#$&* course Phy 201 6/2 11:00 pm ysics Question: `q001. There are two parts to this problem. Reason them out using common sense.
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Given Solution: `aIt will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time: • Will the vehicle require more or less than 10 seconds to reach the lamppost? • Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The car will require less than 10 seconds reaching the lamppost. If the car starts at the same place on the hill, stops that the same lamppost and increases 2 mph each second, but starts at 20 mph then it will take less time than before. Just because the speed is greater than before, this does not mean that the speed is 10mph greater at the lamppost. Since it takes less time to get to the lamppost and the speed is increasing at the same rate, it should actually be less than before. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The car going from 40-90 mph in 20 sec speeds up at the greater rate. - 20 mph to 30 mph/ 5 sec o 10 mph/ 5 sec o 2 mph/ sec - 40-90 mph / 20 sec o 50 mph/20 sec o 2.5 mph/sec confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: OK Self-critique Rating: OK ********************************************* Question: `q004. 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The team that exerts 5000 Newtons on a 2000 kg automobile will win, because they exert more net number of Newtons per kg. Newtons per kg determine how quickly and object will speed up, so the one with the highest number will speed up at a faster rate. - 3000 Newtons/1500 kg o 2 Newtons per kg - 5000 Newtons/ 2000 kg o 2.5 Newtons per kg No, even if someone pulled with a force of 500 Newtons in the opposite direction on the automobile the predicted winner would still win. - 5000 Newtons-500 Newtons= 4500 Newtons o 4500 Newtons/ 2000 kg o 2.25 Newtons per kg Which is still .25 Newtons per kg greater than the other team confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I believe that the 250 lb player moving to 10 feet per second will be moving backwards after the collision. The other player weighs 50 pounds less, but is moving 10 feet per second faster. The combination of the weight and velocity is greater than the other player. First Player ---250 lb*10 ft/sec= 2500 lb ft/sec Second Player---200 lb* 20 ft/sec= 4000 lb ft/sec confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGreater 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There is a 50 lb difference in the climbers and only a 2 oz difference in what they eat. If the climbers had been the same weight, then I would have said the climber that ate more oz would have been able to go further up the mountain. This is based on the common knowledge that the more you eat the more energy your body has to perform movement. However since the climbers were different weights I had to do some simple arithmetic. - 12 oz/ 200 lbs o .06 oz/lbs - 10 oz/ 150 lbs o .067 oz/lbs Based on these calculations the second climber that is lighter and ate 2 oz less will be able to climb further, because they got more energy oz/pound out of the cereal. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT COMMENT I am satisfied with how I worked out the problem, though it would be nice to know what formulas to use in case my instinct is wrong. I should have got the energy used per pound by rereading the question. INSTRUCTOR RESPONSE There are two points to these problems: 1. You can go a long ways with common sense, intuition or instinct, and you often don't need formulas. 2. Common sense, intuition and instinct aren't the easiest things to apply correctly, and it's really easy to get things turned around. A corollary: When we do use formulas it will be important to understand them, as best we can, in terms of common sense and experience. Either way, practice makes the process easier, and one of the great benefits of studying physics is that we get the opportunity to apply common sense in situations where we can get feedback by experimentally testing our thinking. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The car that was moving faster will take more time to stop, because it had more velocity to begin with. I believe the faster car will take twice the distance to stop. The car moving slower will have a greater average coasting velocity, but will be less and twice as great. The distance of the faster car will be twice that of the slower. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIt 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. STUDENT COMMENT: I do not understand why the car would go four times as far as the slower car. INSTRUCTOR RESPONSE: The faster car takes twice as long to come to rest, and have twice the average velocity. If the car traveled at the same average velocity for twice as long it would go twice as far. If it traveled at twice the average velocity for the same length of time it would go twice as far. However it travels at twice the average velocity for twice as long, so it goes four times as far. STUDENT COMMENT: it’s hard to know this stuff without having first discussed it in notes or read it in the book, or have an equation handy. I guess this will all come with the class. INSTRUCTOR RESPONSE One purpose of this and similar exercises is to get students into the habit of thinking for themselves, as opposed to imitating what they see done in a textbook. You're doing some good thinking. When you get to the text and other materials, ideally you'll be better prepared to understand them as a result of this process. This works better for some students than others, but it's beneficial to just about everyone. STUDENT COMMENT I understand, it seems as though it would be easier if there were formulas to apply. I used a little common sense on all but the last one. Reading the responses I somewhat understand the last one. ?????The problem doesn’t indicate the vehicle travels twice the average velocity for twice as long. Should I have known that by reading the problem or should that have become clear to me after working it some????? INSTRUCTOR RESPONSE You did know these things when you thought about the problem. You concluded that the automobile would take twice as long to come to rest, and that it would have twice the coasting velocity. You just didn't put the two conclusions together (don't feel badly; very few students do, and most don't get as close as you did). You should now see how your two correct conclusions, when put together using common sense, lead to the final conclusion that the second automobile travels four times as far. No formula is necessary to do this. In fact if students are given a formula, nearly all will go ahead and use it without ever thinking about or understanding what is going on. In this course we tend to develop an idea first, and then summarize the idea with one or more formulas. Once we've formulated a concept, the formula gives us a condensed expression of our understanding. The formula then becomes a means of remembering the ideas it represents, and gives us a tool to probe even more deeply into the relationships it embodies. There are exceptions in which we start with a formula, but usually by the time we get to the formula we will understand, at least to some extent, what it's about. I suppose this could be put succinctly as 'think before formulating'. STUDENT COMMENT I feel that I did decent on the problem, but I am the student that likes to have formulas. Your insight has opened my eyes to a different way of looking at this problem. I like the comment “Think before Formulate” INSTRUCTOR RESPONSE Your solution was indeed well thought out. I should probably add another comment: 'Think after formulating.' Formulas are essential, but can't be applied reliably without the thinking, which should come first and last. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t realize that air resistance should be factored in to the problem. I also did not realize that air resistance caused so much resistance that it would actually slow the faster car down making it less than exactly 2x the distance. Self-critique Rating: 2 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Based on these figures the person weighing 125lbs will stretch the cord less than 7 ft beyond its initial unstretched length. - 5ft/100 lbs o =.05ft per lbs - 9ft/100 lbs o =.06 ft/lb - 12 ft/ 200 lbs o .06 ft per lb - 125 lbs*.055 ft per lb (.05+.06=.11------------.11/2= .055) o = 6.875 ft per lb confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom 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). STUDENT COMMENT I feel like I nailed this one. Probably just didn’t state things very clearly. INSTRUCTOR RESPONSE You explanation was very good. Remember that I get to refine my statements, semester after semester, year after year. You get one shot and you don't have time to hone it to perfection (not to say that my explanations ever achieve that level). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel like I tried to make this problem more complicated than it should have been. I was multiplying and dividing, instead of just simply looking at the problems information for what it was. I didn’t even think to look at the difference between ft vs. lbs between each problem. Instead I looked at each problem individually and then tried to combine them all. I really feel like I need to stop looking so narrowly at problems and look at the big picture. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The skater will travel more than two times as far as when she was pulled back 4 feet, because she is being pulled back with twice the force and then goes at twice the force. This means that she will probably do quadruple the distance. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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 STUDENT COMMENT: I do not understand the linear proportionality relationship for the force. If the skater is pulled back an extra four feet, does that mean that the amount of pounds propelling her is also doubled? INSTRUCTOR COMMENT: That is so. However the force propelling her isn't the only thing that influences how far she slides. The distance through which the force is applied is also a factor. Doubling the force alone would double the sliding distance. Doubling the distance through which the force is applied would double the sliding distane. Doubling both the applied force and the distance through which it is applied quadruples the sliding distance. STUDENT SOLUTION AND QUESTION She should travel three times as far. The first four feet pulled back yield 20 feet of travel. The second four feet (i.e., feet 5 through 8) will propel her with twice the force as the first four feet. So this interval, by itself, would propel her 40 feet. The 20 feet of the first four-foot interval plus the 40 feet of the second four-foot interval is 60 feet total. But wouldn’t it be the case that by the time the slingshot reaches the four-foot position, the force exerted on the skater would only be half of that exerted when she was eight feet out? I understand why it would be a multiplier of four if the force were the same throughout, but I’m assuming that the force will decrease as the slingshot is contracts. I would appreciate help with this question. Thanks. INSTRUCTOR RESPONSE The average force for the entire 8-foot pull would be double the average force for the 4-foot pull. At this point we don't want to get too mathematical so we'll stick to a numerical plausibility argument. This argument could be made rigorous using calculus (just integrate the force function with respect to position), but the numerical argument should be compelling: Compare the two pulls at the halfway point of each. For a convenient number assume that the 4-foot pull results in a force of 100 lb. Then the 8-foot pull will therefore exert a force of 200 lb. When released at the 4-foot mark, the skater will be halfway back at the 2-foot mark, where she will experience a 50-lb force. When released at the 8-foot mark, the skater will be halfway back at the 4-foot mark, where she will experience a 100-lb force. Since the force is proportional to pullback, the halfway force is in fact the average force. Note that during the second 4 ft of the 8 ft pull the force goes from 100 lb to 200 lb, so the average force for the second 4 ft is 150 lb, three times as great as the average force for the first 4 ft. The max force for the second 4 ft is double that of the first 4 ft, but the second 4 ft starts out with 100 lbs of force, while the first 4 ft starts out with 0 lbs. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To a moth seeking light will see the same brightness, because the two light bulbs will be emitting the same amount of energy. At that distance there should not be a difference in the brightness even though the surface area is greater in the second. The moth on the 2nd sphere will appear to seem like it is less bright, because its surface area is twice as big. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aBoth 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. STUDENT COMMENT: I understand the first part of the problem about the distances. But the second part really confuses me. Looking straight down from the top of the spheres, the bulb is the same intensity and the frosted glass is exactly the same, so why would it seem dimmer? I would think that if a person was standing in front of the spheres, that person would be able to tell a difference, but not extremely close. INSTRUCTOR RESPONSE: Imagine a light bulb inside a frosted glass lamp of typical size. Imagine it outside on a dark night. If you put your eye next to the glass, the light will be bright. Not as bright as if you put your eye right next to the bulb, but certainly bright. The power of the bulb is spread out over the lamp, but the lamp doesn't have that large an area so you detect quite a bit of light. If you put the same bulb inside a stadium with a frosted glass dome over it, and put your eye next to the glass on a dark night, with just the bulb lit, you won't detect much illumination. The power of the bulb is distributed over a much greater area than that of the lamp, and you detect much less light. STUDENT COMMENT: I also didn’t get the second part of the question. I still don’t really see where the ¼ comes from. INSTRUCTOR RESPONSE: First you should address the explanation given in the problem: '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. ' • Do you understand this explanation? • If not, what do you understand about it and what don't you understand? This simple image of a 2x2 square being covered by four 1x1 squares is the most basic reason the larger sphere has four time the area of the smaller. There is, however, an alternative explanation in terms of formulas: • The surface area of a sphere is 4 pi r^2. • If r is doubled, r^2 increases by factor 2^2 = 4. • So a sphere with double the radius has four time the area. • If the same quantity is spread out over the larger sphere, it will be 1/4 as dense on the surface. STUDENT COMMENT: I also have no clue why the extra area doesn’t take away some brightness. INSTRUCTOR RESPONSE: All the light produced by the bulb is passing through either of the spheres. From a distance you see all the light, whichever sphere you're looking at; you see just as much light when looking at one as when looking at the other. From a distance you can't tell whether you're looking at the sphere with larger area but less intensity at its surface, or the sphere with lesser area and greater intensity at its surface. Self-critique : Although I got the first portion of the question right, I did not fully understand the second portion of the question. I did not think that the radius, being twice as big, would make the brightness 4x as bright at the surface. I figured that if the moth was on the light fixture, then the one with fewer diameters would seem brighter, because it would be closer to the light itself. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Melting the ice at its melting point takes the most energy, because even when it is starting to melt the temperature stays the same. Then it takes more energy to increase the ice than the water because water molecules move more quickly, warming up faster. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I answered the first part of the question correctly, but I don’t think I really understood why it was correct. Now I see the 0 C is the melting point and that is why we know that melting the ice at its melting point uses the most energy. I also thought that you could determine which of the others takes more energy, but we do not have enough information in this problem to do so. However, it takes less energy to heat up the ice than the water. Self-critique Rating: 3 ********************************************* Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards, one at either end of the pool, 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the 2 peaks meet at exactly the same time, and they are both 6 inches, then the peak will equal 1 foot. If you want to be in calm water, you need to move to where the peaks and valleys hit you at the same time. confidence rating #$&*: 2 ½ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understood the question, but I was unsure about the math and how to get how far you had to go towards one end to have the peaks and valleys meet. ------------------------------------------------ Self-critique Rating: 2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!