course Phy 242 I now I kind of started going down hill towards the end but I blame the lateness of the night. ¸§õûí’î§âÿÜÍÄØŸá~ú‚ц»ôÁˆÛ§Student Name:
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22:38:48 `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4).
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RESPONSE --> In x-2/x+4, a user would think that the two would need to be divided by six. However, in (x-2)/(x+4), the user would understand the the unit (x-2) needs to be divided by the unit (x+4).
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22:39:43 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 --> Understood
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22:43:55 `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 --> In 2 ^ x + 4, the user would believe that only the x is an exponent. In the equation 2^(x+4) the whole unit (x+4) is the exponent. The answer to 2^x+4 where x = 2 is 8, whereas, in 2^(x+4) the answer is 64.
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22:44:07 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 --> Understood
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22:48:42 `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]. The answer when x = 2 is (95/7).
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22:48:54 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 --> Understood
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22:49:01 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 -->
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22:52:59 `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 in the 4 into (x-5)^2 which equals 1. Then I take and plug the 4 into (x-1) which equals 3. Next I take and plug 4 into 3/x which = 3/4. Then I plug these small problems into the larger problem. 1*3 + (3/4) - 2 3 + (3/4) - 2 1 + (3/4) (4/4) + (3/4) = (7/4)
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22:53:23 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 --> Understood
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˜§„ÍyÜlÔ¯ôÉ°ËçØ|ãå|°ç Student Name: assignment #002 002. Describing Graphs
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22:59:41 `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 --> y=2x+7 X Y -3 1 -2 3 -1 5 0 7 1 9 2 11 3 13 The slope of the graph is 2 which means you go up 2 and over 1. The point at which the line of the graph crosses the y-axis is (0,7). This is know because the y-intercept is 7 and this is know in the equation of the line y=2x+7.
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23:00:19 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 --> Understood. I forgot to mention the x-intercept.
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23:03:04 `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 --> No. The equation is a linear equation which a set slope of 3. This means that the slope constantly increase at a rate of 3 up and 1 over.
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23:03:28 The graph forms a straight line with no change in steepness.
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RESPONSE --> Understood.
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23:04:19 `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 3 which is 3 up and 1 over or (3/1).
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23:05:31 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 --> Understood.
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23:12:35 `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 --> y=x^2 X Y 0 0 1 1 2 4 3 9 The x intercept is (0,0) as is the y intercept. The graph is increasing. The steepness of the graph does change. It is increasing at an increasing rate. For the points from x = 0 to x = 3 the slope of this graph is increasing at an increasing rate.
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23:12:58 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 --> Understood.
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23:15:56 `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 --> y = x^2 X Y -3 9 -2 4 -1 1 0 0 The graph going from x = -3 to x= 0 is decreasing. The steepness of the graph is changing and becoming less steep as it goes from X=-3 to x=0. The graph is decreasing at a decreasing rate because the slope of the graph as it travels from x= -3 to x = 0 is decreasing.
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23:16:16 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 --> Understood.
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23:20:03 `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 --> y = 'sqrt(x) x y 0 0 1 1 2 1.414 3 1.732 The graph is increasing from x = 0 to x = 3. The steepness does change because as x increases the graph is leveling off. The graph is increasing at a decreasing rate.
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23:20:20 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 --> Understood.
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23:24:03 `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 --> y = 5*2^(-x). x y 0 5 1 2.5 2 1.25 3 0.625 The graph is decreasing. The steepness is decreasing as well. The graph is decreasing at a decreasing rate.
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23:24:31 ** 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 --> Understood.
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23:27:00 `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 because the y value would increase as the distance does and the x value will increase at time goes by. Since the car is accelerating this indicates that the slope of the graph would increase as well. This means that the graph would be increasing at an increasing rate.
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23:27:26 ** 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 --> Understood.
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¥p¤V¿ºÌÞ_ʺ㶽ÖÆŽ¶ÓôÇìì¿ Student Name: assignment #006 006. Physics
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23:32:30 `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 --> Vf = Vi + a *t 30 mph = 20 mph + 2mph/s * t 2 mph/s * t = 10 mph t = 5 seconds Vf = 10mph + 2mph/s * 7s Vf = 10mph + 14 mph Vf= 24 mph
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23:32:50 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 --> Understood
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23:37:30 `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 --> Since the speed of the vehicle is double that of the vehicle in the first run the time required to reach the lamppost will be less than 10 seconds given that the acceleration remains constant. Yes, since acceleration is constant and the initial speed is different than the second speed then the final speed will be at least 10mph greater than the final speed from the first trial.
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23:39:02 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 understand that I was incorrect in my asumption of the same increase in speed because I neglected to realize that the time required is different despite the acceleration being constant.
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23:41:44 `q003. The following example shows how we can measure the rate at which an automobile speeds up: If an automobile speeds up from 30 mph to 50 mph as the second hand of a watch moves from the 12-second position to the 16-second position, and its speed changes by 20 mph in 4 seconds. This gives us an average rate of velocity change equal to 20 mph / 4 seconds = 5 mph / second. We wish to compare the rates at which two different automobiles increase their speed: Which automobile speeds up at the greater rate, one which speeds up from 20 mph to 30 mph in five seconds or one which speeds up from 40 mph to 90 mph in 20 seconds?
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RESPONSE --> The first vehicle accelerates at 10mph/5 seconds which = 2 mph/s The second vehicle acclerates at 50mph/20 seconds which = 2.5 mph/s. This means that the second car accelerates at a greater rate then the first car.
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23:41:59 The first automobile's speed changes from 20 mph to 30mph, a 10 mph difference, which occurs in 5 seconds. So the rate of chage in 10 mph / (5 sec) = 2 mph / sec. = rate of change of 2 mph per second. }{The second automobile's speed changes from 40 mph to 90 mph, a 50 mph difference in 20 seconds so the rate of change is 50 mph / (20 sec) = 2.5 mph per second. Therefore, the second auto is increasing its velocity ar a rate which is .5 mph / second greater than that of the first.
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RESPONSE --> Understood.
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23:45:26 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 second team will win because they are pulling at a rate of 2.5 Newtons per kilogram whereas the first team is pulling at a rate of 2 Newtons per kilogram. If someone pulled on car two with a force of 500 Newtons in the opposite direction it would reduce the rate to 2.25 N/kg which is still enough to let team two still win.
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23:45:43 The first team's rate is 3000 Newtons divided by 1500 kg or 2 Newtons per kg, while the second team's rate is 5000 Newtons divided by 2000 kg or 2.5 Newtons per kg. The second team therefore increases velocity more quickly. Since both start at the same velocity, zero, the second team will immediately go ahead and will stay ahead. The second team would still win even if the first team was hampered by the 500 Newton resistance, because 5000 Newtons - 500 Newtons = 4500 Newtons of force divided by 2000 kg of car gives 2.25 Newtons per kg, still more than the 2 Newtons / kg of the first team
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RESPONSE --> ok
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23:48:07 `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 --> The second player weighing 200 lbs and moving at 20 ft/s would be moving backward immediately after the collision because he is moving at 10 ft*lb/s whereas the first football player is moving at 25ft*lb/s.
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23:49:29 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 understand that I divided instead of multiplying though I had the units correct. This change in math also changes the outcome of the answer giving player two greater momentum than player one.
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23:51:32 `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 --> The climber weighing 150 lbs would have more energy per oz of Cheerios at 15 Lb/oz whereas the other climber has a weight to food ratio of 16.67lb/oz.
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23:52:59 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 --> In this case I got the answer correct even though I divided incorrectly. The answer should be in oz/lbs and not lbs/oz.
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23:56:37 `q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop. Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long? Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great? Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?
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RESPONSE --> The automobile with the greater speed will take longer to stop. Due to wind resistance and friction I believe that the faster car would take less than twice as long to slow down. The faster car will have the greater coasting velocity but it will be by less than twice as great. The distance traveled by the faster automobile will be greater than twice that of the slower car.
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23:57:39 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 understand that the distance is wrong .
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23:59:47 `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 --> The rope would be less than 7 ft because the elasticity reaches a maximum at a slower rate at the weight increases.
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00:00:30 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 --> Understood.
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00:05:00 `q009. When given a push of 10 pounds, with the push maintained through a distance of 4 feet, a certain ice skater can coast without further effort across level ice for a distance of 30 feet. When given a push of 20 pounds (double the previous push) through the same distance, the skater will be able to coast twice as far, a distance of 60 feet. When given a push of 10 pounds for a distance of 8 feet (twice the previous distance) the skater will again coast a distance of 60 feet. The same skater is now accelerated by a sort of a slingshot consisting of a bungee-type cord slung between two posts in the ice. The cord, as one might expect, exerts greater and greater force as it is pulled back further and further. Assume that the force increases in direct proportion to pullback (ie.g., twice the pullback implies twice the force). When the skater is pulled back 4 feet and released, she travels 20 feet. When she is pulled back 8 feet and released, will she be expected to travel twice as far, more than twice as far or less than twice as far as when she was pulled back 4 feet?
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RESPONSE --> The skater should be expected to travel twice the distance as before since the rope is pulled back twice as far and this means that the force is twice as great.
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00:05:57 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 understand that the distance is doubled because the initial force is doubled.
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00:09:28 `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 sphere with the larger diameter would appear dimmer since the distance that the light has to travel from the bulb is greater. I would expect the first bulb to seem twice as bright the second bulb.
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00:10:50 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 understand that I forgot to realize a 4 x increase in surface area.
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00:21:47 `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 --> Change from -20 Celsius to -1 Celsius Change from 0 Celsius to 40 Celsius Change from -1 Celsius to 0 Celsius Change from 0 Celsius to 0 Celsius Ice melts at 0 Celsius because that is the temperature that it started at and became water at.
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00:22:37 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 --> Understood
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00:29:05 `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 --> A person would bob up and down around 3 inches. One would need to move 6 inches.
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00:29:53 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 --> Understood.
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