Assignment 2

course Phy 201

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assignment #002

002. Describing Graphs

qa initial problems

05-13-2007

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assignment #002

002. Describing Graphs

qa initial problems

05-13-2007

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17:27:00

`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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RESPONSE -->

The y-intercept is -4. The x-intercept is 4/3. I get this, the easiest way, by plugging in 0 for x and y, respectively, to get the intercepts of the equation (after solving for the remaining variable).

confidence assessment: 3

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17:27:18

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 understand this.

self critique assessment: 3

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17:27:59

`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 constant slope is 3, so the slope/steepness never changes.

confidence assessment: 3

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17:28:09

The graph forms a straight line with no change in steepness.

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RESPONSE -->

I understand this.

self critique assessment: 3

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17:28:35

`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 -->

3 (from the equation y=mx+b, me being the slope constant).

confidence assessment: 3

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17:30:03

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 understand this.

self critique assessment: 3

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17:32:10

`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 -->

The function is merely increasing at an increasing rate, not an increasing constant rate, because the function does not represent a straight line with equal changes (equal slope) between points.

confidence assessment: 2

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17:33:04

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

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

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RESPONSE -->

Okay, I understand how I messed up... one must compare the y values and see if they are increasing (or decreasing, for that matter) in a cohesive sort of pattern.

self critique assessment: 2

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17:35:35

`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 -->

The graph is obviously decreasing. It is steeper between -2 and -3 (y points) than between 0 and -1 and -1 and -2. The graph, I believe, is decreasing at a constant rate, because the numbers of the y form a pattern (difference of 1, 3, 5, etc.).

confidence assessment: 2

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17:39:10

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 -->

Okay. I thought I understood this last time, but I actually do now. For a constant rate, the changes in y must be equal. For decreasing rates they must continue to be getting smaller (""They"" being the difference between y values).

self critique assessment: 2

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17:48:57

`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 -->

The graph is increasing at an increasing rate.

The difference between 0 and 1 y values is less than the difference between the next set of y values.

confidence assessment: 3

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17:52:01

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 guess I am still having a problem understanding ""steepness."" I suppose that ""steepness"" means to what extent the slope is changing--severely verses hardly at all.

As I look at my graph, I see that the SLOPE is positive (increasing), but that the slope, although still positive, begins becoming less steep as the x values get larger. I suppose this makes sense--for some reason the word ""steepness"" was throwing me off.

I feel more confident now (I think!).

Imagine that the graph represents a hill you are climbing. Most people would rather be on the hill described by the square root function, which gets less and less steep and easier to climb, that the y = x^2 hill, which keeps getting steeper and steeper without any limit to its steepness.

self critique assessment: 2

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21:22:03

`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 -->

The graph is decreasing (the slope is negative) and the steepness is decreasing, because the differences in the y values is changing at a decreasing rate (the graph is becoming less steep as the x values increase from 0 to 3). So, the graph is decreasing at a decreasing rate.

confidence assessment: 1

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21:22:25

** 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 understand now.

self critique assessment: 3

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22:31:01

`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 at increasing rates. My rationalization for thinking this is that the rate (from the equation d=rt) is positive, meaning that the equation in general is increasing (the car is moving faster and faster, meaning it is going positive). But, the idea of ""faster and faster"" also indicates that the speed is becoming steeper and steeper as the car drives on.

confidence assessment: 3

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22:31:18

** 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 -->

Yay, got it!

self critique assessment: 3

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assignment #006

006. Physics

qa initial problems

05-13-2007

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22:38:39

`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 -->

It will take 5 seconds to change from 20 mph to 30.

Considering that a car cannot ""go"" negative speed, the mph has to be moving in a positive direction. So, 7 (seconds) x 2 (mph) = 14. 10 (initial) + 14 = 24. So the answer to the second question is 24.

confidence assessment: 3

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22:39:30

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 understand.

self critique assessment: 3

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22:48:54

`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 automobile starts the second time going 20 mph (twice the speed the first time), it obviously will take less time than it did the first time for the car to reach the lampost, meaning that it should take less than 10 seconds for the automobile to reach the lampost the second time (traveling 20 mph starting from the milepost).

Because the speed of the car the second time makes the elapsed seconds between point A and point B less, then one cannot assume that 10 seconds is still the amount of time that it takes the car (starting at 20 mph) to reach the lamppost. Because one cannot assume the time is the same, one, subsequently, cannot guess how fast the car will be going upon reaching the lamppost the second time accurately, because the 2 mph per second increase can only be utilized if one knows for certain the amount of time that a car (or any object) has to get up to speed. So the answer to the second part, simply, is no.

confidence assessment: 2

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22:49:23

If it starts coasting down the same section of road at 20 mph, and if velocity changes by the same amount every second, the automobile should always be traveling faster than if it started at 10 mph, and would therefore take less than 10 seconds.

The conditions here specify equal distances, which implies less time on the second run. The key is that, as observed above, the automobile has less than 10 seconds to increase its speed. Since its speed is changing at the same rate as before and it has less time to change it will therefore change by less.

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RESPONSE -->

I understand this.

self critique assessment: 3

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22:56:28

`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 -->

1. 10 mph/ 5 seconds = 2mph/sec.

2. 50mph/ 20 seconds = 2.5 mph/sec.

So, the first one (increasing 10 mph in 5 seconds) increases speed at the fastest rate.

confidence assessment: 3

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22:57:24

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 -->

Oops! I made a stupid mistake. I figured out the velocities correctly, but simply filled in the wrong numbers in the last part. Obviously 2.5 is greater than 2, meaning it speeds faster. I totally understand my mistake, and it was pure ""in too much of a hurry.""

self critique assessment: 2

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23:06:02

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 team exerting 3000 Newtons on a 2000 kg car would win, b/c their force comes out to be 2.5, whereas the first teams is only 2.

Even after subtracting 500 Newtons from the initial 5000 that the first team has, 4500/2000 still yields 2.25, indicating that the second team, even with the added disadvantage, would still win with 2.25 beating the other team's force of only 2.

self critique assessment: 3

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23:06:24

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 -->

I understand this completely--yay!

self critique assessment: 3

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23:16:58

`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 -->

For the first man, dividing 250 by 10 one gets 25 number force, whereas the second man, 200/20, only has 10 force. So, I assume, from this calculation, that the second man with only 10 force will be moving backward; however, I do not know if that is the way one is supposed to go about solving this equation. I am just guessing from common sense that it is.

confidence assessment: 1

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23:18:30

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 -->

Okay, here I simply did not know the equation. Had I know it, I feel very confident that I would have been able to answer the question effectively. I made a note of the equation (mass x speed = momentum) in my notebook.

self critique assessment: 2

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23:23:15

`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 -->

Common sense leads me to believe that the 150 lb. climber would have the advantage. Again, though, I am not sure what equation to use here.

Perhaps I should divide 200/12=16.7 and 150/10 =15

Obviously, I am trying to find energy yield as the outcome, and I doubt this simple division will give it to me; regardless, I do think the 150 lb. climber has the advantage, all formulas to the wayside.

confidence assessment: 1

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23:25:32

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 -->

Although my answer was right, I do understand now the ""proper"" way of doing the comparison when mere common sense fails.

self critique assessment:

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23:35:34

`q007. Two automobiles are traveling up a long hill with an steepness that doesn't change until the top, which is very far away, is reached. One automobile is moving twice as fast as the other. At the instant the faster automobile overtakes the slower their drivers both take them out of gear and they coast until they stop.

Which automobile will take longer to come to a stop? Will that automobile require about twice as long to stop, more than twice as long or less than twice as long?

Which automobile will have the greater average coasting velocity? Will its average coasting velocity by twice as great as the other, more than twice as great or less than twice as great?

Will the distance traveled by the faster automobile be equal to that of the slower, twice that of the slower or more than twice that of the slower?

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RESPONSE -->

The second automobile, the one with the most speed, will take the longest time to stop. If we use the equation massxspeed=momentum, we see that mass is irrelevant (we are to assume that the cars are of same mass).

So, speed = momentum, and if the speed of the second car is 2x (x being the slower car), then the momentum is also going to be twice, indicating that it will take about twice as long for the second car to stop as the first.

confidence assessment: 2

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23:37:59

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 -->

Wow--I gave the simple answer. Unfortunately, I understand that answer is a ""perfect world"" answer, but I assume we will be taught how to incorporate other factors later. I am intimidated!

self critique assessment: 2

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23:59:37

`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 cord, I think, will stretch a little more than 7. I came to this conclusion by comparing the proportions of each of the three pairs, and I found that the cord stretches less percetage wise with more weight (i.e., it is decreasing in ""steepness""--from 5-9, a difference of four, and then from 9-12, a difference of three).

confidence assessment: 2

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00:00:01

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 -->

I understand.

self critique assessment: 3

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00:12:41

`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 able to travel twice as far given that she is pushed with exactly twice the force. As indicated in the first part, a direct correlation exists between the two.

confidence assessment: 2

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00:16:18

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 think I understand. In the second, the distance and the lbs. exerted are combined into one number--the slingshot number. So, the force exerted would double and be 8 AND the distance is doubled to be 8, both of which we learn in the first example double the distane outcome. So, I think this is how four times is the result. Is this right?

self critique assessment: 2

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00:24:07

`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 -->

At a distance, the two will probably appear the same.

For a moth on the surface, the larger will probably appear to have a brighter light, but not twice as bright.

confidence assessment: 0

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00:30:07

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 the first part; the second, however, I am going to try to explain in my own words.

Because the area of the second light is four times greater, the bulb (the same strength in both lights) is distributed over a much larger area, making it much more difficult for the light to be seen (because the strength of it is dissipated over more area).

I think this is pretty clear now, the concept at least, but I am not sure if I would be able to find out on mine how exactly HOW diluted one lightbult is over an area compared to another light bult.

self critique assessment: 2

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00:47:39

`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 -->

Least to most energy:

1. Increasing the temperature of the ice by 20 degrees to reach its melting point.

2. Increasing the temperature of the water by 20 degrees after all the ice melted.

3. Melting the ice at its melting point.

I simply wrote out the time-line and compared the parts to see which appeared to take the longest.

It appears that ice begins melting at 0 degrees Celcius, because the water was frozen at -1 and began to melt at 0.

confidence assessment: 2

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00:48:37

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 -->

I understand this, only because so much information was given in the problem though. Otherwise, I would have been unable to answer the question.

self critique assessment: 3

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00:51:59

`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 -->

Wow. I have absolutely no idea how to answer this question. I am going to go ahead and look at the answer and try to decipher.

confidence assessment: 0

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01:04:21

If the two 6-inch peaks meet and reinforce one another completely, the height of the 'combined' peak will be 6 in + 6 in = 12 in.

If for example you move 3 ft closer to one end you move 3 ft further from the other and peaks, which are 6 ft apart, will still be meeting peaks. [ Think of it this way: If you move 3 ft closer to one end you move 3 ft further from the other. This shifts your relative position to the two waves by 6 feet (3 feet closer to the one you're moving toward, 3 feet further from the other). So if you were meeting peaks at the original position, someone at your new position would at the same time be meeting valleys, with two peaks closing in from opposite directions. A short time later the two peaks would meet at that point. ]

However if you move 1.5 ft the net 'shift' will be 3 ft and peaks will be meeting valleys so you will be in the calmest water.

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RESPONSE -->

On the first part, I understand that you merely add the two peaks together to see how high the combination of the two would be.

The other part, basically, however far you move you must double the value, because you are moving that far away from one way creator and that much closer to another. If the number that originates from doubling the distance that you move is even (since there are two sets of waves), you will still be in ""stormy"" water; however, if the number is odd, then you will be caught primarily where the troughs and peaks meets, creating calm water in that area.

self critique assessment: 2

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Your work looks very good. You are reasoning very well and expressing your reasoning clearly and insightfully.

Let me know if you have questions.