course Phy 122 end programn????x??~???p?~????assignment #006
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12:30:37 `q001. There are two parts to this problem. Reason them out using common sense. If the speed of an automobile changes by 2 mph every second, then how long will it take the speedometer to move from the 20 mph mark to the 30 mph mark? Given the same rate of change of speed, if the speedometer initially reads 10 mph, what will it read 7 seconds later?
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RESPONSE --> First the difference in speed of the vehicle is 30mph - 20mph= 10 mph in which the speed increased in the questionable amount of time. Every second is worth 2 mph so 10mph/2s= 5 seconds that it took the car to reach 30mph. The second part is what mph the car is in 7s starting at 10mph. In seven seconds the car increased its speed by 2mph every second so 7s x 2mph/s= 14mph. Plus the 10mph it had started off at would be 10mph + 14mph= 24mph 7 seconds later from the initial 10mph. confidence assessment: 3
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12:31:03 It will take 5 seconds to complete the change. 30 mph - 20 mph = 10 mph change at 2 mph per second (i.e., 2 mph every second) implies 5 seconds to go from 20 mph to 30 mph Change in speed is 2 mph/second * 7 seconds = 14 mph Add this to the initial 10 mph and the speedometer now reads 24 mph.
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RESPONSE --> self critique assessment: 3
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12:40:43 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> The vehicle would need less than 10s to reach the lamppost this time because its inital speed is twice of that of the original. At the end of the coast down the hill when the first time it reached the lamppost, the car was going 15mph (10s / 2mph/s=5mph + 10mph=15mph). The second car was already going more than that to start off and was still increasing by 2mph/s. No, because since the car is going faster and covering a greater distance in less time, the car will be able to travel to the lamppost quicker than the first time. Since less seconds passed, the car was not able to increase as many mph as before (less seconds to increase by 2mph/s). confidence assessment: 2
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12:41:23 If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds. The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.
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RESPONSE --> self critique assessment: 3
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12:46:37 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> The first vehicle speed changes by 10mph in 5s which gives me an average rate of velocity change of 10mph / 5s = 2mph/s. The other vehicla changes by 50mph in 20s which is an average rate of velocity change of 50mph / 20s = 2.5mph/s. The second vehicle is speeding up at a greater rate with a difference of 0.5mph/s. confidence assessment: 3
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12:46:50 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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RESPONSE --> self critique assessment: 3
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12:53:50 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The first team is exerting 3000N / 1500kg = 2N per kg whereas the second team is exerting 5000N / 2000kg = 2.5N per kg The second team will win because they are exerting 0.5 more Newtons per kg then the first team. If someone pulled in the opposite direction with 500 Newtons then it would be (5000-500)N / 2000kg = 2.25 N per kg which would still allow the second team to win by 0.25 more N per kg. self critique assessment: 3
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12:54:15 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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RESPONSE --> self critique assessment: 3
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12:56:51 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> I really can not remember how to do this simple problem. What equation would I need in order to solve this? confidence assessment: 0
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12:59:12 Greater speed and greater mass both provide advantages. In this case the player with the greater mass has less speed, so we have to use some combination of speed and mass to arrive at a conclusion. It turns out that if we multiply speed by mass we get the determining quantity, which is called momentum. 250 lb * 10 ft/sec = 2500 lb ft / sec and 200 lb * 20 ft/sec = 4000 lb ft / sec, so the second player will dominate the collision. In this course we won't use pounds as units, and in a sense that will become apparent later on pounds aren't even valid units to use here. However that's a distinction we'll worry about when we come to it.
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RESPONSE --> So momentum in this problem can be calculated by multiplying the weight of the player times the ft traveled per second. 4000 lb ft /s would be much more than 2500 lb ft / s.
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13:04:37 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> How would you be able to determine which climber got the most energy out of the meal? The first climber is 200lb and ate 12 ounces which would be 200lb / 12 ounces =16.67lb/ounce The second climber is 150lb and ate 10 ounces which would be 150lb/10ounces = 15lb/ounce. I would say that the first climber would be able to climb further. confidence assessment: 0
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13:09:58 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> I had gotten the equation backwards, I had wanted oz/lb fro the energy per pound of body weight. self critique assessment: 1
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13:14:15 `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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RESPONSE --> The second automobile will take longer to stop because its initial speed was faster of that of the first car and since they are continuing up the same steep hill it will be longer before the second stops. How would I determine how long it would take for the second vehicle to stop? I would say that it would take less than twice as long because coasting up a steep hill would still decrease its speed continually as time passes. confidence assessment: 1
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13:16:22 It turns out that, neglecting air resistance, since the slope is the same for both, both automobiles will change velocity at the same rate. So in this case the second would require exactly twice as long. If you include air resistance the faster car experiences more so it actually takes a bit less than twice as long as the slower. For the same reasons as before, and because velocity would change at a constant rate (neglecting air resistance) it would be exactly twice as great if air resistance is neglected. Interestingly if it takes twice as much time and the average velocity is twice as great the faster car travels four times as far. If there is air resistance then it slows the faster car down more at the beginning than at the end and the average velocity will be a bit less than twice as great and the coasting distance less than four times as far.
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RESPONSE --> The change of velocity is at the same rate for both the vehicles so the car would travel twice as long as expected for the first at least close to if air resistance is included. self critique assessment: 2
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13:24:35 `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?
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RESPONSE --> I would expect it to be a little less or equal to 7 ft beyond its initial point because the examples are changing by 50 lbs and the ft changes go from 4ft to 3ft of difference which means the difference decreases slightly as more and more weight is applied. 125lbs is exactly between 100lb and 150lb with a difference of feet of 4. 4/s is 2ft + 5 ft would be about 7ft. confidence assessment: 2
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13:41:08 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> Ok so the increase in the bungee cord would still be greater than 7ft because the increase is greater between 100lb to 125lb than 125lb to 150lb. This makes sense to me now.
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13:44:44 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> The skater would be expected to go twice the distance as before as twice the pullback equals twice the force. 20ft x2 would be 40ft that she would be expected to travel. confidence assessment: 2
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13:46:21 The distance through which the force acts will be twice as great, which alone would double the distance; because of the doubled pullback and the linear proportionality relationship for the force the average force is also twice as great, which alone would double the distance. So we have to double the doubling; she will go 4 times as far
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RESPONSE --> Ok, I understand now. self critique assessment: 3
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13:50:57 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> I am not sure on how to find out which sphere would appear to be brighter or dimmer than the other. My guess would be that the larger one would appear to be dimmer because of the closeness of the bulb to the surface of the frosted glass and the distortion of light. Therefore the larger sphere would appear to have half the brightness of the first because the light waves would have a longer distance to travel in which to get distorted and lost. confidence assessment: 1
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13:55:11 Both bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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RESPONSE --> I did not know how to get to this conclusion before I read the response, but now that I see it, I understand how I can get to that conclusion. The two lamps will look the same size at a distance to the insect because of the intensity of the light bulbs are the same. However, because of the 4x area of the second lamp there would be 1/4 less light on the surface of the lamp then the littler lamp because of the lights distribution across the surface. self critique assessment: 2
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14:05:29 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> Increasing the temperature of the ice by 20 degrees to reach its melting point took the least energy because it was only in the microwave for 10s to do so (600J/s x 10 = 6000J). Then increasing the temperature of the water by 20 degrees after all the ice is melted because that only took less than a minute in the same microwave using the same 600 J/s. Lastly, Melting the ice at its melting point took. It looks as though ice melts at 0 degrees Celcius because it stayed at 0 degrees as the ice melted slowly. confidence assessment: 1
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14:06:48 Since the temperature is the same when a little of the ice is melted as when most of it is melted, melting takes place at this temperature, which is 0 Celsius. The time required to melt the ice is greater than any of the other times so melting at 0 C takes the most energy. Since we don't know how much ice remains unmelted before the final minute, it is impossible to distinguish between the other two quantities, but it turns out that it takes less energy to increase the temperature of ice than of liquid water.
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RESPONSE --> Ok I understand it better when it is put in those words. self critique assessment: 3
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14:25:26 `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fll in' the valley, with the result that you won't bob up and down very much. If the peaks of the approaching waves are each 6 inches high, how far would you expect to bob up and down when you are at the center point? How far would you have to move toward one end or the other in order for peaks to meet valleys, placing you in relatively calm water?
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RESPONSE --> 12 inches because they combine in the center and increase 2x as much as they originally were making them 12in. I would expect to bob 12 inches higher than the surface of the pool and 12 inches lower than the center point of the pool. I would predict between half way of the one friend and the middle where I was located, although I am not sure if this is right or not and have nothing really to back up my answer. confidence assessment: 0
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14:28:30 If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in. If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ] However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.
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RESPONSE --> Ok I understand now. Since the distance between the crests are 6ft coming from both sides, it would be 3ft where crests would be meeting valleys and making calm water. To move only 1.5ft in one direction means that you are moving toward one source by 1.5ft and away from another at 1.5ft making it 3 ft total of a move. self critique assessment: 2
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