course PHY 201 I still don't know if I am doing what you want me to for these assessments or what I need to be doing in order to get full credit, but I think it is because I haven't gotten my other ones back from you yet. I'm also not getting all of these right- is that ok? J䍎}z{assignment #006
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15:33:19 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> confidence assessment:
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kzՉz|Qڡc assignment #006 006. Physics qa initial problems 06-03-2008
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15:37:36 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> The speed will be changing by 10 mph and goes up 2 mph each second. Because of this 10 is to be divided by 2 which would give 5 or in other words, it will take 5 seconds for the speed to increase from 20 to 30 mph. The speed changes 2 mph every sec. To get the speed 7 seconds later, 7 is multiplied by 2 which gives 14. the original 10 mph is added to the 14 mph that the car will travel in 7 seconds. This gives 24mph as the final speed. confidence assessment: 3
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15:38:28 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> After reading and comparing with my answers I see that I was correct and thinking about this problem in the right way. self critique assessment: 3
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15:48:10 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> It will take less time for the vehicle to reach the lampost the second time because of its increased initial speed. The speed would still increase 2 mph each second, but since the vehicle is starting out at 20 mph, it will not take as long to reach the second lampost. No, it does not. The vehicle may start out faster but since it will not take as much time for the vehicle to reach the second lampost, it will not be able to increase in as much speed as it had before. confidence assessment: 2
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15:49:30 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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RESPONSE --> I was correct in my way of thinking, especially in the second part of the problem. The speed will increase by less due to the fact that the same amount of change in speed will occur over a shorter time span. self critique assessment: 2
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15:57:21 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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RESPONSE --> I hit the wrong button again- I got that: An automobile that speeds up from 40 to 90 mph in 20 sec. would increase its rate faser than one that started at 20 mph and increased to 30 in 5 sec. This is beacuse 90-40=a difference of 50 mph which is divided by 20 to obtain 2.5 mph/sec that is larger than the difference of 10 mph divided by 5 sec to obtain a rate of 2 mph/sec, which is stated here. Anyway, I understand how it was obtained and was thinking correctly but just hit the wrong button. self critique assessment: 2
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16:06:43 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The team with 5000 net Newtons with a 2000kg automobile would pull with a rate of 2.5 Newton/kg (5000/2000) as opposed to the 3000 net Newtons with a 1500kg automobile whic would pull at a rate of 2.0 Newtons/kg (3000/1500). Since they would be moving with a greater rate, the automobile would accelerate faster and allow them to win. Should someone pull in the opposite direction at 500 Newtons, the team would still win beacause an overall net of 4500 Newtons with the 200kg automobile would still move a little faster rate of 2.25 Newtons/kg (4500/2000). self critique assessment: 2
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16:08:13 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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RESPONSE --> I was correct in my way of thinking. The second team will win both pulling matches. self critique assessment: 2
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16:11:35 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> I would predict that the 200 lb player would go backwards because he weighs less and is traveling at a faster speed, which makes him harder to be stopped as opposed to the other player that is heavier and moving at a lower speed which makes it easier for him to be stopped. confidence assessment: 2
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16:28:11 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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RESPONSE --> I now see what I was supposed to do. I was just basing my answer on ""common sense"" as was mentioned to do earlier but really wasn't sure. After seeing these calculations, it does make sense that the second player would cause the first to fall. self critique assessment: 1
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16:33:52 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> At this point, I don't think it would make sense to multiply the amount each climber ate by their weight and think that the obtained value would have anything to do with their climbing abitlity. It makes the most sense to me, however, to think that the 150 lb climber would be able to go further up the mountain than the 200 lb one not because of their weight difference, but because the mountain is so steep and it seems as if it would take more energy for the heavier climber to pull an addtional 50 lbs up the mountain and this extry weight would cause a quicker dwindling of energy. confidence assessment: 1
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16:37:49 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> I did have the correct answer but I did not understand how the calculations really went into it. I was on the right track however, because I did realise that the lighter climber would not drain its energy as quickly as the heavier. Now i see how this can be seen in calculatoins though. self critique assessment: 2
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16:47:06 `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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RESPONSE --> As was statest from ""common sense"", I think that the faster automobile will take longer to come to a stop because of its increased initail speed, so it seems likely that it will take twice as long to stop it as the other automobile. It makes sense that the car with the greater coasting velocity would be the faster car too because it started at a higher rate and will take longer to stop and will therefore be coasting for a longer period of time. It also seems as if it will go twice as far as the slower one to stop because it will need at least twice as much area to come to a stop due to its greater speed. confidence assessment: 1
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16:52:21 It turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long. If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower. For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected. Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far. If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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RESPONSE --> I think that this is kind of tricky to understand. I am not quite sure how the calculations work out to this, but I think I would be better able to see it with real speeds. I am still a little uncertain of this but I think that I was starting to see this in the right way. self critique assessment: 1
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17:01:57 `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length. Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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RESPONSE --> Judging from these figures, I think that the cord would stretch a little less than 7 feet beyond its original length. I came to this by taking 5 ft and dividing it by the 100 lb weight, which is 0.05 ft/lb. I did this similarly for the 150 and 200 lb persons and got 0.06 ft/lb for each. In other words, the stress of foot per pound seems to be leveling off the more the person weighs. When I did this for the 125 lb person, I got 0.056 ft/lb the cord will stretch. Since this value didn't seem to have the same leveling off as the others did, it lead me to believe that not as much stress was placed on the cord to stretch so it would not stretch more than 7 ft beyond its original lenght but would stretch a little less than this. confidence assessment: 1
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17:06:38 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> I was completely wrong, but I now see how this was obtained. The graphing of these values especially shows this difference. If I had thought about this in a different way I think I would have seen it. self critique assessment: 1
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17:16:28 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> Since the force increases in proportion to the pullback, I think that the skater which is pulled back 8 ft (eight time the force) would be expected to be launched eight times as far. In other words, I think she will go more than twice as far as she did when pulled back 4 ft. confidence assessment: 1
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17:19:23 The distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far
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RESPONSE --> I was mostly correct but overshot the ice skater by alot. I failed to really understand the linear proportionality relationship, but I think that now it makes more sense. self critique assessment: 1
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17:27:16 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> The only difference between the light bulbs is there size, therefore they will have the same brightness even though the moth cannot tell a difference in their size from the long distance. The moth walking on the spheres would think that the second (larger) light was halfas bright because it has more surface area and the light would appear to be less intense than that of the first (smaller) sphere. confidence assessment: 0
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17:31:01 Both bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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RESPONSE --> I was sort of right, but not quite there. I better understand now why the lights appear to be the same size and how the surface area plays into how bright the bulb appears. self critique assessment: 1
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17:44:04 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> Least energy to most energy: degrees to reach its melting point degrees after all the ice melted. degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the ice by 20 -It will take the most to increase the temp of the ice because of the phase changes that the ice must make, then less energy to melt the ice though this still takes alot because of the phase change. If the ice is at a temperature, it is slightly easier to reach its melting point, after the ice has melted it is there is no energy being used to make the phase change. It takes the least energy to reach its melting point than the other changes. The ice melts at 0 degrees Celsius. You can see this because it was at this temperature that the most change to the ice (turning to water), takes place. confidence assessment: 0
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17:45:41 Since the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius. The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.
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RESPONSE --> I see why it would take more energy to melt the ice now, because this is where all the phase changes are taking place it does make sense that more energy would be required for these processes to occur. self critique assessment: 1
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17:51:05 `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much. If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point? How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?
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RESPONSE --> Since you are in the center point and are equal distant from each person, you would bob 12 inches. In order to experience no bob, you would have to move closer to one person by a 1/2 of the earlier distance you were from each person, that way one person's wave would come at you sooner, and the other person's wave would come at you at the same time. confidence assessment: 1
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17:53:40 If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in. If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ] However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.
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RESPONSE --> I got this mostly right, but I see where I messed up. Moving 3 feet would just get you hammered at first and then the peaks would be similar to the first case. But moving the net distance of 1.5 ft would cause a shift in the flow of the peak and valley formation and keep you in calm water. self critique assessment: 2
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