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12:55:48 `q001. There are 9 questions and 4 summary questions in this assignment.
What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?......!!!!!!!!...................................
RESPONSE --> V = lwh = 3cm * 5cm * 7cm Volume of a rectangular solid = 105cm^3
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12:57:07 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2.
Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.......!!!!!!!!...................................
RESPONSE --> ok.
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12:58:02 `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?
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RESPONSE --> V = A * h = 48 * 2 = 96m^3
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13:01:06 Using the idea that V = A * h we find that the volume of this solid is
V = A * h = 48 m^2 * 2 m = 96 m^3. Note that m * m^2 means m * (m * m) = m * m * m = m^2.......!!!!!!!!...................................
RESPONSE --> Do you mean to say that m*m*m = m^3, not m^2?
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13:09:21 `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?
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RESPONSE --> V = pi r^2 l is the equation for the volume of a cylinder. Since I don't have the radius, I used A = pi r^2 to solve for r. 20 = pi r^2 = 20/pi = r^2 = sqrt(20/pi) = r r = 2.523m
So the volume should be V = pi 2.523^2 * 40 = 317m^3 Not sure if I went about this the right way or if I could just multiply the base area and altitude as done before with the rectangular solid....................................................
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13:10:54 V = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that
V = A * h = 20 m^2 * 40 m = 800 m^3. The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.......!!!!!!!!...................................
RESPONSE --> Ok I should have gone with my initial idea of using the same V = Ah formula as before. Can't say my attempt wasn't creative but I can say it was blatantly wrong.
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13:14:32 `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?
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RESPONSE --> Now I can use the previous formula. V = pi r^2*l = pi 5^2 * 30 = 2356.2
I just realized that on my previous answer, I actually had the right concept with figuring out the radius and using this V = pi r^2*l formula, but that when I did the calculations, I forgot to square the radius. If I had, my answer would be 799.9..................................................
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13:17:12 The cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies.
The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2. Since the altitude is 30 cm the volume is therefore V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3. Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.......!!!!!!!!...................................
RESPONSE --> Ok. I used my calculater and multiplied pi into the equation as opposed to just leaving the answer at 750 pi cm^3. Is that ok or from now on should I now include pi in the calculations.
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13:26:16 `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?
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RESPONSE --> I measured a height of 12.8cm and a diameter of 8cm. The area is A = pi r^2 = 50.3cm^2 The volume is V = A h V = 50.3 * 12.8 = 643cm^3
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13:27:38 People will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using.
A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3. Approximating, this comes out to around 35 in^3. Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^2.......!!!!!!!!...................................
RESPONSE --> ok.
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13:30:57 `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?
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RESPONSE --> I'm pretty sure the notes before mentioned this V = Ah equation working for uniform prisms also, so V = 50cm^2 * 60cm = 3000cm^3
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13:34:24 We can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box.
So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.......!!!!!!!!...................................
RESPONSE --> As usual, I tried to decide which equation to use (either the V=Ah or V=1/3 pi r^2 h) and I chose the wrong one.
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13:35:21 `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?
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RESPONSE --> V = 1/3Ah V = 1/3 * 20m^2 * 9m = 60m^3
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13:36:08 Just as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone.
In this case the base area and altitude are given, so the volume of the cone is V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.......!!!!!!!!...................................
RESPONSE --> ok
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13:38:35 `q008. What is a volume of a sphere whose radius is 4 meters?
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RESPONSE --> V = 4/3 pi r^3 is the equation for the volume of a sphere. At a radius of 4m, the volume = 4/3 pi 4^3. = 268m^3 or 85.3 pi m^3.
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13:39:14 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so
V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.......!!!!!!!!...................................
RESPONSE --> ok
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13:40:38 `q009. What is the volume of a planet whose diameter is 14,000 km?
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RESPONSE --> V = 4/3 pi r^3 V = 4/3 pi 14000km^3 = 18666 pi km^3 or 58643km^3.
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13:42:07 The planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is
V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3. This result can be approximated to an appropriate number of significant figures.......!!!!!!!!...................................
RESPONSE --> Oh I forgot to find the radius. I screwed up and used the diameter instead. Come to think of it, I forgot to cube the number also. I think I need a break.
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13:44:08 `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?
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RESPONSE --> The basic principle is that we can take the Area of the object and multiply it by the height or altitude to get the volume. So in the case of a cylinder, The area is A = pi r^2, so the V = A h.
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13:44:18 The principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.
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RESPONSE --> ok
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13:45:57 `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?
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RESPONSE --> A pyramid covers 1/3 the area of a rectangle, so to get the volume, we multiply 1/3 A h, where A is the area (pi r^2) and h is the altitude.
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13:46:05 The volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.
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RESPONSE --> ok
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13:46:57 `q012. Summary Question 3: What is the formula for the volume of a sphere?
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RESPONSE --> V = 4/3 pi r^3 is the formula for the voume of a sphere.
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13:47:03 The volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.
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RESPONSE --> ok
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13:49:20 `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> I mainly learned that I really need to pay more attention to my calculations and to make sure I remember to square or cube everything in order to avoid simple mistakes. Also knowing that the Volume of the objects is just Area * height helps reinforce the previous exercise on areas because all the concepts have to be used in order to find the volume.
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13:49:25 This ends the second assignment.
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RESPONSE --> ok"
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13:50:05 ** Questions about velocity, average velocity, acceleration, etc. are very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.
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RESPONSE --> ok
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13:50:27 It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.
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RESPONSE --> ok
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13:50:34 That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **
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RESPONSE --> ok
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14:08:29 Query You should write up Video Experiment 2 as directed and submit it. What were your final results? Give your slopes and your rates of velocity change as rate vs. slope ordered pairs according to the y vs. x convention (slope first, rate second), and specify the slope of the straight line you got at the end.
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RESPONSE --> I've submitted Video Exp. #2 but had a problem with my average rate of change of velocity calculations. I've tried to figure out what to use to get the average rate, but I'm not sure if I do the change in distance / change in time, and if so, is that that final vel. - vel / time or is it just the slope or what? I looked at the Idea 1: Rates notes but it seems like any of them could apply.
The slopes I have are: Slope (cm) 0.028636364 0.046363636 0.038409091 0.021590909 As for the rates of vel. change, I still have no idea. 55.73226626 (cm) was the slope of the line but it's not right since my rates aren't correct..................................................
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14:15:59 ** Common Error: A common error is to divide ave velolcity by the time required, which does not give acceleration. If you did these calculations correctly but didn't justify them on the experiment, justify your calculations now (just show one calculation in detail). If you did the calculations incorrectly then revise your calculations and give the same sort of justification.
Note that only a Summary is required in the Query: This query just calls for a few summary items; if the experiment has been submitted and returned at OK there's no need to go into great detail in the Query. **......!!!!!!!!...................................
RESPONSE --> The book says that acceleration is the change of velocity / time elapsed... so I tried Velocity / time and got Time (sec) Velocity (cm/s) Acceleration 2.097617696 20.97617696 10 2.039607805 21.57277487 10.57692308 2.078460969 21.16950987 10.18518519 3.186220331 13.80946558 4.334121355
I don't know if that's right either but I'm not sure what else I should try..................................................
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14:17:06 Explain briefly how you determined your accelerations.
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RESPONSE --> I tried dividing the velocity by the time it took the ball to roll down the ramp. I don't think it's right though.
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14:27:07 **To determine acceleration we first find the average velocity by dividing displacement by elapsed time: vAve = `ds / `dt.
Then since acceleration is uniform and init velocity is zero we can double the average velocity to get the final velocity. Finally we find acceleration, which is rate of velocity change. The rate of velocity change is (change in velocity) / (change in clock time). Change in velocity is the change from initial to final velocity: `dv = vf - v0. Change in clock time is the time required to accelerate down the ramp, which is what you observed. **......!!!!!!!!...................................
RESPONSE --> I thought about subtracting the final velocity and the velocity, but since we doubled the velocity to get the final in the first place, that leaves us right back at the original velocity / clock time. Which is what I did. So I'm assuming I'm right and I'll regraph that now and see if I can figure out the slopes.
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14:28:14 What were the units of the slope of your straight line? Hint: The slope of a ramp has no units; what were the units of the rise between two points? (you won't be timed on this one, so don't worry about taking a few minutes if you need it).
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RESPONSE --> I think the units for the rise between the points would be the same units as the acceleration since it's on the y axis... so cm/s/s...
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14:28:38 ** The rise measures change in acceleration in cm/s while the run measures change in clock time in s, so the units of the graph of acceleration vs. clock time are (cm/s) / s = cm/s^2. **
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RESPONSE --> ok
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14:29:49 How is acceleration an example of a rate of change?
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RESPONSE --> It measures the rate of change in velocity.
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14:32:29 ** Velocity is the rate of change of position. Acceleration is rate of change of velocity--change in velocity divided by the time period.
To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in time. The average rate of change of velocity with respect to clock time is the same as the acceleration **......!!!!!!!!...................................
RESPONSE --> ok
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14:33:52 If you know average acceleration and time interval what can you find?
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RESPONSE --> I think you can get the change in velocity if you know the avg. acceleration and the time interval.
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14:34:19 ** Accel = change in vel / change in clock time, so if you know accel and time interval (i.e., change in clock time) you can find change in vel = accel * change in clock time.
In this case you don't know anything about how fast the object is traveling. You can only find the change in its velocity. COMMON ERROR (and response): Average acceleration is the average velocity divided by the time (for the change in the average velocity)so you would be able to find the average velocity by multiplying the average acceleration by the change in time. INSTRUCTOR RESPONSE: Acceleration is rate of change of velocity--change in velocity divided by the change in clock time. It is not average velocity / change in clock time. COUNTEREXAMPLE TO COMMON ERROR: Moving at a constant 60 mph for 3 hours, there is no change in velocity so acceleration = rate of change of velocity is zero. However average velocity / change in clock time = 60 mph / (3 hr) = 20 mile / hr^2, which is not zero. This shows that acceleration is not ave vel / change in clock time. COMMON ERROR and response: You can find displacement INSTRUCTOR RESPONSE: From average velocity and time interval you can find displacement. However from average acceleration and time interval you can find only change in velocity. Acceleration is the rate at which velocity changes so average acceleration is change in velocity/change in clock time. From this it follows that change in velocity = acceleration*change in clock time. **......!!!!!!!!...................................
RESPONSE --> ok
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14:34:49 Can you find velocity from average acceleration and time interval?
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RESPONSE --> No I think with velocity you need the distance traveled and the time interval, not the avg. acceleration.
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14:36:16 ** Ave accel = change in vel / change in clock time. If acceleration is constant, then this relationship becomes acceleration = change in velocity/change in clock time.
Change in clock time is the time interval, so if we know time interval and acceleration we can find change in velocity = acceleration * change in clock time = acceleration * change in clock time. We cannot find velocity, only change in velocity. We would need additional information (e.g., initial velocity, average velocity or final velocity) to find an actual velocity. For example if we know that the velocity of a car is changing at 2 (mi/hr) / sec then we know that in 5 seconds the speed will change by 2 (mi/hr)/s * 5 s = 10 mi/hr. But we don't know how fast the car is going in the first place, so we have no information about its actual velocity. If this car had originally been going 20 mi/hr, it would have ended up at 30 miles/hr. If it had originally been going 70 mi/hr, it would have ended up at 80 miles/hr. Similarly if an object is accelerating at 30 m/s^2 (i.e., 30 (m/s) / s) for eight seconds, its velocity will change by 30 meters/second^2 * 8 seconds = 240 m/s. Again we don't know what the actual velocity will be because we don't know what velocity the object was originally moving. ANOTHER SOLUTION: The answer is 'No'. You can divide `ds (change in position) by `dt (change in clock time) to get vAve = `ds / `dt. Or you can divide `dv (change in vel) by `dt to get aAve. So from aAve and `dt you can get `dv, the change in v. But you can't get v itself. EXAMPLE: You can find the change in a quantity from a rate and a time interval, but you can't find the actual value of the quantity. For example, accelerating for 2 sec at 3 mph / sec, your velocity changes by 6 mph, but that's all you know. You don't know how fast you were going in the first place. Could be from 5 mph to 12 mph, or 200 mph to 206 mph (hopefully not down the Interstate). COMMON ERROR: Yes. Final velocity is average velocity multiplied by 2. INSTRUCTOR RESPONSE: We aren't given ave velocity and time interval, we're give ave accel and time interval, so this answer is not valid. Note also that final velocity is average velocity multiplied by 2 ONLY when init vel is zero. Be sure you always state it this way. ANOTHER EXAMPLE: You can't find velocity from ave accel and time interval--you can only find change in velocity from this information. For example a velocity change of 10 mph would result from ave accel 2 m/s^2 for 2 seconds; this change could be between 10 and 20 mph or between 180 and 190 mph, and if all we know is ave accel and time interval we couldn't tell the difference. ONE MORE RESPONSE: You can find the change in velocity. The actual velocity cannot be found from ave accel and time interval. For example you would get the same result for acceleration if a car went from 10 mph to 20 mph in 5 sec as you would if it went from 200 mph to 220 mph in 10 sec. **......!!!!!!!!...................................
RESPONSE --> ok
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14:36:57 Can you find change in velocity from average acceleration and time interval?
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RESPONSE --> Yes. As mentioned before, just multiply avg. acceleration * time interval and you get the change in velocity.
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14:37:18 **Good student response:
Yes, the answer that I provided previously is wrong, I didn't consider the 'change in velocity' I only considered the velocity as being the same as the change in velocity and that was not correct. Change in velocity is average accel * `dt. CALCULUS-RELATED ANSWER WITH INSTRUCTOR NOTE(relevant mostly to University Physics students) Yes, you take the integral with respect to time INSTRUCTOR NOTE: That's essentially what you're doing if you multiply average acceleration by time interval. In calculus terms the reason you can't get actual velocity from acceleration information alone is that when you integrate acceleration you get an arbitrary integration constant. You don't have any information in those questions to evaluate c. **......!!!!!!!!...................................
RESPONSE --> ok
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14:38:25 Can you find average velocity from average acceleration and time interval?
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RESPONSE --> No. I think you can only find the velocity. To get the average velocity, you need the change in distance, not the avg. acceleration.
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14:41:59 ** CORRECT STATEMENT BUT NOT AN ANSWER TO THIS QUESTION:
The average acceleration would be multiplied by the time interval to find the change in the velocity INSTRUCTOR RESPONSE: Your statement is correct, but as you say you can find change in vel, which is not the same thing as ave vel. You cannot find ave vel. from just accel and time interval. There is for example nothing in accel and time interval that tells you how fast the object was going initially. The same acceleration and time interval could apply as well to an object starting from rest as to an object starting at 100 m/s; the average velocity would not be the same in both cases. So accel and time interval cannot determine average velocity. CALCULUS-RELATED ERRONEOUS ANSWER AND INSTRUCTOR CLAIRIFICATION(relevant mostly to University Physics students: Yes, you take the integral and the limits of integration at the time intervals CLARIFICATION BY INSTRUCTOR: A definite integral of acceleration with respect to t gives you only the change in v, not v itself. You need an initial condition to evaluate the integration constant in the indefinite integral. To find the average velocity you would have to integrate velocity (definite integral over the time interval) and divide by the time interval. **......!!!!!!!!...................................
RESPONSE --> I thought my statement did answer the question. I said that you couldn't find the avg. velocity with the avg. acceleration and time interval, they only give you the change in velocity...
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14:44:28 You can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval?
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RESPONSE --> In order to find the average velocity, you need to know the difference or change in the velocity / time interval. Using acceleration and time interval only gives you the change in velocity, not the average.
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14:44:41 ** Average velocity is change in position/change in clock time. Average velocity has no direct relationship with acceleration.
CALCULUS-RELATED ANSWER you dont know the inital velocity or the final velocity INSTRUCTOR COMMENT: . . . i.e., you can't evaluate the integration constant. **......!!!!!!!!...................................
RESPONSE --> ok
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14:51:43 Introductory Force Experiment
Give your data for the number of 1/4 cups of water vs. the length of the rubber band in ordered-pair form (i.e., (13,1), (14,2), (14.8,3), ... would indicate lengths of 13 cm, 14 cm and 14.8 cm for 1, 2 and 3 1/4-cups).......!!!!!!!!...................................
RESPONSE --> Rubber band #1: (7,0), (7.4,1), (7.6,2), (7.9,3), (8.3,4), (8.7,5), (9.1,6) Rubber band #2: (7.6,0), (7.6,1), (8.1,2), (8.5,3), (8.7,4), (9.0,5), (9.5, 6) Rubber band #3: (7.5,0), (7.7,1), (8.1,2), (8.4,3), (8.7,4), (9.2,5), (9.7,6) Rubber band #5: (7.7,0), (7.9,1), (8.1,2), (8.1,3), (8.9,4), (9.1,5), (9.5,6)
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14:52:06 **COMMON ERROR:
(0, 8.5) (1, 9) (2, 9.4) (3, 9.8) (4, 10.4) (5, 10.8) INSTRUCTOR RESPONSE: You have your coordinates backwards. The numbers of quarter-cups are 0, 1, 2, 3, 4. The rubber band lengths obtained for your specific rubber band, as you report them, are 8.5 cm, 8.8 cm, etc. Different students might get different results. Remember, the order is y vs. x. The y coordinate comes before 'vs.', the x coordinate after. When listing ordered pairs the x coordinate is the first and the y coordinate the second. So a graph of quarter-cups vs. length in cm. might contain the points (8.5, 0), (8.8, 1), etc. , etc.; but it wouldn't contain the points (0, 8.5), (1, 8.8) etc. **......!!!!!!!!...................................
RESPONSE --> ok
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14:58:35 Give your results for weight vs. length in ordered-pair form. [ Note that the requested order of the variables seems to violate our convention for independent vs. dependent variable. However, we will later exert forces by stretching the rubber band, and will regard the stretch as the variable we control with force being dependent on stretch. So in this experiment we switch what seems to be the logical order of the variables. ]
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RESPONSE --> (weight,length)
Rubber band #1: (0,7), (.21,7.4), (.42,7.6), (.63,7.9), (.84,8.3), (1.05,8.7), (1.26,9.1) Rubber band #2: (0,7.6), (.21,7.6), (.42,8.1), (.63,8.5), (.84,8.7), (1.05,9), (1.26,9.5) Rubber band #3: (0,7.5), (.21,7.7), (.42,8.1), (.63,8.4), (.84,8.7), (1.05,9.2), (1.26,9.7) Rubber band #5: (0,7.7), (.21,7.9), (.42, 8.1), (.63,8.1), (.84,8.9), (1.05,9.1), (1.26,9.5).................................................
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15:01:47 ** Each quarter-cup weighs about .55 Newtons, as given in the experiment, so that 0, 1, 2, . . . quarter-cups weigh 0, .55 N, 1.1 N, etc.. If the graph points for the preceding question include (0, 8.5), (1, 8.8), (2, 9.4) etc. then the graph of weight (in Newtons) vs. length (in cm) will include the points(8.5, 0), (8.8, .55), (9.4, 1.1), etc.. Your results will likely differ a bit but this example should show you whether you did the graph correctly. **
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RESPONSE --> ok
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15:03:56 What length you would expect for .8 Newtons of weight and for 2 Newtons of weight?
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RESPONSE --> Since rubber band #1 was the first one to reach it's maximum stretch o 30%, I'm using it's data to answer this question. At 0.8newtons, the length will be approximately 8.2cm. At 2newtons, the length will approx. be 11.6cm.
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15:04:27 ** You should look at your graph of weight vs. length. .8 Newtons will occur as the 'y' coordinate. Look straight across from that coordinate on the y axis to find the point on the graph where y = .8 Newtons. Then look straight down to find the corresponding 'x' coordinate, which will represent the stretch.
If the weight vs. stretch points include (8.5, 0), (8.8, .55), (9.4, 1.1) then the .8 Newtons weight would be between .55 N and 1.1 N, so that the length would likely lie between 8.8 cm and 9.4 cm. The precise estimate of the length would depend on exactly how the graph is drawn. **......!!!!!!!!...................................
RESPONSE --> ok
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15:10:37 What weight would you expect to stretch the rubber band by 1, 2, 3, 4, and 5 cm?
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RESPONSE --> 0.63 newtons stretched the rubber band by 1cm, 1.2newtons stretched it 2cm, 1.85newtons stretched it 3cm, 2.31newtons stretched it 4cm, and 2.9newtons stretched it 5cm.
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15:13:30 ** You should look at your graph of weight vs. length. This graph can be relabeled to give you weight vs. stretch: Simply subtract the 0-weight length from each of the given lengths and write this number beneath the length on the x axis.
Each stretch will occur as a newly labeled 'x' coordinate. Locate on the x axis the points corresponding to stretches of 1, 2, 3, 4, and 5 cm. Look straight up from each coordinate on the x axis to find the corresponding point on the graph. Then look straight across to the 'y' axis to find the corresponding 'y' coordinate, which will represent the weight necessary for that stretch. **......!!!!!!!!...................................
RESPONSE --> My rubber band did not stretch more than 2.5in to reach it's 30% stretch that was allowed. So in order to get the weights needed to stretch it by a cm at a time, I just extended my chart, adding .21 newtons (the weight of the cup of water) and adding 0.4cm to each weight change because that was the average rate that the rubber band lengthened each time water was added. That's how I got my estimates.
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15:14:38 Describe your force vs. stretch graph (i.e.: Do the points seem to lie on a straight line or a curve? If it is a curve for what stretches is the curve steepest, and for what stretches is it least steep? Does the direction of the curvature change and if so how?).
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RESPONSE --> Because of how I did the estimates, my graph is a straight line.
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15:15:28 **The graph will generally increase but at a decreasing rate. If you stretched the rubber band further than you should have, the graph might eventually straighten out for an instant then begin increasing at an increasing rate. **
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RESPONSE --> I understand why the graph should increase at a decreasing rate, but I don't really know of a way that I can safely extend my data without it being horribly inaccurate.
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15:16:01 What do you think makes the rubber band stretch?
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RESPONSE --> The additional force applied to rubber band is what stretches it. In this case, adding water.
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15:16:15 ** The gravitational force between Earth and the water, which is what we mean by the 'weight' of the water, pulls the water down. The further the rubber band is stretched the harder it pulls back up, until the pull of the rubber band is equal and opposite to the gravitational force on the water. **
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RESPONSE --> ok
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15:17:06 Is there a force on the bag and water? If so what is the source of the force? What is the direction of this force?
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RESPONSE --> Yes there is a force on the bag and the water. The gravitational pull forcing it downward and the pull of the rubberbands. The direction of the force is downward.
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15:17:15 ** The source of the force is the gravitational attraction between water and Earth **
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RESPONSE --> ok
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15:18:05 Is there more than one force on the bag and water? If so what are the sources of these forces? What are the directions of these forces?
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RESPONSE --> As mentioned before, I think the rubber bands are a force also. I'm guessing that the direction of the force from the rubber bands would be upward.
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15:20:50 ** gravity pulls down, the hook pulls up, and the two are in balance so they must be equal and opposite
GOOD (but slightly flawed) STUDENT ANSWER: There is more than one force on the bag and the water. the rubber band would be exerting an upward force on the bag and the water although the force of the water in the bag pulling down on the rubber band is greater than the force of the rubber band pulling upward on the bag and water. INSTRUCTOR RESPONSE: Good, but since the bag isn't accelerating in either the upward or downward direction the two forces must be equal and opposite. If one was greater the bag would accelerate in the direction of that force. The rubber band stretches until it exerts exactly enough force to create this equilibrium with the gravitational force. **......!!!!!!!!...................................
RESPONSE --> Ok so the forces between the rubber bands and gravity have to be equal. I know the bag isn't considered to be accelerating, but it does move downward by a little each time more weight is added, so I got confused by that and thought that the gravitational pull was going to be greater. Now that I think of it, if gravity was greater than the bag would continue to fall... I see what the explanation's saying now.
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15:28:06 General College Physics only: Problem #10 Summarize your solution Problem 1.10 (approx. uncertainty in area of circle given radius 2.8 * 10^4 cm).
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RESPONSE --> I'm not sure if I have to do this or not, but just in case... A = pi r^2 A = pi (3.8*10^4)^2 = 1.4*10^-9 pi
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15:36:18 ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.
This means that the area is between pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. The difference is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2, which is the uncertainty in the area. Note that the .1 * 10^4 cm uncertainty in radius is about 4% of the radius, which the .176 * 10^9 cm uncertainty in area is about 8% of the area. This is because the area is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. **......!!!!!!!!...................................
RESPONSE --> I see how I didn't take into account the uncertainty, and my calculater was in radians and not degrees so I think that was giving me wrong answers. It took me a few, but I see what it's asking for and how to get it now.
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15:37:41 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I've always had trouble making sure which direction to go if it's 4*10^5 or 4*10^-5. Other than that, everything made sense to me after I got my calculator in the ride mode.
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15:40:07 ** COMMON STUDENT COMMENT AND INSTRUCTOR RESPONSE: I am really confused about velocity, acceleration changes in velocity and acceleration, etc. I guess I am the type that works with a formula and plugs in a number. I have went back to the class notes and the problem sets to summarize formulas. Any suggestions?
RESPONSE: I note that you are expressing most of your answers in the form of formulas. Ability to use formulas and plug in numbers is useful, but it doesn't involve understanding the concepts, and without an understanding of the concepts we tend to plug our numbers into equations that don't apply. So we deal first with concepts. However formulas do come along fairly soon. The concepts of velocity, acceleration, etc. are very fundamental, but they are tricky and they take awhile to master. You are doing OK at this point. You'll see plenty more over the next few assignments. If you look at the Linked Outline (on the main Physics 1 page--the one where you click on the Assts button--click on the Overviews button, then on the Linked Outline. You will see a table with a bunch of formulas and links to explanations. You might find this page very useful. Also the Introductory Problem Sets give you formulas in the Generalized Solutions. **......!!!!!!!!...................................
RESPONSE --> As for the velocity and acceleration changes, I feel like I understand it somewhat, but I'm horrible at testing so what I feel I may understand, I always second guess myself. I'll definately check out that link you mentioned. And I'm making flash cards for the definitions so I have them down word for word.
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