course Phy 121 ??????????O????q?assignment #006006. Physics
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01:55:16 `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 --> If you were to count by two up to 30, it would have taken 5 seconds for it to change. If it starts at 10, then by using the same method it will have gained 14 mph since the original 10. You would just add them together to get 24mph. confidence assessment: 2
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01:55:34 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 --> ok self critique assessment: 2
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01:57:09 `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. Will the vehicle require more or less than 10 seconds to reach the lamppost? Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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RESPONSE --> It will require less than 10 seconds as it is going faster. It would make sense for it to be 10mph greater at the lamppost as well. confidence assessment: 2
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01:57:39 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 --> ok self critique assessment: 2
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01:59:48 `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 car would be traveling at 2mph/second and the second car would be traveling 2.5mph/second. Therefore, the second car would be speeding up at a greater rate. confidence assessment: 2
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01:59:59 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 --> ok self critique assessment: 2
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02:03:43 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The force that the first team exerts in Newtons per kg is 3000/1500 = 2 and the force that the second team exerts in Newtons per kg is 5000/2000 = 2.5. Therefore, the second team will win because they are exerting more force. If someone were pulling with a force of 500 Newtons in the opposite direction, the second team would still beat the first team because the force they are exerting, which is 2.25 is still greater than 2. self critique assessment: 2
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02:03:54 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 --> ok self critique assessment: 2
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02:05:24 `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you precidt will be moving backward immediately after the collision, and why?
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RESPONSE --> I would predict that the 200lb player would be pushed backwards. confidence assessment: 2
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02:06:08 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 --> ok self critique assessment: 1
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02:10:06 `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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RESPONSE --> I would assume the 150lb climber would use up less energy because he wouldn't have to exert as much energy as the 200lb climber as he is lighter. confidence assessment: 1
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02:10:40 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 --> ok self critique assessment: 1
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02:32:13 `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?
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RESPONSE --> The automobile that is traveling faster will take longer to come to a stop. It should take about twice as long. The faster vehicle will have the greater average coaster velocity. It too should be about twice as great. The distance should be twice as the slower vehicle. confidence assessment: 0
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02:37:17 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 --> ok self critique assessment: 2
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02:41:02 `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 less than 7. confidence assessment: 2
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02:41:40 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 self critique assessment: 0
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02:43:32 `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 --> She should travel twice as far as it's stated that twice the pullback implies twice the force. confidence assessment: 2
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02:44:11 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 self critique assessment: 1
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02:46:30 `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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RESPONSE --> The larger sphere will appear dimmer than the first as there is twice as much space inside. In the second part of the question, the larger light will appear to have half of the brightness of the first. confidence assessment: 2
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02:47:22 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 --> ok self critique assessment: 2
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02:52:27 `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 --> The order in least amount of energy to most amount of energy required are: 1)Increasing the temperature of the ice by 20 degrees to reach its melting point. 2)Increasing the temperature of the water by 20 degrees after all the ice melted. 3)Melting the ice at its melting point. Ice melts at 0 Celsius. The evidence is at that temperature, the ice started to melt and it stayed at that temperature until all of the ice was melted. confidence assessment: 2
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02:52:57 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 self critique assessment: 2
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02:57:34 `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 --> You would move about 3 feet to either side for the valleys and peaks to meet if the waves are 6 feet apart. confidence assessment: 0
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02:58:14 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 self critique assessment: 2
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???a?????????? assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density qa areas volumes misc 01-17-2009
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03:13:53 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> The formula for surface area of a rectangular prism is SA = 2(lw + hl + hw) You would just plug in the values for the dimensions. SA = 2((3m*4m) + (6m*3m) + (6m*4m) SA = 2(12m^2+18m^2+24m^2) SA = 2(54m^2) SA = 108m^2 confidence assessment: 2
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03:13:59 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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RESPONSE --> ok self critique assessment: 2
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03:17:53 `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?
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RESPONSE --> The formula for surface area of a cylinder is SA = 2pi*r(r+h) SA = 2pi*5m(5m+12m) SA = 2pi*5m(17m) SA = 2pi*85m^2 SA = 2(266.9m^2) SA = 533.8m^2 confidence assessment: 2
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03:19:03 The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2. If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.
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RESPONSE --> ok confidence assessment: 2
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03:20:51 `q003. What is surface area of a sphere of diameter three cm?
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RESPONSE --> The formula is SA = 4pi*r^2. To get r, you would need to find the radius by dividing 3 by 2 to get 1.5. SA = 4pi*(1.5cm^2) SA = 4pi(2.25cm^2) SA = 28.26cm^2 confidence assessment: 2
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03:21:38 The surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.
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RESPONSE --> ok self critique assessment: 2
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03:23:29 `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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RESPONSE --> You would use the pythagorean theorem. a^2 + b^2 = c^2 5^2 + 9^2 = c^2 25 + 81 = c^2 106 = c^2 c= sqrt(106) confidence assessment: 2
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03:24:04 The Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx.. Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.
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RESPONSE --> ok self critique assessment: 2
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03:25:49 `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
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RESPONSE --> The pythagorean theorem would be used again. a^2 + b^2 = c^2 4^2 + b^2 = 6^2 16 + b^2 = 36 b^2 = 20 b = sqrt(20) confidence assessment: 2
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03:26:25 If c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg: a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m, or approximately 4.4 m.
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RESPONSE --> ok self critique assessment: 2
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03:29:37 `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?
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RESPONSE --> Density is found by dividing the mass by the volume. To find the volume of the rectangle, we need to multiply the dimensions together. V = 4cm * 7cm * 12cm V = 336cm^3 Then we find the density. D = 700g/336cm^3 D = 2.083g/cm^3 confidence assessment: 2
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03:30:24 The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3. Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that density = 700 grams / (336 cm^3) = 2.06 grams / cm^3. Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams. Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).
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RESPONSE --> ok self critique assessment: 2
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03:40:17 `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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RESPONSE --> confidence assessment: 0
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03:49:58 A average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg. The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg. This result can be approximated to an appropriate number of significant figures.
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RESPONSE --> ok self critique assessment: 2
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04:04:40 `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?
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RESPONSE --> The average density would be found by finding the total density then averaging them. (6cm^3 at 4g/cm^3 + 10cm^3 at 2g/cm^3)/2 = (16cm^3 at 6g/cm^3)/2 = 8cm^3 at 3g/cm^3 confidence assessment: 1
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04:06:13 The first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams. The average density of this object is average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.
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RESPONSE --> ok self critique assessment: 2
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04:31:12 `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
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RESPONSE --> To find the average density, we need to find the total mass and total volume. The total mass of the sand is found by multiplying 27 cubic meters and 2100 kg/cubic meter. The mass of the sand is 56700kg. The total mass of the cannonballs is found by multiplying 3 cubic meters and 8000kg/cubic meter. The total mass would be 80700kg Then to find the total volume you would add 27cm^3 and 3cm^3. Next, you would divide the total mass by the total volume to find the average density. 80700kg / 30cm^3 = 2690kg/cm^3 confidence assessment: 1
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04:31:26 We find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg. The average density is therefore average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..
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RESPONSE --> ok self critique assessment: 2
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04:40:59 `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?
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RESPONSE --> To find how many cubic meters it is, you would multiply 1700000 m^2 by .015m. It turns out to be 25500m^3. To find the density of the oil, you would multiply 22500m^3 by 860kg/cm^3. You would end up getting 19350000kg. confidence assessment: 1
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04:41:30 The volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3. The mass of the slick is therefore mass = density * volume = 860 kg / m^3 * 24,400 m^3 = 2,193,000 kg. This result should be rounded according to the number of significant figures in the given information.
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RESPONSE --> ok self critique assessment: 2
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04:42:33 `q011. Summary Question 1: How do we find the surface area of a cylinder?
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RESPONSE --> We find the circumference of the top and bottom then multiply it by the height. confidence assessment: 2
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04:42:49 The curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude. The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2. {]The total surface area is therefore Acylinder = 2 pi r h + 2 pi r^2.
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RESPONSE --> ok self critique assessment: 2
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04:43:16 `q012. Summary Question 2: What is the formula for the surface area of a sphere?
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RESPONSE --> The formula is SA = 4pi r^2. confidence assessment: 2
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04:43:20 The surface area of a sphere is A = 4 pi r^2, where r is the radius of the sphere.
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RESPONSE --> ok self critique assessment: 2
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04:44:16 `q013. Summary Question 3: What is the meaning of the term 'density'.
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RESPONSE --> Density is how many units of mass are in each unit of volume. confidence assessment: 2
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04:44:30 The average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'
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RESPONSE --> ok self critique assessment: 2
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04:46:48 `q014. Summary Question 4: If we know average density and mass, how can we find volume?
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RESPONSE --> You would find it by setting it up like the formula D = m/v D = M/v Dv = M v = M/D confidence assessment: 2
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04:46:59 Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.
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RESPONSE --> ok self critique assessment: 2
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04:47:30 `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I thought about each question logically and figured out what I had to do. confidence assessment: 2
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