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course Phy 121
001. Typewriter notation Note that there are six questions in this exercise. Be sure to continue scrolling down until you get to the end of the exercise.
Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.
Your solution: The first one you would divide and then you would work on the addition and subtraction parts. You substitute 2 for x and then it would look like 2-2/2+4= 2-1+4 and your answer would equal 5. For the second problem you would do the parentheses first and then division and lastly do the subtraction and addition. You substitute 2 for x (x-2)/ (x+4) = (2-2)/(2+4)+ 0/6= 0. The answer would equal 0 because you would add and subtract the numbers in the parentheses first then divide.
confidence rating #$&*:: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2. Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.
Your solution: The first problem you would perform exponentiation first then perform the addition part. You substitute 2 for x for 2^x+4 = 2^2+4 = 4+4 = 8. The second problem you would perform the parentheses first then the exponentiation. You substitute 2 for x for 2^(x+4) = 2^(2+4) = 2^6 = 64.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. What is the numerator of the fraction in the expression x - 3 / [(2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?
Your solution: The numerator is the 3 and the denominator is[(2x-5)^2*3x+1]. When you substitute 2 for x for x-3/[(2x-5)^2*3x+1]-2+7x = 2-3/ [(2(2)-5)^2*3(2)+1]-2+7x = first you perform parentheses 2-3/[(4-5)^2*6+1]-2+14 = second you do multiplication outside the parentheses 2-3/[(-1)^2*6+1 ]-2+14 = add inside the parentheses 2-3/[1*6+1]-2+14 = the exponents in the bracket 2-3/ 7-2+14 = evaluate in bracket, 2-2+14 -3/7. 14 - 3/7= then you get the common denominator would be 7 so in order to get 7 for the denominator for 14 you need to multiply 14 by 7 = 98/7. Then the last step is 98/7 -3/7 = 95/7 is the answer.
confidence rating #$&*: 2
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Self-critique (if necessary): This question took me a while because at the last steps I forgot about the x not being the numerator it was only 3. Also I forgot that you needed to find a common denominator in order to complete the problem. After I read the students and instructor responses I remembered this step. It showed me that I need to review these types of problems.
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Self-critique Rating: 2
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Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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Your solution: First you evaluate the parentheses (4-5)^2 *4 -1 +3/ 4-2 = (-1)^2 *4 -1 +3/ 4-2. Second you would evaluate your exponents 1 * 4 - 1 +3/4 -2 = then you evaluate multiplication and division 4 - 1 + ¾ -2= then you switch the order to where it is 4 - 1 - 2+3/4 = 1 + ¾ = 1 ¾ = 7/4 is the answer.
confidence rating #$&*: 2
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Self-critique (if necessary): I understand every step until I get to the last step where you switch the order and determine the correct answer. After I read the comments made from the student and instructor it makes a little bit more sense but I still get confused when I try to complete it by myself.
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Self-critique Rating: Ok
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Question: `q005. At the link http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm. (copy this path into the Address box of your Internet browser; alternatively use the path http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples and you will find a page containing a number of additional exercises and/or examples of typewriter notation. Locate this site, click on a few of the links, and describe what you see there.
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Your solution: It gives you different examples of these problems and it gives you practice problems to where you can help make your understanding of these problems more clearly and easy. Also there includes about 35 problems to complete and underneath the problem if you click on the picture it will show you what it should look like.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions. However it's very important to understand order of operations, and students do get used to this way of doing it. You should of course write everything out in standard notation when you work it on paper. It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation. Indicate your understanding of why it is important to understand this notation.
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Your solution: You need to make sure you understand this notation to help make sure you complete the problems correctly and if you understand and follow the steps of order of operations then you should be able to get the correct answer every time.
confidence rating #$&*: 3
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002. Describing Graphs
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Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.
Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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Your solution: You first try to figure out the x value so you substitute the y value to 0 = 3x-4, 4=3x, 4/3=x So the x intercept will be (4/3, 0). Then to figure out the y value you substitute the x value to 0. Y=3(0)-4, Y=0-4, Y= -4. So your y intercept will be (0, -4). If you check your graph is should confirm the x and y interceptions.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.
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Your solution: No because the graph is a straight line so the steepness never changes. But the line itself is looks like a steep hill but the steepness of the line stays the same throughout the line. So I would say the line is steep but the steepness does not change.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4; slope is rise / run between two points of the graph)?
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Your solution: I took the x values of 3 and 5. When x = 3 then y =3*3-4 = 5 and when x=5 then y = 3 *5 - 4 = 11. Then you take y2-y1/x2-x1 which gives you rise over run. So the formula would be 11-5/5-3 = 6/2 = 3. So the slope of the graph would be 3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at a decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: I would say that the graph is increasing and the steepness would also increase as the numbers increase. I would say the graph is increasing at an increasing rate because each increased number makes the graph increase more. Some of the points on the graph are (0,0), (1,1), (2,4), and (3,9). The explanation for how I got my points on the graph Y=(3)^2 =9, Y=(2)^2=4, Y=(1)^2=1, and Y=(0)^2=0.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: The graph would be decreasing left to right (-3,9), (-2,4), (-1,1), and (0,0). The steepness would also be decreasing with each x value coming from left to right. I say the graph would be decreasing at a decreasing rate. The explanation of how I got my points on my graph is Y=(-3)^2 =9, Y=(-2)^2=4, Y=(-1)^2=1 and Y=(0)^2=0.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at a decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: The points on the graph included (0,0), (1,1), (2,1.41), and (3, 1.73). I would say that the graph increases at a decreasing rate and the steepness of the graph is decreasing also because at the beginning the steepness was 1 then as the numbers increased the steepness decreased. The explanation of how I got my points on the graph is Y= sqrt 0=0,Y sqrt 1=1,Y= sqrt 2=1.41,and Y= sqrt 3=1.73.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `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 a decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: I would say the graph was decreasing at a decreasing rate with the points on the graph (0,5), (1,2.5), (2,1.25), and (3,.625). The steepness decreases at a decreasing rate. The explanation for the points on the graph is Y=5*2^(-0)=5, Y=5*2^(-1)=2.5, Y=5*2^(-2)=1.25, and Y=5*2^(-3)=1.73.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at a decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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Your solution: I say the graph would increase because the longer time the car is moving the further the distance the car is away from you. It would also be increasing at an increasing rate because of the previous statement.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
006. Physics
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Question: `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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Your solution: It will take 5 seconds to move the speedometer from 20 mph to 30 mph. You take 2 * 5 = 10 and that’s the difference between 20 and 30 mph. If you start out at 10 mph and 7 seconds later it would end up at 24 mph. 7 * 2= 14 then you add 14 to 10 and get 24 mph.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q002. An automobile traveling down a hill passes a certain milepost traveling at a speed of 10 mph, and proceeds to coast to a certain lamppost further down the hill, with its speed increasing by 2 mph every second. The time required to reach the lamppost is 10 seconds. It then repeats the process, this time passing the milepost at a speed of 20 mph. This time:
Will the vehicle require more or less than 10 seconds to reach the lamppost?
Since its initial speed was 10 mph greater than before, does it follow that its speed at the lamppost will be 10 mph greater than before?
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Your solution: If I’m traveling down a hill increasing 2 mph every second and 10 seconds pass then I would have increased up to 20 mph and add that to the 10 mph I was already traveling I would reach up to 30 mph. The vehicle will require less than 10 seconds because it is already traveling 5 seconds faster than the vehicle before. The change will be less because instead of 10 seconds it only has 5 seconds to increase its speed.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed:Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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Your solution: 10mph /5 sec = 2mph/sec for the one that speeds up from 20 to 30mph and the one that speeds up from 40 mph to 90 mph equals 50mph/20 sec. = 2.5 mph/sec. So the vehicle that went from 40 to 90 mph speeds up at a faster rate by a half a sec compared to the vehicle that changed from 20 to 30 mph.
confidence rating #$&*: 3
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Self-critique:
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Self-critique Rating: Ok
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Question: `q004. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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Your solution: The first team’s rate would be 3000 Newtons/ 1500 kg = 2 Newtons/kg. The second team’s rate would be 5000 Newtons / 2000 kg = 2.5 Newtons/kg. The second team would win because the can accelerate a 0.5 second faster than the first team.
The second team’s rate with the opposite force would be 5000 Newtons - 500 Newtons/ 2000 kg = 2.25 Newtons/ kg. So the second team would still win because there ending result would still be higher than the first team.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q005. Both the mass and velocity of an object contribute to its effectiveness in a collision. If a 250-lb football player moving at 10 feet per second collides head-on with a 200-lb player moving at 20 feet per second in the opposite direction, which player do you predict will be moving backward immediately after the collision, and why?
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Your solution: First you need to figure out the momentum of each football player:
250 lb. * 10 ft/sec. = 2500 lb ft/sec
200 lb * 20 ft/sec = 4000 lb ft/sec
So the 200 lb football player will push the 250 lb football player backwards because the 200 lb football player has a greater momentum than the 250 lb football player.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006. Two climbers eat Cheerios for breakfast and then climb up a steep mountain as far as they can until they use up all their energy from the meal. All other things being equal, who should be able to climb further up the mountain, the 200-lb climber who has eaten 12 ounces of Cheerios or the 150-lb climber who has eaten 10 ounces of Cheerios?
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Your solution: First climber 12 oz/ 200 lb = 0.06 oz / lb; Second climber 10 oz / 150 lb = 0.67 oz / lb
So the second climber would have more energy per pound of body weight than the first climber.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q007. Two automobiles are traveling up a long hill with a 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 them 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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Your solution: The faster automobile will take longer to stop. It will prolly take twice as long to stop because its speed is twice as fast as the slower automobile so you have force to stop. The faster automobile will have an average coasting velocity twice as much as the slower automobile. Since the faster automobile will be going more than twice as fast it will be able to travel twice as much distance as the slower automobile.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q008. When a 100 lb person hangs from a certain bungee cord, the cord stretches by 5 feet beyond its initial unstretched length. When a person weighing 150 lbs hangs from the same cord, the cord is stretched by 9 feet beyond its initial unstretched length. When a person weighing 200 lbs hangs from the same cord, the cord is stretched by 12 feet beyond its initial unstretched length. Based on these figures, would you expect that a person of weight 125 lbs would stretch the cord more or less than 7 feet beyond its initial unstretched length?
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Your solution: From 100 to 150 lbs the bungee cord stretches from 5 ft to 9 ft which is a total difference 4 ft. From 150 to 200 lbs the bungee cord stretches from 9 to 12 ft which is a total difference of 3 ft. So if you go from 100 to 125 lbs the bungee cord will probably stretch 2 ft because it is the half distance the bungee cord was stretched from 100 to 150 lbs. So I think the 125 lb person will stretch the bungee cord to 7 ft. or more.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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Your solution: The skater would travel more than twice as far because the distance the skater gets pulled back would be twice as much and the distance traveled is also be doubled. The pullback force went from 4 to 8 ft and the distance traveled went from 20 to 80 ft. I got 80 by taking 20 and multiplying it by 4. I think the skater would go 4 times a far as the skater that only was pullback 4 ft.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q010. Two identical light bulbs are placed at the centers of large and identically frosted glass spheres, one of diameter 1 foot and the other of diameter 2 feet. To a moth seeking light from half a mile away, unable to distinguish the difference in size between the spheres, will the larger sphere appear brighter, dimmer or of the same brightness as the first? To a small moth walking on the surface of the spheres, able to detect from there only the light coming from 1 square inch of the sphere, will the second sphere appear to have the same brightness as the first, twice the brightness of the first, half the brightness of the first, more than twice the brightness of the first, or less than half the brightness of the first?
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Your solution: The first bulb will be brighter than the second because its diameter is smaller and does not have as much space to light up. If you are a certain distance away you would not be able to tell which light is brighter and they would look the same. They also would have the same amount of energy per second. I think the second sphere will appear to have twice the brightness of the first bulb.
confidence rating #$&*: 2
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Self-critique (if necessary): The second part of this question confuses me just like the student and instructor comments above. I don’t quite understand how you would get the second part of this question even with the explanations.
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Self-critique Rating: 2
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Question: `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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Your solution: Ice melts at 0 degrees Celsius because it states above that 10 seconds after -1 degrees Celsius the ice starts to melt at 0 degrees Celsius and then after a minute most of the ice is melted but there is still a little bit of ice left. At -1 degrees the ice is still solid and after the temperature reaches 40 degrees the ice is melted and the water temperature has risen.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q012. Suppose you are in the center of a long, narrow swimming pool (e.g., a lap pool). Two friends with kickboards, one at either end of the pool, are using them to push waves in your direction. Their pushes are synchronized, and the crests of the waves are six feet apart as they travel toward you, with a 'valley' between each pair of crests. Since your friends are at equal distances from you the crests from both directions always reach you at the same instant, so every time the crests reach you the waves combine to create a larger crest. Similarly when the valleys meet you experience a larger valley, and as a result you bob up and down further than you would if just one person was pushing waves at you. Now if you move a bit closer to one end of the pool the peak from that end will reach you a bit earlier, and the peak from the other end will reach you a little later. So the peaks won't quite be reaching you simultaneously, nor will the valleys, and you won't bob up and down as much. If you move far enough, in fact, the peak from one end will reach you at the same time as the valley from the other end and the peak will 'fill 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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Your solution: You would add the peaks of the waves together to get the combined peak which would equal 12 inches. If you moved 1.5 ft from each end you would still be in the middle where the peaks meet the valleys and that would place you in calm water.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
001. Rates
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. Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.
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Question: If you make $50 in 5 hr, then at what rate are you earning money?
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Your solution: You would take 50 and divide it by 5 and you would get 10 dollars/hour.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003.If you make $60,000 per year then how much do you make per month?
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Your solution: You take 60,000 divided by 12 months and you would get 5,000 dollars a month.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `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?
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Your solution: I would say that the business makes an average of $5000 per month because you don’t know if they will make the same amount every year.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `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?
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Your solution: You would take 300 miles and divide it by 6 hours to figure out the average rate of distance. The answer would be 50 miles/hour.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?
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Your solution: You would take 60 gallons of gasoline and divide it by 1200 miles and that would get you the average rate of 0.5 gallons of gasoline/mile.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
Question: `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?
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Your solution: Because the values have already been adding up and we are trying to find out how many it takes of one thing to get a certain amount of another thing.
confidence rating #$&*:: 2
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Self-critique (if necessary):
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Self-critique Rating: OK
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Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?
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Your solution: the second group did 40 more pushups then the first group and at the end the second group only did 15 more pounds of lifting strength than the first. So I would take 15 pounds/ 40 pushups and you would get 0.374 pounds/pushup.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `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?
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Your solution: there are a 20 more pushups and 17 more pounds. The average rate is 17 pounds/20 pushups. The answer is 0.85 pounds/ pushups.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?
Your solution: 100 meters/10 seconds equals 10 meters/second. The runner is running at a speed of 10 meters per second.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
Question: `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 intervening 100 meter distance?
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Your solution: It took 10 seconds to run 100 meters but the runner slowed down to only being able to run 9 meters per second. 10 meters/second + 9 meters/second and then you divide by 2. The answer is 9.5 meters/second.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now?
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Your solution: Because before we were only given one rate to figure out and the last question there was two rates that we had to determine the average rate between the 2 of them.
confidence rating #$&*: 2
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Self-critique (if necessary):
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Self-critique Rating: Ok
001. Areas
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Question: `q001. There are 11 questions and 7 summary questions in this assignment. What is the area of a rectangle whose dimensions are 4 m by 3 meters.
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Your solution: Area = length times width. You would take 4m and multiply it by 3 m. The answer would be 12 meters squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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Your solution: Area = ½ * base * height. 4.0 meters * 3.0 meters * ½ = 6.0 meters squared then you take 12.0 meters squared and divide it by 2.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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Your solution: Area = base * height: 5.0 meters * 2.0 meters = 10 meters squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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Your solution: Area = ½ * base * height : ½ * 5.0 cm * 2.0 cm = 5 cm squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: ok
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Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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Your solution: Area = base * height: 4.0 km * 5.0 km = 20.0 km squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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Your solution: Average altitude = (3 cm + 8 cm)/2 = 5.5 cm so then you would take 4 cm * 5.5 cm = 22 cm squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q007. What is the area of a circle whose radius is 3.00 cm?
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Your solution: A =∏ * r^2. A = ∏*3.00 cm ^ 2= 28.27 cm squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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Your solution: Circumference is C = 2 ∏ and r= 2∏ * 3 = 18.8 cm.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?
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Your solution: the radius would be half of the diameter so the radius would be 6 meters. The area would be A= ∏r^2. ∏*(6)^2 = ∏ * 36 m^2 = 113.10 meters squared.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?
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Your solution: A = ∏ r^2. A = ∏ 7^2= 153.94 meters squared.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q011. What is the radius of circle whose area is 78 square meters?
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Your solution: A= ∏ * r^2: We need to solve for r: sqrt r=(78meters^2)/∏. Square root of r = 24.828 meters ^2 = r= 4.98 meters ^= 5.0 meters.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?
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Your solution: We find the area by multiplying length * width. How long it is compared to how wide it is.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?
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Your solution: We multiply the base and height and by ½ because of where it is a triangle and not a rectangle. But it is solved the same way you would solve a rectangle you just have to half that because a triangle is a lot smaller than a rectangle.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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Your solution: you use the same equation as a rectangle area = base times height but you use the altitude inside of height.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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Your solution: We use the same equation area = base times height.
confidence rating #$&*: 2
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q016. Summary Question 5: How do we calculate the area of a circle?
Your solution: You would use Area = ∏* r^2.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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Your solution: Circumference is solved by 2 times ∏ times radius. To find the area you have to square the radius unlike the circumference.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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Self-critique (if necessary): This assignment has helped me review the formulas for calculating area for the different shapes like triangle, rectangle, circle, etc.
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Self-critique Rating: ok
004. Units of volume measure
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Question: `q001. There are 10 questions and 5 summary questions in this assignment. How many cubic centimeters of fluid would require to fill a cubic container 10 cm on a side?
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Your solution: 10cm * 10 cm * 10 cm = 1000 cm^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q002. How many cubes each 10 cm on a side would it take to build a solid cube one meter on a side?
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Your solution: I think that you would need 10 cubes to make it one meter on a side.
confidence rating #$&*: 2
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. How many square tiles each one meter on each side would it take to cover a square one km on the side?
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Your solution: It would take 1,000,000 squares to cover a km on a side. A square km is 1 km^2 = 1000m^2 = 1 km.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q004. How many cubic centimeters are there in a liter?
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Your solution: A liter is equal to 1000 cm^3. So a cubes volume would have to be 10 cm * 10 cm * 10 cm = 1000 cm^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q005. How many liters are there in a cubic meter?
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Your solution: There is a 1000 liter in a cubic meter.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q006. How many cm^3 are there in a cubic meter?
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Your solution: There is 1000 cm^3 in a liter and 1000 liters in m^3 so you would multiply 1000 * 1000 = 1,000,000 cm^3 in a m^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q007. If a liter of water has a mass of 1 kg then what is the mass of a cubic meter of water?
Your solution: A 1000 liter is in a cubic meter that would mean the mass of the water would be 1000 kg.
confidence rating #$&*:: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q008. What is the mass of a cubic km of water?
Your solution: (1000 m)^3 = 1,000,000,000 m^3 then you would multiply 1,000,000,000 m^3 * 1000 kg = 1,000,000,000,000 kg and you can simplify it to 10^12kg.
confidence rating #$&*:: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q009. If each of 5 billion people drinks two liters of water per day then how long would it take these people to drink a cubic km of water?
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Your solution: 5 billion people * 2 liters of water = 10 billion liters of water a day. 1,000,000,000,000 liters/ 10,000,000,000 liters per day = 100 days.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q010. The radius of the Earth is approximately 6400 kilometers. What is the surface area of the Earth? If the surface of the Earth was covered to a depth of 2 km with water that what would be the approximate volume of all this water?
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Your solution: 1,020,000,000,000 km^3.
confidence rating #$&*: 1
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Self-critique (if necessary): I don’t quite understand how to get the answer to this question.
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Self-critique Rating: 2
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Question: `q011. Summary Question 1: How can we visualize the number of cubic centimeters in a liter?
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Your solution: A liter is a cube 10 cm so it would be 10 cm for the length and 10 cm for the width and 10 cm height which would equal 1000 cubic cm in a liter.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q012. Summary Question 2: How can we visualize the number of liters in a cubic meter?
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Your solution: we could see 10 cm length and 10 cm width and 10 cm height. This equals 1000 (10 cm) cubes.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q013. Summary Question 3: How can we calculate the number of cubic centimeters in a cubic meter?
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Your solution: It takes 100 cm to make a meter which is (100 cm)^3 for a cubic meter which equals 1,000,000 cm^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q014. Summary Question 4: There are 1000 meters in a kilometer. So why aren't there 1000 cubic meters in a cubic kilometer? Or are there?
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Your solution: There is a 1000 cubic meter is a cubic kilometer.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q015. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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Self-critique (if necessary): It has helped me review the different volumes equations and give me a better understanding at the subject.
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Self-critique Rating: Ok
002. Volumes
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Question: `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 meters by 5 meters by 7 meters?
Your solution: V= 3 meters * 5 meters* 7 meters = 105 meters^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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Your solution: V= a*h : V= 2 m * 48 m^2 = 96 m^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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Your solution: V= 20 m^2 * 40 m = 800 m^3.
confidence rating #$&*: 3
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Self-critique (if necessary):
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Self-critique Rating: Ok
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Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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Your solution: first find the area = pi r^2 = pi (5cm)^2 V = 25 pi cm^2 * 30 cm = 750 pi cm^3. The final answer is 2355 cm^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q005. Estimate the dimensions of a metal can contain food. What is its volume, as indicated by your estimates?
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Your solution: A can’s estimated dimensions is with the diameters at 3 inches and altitude at 5 inches. V= A * H = 9pi/4 in^2 *5ic = 45 pi/4 in^3 = 35 in^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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Your solution: V=1/3 *A*H = 1/3 *50cm^2 * 60cm = 1000 cm^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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Your solution: V=1/3*A*H = 1/3*20m^2*9m = 60 m^3
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q008. What is a volume of a sphere whose radius is 4 meters?
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Your solution: V=4/3pi r^3= 4/3 pi (4m) ^3 = 4/3 pi *64 m^3 = 256/3 pi m^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q009. What is the volume of a planet whose diameter is 14,000 km?
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Your solution: V= 4/3 pi r^3 = 4/3 pi (7,000 km)^3 = 4/3 pi *343,000,000,000 km^3 = 1,372,000,000,000/ 3* pi km^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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Your solution: V= A* h.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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Your solution: V=1/3 A * h
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q012. Summary Question 3: What is the formula for the volume of a sphere?
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Your solution: V= 4/3 pi r^3
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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Self-critique (if necessary):
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Self-critique Rating: It has helped me review how to solve a problem related to volumes.
003. Misc: Surface Area, Pythagorean Theorem, Density
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Question: `q001. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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Your solution: 3 m by 4 m, 3 m by 6m, and 4 m by 6 m, areas 12 m^2, 18 m^2, and 24 m^2. Total area is 2*12 m^2+2*18 m^2+2*24 m^2 = 108 m^2.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `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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Your solution: the circumference would be 10 pi m = 2 pi * 5m. A = 10 pi m * 12 m = 120 pi m^2. Total area = 120 pi m^2 +2* 25 pi m^2 = 170 pi m^2.
confidence rating #$&*: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q003. What is surface area of a sphere of diameter three cm?
Your solution: Surface A=4 pi r^2. Radius = 3: 4 pi (3/2)^2 = 9 pi cm^2.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?
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Your solution: c^2= a^2 +b^2 = 5m^2 9m^2 = 25m^2 +81 m^2 = 106m^2; c^2 = 106m^2, c = square root of 106 m^2.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?
Your solution: a^2= 6m^2 - 4m^2 = 36m^2 - 16 m^2= 52m^2 so a = would equal the square root of 20 = 4.4 m
confidence rating #$&*:: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `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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Your solution: V = 4 cm * 7 cm * 12 cm = 336 cm^3. Density = 700 grams/336 cm^3 = 2.08 g/cm^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?
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Your solution: V= 4/3 pi r^3 = 4/3 pi (4m) ^3 = 4/3 pi *64m^3 = 256/3 pi m^3. D= 256/3 pi m^3 * 3000/m^3 = 256,000 * pi kg.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q008. If we build 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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Your solution: Mass of the first one = 4g/cm^3 *6 cm^3 = 24 grams and the second mass is 2 g/cm^3 * 10 cm^3 = 20 grams. The total mass would be 24 g + 20 g = 44g. So the average density is (24g + 20 g)/(6 cm^3 +10 cm^3) = 44 g/16 cm^3 = 2.75 g/cm^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?
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Your solution: Mass of sand= 27m^3 *2100 kg/m^3 = 56700kg. Mass of cannonballs is 3 m^3 * 8000kg/m^3 = 24000 kg. Average density = (56700 kg +24000 kg)/(27m^3 + 3m^3) = 80700 kg/30m^3 = 2700 kg/m^3.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `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 what is the mass of the oil slick?
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Your solution: V= A*h = 1700000m^2 *.015 m = 25500m^3. Mass = density * volume = 860 kg/m^3 *25500 m^3 = 21,930,000 kg.
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder?
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Your solution: Surface area = 2 pi r h +2 pi r^2.
confidence rating #$&*: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere?
Your solution: Surface area is A = 4 pi r^2.
confidence rating #$&*:: 3
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Self-critique (if necessary):
Self-critique Rating: Ok
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Question: `q013. Part 1 Summary Question 3: What is the meaning of the term ‘density’?
Your solution: Density is mass over volume.
confidence rating #$&*:: 3
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Self-critique Rating: Ok
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Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume?
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Your solution: Volume = mass/ average density
confidence rating #$&*: 3
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Self-critique Rating: Ok
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Question: `q015. Part 1 Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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Self-critique (if necessary): It helped me review Density problems along with mass and volume and a little of surface area problems.
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Self-critique Rating: Ok
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#*&!
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#*&!#*&!
@& Good, but you appear to have deleted all the given solutions.
No problem here, but in the future, never delete anything from the original document. Insert your solutions, but don't delete anything that was there before.*@