course Phy 201 I completed part of this assignment before 12:00 last night and the other part this morning. I copied and pasted each send file together and pasted them below. O֧걏w攨assignment #002
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19:25:46 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> I made a table with an x and y column. I used the numbers -3, -2, -1, 0, 1, 2, and 3 down the x column. I filled in -3 for x, and got (3*-3) - 4 = -13. When I substituted the rest of the numbers, I got -10, -7, -4, 1, 2, and 5. I then solved for x and got the following equation: x = (y+4)/3. I filled in zero for y and got x= (0+4)/3 = 4/3 or 1.3333. Now i have my intercepts: (0, -4) (4/3, 0) The graph is a straight line with a positive slope, passing through the axis at x=4/3 and y=-4. confidence assessment: 3
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19:29:55 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> My graph confirmed both intercepts correctly. I was unsure exactly which intercept was which, but I understand now that the x intercept is when y=0, and the x=0 at the y intercept. self critique assessment: 2
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19:45:37 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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RESPONSE --> I made a table for y=3x-4 using the numbers -2,-1,0-1,2 for the x's. When I substituted -2 for x, I got y=-6-4 = -10. I got -7, -4, -1, and 2 for the following y's. To find the ""steepness"" I used the slope formula: m=(difference in x's)/(difference in y's). First, I did m = (-2+1)/(-10+7) = -1/-3 = 1/3. Then I used different points: m = (1-2)/(-1-2) = -1/-3 = 1/3. Also, the equation y=3x-4 is a straight positive line, so the slope can not change in a straight line. confidence assessment: 3
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19:45:52 The graph forms a straight line with no change in steepness.
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RESPONSE --> self critique assessment: 3
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19:49:59 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> I found the slope in the previous question with the points (-2, -10) and (-1, -7), and got the slope = 1/3 confidence assessment: 3
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19:54:37 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> I should have used (difference in the y's)/(difference in the x's), or (rise)/(run), instead I switched them and got the wrong answer. I understand what I did incorrectly. self critique assessment: 2
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20:03:10 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph of y=x^2 is a parabola. I used the x values 0, 1, 2, and 3, and got the following y values: 0, 1, 4, and 9. Between the points (0,0) and (3,9), the graph is increasing, but not at a steady rate. The steepness of the graph is changing, increasing, as the x values increase. (Increasing at an increasing rate) confidence assessment: 3
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20:03:53 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.
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RESPONSE --> self critique assessment: 3
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20:09:43 `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> When I used the x values -3, -2, -1, and 0, I got the following y values: 9, 4, 1, and 0. This graph is decreasing at a decreasing rate because while the x values are decreasing at a constant rate, the y values are are decreasing at a decreasing rate. confidence assessment:
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21:20:57 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> self critique assessment: 3
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21:25:41 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> When using the x values: 0, 1, 2, and 3, the corresponding y values are: 0, 1, 1.41, and 1.73. The graph is increasing. The steepness is increasing at a decreasing rate because it is increasing, but more and more slowly each time. confidence assessment: 3
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21:28:54 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> I probably should have used more points (I did not use the point (4,2) on my graph), but I understand the concept and my graph has the same basic shape. self critique assessment: 2
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21:38:06 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> I used the x values 0, 1, 2, and 3. The corresponding y values are 5, 2.5, 1.25, and 0.625. The graph is decreasing. The steepness of the graph is decreasing at a decreasing rate because the y values are becoming smaller in smaller increments each time. confidence assessment: 3
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21:38:34 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> self critique assessment: 3
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21:43:37 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph would be increasing if distance was on the y axis and time was on the x axis. The graph would increase at an increasing rate because the car woud speed up at an increasing rate each second. confidence assessment: 2
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21:45:23 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> self critique assessment: 3
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ԑ}nm}FWfƞ assignment #006 006. Physics qa initial problems 06-06-2007
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21:59:42 `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 --> It should take 5 seconds for the speedometer to move from 20 mph to 30 mph if the speed of an automobile changes by 2 mph in one second. (30-20=10mph) If the speed changes 2mph every second, 10/2 = 5 seconds. If the rate of change of speed is a constant 2 mph/second, 7*2 = 14mph. 10mp+14mph = 24 mph confidence assessment: 2
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22:00:01 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 --> self critique assessment: 3
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22:08:42 `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 --> If the vehicle is passing the milepost at a speed of 20 mph, it will require less time to reach the lamppost. It is starting out at a faster speed and the speed is increasing at a rate of 2 mph every second. No, the speed is not a constant rate, it is increasing at 2 mph every second so it will be greater than 10 mph greater than before. confidence assessment: 3
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22:14:12 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 misunderstood what the second part of the question was asking. I understand that if it takes less than 10 seconds to reach the post, it will have less time to change, so it will in turn change by less. self critique assessment: 2
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22:27:08 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> The automobile that speeds up from 40 mph to 90 mph in 20 seconds is speeding up at a greater rate because (90-40)/20=2.5 and (30-20)/5=2, and 2.5>2. confidence assessment: 3
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22:41:09 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 am sorry, I accidentally skipped question #4. I think I clicked the next button twice and it skipped to the answer. self critique assessment: 2
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23:02:37 `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 predict the 250lbplayer to move backward immediately after the colision because velocity/mass is (10ft/s)/(250lbs)=1/25 or 0.04 and (20ft/s)/(200lbs)=1/10 or 0.1, and 0.04<0.1 confidence assessment: 1
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23:06:22 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 --> Well, I had the right answer, but I went about it the wrong way. I should have multiplied the speed and mass to get momentum instead of divide. self critique assessment: 2
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23:12:32 `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 --> I multiplied the amount of Cheerios each climber had with their weight. 200*12=2400lbsoz 150*10=1500lbsoz The climber who has eaten 12 oz of Cheerios should have more energy. confidence assessment: 1
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23:18:42 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 should have divided the oz of food each climber consumed by their weight in pounds. I multiplied, and therefore my answer was incorrect. I understand that to calculate energy, in this case, I must use the (oz of food) / (total body weight) self critique assessment: 2
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23:35:38 `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 --> The automobile that was traveling at a speed twice as fast should take longer to stop because it has more momentum. I am guessing that it will take less than twice as long to stop because, even though it is traveling faster, it will experience the same amount of friction and it will no longer have anything forcing it's speed to increase at a faster rate. confidence assessment: 1
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23:42:49 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 understand that both automobiles change velocicty at the same rate, so the second car would take twice as long. Disregarding air resistance, it would take exactly twice as long for the faster car to come to a stop. self critique assessment: 2
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23:51:28 `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 --> I would expect the cord to stretch less than 7 feet beyond. Based on the figures, 125 is exactly in the middle of 100 and 150, and 5/100=.05, 9/150=.06. .055 is in the middle of .05 and .06, and .055*125=6.875 which is less than 7. confidence assessment: 1
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23:57:22 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 --> The cord should actually stretch more than 7 ft because from 100lbs to 150lbs, the cord stretches 4 more feet, and from 150lbs to 200lbs, the cord stretches only 3 ft more. With each increase in pounds, the cord stretches less proportionally, so between 100lbs and 125lbs, the cord would be expected to stretch a little more than 2 more feet (or 7ft total). self critique assessment: 2
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08:05:58 `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 --> If the skater was pulled back twice as far, she should travel twice as far. confidence assessment: 1
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08:09:16 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 --> The skater would actually go 4 times as far because when she is pulled back in the slingshot she is pushed with double force. Pulling the slingshot back twice as far would result in (2*double the distance) or 4 times as far. self critique assessment: 2
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08:13:52 `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 --> At first, the larger sphere would appear brighter to the moth, if it was seeing them both from a distance. If the moth was on the surface, both spheres would appear to be producing the same brightness per square inch, regardless of size. confidence assessment: 2
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08:23:03 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 --> From a great distance, both bulbs would appear the same size. The larger sphere may not appear as bright. The light on the larger sphere is producing 1/4 the illumination, because it has 4 times the surface area to cover with the same amount of light as the smaller bulb. self critique assessment: 2
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08:31:49 `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 --> 1. Increasing the temperature of the water by 20 degrees after all the ice melted. 2. Melting the ice at its melting point. 3. Increasing the temperature of the ice by 20 degrees to reach its melting point. The ice melts at 0 degrees Celsius because at -1 degrees, the ice is still solid, and at 0 degrees it begins to melt and continues to melt almost completely at that temperature. confidence assessment: 3
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08:34:08 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 --> It actually takes less energy to increase the temperature of ice than water. self critique assessment: 2
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08:41:10 `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 --> If the peaks of both waves are 6 inches high, I would expect to rise twice as high (12 inches) if I was directly in the center. I would have to move where a peak from one end and a valley from the other end will reach me at the same time to be in calm water. confidence assessment: 2
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08:44:07 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 would have to move half of the distance of a full wave. If each wave took 3 ft to complete, I could move 1.5 ft towards one end and the peaks will meet the valleys. self critique assessment: 2
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