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09:56:45 What do we mean by velocity?
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RESPONSE --> Velocity is the rate in which an object changes its position.
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09:57:02 ** STUDENT RESPONSE:Velocity is the speed and direction an object is moving. INSTRUCTOR COMMENT: Good. More succinctly and precisely velocity is the rate at which position is changing. obtained by dividing change in position by change in clock time **
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RESPONSE --> ok
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10:03:19 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> In terms of average velocity we can take the displacement (change in the x value) and divide it by the time associated with that displacement. Ex: A ball takes 2 seconds to travel down a 10 cm incline. The velocity would be 10cm/2seconds = 5cm/s
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10:03:31 ** We divide displacement by the time interval to get average velocity. If the ball happens to be starting from rest and acceleration is uniform we can also conclude that the final velocity attained on the incline is double the average velocity. **
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RESPONSE --> ok
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10:07:46 We anticipate from our experience that a ball traveling down a greater incline, starting from rest, will experience a greater change in velocity. How can we determine whether the velocity actually changes, and whether the velocity increases in the manner that we expect?
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RESPONSE --> Velocity is the displacement divided by the time..so we can time a ball on two seprate ramps one at a greater incline than the other. Comparing the two velocities will show us that the ball on the greater inlcine will experience a greater velocity
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10:08:01 ** We divide displacement by the time interval to get average velocity. We time the ball down one incline, then down the other and determine average velocity for each.
We then infer that since both balls started from rest, the greater average velocity implies a greater change in velocity. **......!!!!!!!!...................................
RESPONSE --> ok
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10:15:33 How could we determine the velocity of the ball at a specific point? The specific points are measured for distance and the ball is timed when it reaches these specific points. The distance is then divided by the time.
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RESPONSE --> Instantaneous velocity is the limit of the average velocity as the time invterval approaches zero; it equals the intantaneous rate of change of position with time. Ex. To find the instantaneous velocity of an object at Point P1, I would imagine moving the second point P2 closer and closer to the first point P1. I would compute the average velocity 'dx/'dt over these shorter and shorter displacements and time intervals.
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10:15:56 ** Short answer: The question concerned one specific point. We can't really measure this precisely. The best we can do is use two points close together near the point we are interested in, but not so close we can't measure the time accurately enough to trust our result.
More detailed answer: The question really asks how we determine the velocity at a given point, for an object in the real world. Assuming that the velocity is always changing, how can we ever know the velocity at an instant? This involves a limiting process, thinking of shorter and shorter time intervals and shorter and shorter position changes. If we know the velocity function, or if we can accurately infer the velocity function from our data, then the velocity of a ball at a specific point is obtained by finding the slope of the tangent line of the position vs. clock time graph at that point, which calculus-literate students will recognize as the derivative of the velocity function. **......!!!!!!!!...................................
RESPONSE --> ok
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10:26:05 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> Acceleration describes the rate of change of velocity with time. Given a time range and a displacement over that time we can caculate the Average Acceleration by 'dv/'dt in which we divide the change in velocity by the change in time for a specific data set.
Your analogy is good. A movement in a cars speedometer dictates a change in velocity. The rate in which the speedometer moves would be the rate at which the velocity of the car is changing..................................................
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10:26:20 ** We find the change in velocity then divide by the change in the clock time. Any rate consists of the change in one quantity divided by the change in another. **
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RESPONSE --> ok
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10:35:07 It is essential to understand what a trapezoid on a v vs. t graph represents. Give the meaning of the rise and run between two points, and the meaning of the area of a trapezoid defined by a v vs. t graph.
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RESPONSE --> the rise in the graph (y) is the velocity and the run of the graph (x) is the time
We can calculate the position change as an area of the trapezoid defined by a v vs. t graph..................................................
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10:35:25 ** Since the rise represents the change in velocity and the run represents the change in clock time, slope represents `dv / `dt = vAve, the average velocity over the corresponding time interval.
Since the average altitude represents the average velocity and the width of the trapezoid represents the time interval the area of the trapezoid represents vAve * `dt, which is the displacement `ds. **......!!!!!!!!...................................
RESPONSE --> ok
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10:36:18 What does the graph of position vs. clock time look like for constant-acceleration motion?
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RESPONSE --> The graph would be increasing at an increasing rate
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10:36:28 ** For constant positive acceleration velocity is increasing. The greater the velocity the steeper the position vs. clock time graph. So increasing velocity would be associated with a position vs. clock time graph which is increasing at an increasing rate.
The reason velocity is the slope of the position vs. clock time graph is that the rise between two points of the position vs. clock time graph is change in position, `ds, and run is change in clock time, `dt. Slope therefore represents `ds / `dt, which is velocity. Other shapes are possible, depending on whether initial velocity and acceleration are positive, negative or zero. For example if acceleration was negative and initial velocity positive we could have a graph that's increasing at a decreasing rate. Negative initial velocity and positive acceleration could give us a graph that's decreasing at a decreasing rate. **......!!!!!!!!...................................
RESPONSE --> ok
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10:47:53 How can we obtain a graph of velocity vs. clock time from a position vs. clock time graph?
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RESPONSE --> The slope of the position vs. time graph represents average velocity and will give back the average velocity values
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10:48:05 ** We can find the slope of the position vs. clock time graph at a series of clock times, which will give us the velocities at those clock times. We can put this information into a velocity vs. clock time table then plot the velocities vs. clock time as a 'guidepost points', and fill in the connecting curve in such a way as to be consistent with the trend of the slopes of the position vs. clock time graph.
COMMON MISCONCEPTION: To get velocity vs. clock time find average velocity, which is position (m) divided by time (s). Plot these points of vAvg on the velocity vs. time graph. INSTRUCTOR RESPONSE: Ave velocity is change in position divided by change in clock time. It is not position divided by time. Position can be measured from any reference point, which would affect a position/time result, but which would not affect change in position/time. Graphically velocity is the slope of the position vs. clock time graph. If it was just position divided by time, it would be the slope of a line from the origin to the graph point. **......!!!!!!!!...................................
RESPONSE --> OK
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10:52:47 How can we obtain a graph of position vs. clock time from a velocity vs. clock time graph
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RESPONSE --> Given that the change of position is represented by the area under the velocity vs.clock time graph, it can be determined the velocities at each second intervals using similar triangles. We can determine the acceleration by the slope of the graph and plot the position changes based on the rate of the constant acceleration
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10:53:10 ** We can divide the graph of v vs. t into small strips, each forming an approximate trapezoid. The area of each strip will represent ave vel * time interval and will therefore represent the change in position during that time interval. Starting from the initial clock time and position on the position vs. clock time graph, we add each subsequent time increment to the clock time and the corresponding position change to the position to get our new position.
When the graph is constructed the slopes of the position vs. clock time graph will indicate the corresponding velocities on the v vs. t graph. **......!!!!!!!!...................................
RESPONSE --> ok
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10:55:05 How can we obtain a graph of acceleration vs. clock time from a velocity vs. clock time graph?
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RESPONSE --> We can obtain the acceleration which is the slope of the velocity vs. clock time graph and plot accordingly
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10:55:24 ** Accel is the rate of change of velocity, represented by the slope of the v vs. t graph. So we would plot the slope of the v vs. t graph vs. t, in much the same way as we plotted slopes of the position vs. clock time graph to get the v vs. t graph.
}University Physics Students note: Acceleration is the derivative of the velocity. COMMON MISCONCEPTION: Take speed/ time to find the acceleration per second. The form an acceleration v. time graph and draw a straight line out from the number calculated for acceleration above. INSTRUCTOR RESPONSE: Ave acceleration is change in velocity divided by change in clock time. (note that this is different from velocity divided by time--we must use changes in velocity and clock time). (Advanced note: Velocity is always measured with respect to some reference frame, and the velocity of the reference frame itself affects a velocity/time result, but which would not affect change in velocity/time). Graphically acceleration is the slope of the velocity vs. clock time graph. If it was velocity divided by time, it would be the slope of a line from the origin to the graph point. **......!!!!!!!!...................................
RESPONSE --> ok
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11:11:26 How can we obtain a graph of velocity vs. clock time from an acceleration vs. clock time graph
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RESPONSE --> The area under the graph represents the total displacement, so for each t interval we can individually calculate the area of the trapezoid between each second interval and we have the displacement over 1 second which can be used to calculate the velocity of the object and graph accordingly
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11:14:32 ** STUDENT RESPONSE: Take your acceleration and multiply by time to find the change in velocity. Start with initial velocity and graph your velocity by increasing initial velocity by the slope, or change in velocity. INSTRUCTOR COMMENT: Good. More precisely we can approximate change in velocity during a given time interval by finding the approximate area under the acceleration vs. clock time graph for the interval. We can then add each change in velocity to the existing velocity, constructing the velocity vs. clock time graph interval by interval.
A velocity vs. clock time graph has slopes which are equal at every point to the vertical coordinate of the acceleration vs. clock time graph. University Physics students note: These two statements are equivalent, and the reason they are is at the heart of the Fundamental Theorem of Calculus. **......!!!!!!!!...................................
RESPONSE --> I think I misinterpreted the question. I agree that to obtain the the velocity for the graph by accleration multiplied by time and starting from the intital velocity I need to increase the initial velocity by the slope
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ٚhyy⪞ assignment #001 ؚmfoVЮ Physics I Vid Clips 09-10-2005
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18:14:32 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> If the ball is on a completely level incline I would suspect the balls velocity to be increasing at a decreasing rate.
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18:16:33 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases.
A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **......!!!!!!!!...................................
RESPONSE --> If the ball were on a completely level incline...I would think that the ball would eventually stop due to friction of the rail...in this case the velocity would decrease at a increasing rate. If the ball were on a slightly straight incline then the ball would in fact increase in velocity at an increasing rate
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18:17:07 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> the rate that the speedometer is changing..either slowing down or speeding up.
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18:17:23 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity.
INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **......!!!!!!!!...................................
RESPONSE --> ok
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18:17:59 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> The graph would be increasing at an increasing rate
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18:18:08 ** If the ramp has an increasing slope, the velocity would increase at an increasing rate and the graph would curve upward, increasing at an increasing rate. If the ramp has a decreasing slope, like a hill that gradually levels off, the graph would be increasing but at a decreasing rate. On a straight incline it turns out that the graph would be linear, increasing at a constant rate, though you aren't expected to know this at this point. All of these answers assume an absence of significant frictional forces such as air resistance. **
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RESPONSE --> ok
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18:20:06 A ball rolls down ramp which curves upward at the starting end and otherwise rests on a level table. What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> The initial graph would increase at an increasing rate until the incline becomes level then I would suspect that due to the level ramp the velocity of the ball would either decrease at an increasing rate or be constant
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18:20:23 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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RESPONSE -->
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18:20:56 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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RESPONSE --> Velocity would be increasing at an increasing rate
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18:23:59 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right.
Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**......!!!!!!!!...................................
RESPONSE --> Again I am confused by what you mean by straight incline....but in either case I do understand that the ball would increase in velcocity at a constant rate due to the straight incline applying the same force of gravity at each interval...thus the velocity would change at a constant rate
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{l{ Уצ assignment #001 ؚmfoVЮ Physics I 09-10-2005
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18:34:45 Briefly state what you think velocity is and how you think it is an example of a rate.
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RESPONSE --> Velocity is the rate in which an object changes is position. Ex. 40m/s South
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18:34:52 ** A rate is a change in something divided by a change in something else.
This question concerns velocity, which is the rate of change of position: change in position divided by change in clock time. **......!!!!!!!!...................................
RESPONSE --> ok
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18:37:39 Given average speed and time interval how do you find distance moved?
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RESPONSE --> Given the average speed of 10m/s and we know the object traveled for 10s then;
10*10 = 100meters.................................................
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18:37:49 ** You multiply average speed * time interval to find distance moved.
For example, 50 miles / hour * 3 hours = 150 miles. **......!!!!!!!!...................................
RESPONSE --> ok
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18:39:49 Given average speed and distance moved how do you find the corresponding time interval?
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RESPONSE --> Given the average speed of 10m/s and the object moved a distance of 100m then;
(100m)/10m/s = 10 seconds divide the distance by the average speed to get the time interval.................................................
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18:40:08 ** time interval = distance / average speed. For example if we travel 100 miles at 50 mph it takes 2 hours--we divide the distance by the speed.
In symbols, if `ds = vAve * `dt then `dt = `ds/vAve. Also note that (cm/s ) / s = cm/s^2, not sec, whereas cm / (cm/s) = cm * s / cm = s, as appropriate in a calculation of `dt. **......!!!!!!!!...................................
RESPONSE --> ok
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18:42:49 Given time interval and distance moved how do you get average speed?
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RESPONSE --> The average speed is distance divided by the time time Ex. If the object moved 10meters in 5 seconds then the average speed would be 10/5 m/s or 2m/s
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18:42:56 ** Average speed = distance / change in clock time. This is the definition of average speed.
For example if we travel 300 miles in 5 hours we have been traveling at an average speed of 300 miles / 5 hours = 60 miles / hour. **......!!!!!!!!...................................
RESPONSE --> ok
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18:43:18 You should have written up Video Experiment 2 as directed and submitted it. If you have done so within a couple of days of submitting this query then you need not answer any question here that was answered on your writeup. This is a general principle for experiments and can be applied on all queries. 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 --> ok
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18:43:36 ** It's very important to express all quantities in terms of the correct units.
Slope, being rise / run in units of cm / cm, ends up unitless (the cm 'cancel'). Velocities are expressed in cm / sec. Rates of velocity change are expressed in (cm / sec) / sec or cm / sec^2. Slopes are expressed in units of rise / units of run, in this case giving just cm / sec^2. **......!!!!!!!!...................................
RESPONSE --> ok
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18:44:27 Explain briefly how you calculated your slopes.
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RESPONSE --> Slope is the 'd rise (y value) / 'd run (x value)
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18:44:39 ** slopes are caclulated by using the basic rise over run formula.
For instance one student reports that when the paper stack was 1.26 cm tall and the desk was 44 cm long the slope was 1.26/44 which is equivalent to 0.0286 **......!!!!!!!!...................................
RESPONSE --> ok
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18:47:10 Explain briefly how you determined your rates of velocity change.
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RESPONSE --> acceleration is the 'dx/'dt
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18:47:18 ** rates of velocity change are calculated by dividing change in velocity by change in clock time.
From your timing you got average velocity. Since accel is uniform and initial velocity is zero, final velocity must be double average velocity. Change in velocity is obtained by subtracting init vel from final vel. Change in clock time is the time required to accelerate down the incline. Note units: when you divide change in velocity by change in clock time you are dividing cm/sec by sec, giving cm / sec * 1/sec = cm/sec^2. **......!!!!!!!!...................................
RESPONSE --> ok
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18:49:46 What were the units of the slope of your straight line on the graph of rate of velocity change vs. slope? 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 --> m
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18:50:57 ** Rise is change in the rate of velocity change, in cm/sec^2. Ramp slope is unitless. So graph slope is in units of cm/sec^2.
The ideal value would be very close to the acceleration of gravity, which is 980 cm/sec/sec. Typical range when timing with the pendulum is from 600 to 1500 cm/sec/sec. Using very accurate timing (electronic equipment in the lab) with these clips we still end up with an error of about 8% on the high side, which shouldn't happen. The experiment done with accurate timing should be quite accurate. **......!!!!!!!!...................................
RESPONSE --> I sorry I misunderstood the question...the y value of the graph was the change in the rate of velocity which was measured in cm/s/s or cm/s^2
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́[z Student Name: assignment #001
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09:11:42 This exercise was to have included a simulation of the pendulum as a timer. However due to delays in the startup of the course this exercise will not be included this semester. Your real first assignment with this program will be #2. The program will exit after you enter your response (you can enter a blank response or any other response you wish). Then click on Quit Program (upper right-hand corner).
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RESPONSE --> ok
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ߺx}l Student Name: assignment #001 001. Areas
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11:21:39 `q001. There are 11 questions and 7 summary questions in this assignment.
What is the area of a rectangle whose dimensions are 4 m by 3 meters.......!!!!!!!!...................................
RESPONSE --> 4m*3m=12 meters
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11:23:02 A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.
The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2. Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.......!!!!!!!!...................................
RESPONSE --> Sorry I forgot to type in ^2 since it is an area of 12 square meters
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11:29:24 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?
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RESPONSE --> 1/2bh = 1/2*4*3 = 6 m^2
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11:29:33 A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.
The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2. The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.......!!!!!!!!...................................
RESPONSE --> OK
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11:30:48 `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?
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RESPONSE --> area of parallelogram = b * h = 5*2 = 10 m^2
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11:30:54 A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.
The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.......!!!!!!!!...................................
RESPONSE --> OK
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11:31:37 `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?
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RESPONSE --> 1/2 b*h = 1/2*5*2 = 5m^2
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11:39:14 It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.
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RESPONSE --> I need to slow down a bit and pay attention to the units which are in cm and not m.
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11:41:57 `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?
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RESPONSE --> 1/2*(b1+b2)*h = 1/2*(4km+4km)*5km 1/2*8km*5km = 20 km^2
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11:42:02 Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.
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RESPONSE --> ok
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17:35:57 `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?
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RESPONSE --> Per the previous formula.. I will assume
1/2*(3cm+8cm)*4cm= 22cm^2.................................................
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17:36:14 The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.
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RESPONSE --> ok
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17:39:14 `q007. What is the area of a circle whose radius is 3.00 cm?
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RESPONSE --> Pi*r^2
3.14*3cm^2 28.26cm^2.................................................
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17:39:51 The area of a circle is A = pi * r^2, where r is the radius. Thus
A = pi * (3 cm)^2 = 9 pi cm^2. Note that the units are cm^2, since the cm unit is part r, which is squared. The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius. Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.......!!!!!!!!...................................
RESPONSE --> **I needed to round to significant figures.
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17:42:18 `q008. What is the circumference of a circle whose radius is exactly 3 cm?
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RESPONSE --> C=pi*d C=pi*(3cm + 3cm) C=3.14*6cm C= 18.8cm
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17:42:27 The circumference of this circle is
C = 2 pi r = 2 pi * 3 cm = 6 pi cm. This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm. Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.......!!!!!!!!...................................
RESPONSE --> ok
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17:43:42 `q009. What is the area of a circle whose diameter is exactly 12 meters?
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RESPONSE --> Area = pi*r^2
area = 3.14*6m^2 Area = 113.04m^2.................................................
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17:43:56 The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is
A = pi ( 6 m )^2 = 36 pi m^2. This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.......!!!!!!!!...................................
RESPONSE --> ok
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17:48:46 `q010. What is the area of a circle whose circumference is 14 `pi meters?
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RESPONSE --> Area = Pi*r^2
**Assuming that the equations is stating that C=14*Pi meters Then 14 would be the diameter....1/2 of the diameter is 7 thus; Area=3.14*7^2 = 153.86m.................................................
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17:50:54 We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.
We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m. We use this to find the area A = pi * (7 m)^2 = 49 pi m^2.......!!!!!!!!...................................
RESPONSE --> I misinterpreted the typewriter notation...I do see my mistake in this. I assumed that the notation meant the equation C=Pi*d with d being 14 meters.
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17:55:30 `q011. What is the radius of circle whose area is 78 square meters?
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RESPONSE --> A=Pi*r^2 78m^2=3.14*r^2 24.84=r^2
r=4.99m.................................................
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17:55:41 Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).
Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution. Now we substitute A = 78 m^2 to obtain r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{} Approximating this quantity to 2 significant figures we obtain r = 5.0 m.......!!!!!!!!...................................
RESPONSE --> ok
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17:56:48 `q012. Summary Question 1: How do we visualize the area of a rectangle?
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RESPONSE --> Area = l*w
mutiply the unit of lenght times the unit of width which would give us the number of square units..................................................
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17:56:55 We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.
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RESPONSE --> ok
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17:58:42 `q013. Summary Question 2: How do we visualize the area of a right triangle?
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RESPONSE --> The area of a right triangle is visualized by 1/2 a square thus the (l*w)/2 would give half the grid of a square of the area
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17:58:49 We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.
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RESPONSE --> ok
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18:00:17 `q014. Summary Question 3: How do we calculate the area of a parallelogram?
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RESPONSE --> Area = base * height
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18:00:24 The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.
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RESPONSE --> ok
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18:01:38 `q015. Summary Question 4: How do we calculate the area of a trapezoid?
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RESPONSE --> Area = (1/2)*(B1+B2)*Height
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18:01:48 We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.
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RESPONSE --> ok
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18:02:01 `q016. Summary Question 5: How do we calculate the area of a circle?
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RESPONSE --> Area = Pi r^2
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18:02:07 We use the formula A = pi r^2, where r is the radius of the circle.
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RESPONSE --> ok
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18:03:22 `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?
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RESPONSE --> C=Pi * d
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18:03:33 We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.
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RESPONSE --> ok
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18:04:17 `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.
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RESPONSE --> OK
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18:04:25 This ends the first assignment.
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RESPONSE --> ok
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