top of the Hoover dam is roughly equal to 97kJ .versus a bowling ball a half a mile up rolling down a 90 degree angle slope producing roughly 48KJ
Are these numbers completely accurate? (of course not I never used collisions in two dimensions for the water to dam nor did I use Ideal fluids in Motion). But I do believe they give a good representation of rather or not Water on top of a dam could be used as an alternative answer.
Given this information wouldn’t it seem reasonable that Water at the top of a dam possesses kinetic energy? If I’m incorrect on this information by all means please show me how I am wrong.
Sincerely
AJ