course PHY 201 Contents include 2 days of SEND files. ???????????Student Name:
......!!!!!!!!...................................
00:47:06 `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. 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.
......!!!!!!!!...................................
RESPONSE --> The x-intercept, or where the line crosses the ""0"" value of ""x"", is ""y=-4"" (0, -4). The y-intercept, or where the line crosses the ""0"" value of ""y"" is somewhere between ""x=1"" and ""x=2"". I didn't make the table with anything more than whole numbers. It's somewhere around ""1.3"".
.................................................
......!!!!!!!!...................................
00:47:24 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.
......!!!!!!!!...................................
RESPONSE --> Yes, it does. And I always try and turn the numbers to decimals...something about leaving the number 4/3 just never sat well with me. Mathematically, I probably should, though.
.................................................
......!!!!!!!!...................................
00:47:54 `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
......!!!!!!!!...................................
RESPONSE --> The steepness will remain constant, since it is a linear function. I will attempt to explain: the value the ""y"" changes is in direct proportion to the value that the ""x"" is changing, resulting in a straight line that maintains the same slope over its run.
.................................................
......!!!!!!!!...................................
00:48:22 The graph forms a straight line with no change in steepness.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Just an OK, although I am going to try and change my mindset to allow answers like ""4/3"".
.................................................
......!!!!!!!!...................................
00:48:42 `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)?
......!!!!!!!!...................................
RESPONSE --> The slope is ""3"" (3/1), or the line rises along the ""y"" axis 3 units for every 1 unite of ""x"".
.................................................
......!!!!!!!!...................................
00:49:07 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.
......!!!!!!!!...................................
RESPONSE --> I think I'm OK up to this point.
.................................................
......!!!!!!!!...................................
00:49:46 `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?
......!!!!!!!!...................................
RESPONSE --> Using the definition in the last problem, I would say that the function ""y=x^2"" for values of x between 0 and 3 is increasing, since as we move from left to right the line in moving ""up"". The steepness of the graph does change: the steepness is increasing at an increasing rate, although it gets hard to see after a certain point.
.................................................
......!!!!!!!!...................................
00:50:17 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:51:28 `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?
......!!!!!!!!...................................
RESPONSE --> I'm not sure what I did, but this is the same question I just answered. Anyway, graph is increasing and increasing at an increasing rate.
.................................................
......!!!!!!!!...................................
00:53:14 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.
......!!!!!!!!...................................
RESPONSE --> OK, I think I messed up. Apparently this referred to the other side of the parabola. My mistake. I do understand, though, that this side of the function is decreasing and decreasing at a decreasing rate.
.................................................
......!!!!!!!!...................................
00:54:42 `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?
......!!!!!!!!...................................
RESPONSE --> I would say that, for a graph a y='sqrt(x), that the graph is increasing, and that the steepness of the graph is increasing at a decreasing rate.
.................................................
......!!!!!!!!...................................
00:55:06 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.
......!!!!!!!!...................................
RESPONSE --> Got it right this time.
.................................................
......!!!!!!!!...................................
01:00:02 `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?
......!!!!!!!!...................................
RESPONSE --> For the function ""y-5*2^(-x) between x=0 and x=3, I would say that the graph is decreasing, and it is decreasing at a decreasing rate. I mean that the rate of change (decreasing) is lessening as the line moves from left to right.
.................................................
......!!!!!!!!...................................
01:00:25 ** 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. **
......!!!!!!!!...................................
RESPONSE --> OK
.................................................
......!!!!!!!!...................................
01:02:44 `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?
......!!!!!!!!...................................
RESPONSE --> I would say the the graph of y vs. t would be increasing and that it would be increasing at an increasing rate, since the car would be accelerating as it moves away, and the rate of accelerating would be ever-increasing.
.................................................
......!!!!!!!!...................................
01:02:53 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Alright.
.................................................
?????????? ?Student Name: assignment #006 006. Physics
......!!!!!!!!...................................
01:12:07 `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?
......!!!!!!!!...................................
RESPONSE --> For a change of 10 mph (moving from 20 mph to 30 mph) and a rate of change of 2 mph per second, it would take the car 5 seconds, or 10 mph = 2 mph/second * 5 seconds. For every second, the car increased 2 mph. For a total of 10 mph, it would take 5 seconds. If the speedometer initially reads 10 mph, then 7 seconds later it will read 24 mph, since 2 mph/second multiplied by 7 seconds would be a 14 mph greater than the original value of 10 mph.
.................................................
......!!!!!!!!...................................
01:12:24 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.
......!!!!!!!!...................................
RESPONSE --> Hopefully I explained that well.
.................................................
......!!!!!!!!...................................
01:21:47 `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?
......!!!!!!!!...................................
RESPONSE --> The vehicle should require less than 20 seconds to reach the lamppost, since everything remained constant (the rate of increase in speed, the distance to travel) except the initial speed. Since speed is distance over time, if the speed increases and distance remains the same, the time should decrease. I don't think it follows that the speed at the lamppost will be increase by the same amount at the two different speeds, and I'm basing it on an assumption that the inertia or momentum that carries the car through its coasting phase will vary with the initial speed of the car.
.................................................
......!!!!!!!!...................................
01:22:58 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.
......!!!!!!!!...................................
RESPONSE --> OK, I see where I missed that one: I didn't take into account that the distance between the sign and lamppost was constant.
.................................................
......!!!!!!!!...................................
01:26:01 `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?
......!!!!!!!!...................................
RESPONSE --> The first auto increases at 10 mph / 5 seconds, or 2 mph/second. The second auto increases at 50 mph/20 seconds, or 2.5 mph/second. So I'd say that the second auto speeds up at the greater rate of 2.5 mph/second.
.................................................
......!!!!!!!!...................................
01:26:13 09-12-2006 01:26:13 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.
......!!!!!!!!...................................
NOTES -------> Got that one.
.......................................................!!!!!!!!...................................
01:34:24 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?
......!!!!!!!!...................................
RESPONSE --> The first team has a mass of 1500 kg and a net force of 3000 Newtons, so its rate is determined by 3000/1500 = 2 Newtons/kg. The second team has a mass of 2000 kg and a force of 5000 Newtons, so its rate is 5000/2000 = 2.5 Newtons/kg. So, the second team should win, since it will be able to accelerate its load faster than the other. If there was a 500 Newton force opposing the winning team, then its new rate would be (5000-500)/2000 = 2.25 Newtons/kg, which is still greater than the first team, so it should still win.
.................................................
......!!!!!!!!...................................
01:34:38 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
......!!!!!!!!...................................
RESPONSE --> So far so good.
.................................................
......!!!!!!!!...................................
01:42:11 `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?
......!!!!!!!!...................................
RESPONSE --> I would predict that the 250-lb player moving at 10 feet per second would be moving backward after the hit, because both mass and velocity contribute to the effectiveness of the collision and the 250-lb player at 10 feet per second has 2500 wham-o units, while the 200-lb player moving at 20 feet per second has 4000 wham-0 units.
.................................................
......!!!!!!!!...................................
01:42:41 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.
......!!!!!!!!...................................
RESPONSE --> I liked wham-o units as a unit of measurement.
.................................................
......!!!!!!!!...................................
01:46:01 `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?
......!!!!!!!!...................................
RESPONSE --> I think the 150-lb climber would go further, since his 10 ounces of cheerios are spread across 150 lbs, givng him a cheerios energy quotient of .067, or more energy available for him to use. His friend, being 200-lbs and eating 12 ounces, has a cheerios energry quotient of .06, or less fuel available to burn.
.................................................
......!!!!!!!!...................................
01:46:21 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.
......!!!!!!!!...................................
RESPONSE --> So far so good.
.................................................
......!!!!!!!!...................................
01:51:30 `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?
......!!!!!!!!...................................
RESPONSE --> The twice-as-fast car should take longer, since its energy is twice that of the other car and it is being shed across the same load, or hill. It should take about twice as long to stop. The twice-as-fast car should have a greater average coasting velocity, since it has more velocity to average out. The distance traveled by the faster car should be twice that of the slower.
.................................................
......!!!!!!!!...................................
01:53:16 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.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. I didn't get that ""four times as far"" thing, but I suppose it would have come out it we were working out the formula. More time for that later.
.................................................
......!!!!!!!!...................................
02:01:14 `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?
......!!!!!!!!...................................
RESPONSE --> I would expect a 125 lb person to stretch the cord more than 7 feet beyond initial. The amount the cord stretches seems to be decreasing as the weight increases, so I would think it was greater at the lighter end and then tapering off.
.................................................
......!!!!!!!!...................................
02:02:05 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).
......!!!!!!!!...................................
RESPONSE --> Yes, I was sort of picturing the graph when I came up with that answer. (Sort of...I didn't actually graph it.)
.................................................
......!!!!!!!!...................................
02:05:20 `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?
......!!!!!!!!...................................
RESPONSE --> When pulled back 8 feet, she should be able to travel twice as far as the cord was pulled back 4 feet, since it keeps with the linear examples shown in the beginning.
.................................................
......!!!!!!!!...................................
02:06:36 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
......!!!!!!!!...................................
RESPONSE --> Aah, right: double the distance pushed AND double the weight behind the pushing.
.................................................
......!!!!!!!!...................................
02:10:11 `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?
......!!!!!!!!...................................
RESPONSE --> The larger sphere should appear dimmer, since the moth is getting less light because it is diffused across more surface area. The small moth walking on the surface shouldbe seeing less light than she would with the first, since the light is diffused across more surface area.
.................................................
......!!!!!!!!...................................
02:10:34 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.
......!!!!!!!!...................................
RESPONSE --> OK.
.................................................
a{???????Q? Student Name: assignment #006 006. Physics ?????????x??Student Name: assignment #006 006. Physics ?????H??????Student Name: assignment #006 006. Physics
......!!!!!!!!...................................
21:58:02 `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?
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. .
.................................................
......!!!!!!!!...................................
21:58:11 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.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK..
.................................................
......!!!!!!!!...................................
21:58:13 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:16 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:19 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:21 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:23 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:25 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
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:26 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:28 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:30 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:58:32 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
??????}?[?x? Student Name: assignment #006 006. Physics
......!!!!!!!!...................................
21:59:11 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:13 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:15 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:17 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:18 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:20 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:21 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:23 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
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:25 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:26 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:28 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:30 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:32 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:33 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:35 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:37 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).
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:39 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
21:59:41 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
......!!!!!!!!...................................
RESPONSE -->
.................................................
???\???{??????? Student Name: assignment #006 006. Physics
......!!!!!!!!...................................
22:00:51 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:00:53 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:00:55 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:00:57 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:00:58 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:00 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:01 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:03 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
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:04 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:06 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:08 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:10 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:13 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:15 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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:18 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:20 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).
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:26 `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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:01:29 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
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
22:06:06 `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?
......!!!!!!!!...................................
RESPONSE --> To the moth a half a mile away, the larger sphere should appear dimmer than the smaller sphere. The light bulbs put out the same amount of energy, but inside the larger sphere the energy is diffused across a larger surface, so a smaller amount of energy would be visible along the line of sight of the moth. To the moth walking along the surface of the sphere, the light coming from 1 square inch should be half as bright as the light coming from 1 square inch on the smaller sphere, again using the constant amount of energy being spread across a surface of the sphere that is twice as large as the first sphere's.
.................................................
......!!!!!!!!...................................
22:11:14 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.
......!!!!!!!!...................................
RESPONSE --> OK, I didn't account for the surface area properly, thinking twice the surface area instead of four times the area.
.................................................
......!!!!!!!!...................................
22:25:52 `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?
......!!!!!!!!...................................
RESPONSE --> 1. Increasing temp of ice by 20 deg to reach melting point (10 sec + 10 sec = 20 sec * 600 Joules/sec = 12,000 Joules). 2. Increasing temp of water by 20 deg after ice has melted (30 sec * 600 Joules/sec = 18,000 Joules). 3. Melting the ice at its melting point (at least 70 sec * 600 Joules/sec = 42,000 Joules). It appears that the ice melts at 0 deg C, since no melting took place at -1 deg C.
.................................................
......!!!!!!!!...................................
22:27:01 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.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. OK, I did some calculation to come up with Joules spent that couldn't be totally accurate, but they helped me to visualize the energy being spent, so I don't think it was a wasted exercise.
.................................................
......!!!!!!!!...................................
22:41:40 `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?
......!!!!!!!!...................................
RESPONSE --> If the peaks are 6 inches high at the center, when the peaks reach simultaneously I would expect to bob up 12 inches, fall back to the original waterline, then bob down 12 inches. So 24 inches in total. Since the wave crests are 6 feet apart, moving 3 feet in either direction should put you in calmer water.
.................................................
......!!!!!!!!...................................
22:43:51 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. 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.
......!!!!!!!!...................................
RESPONSE --> I actually thought about that 3 feet thing causing a net zero change, but I was having trouble visualizing the timing of the waves.
.................................................
??w??????l???Student Name: assignment #001
.................................................
......!!!!!!!!...................................
22:59:24 `q001. It will be very important in this course for your instructor to see and understand the process of visualization and reasoning you use when you solve problems. This exercise is designed to give you a first experience with these ideas, and your instructor a first look at your work. Answer the following questions and explain in commonsense terms why your answer makes sense.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. OK
.................................................
......!!!!!!!!...................................
23:07:18 For each question draw a picture to make sense out of the situation, and include a description of the picture. Samples Sample question and response Question: If a bundle of shingles covers 30 square feet, how many bundles are required to cover a 600 square foot roof? Response: We might draw a picture of a rectangle representing the area, dividing the rectangle into a number of smaller rectangles each representing the area covered by a single bundle. This makes it clear that we are dividing the roof area into 1-bundle areas, and makes it clear why we are going to have to divide. Reasoning this problem out in words, we can say that a single bundle would cover 30 square feet. Two bundles would cover 60 square feet. Three bundles would cover 90 square feet. We could continue in this manner until we reach 600 square feet. However, this would be cumbersome. It is more efficient to use the ideas of multiplication and division. We imagine grouping the 600 square feet into 30 square foot patches. There will be 600 / 30 patches and each will require exactly one bundle. We therefore require 600 / 30 bundles = 20 bundles. {}Your responses might not be as clear as the above, though they might be even more clear. I won't be looking for perfection, though I wouldn't object to it, but for a first effort at visualizing a situation and communicating a reasoning process. This is not something you are used to doing and it might take a few attempts before you can achieve good results, but you will get better every time you try. {}You might be unsure of what to do on a specific question. In such a case specific questions and expressions of confusion are also acceptable responses. Such a response must include your attempts to come up with a picture and reason out an explanation. For example your response might be Sample expression of confusion: I've drawn a picture of a pile of bundles and a roof but I'm not sure how to connect the two. I tried multiplying the number of bundles by the square feet of the roof but I got 18,000, and I know it won't take 18,000 bundles to cover the roof. How do you put the area covered by a bundle together with the roof area to get the number of bundles required? A poor response would be something like 'I don't know how to do #17'. This response reveals nothing of your attempt to understand the question and the situation. Nor does it ask a specific question. Incidentally, you might be tempted to quote rules or formulas about rates and velocities in answering these questions. Don't. This exercise isn't about being able to memorize rules and quote them. It is about expanding your ability to visualize, reason and communicate.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. OK
.................................................
......!!!!!!!!...................................
23:09:37 In your own words briefly summarize the instructions and the intent of this exercise.
......!!!!!!!!...................................
RESPONSE --> I am to create a picture to illustrate the solution to a problem and describe that picture in detail. The point is to show my visualization methods and ability to communicate either my understanding or confusion regarding the problem. Quoting formulas is not the point...it's about visualizing solutions and communicating ideas.
.................................................
......!!!!!!!!...................................
23:12:03 `q001. If you earn 50 dollars in 5 hours, at what average rate are you earning money, in dollars per hour?
......!!!!!!!!...................................
RESPONSE --> Based on the fact the it was neatly 50 dollars in 5 hours, I pictured 5 $10 bills laid out in a line, with hour 1 above the first, hour 2 above the second, etc. Since all five hours had equal quantities in them, I just circled the first hour and came up with the solution of the average rate = $10 per hour.
.................................................
......!!!!!!!!...................................
23:15:06 If you travel 300 miles in 6 hours, at what average rate are you traveling, in miles per hour?
......!!!!!!!!...................................
RESPONSE --> I pictured a line 300 ""miles"" long. Since the length of time to travel was 6 hours, I divided the line into 6 equal sections. Now, all I have to do is to measure one of the sections, at which point I find out that it is 50 miles long. Since each section equaled 1 hour of traveling, the answer would be 50 miles per hour of travel.
.................................................
......!!!!!!!!...................................
23:17:05 `q002. If a ball rolling down a grooved track travels 40 centimeters in 5 seconds, at what average rate is the ball moving, in centimeters per second?
......!!!!!!!!...................................
RESPONSE --> Same process as the previous question: the track is represented as a line 40 cm long. That line is divided into 5 equal sections. Measuring a section, I find that it is 8 cm long. Since each section represented 1 second of travel time, the average rate the ball is moving is 8 cm per second.
.................................................
......!!!!!!!!...................................
23:19:57 The preceding three questions illustrate the concept of a rate. In each case, to find the rate we divided the change in some quantity (the number of dollars or the distance, in these examples) by the time required for the change (the number of hours or seconds, in these examples). Explain in your own words what is meant by the idea of a rate.
......!!!!!!!!...................................
RESPONSE --> Rate is the speed at which a task is peformed with regards to a quantity...if it takes 1 hour to travel 60 miles, the speed that the task of traveling 60 miles is performed at is 60 miles per hour.
.................................................
......!!!!!!!!...................................
23:21:39 `q003. If you are earning money at the average rate of 15 dollars per hour, how much do you earn in 6 hours?
......!!!!!!!!...................................
RESPONSE --> Picturing this involved 6 stacks of bills, each containing 15 dollars. Then, counting up the stacks, the answer is $90.
.................................................
......!!!!!!!!...................................
23:23:12 If you are traveling at an average rate of 60 miles per hour, how far do you travel in 9 hours?
......!!!!!!!!...................................
RESPONSE --> I pictured a line segment 60 ""miles"" long. Then, piecing together 9 of those segments end-to-end. At this point, the line would be measured and found to be 540 ""miles"" long.
.................................................
......!!!!!!!!...................................
23:24:58 `q004. If a ball travels at and average rate of 13 centimeters per second, how far does it travel in 3 seconds?
......!!!!!!!!...................................
RESPONSE --> Same as the previous question: I pictured a line segment, 13 cm long, representing the average distance traveled in 1 second. Then, piecing 3 of those segments together end-to-end, I came up with a line that is 39 cm long.
.................................................
......!!!!!!!!...................................
23:26:33 In the preceding three exercises you turned the concept of a rate around. You were given the rate and the change in the clock time, and you calculated the change in the quantity. Explain in your own words how this increases your understanding of the concept of a rate.
......!!!!!!!!...................................
RESPONSE --> This helps because it shows the quantity being calculated in an additive way, or work being performed sequentially, like it would in real life and real time. You can see the progression of work and the cumulative result of the work.
.................................................
......!!!!!!!!...................................
23:30:19 `q005. How long does it take to earn 100 dollars at an average rate of 4 dollars per hour?
......!!!!!!!!...................................
RESPONSE --> I pictured the 100 dollars as vertical tank with a level gage along side. Then, to represent the $4/hour rate, I begin filling the tank in increments of $4, counting how many increments it takes to reach the $100 mark. This turns out to be 25 increments, and since each increment respresented an hour, the answer would be 25 hours.
.................................................
......!!!!!!!!...................................
23:36:37 How long does it take to travel 500 miles at an average rate of 25 miles per hour?
......!!!!!!!!...................................
RESPONSE --> To visualize this, just to be different this time, I pictured a track that was 25 miles around. Each time a car completed one lap, the scoreboard displays a count of the completed lap and a running total of miles traveled: first lap shows a ""1"" and ""25 miles; second lap shows a ""2"" and ""50 miles""; and so on until the scoreboard shows ""500 miles"", which would be after the 20th lap. Since each lap was 25 miles and the rate was 25 miles per hour, the length of time would be the number of laps = hours, or 20 hours.
.................................................
......!!!!!!!!...................................
23:38:41 `q006. How long does it take a rolling ball to travel 80 centimeters at an average rate of 16 centimeters per second?
......!!!!!!!!...................................
RESPONSE --> I pictured a line 80 cm long, and I had a ruler that is 16 cm long. Marking off how many lengths of the ruler it takes to reach the end of the line, I come up with 5 ruler lengths, and since each ruler represented the rate of 16 cm per second, the answer would be 5 seconds at 16 cm per second to travel 80 cm.
.................................................
......!!!!!!!!...................................
23:40:14 In the preceding three exercises you again expanded your concept of the idea of a rate. Explain how these problems illustrate the concept of a rate.
......!!!!!!!!...................................
RESPONSE --> I think they (or at least the way I decided to visualize them) showed the rate of work being done per unit of time as a building block. These blocks, when counted up, would indicate how much time was required to complete work.
.................................................
???O??????Student Name: assignment #001 001. Rates
......!!!!!!!!...................................
23:44:55 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
......!!!!!!!!...................................
RESPONSE --> I understand...especially since this isn't the first time I've used this. I probably will try the save as notes button, although I prefer using actual paper for notes.
.................................................
......!!!!!!!!...................................
23:45:06 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
......!!!!!!!!...................................
RESPONSE --> OK
.................................................
......!!!!!!!!...................................
23:46:24 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
......!!!!!!!!...................................
RESPONSE --> This question sounds familiar: $50 divided into 5 equal ""hours"" would result in there being $10 in each hour, so the rate is $10 per hour.
.................................................
......!!!!!!!!...................................
23:47:21 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. Understood. And since this assignment isn't about visualization, I may shorten my answers.
.................................................
......!!!!!!!!...................................
23:49:13 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE --> $60,000 / 1 year = $60,000 / year $60,000 / 12 months = $5,000 / month
.................................................
......!!!!!!!!...................................
23:49:32 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
23:51:35 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
......!!!!!!!!...................................
RESPONSE --> It would be more appropriate to say is made an average of $5,000 per month, since the only quantity stated definitively was the yearly amount. We don't know what was earned in each month...one month could have been $7,500 and the next month $2,500, but they would average out to $5,000 month.
.................................................
......!!!!!!!!...................................
23:51:47 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
23:56:15 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
......!!!!!!!!...................................
RESPONSE --> 300 miles = 6 hours x miles traveled per hour Move the 6 to the other side of the equal sign by dividing both sides by 6. 50 = miles traveled per hour Again, average rate is more accurate because the beginning of the trip may include acceleration from a standstill, while the end may include deceleration to a stop.
.................................................
......!!!!!!!!...................................
23:56:16 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
.................................................
Here begins the second SEND file..... ??g?????n?????K?Student Name: assignment #001 001. Rates
......!!!!!!!!...................................
00:06:41 `q001. You should copy and paste these instructions to a word processor for reference. However you can always view them, as well as everything else that has appeared in this box, by clicking the 'Display Everything' button. 1. For the next question or answer, you click on 'Next Question / Answer' button above the box at top left until a question has been posed. Once a question has been posed you are to answer before you click again on this button. 2. Before clicking for an answer, type your best answer to the current question into the box to the right, then clip on the 'Enter Answer' button. 3. After entering your answer you will click on 'Next Question / Answer' to view the answer to the question. Do not tamper with the information displayed in the left-hand box. 4. If your answer was incorrect, incomplete or would otherwise require revision, you will enter a self-critique. If you learned something from the answer, you need to restate it in your own words in order to reinforce your learning. If there is something you feel you should note for future reference, you should make a note in your own words. Go to the response box (the right-hand box) and type in a self-critique and/or notes, as appropriate. Do not copy and paste anything from the left-hand box, since that information will be saved in any case. 5. If you wish to save your response to your Notes file you may choose to click on the 'Save As Notes' button rather than the 'Enter Answer' button. Doing so will save your work for your future reference. Your work will be saved in a Notes file in the c:\vhmthphy folder. The title of the Notes file will also include the name you gave when you started the program. 6. After clicking either the 'Enter Response' or the 'Save as Notes' button, click on 'Next Question / Answer' and proceed in a similar manner. In the right-hand box briefly describe your understanding of these instructions, then click 'Enter Answer'.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:43 Your answer has been noted. Enter 'ok' in the Response Box and click on Enter Response, then click on Next Question/Answer for the first real question.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:45 `q002. Note that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Continue as far as you can until you are completely lost. Students who are prepared for the highest-level math courses might not ever get lost. If you make $50 in 5 hr, then at what rate are you earning money?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:47 The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:49 `q003.If you make $60,000 per year then how much do you make per month?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:51 Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:52 `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:54 Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:06:56 `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:07:02 The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:09:37 `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
......!!!!!!!!...................................
RESPONSE --> The rate that gas is being used would be 60 gallons per 1200 miles, or simplified down to 60 / 1200 = 0.05 gallons used per each mile traveled. This question has burned me before, and burned me once on paper before I put the answer down...I first did miles per gallon, but went back because it seemed wrong.
.................................................
......!!!!!!!!...................................
00:09:54 The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover those miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that require the use of more fuel on some miles than on others. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:14:04 `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?
......!!!!!!!!...................................
RESPONSE --> We're starting with the end quantity and one of the two factors needed to calculate it. So naturally we divide to find the missing factor...we're trying to derive an average after the fact, when we know the end result but not the individual quantities that added up to make that end result.
.................................................
......!!!!!!!!...................................
00:14:20 The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:19:56 `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
......!!!!!!!!...................................
RESPONSE --> Lifting strength is in pounds. The second group did more pushups and lifted more weight than the first. So the difference in weight was 162 - 147 = 15 pounds, and the difference in pushups per day was 50 - 10 = 40 pushups. So the rate increased 15 pounds by doing 40 more pushups per day. 15 lbs / 40 daily pushups = .375 lbs increase per daily pushup.
.................................................
......!!!!!!!!...................................
00:20:05 The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:25:25 `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?
......!!!!!!!!...................................
RESPONSE --> The second group's lifting strength was 188 lbs, while the first group's was 171, for a difference of 17 pounds. The second group used 30 pound shouder weights, while the first group used 10 pound weights for a difference of 20 lbs. So, the average lifting strength increase per added shoulder weight would be 17 lbs lifting strength / 20 lbs shoulder weight = .85 lbs increase in lifting strength for each added pound of shoulder weight.
.................................................
......!!!!!!!!...................................
00:25:35 The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:29:51 `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
......!!!!!!!!...................................
RESPONSE --> The runner traveled between the 100 m and 200 m marks, for a distance of 100 m. The difference in time between those marks was 22 sec - 12 sec for a total of 10 sec. So the average rate would be 100 m / 10 sec = 10 meters per second.
.................................................
......!!!!!!!!...................................
00:30:00 The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:33:30 `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the 100 meter distance?
......!!!!!!!!...................................
RESPONSE --> I'd say if the runner was traveling at 10 m/sec at the beginning and 9 m/sec at the end, a reasonable assumption of average speed would be 9.5 m/sec. Since the distance traveled at this average rate was 100 meters, it would take the runner 100 m / 9.5 meters per second = 10.53 m/sec.
.................................................
......!!!!!!!!...................................
00:33:46 At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................
......!!!!!!!!...................................
00:36:01 `q012. We just averaged two quantities, adding them in dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
......!!!!!!!!...................................
RESPONSE --> In the earlier questions we were given the average rate or the average speed. In this one, we were given 2 speeds taken at two points, so we had to calculate the average rate ourselves.
.................................................
......!!!!!!!!...................................
00:36:13 In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.
......!!!!!!!!...................................
RESPONSE --> Understood
.................................................