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14:18:36 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> The order of operations on x-2/x+4 would dived 2 by x and then add 4 to that answer and then subtract that answer from 2. Where as the second problem (x - 2) / (x + 4) subtracts 2 from x on the top, adds 4 to x on the bottom and then divides the top by the bottom.
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14:20:27 The order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction. It makes a big difference whether you subtract the 2 from the 2 or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract: 2 - 2 / 2 + 4 = 2 - 1 + 4 (do multiplications and divisions before additions and subtractions) = 5 (add and subtract in indicated order) If there are parentheses you evaluate the grouped expressions first: (x - 2) / (x - 4) = (2 - 2) / ( 4 - 2) = 0 / 2 = 0.
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RESPONSE --> I was correct in my thinking on the order of operations. I knew that parentheses came before dividing and that dividing came before adding or subtracting.
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14:25:43 `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.
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RESPONSE --> Again the order of operations state that exponets come before adding. The expression 2 ^ x + 4 for x =2 is: 2^2 + 4 or 4 + 4 = 8 Because the exponet of 2 is calculated before teh 4 is added. The expression 2 ^ (x + 4) is evaluated: 2^ (2+4) = 2^6 = 64 Because the paranthesese are added before they are used as an exponent.
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14:27:14 2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4. 2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power. If x = 2, then 2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8. and 2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.
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RESPONSE --> Again I was correct in my thinking of order of operations in that I knew that 2^2+4 is different than 2^(2+4) because of the way the exponet was grouped.
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14:41:57 `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?
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RESPONSE --> The numerator is 3 The denominator is [ (2x-5)^2 * 3x + 1 ] When we evaluate this expression for x = 2 we get: 2 - 3/[ (2*2-5)^2 * 3*2 + 1 ] -2 +7*2 (full problem) = 2- 3/((-1)^2 * 6 +1) - 2 +14 (initial para. solved. with outside multiplication) = -3/ (1*6+1) + 14 (exponet solved along with outside addition and subtraction). = 3/7 + 14 (reduced) = 101/7~ 14.43 ( reduced to final answer) *sig digits not used.
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14:43:19 The numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3. If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way. The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ] Evaluating the expression for x = 2: - 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 = 2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses 2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses 2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term; 2 - 3 / 7 - 2 + 14 = evaluate in brackets 13 4/7 or 95/7 or about 13.57 add and subtract in order. The details of the calculation 2 - 3 / 7 - 2 + 14: Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have 2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.
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RESPONSE --> Unfortunatily I forgot, as I do commonly, to notice to negitive sign infront of the 3/7 and added this to the 98/7 wich gave me the incorrect 101\7. I see my mistake.
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14:45:51 `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.
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RESPONSE --> First I plug 4 in for all the x's Next I would solve for the Para. Then I would solve all the multiplication and division. I would then solve the exponet portion of the problem. Finally I would add and subtract the remaining number to achieve a final answer.
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14:49:02 We get (4-5)^2 * 4 - 1 + 3 / 1 - 4 = (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses = 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4 = 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get = 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4). COMMON ERROR: (4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 = -1 ^ 2*4 - 1 + 3 / 4-2 = -1 ^ 8 -1 + 3 / 4 - 2. INSTRUCTOR COMMENTS: There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication. Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4). Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.
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RESPONSE --> Again I was wrong because I multiplied the exponet and I was not suppose to. I did not fully type out each expression as I thought I was only to list the steps. I made the listed mistake.
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۶ʋzYLwΞ Student Name: assignment #002 002. Describing Graphs
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14:58:35 `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points. Now make a table for and graph the function y = 3x - 4. Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.
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RESPONSE --> x y -3 -13 -2 -10 -1 -7 0 -4 1 -1 2 2 3 5 Y intercept is at -4 X intercept is at aprox 1.33
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15:00:17 The graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this.
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RESPONSE --> I did not give the exact point coordinates, I just gave the location along the given axis where the point would be located. I should give the (x,y) coordinate where this point is located.
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15:01:43 `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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RESPONSE --> The steepness stays the same throughtout the entire graph, that meaning the slope is constant or linear.
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15:01:57 The graph forms a straight line with no change in steepness.
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RESPONSE --> I was correct in this thinking.
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15:03:04 `q003. What is the slope of the graph of the preceding two exercises (the function ia y = 3x - 4;slope is rise / run between two points of the graph)?
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RESPONSE --> The slope is derived from the 3x. We say that the slope is 3\1x or that the slope has a rise of 3 and a run of 1.
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15:04:25 Between any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.
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RESPONSE --> I did not use two points to calculate the slope, but I could have. I recal from prior math classes that the slope is the number multiplied by x in this case it is 3.
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15:10:35 `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> x y 0 0 1 1 2 4 3 9 This graph is increasing because it is climbing toward +infinity. The steapness is changing at the rate x^2, for example if x is 4 y is 16 and the there is a grater steepness than beween the points (1,1) and (2,4) and (2,4) adn (4,16). This graph is increasing at a constant rate.
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15:12:20 Graph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate.
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RESPONSE --> I was incorrect in my thinking that it was a constant rate. a constant rate would be steady, an increasing rate would be steeper as the numbers get larger. It has been along time on graphing.
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15:15:32 `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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RESPONSE --> table x y -3 9 -2 4 -1 1 0 0 This graph is decreasing as we go from right to left. The steepness changes and gets less steep as we approach 0. The graph is Decreasing at a Decreasing rate because the amount the steepness is decreasing is less than it was the moment before.
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15:16:59 From left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph is decreasing at a decreasing rate.
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RESPONSE --> I was correct in my thinking that the graph was decreasing from right to left. I was also correct in thinking the graph was decreasing at a decreasing rate. This is because as the graph approaches 0 the steepness is much less.
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15:20:59 `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> table x y 0 0 1 1 2 1.4 3 1.7 This graph is increasing from left to right. The graph gets less steep as it approache +infinity The graph is increasing at a decreasing rate because the closer to +infinity we go the smaller the steepness gets.
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15:24:42 If you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate.{}{} If the graph respresents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. {}{}If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as desribed take another look at your plot and make a note in your response indicating any difficulties.
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RESPONSE --> I misread the question about the steepness. I understand that the steepness is decreasing. I confused the steepness and the direction of change. I now understand the difference. My graph is correct.
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15:29:25 `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph is decreasing as the steepness gets less from left to right. The speepness gets less as y approaches 0. The graph is decreasing at a decreasing rate
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15:30:00 ** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. **
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RESPONSE --> I was correct in my thinking and graphing in this case.
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15:33:34 `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?
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RESPONSE --> The graph would be increasing based on the fact that as time goes on the car gets farther and farther away each seccond. This graph would be increasing at and increasing rate because the distance the car travels a seccond gets larger and larger.
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15:34:31 ** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **
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RESPONSE --> I was correct in my thinking that the car travels faster and faster per seccond. Therefore creating a graph that is increasing at an increasing rate.
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͐D䂃{y Student Name: assignment #006 006. Physics
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15:38:49 `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 --> the rate of change is 2mph per seccond so the difference between the two speeds 20mph and 30 mph which is 10mph would be divided by the 2mph/sec which would give us 10mph multiplied by 1/2mph/sec which gives us 5 seconds. Given the fact that the speed increases at 2mph/sec we would multiply the number of secconds 7 by the rate of increase 2mph/sec and add the 14 mph to the 10 initial mph which is 24 mph.
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15:39:31 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 --> I was correct not only in the answers of each problem but also in the unit conversions as well.
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15:51:49 `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 --> Based on the fact that the automobile has a greater initial speed the time would be lesss than 10 seconds to reach the lamppost. Based on the given information we can figure that the speed of the car at the lamp post was 10mph plus the amount of time 10 sec multiplied by the increase in speed of 2mph/sec is 10 mph + 20 mph or 30 mph. The seccond car would have an initial speed of 20 mph. In order for the car to have a final speed of 40 mph( the previous 30 mph at the lamppost plus the suggested 10 mph more) would have to take the same amount of time to travel between the milepost and the lamppost. Considereing the process is completed with the same distance between the milepost and the lamppost, the time taken to complete the process the second time would be less that the required 10 seconds, so the speed at the lamppost will not be 10 miles greater than before.
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15:53:03 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 --> I was correct. I just did not call the velocity change as I should have. I also noted the conditions.
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15:57:36 `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 --> We take the difference between the speed of the first automobile which is 30mph -20mph or 10mph. We then divide this 10mph by the 5sec given. We have 10mph/5sec or 2mph/sec The second car the speed difference is 90mph -40mph or 50mph we divide this amount by the given 20sec. This is 50mph/20sec or 2.5mph/sec. Based on these calculations the seccond car has a higher rate of change or speeds up faster.
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15:58:31 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 --> I was correct in my calculations which proved the second car's speed was increasing faster.
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16:08:05 4. If an automobile of mass 1200 kg is pulled by a net force of 1800 Newtons, then the number of Newtons per kg is 1800 / 1200 = 1.5. The rate at which an automobile speeds up is determined by the net number of Newtons per kg. Two teams pulling on ropes are competing to see which can most quickly accelerate their initially stationary automobile to 5 mph. One team exerts a net force of 3000 Newtons on a 1500 kg automobile while another exerts a net force of 5000 Newtons on a 2000 kg automobile. Which team will win and why? If someone pulled with a force of 500 Newtons in the opposite direction on the automobile predicted to win, would the other team then win?
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RESPONSE --> The first team has chose to exert a force of 3000 N on a 1500 kg automobile, the N/kg are 3000N/1500kg or 2N/kg. The Second team has chose to exert a force of 5000N on a 2000kg automobile. The N/kg are 5000N/2000kg or 2.5 N/kg. Because the seccond group is exerting a larger number of N/kg their car will reach the speed of 5mph quicker. In the case that someone pulled on an automoblie with 500N in the oppisit direction the opposing force would be 500N/2000kg or .25N/kg. We would subtract this force from the force calculated first so 2.5N/kg-.25N/kg= 2.25N/kg. So the second team would still win the race to 5mph.
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16:10:06 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 --> My method of calculating who would win the first part of the problem was correct. I however did more work on the second part than was required because I could have initially subtracted the Newtons and saved a whole group of calculations. However, I did arive at the correct answer.
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16:13:33 `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 am not completely sure how to solve this problem. But based on what I can remember the force of gravity and the speed at whcih the 250lb player would outdue the force of gravity and speed of the 200lb player. So the smaller player would immediately move backward.
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16:15:08 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 --> I was incorrect in my thinking. I had forgotten about momentum which is mass*speed. I now see that the second player has a greater momentum.
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16:18:46 `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 --> Again I am not sure how to solve this problem. Based on the knowledge that it would take more energe to propel a heavier person up the mountain and the difference of weight per ounce is: 16.66lb/oz for the first 15lb/oz for the second. I would say the second would make it farther up the mountain.
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16:20:34 The comparison we make here is the number of ounces of Cheerios per pound of body weight. We see that the first climber has 12 oz / (200 lb) = .06 oz / lb of weight, while the second has 10 0z / (150 lb) = .067 oz / lb. The second climber therefore has more energy per pound of body weight. It's the ounces of Cheerios that supply energy to lift the pounds of climber. The climber with the fewer pounds to lift for each ounce of energy-producing Cheerios will climb further.
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RESPONSE --> I was almost correct in my thinking. However, I reversed the calculations to come up with my answer. I calculated the lb of body weight per oz of cheerios.
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16:28:02 `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 automobie that was traveling twice as fast would take longer to stop because it is going twice the speed. The faster automobile will take about twice as long to stop because the slope of the hill is constant so the decrease in speed would be equal on both automobiles. The second faster automobile will have the greater costing velocity because it had the faster initial speed and both will end up at 0. The average velocity should be twice as great. The distance traveled by the slower vehicle should, based on the constant slope, be twice that of the slower because the speed was exactly twice that of the slower.
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16:34:31 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 --> I did not specify the neglect of air resistance but my therories were all correct except the last on distance. How is this distance calculated? Is it based on the idea that the faster car traveles the same distance of the slower car in half the time the slow car traveles. And because the faster is moving twice as fast and twice the amount of time the distance would be double in the same amount of time and 4 times in double the time?
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16:45:05 `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 --> A person 100 lb streches the cord 5 ft A person 150 lb streches the cord 9 ft A person 200 lb streches teh cord 12ft The rate of change between the 100 lb and the 150 lb is 4ft The rate of change between the 150 lb and the 200 lb is 3 ft. The So based on these calculations the rate of chang in the amound the cord stretches decreases as the weight gets hevier. Even though the distance stretched is greater. So a 125 lb person would be between 100 and 125 which because of the increase in rate of change would be less than 7 ft.
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16:47:08 From 100 lbs to 150 lbs the stretch increased by 4 feet, from 150 lbs to 200 lbs the increase was only 3 feet. Thus it appears that at least in the 100 lb - 200 lb rands each additional pound results in less increase in length than the last and that there would be more increase between 100 lb and 125 lb than between 125 lb and 150 lb. This leads to the conclusion that the stretch for 125 lb would be more than halfway from 5 ft to 9 ft, or more than 7 ft. A graph of stretch vs. weight would visually reveal the nature of the nonlinearity of this graph and would also show that the stretch at 125 lb must be more than 7 feet (the graph would be concave downward, or increasing at a decreasing rate, so the midway stretch would be higher than expected by a linear approximation).
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RESPONSE --> My thinking was correct but I could not put together all the thoughts to come up with the correct answer I realized all of the information but did not corectly analyse it. The graph is a good idea to look at this problem.
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16:51:44 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> Based on the information above when the force was doubled the skater traveled twice as far. And when the distance the force was exerted was twice that of the original the skater traveled twice as far. Because the problem states that twice the pullback = twice the force, the skater would travel twice the original distance based on a pull of double the initial length.
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16:53:09 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 --> I forgot the distance was twice as great. I see that twice the force used on twice the distance would multiply the distance by 4.
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16:57:38 `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 would appear dimmer than the first because the surface of the sphere is farther from the source of light. The seccond sphere being considered the small sphere would be more than twice as bright as the first.
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17:00:18 Both bulbs send out the same energy per second. The surface of the second bulb will indeed be dimmer than the first, as we will see below. However the same total energy per second reaches the eye (identically frosted bulbs will dissipate the same percent of the bulb energy) and from a great distance you can't tell the difference in size, so both will appear the same. The second sphere, while not as bright at its surface because it has proportionally more area, does have the extra area, and that exactly compensates for the difference in brightness. Specifically the brightness at the surface will be 1/4 as great (twice the radius implies 4 times the area which results in 1/4 the illumination at the surface) but there will be 4 times the surface area. Just as a 2' x 2' square has four times the area of a 1' x 1' square, a sphere with twice the diameter will have four times the surface area and will appear 1 / 4 as bright at its surface. Putting it another way, the second sphere distributes the intensity over four times the area, so the light on 1 square inch has only 1 / 4 the illumination.
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RESPONSE --> I think I was correct in my thinking on which bulb was the brightest from the surface, meaning the smaller one. I did not realize that they would look the same from a distance.
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17:05:55 `q011. The water in a small container is frozen in a freezer until its temperature reaches -20 Celsius. The container is then placed in a microwave oven, which proceeds to deliver energy at a constant rate of 600 Joules per second. After 10 seconds the ice is still solid and its temperature is -1 Celsius. After another 10 seconds a little bit of the cube is melted and the temperature is 0 Celsius. After another minute most of the ice is melted but there is still a good bit of ice left, and the ice and water combination is still at 0 Celsius. After another minute all the ice is melted and the temperature of the water has risen to 40 degrees Celsius. Place the following in order, from the one requiring the least energy to the one requiring the most: Increasing the temperature of the ice by 20 degrees to reach its melting point. Melting the ice at its melting point. Increasing the temperature of the water by 20 degrees after all the ice melted. At what temperature does it appear ice melts, and what is the evidence for your conclusion?
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RESPONSE --> Increasing the temperature of the water by 20 degrees. Increasing the temperature by 20 degreese to reach it's melting point. Melting the ice at it's melting point seemed to take the most energy because the temperature did not change that much whcih shows that most of the energy was used melting the ice. Ice seems to melt around 0 degrees C Because it had just started to melt when the temperature was 0 and had not melted at all at -1 degreese and at 40 degreese it was completely water.
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17:07:33 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 --> I was almost correct. I guessed that the water heated quicker because we were not told when in that minute the Ice was completely water. we were told that in the first minute the Ice changed from -40 degrees to -1 degrees.
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17:10:59 `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 --> At the center point I would expect to bob 12 inches because the distance of the two would combine. I would expect to move abot 1/4 the distance of the pool because one wave would be 3/4 as high as the valley.
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17:12:00 If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in. If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.
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RESPONSE --> I was correct in my first thoughts, but the second were not even close.
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