Motions Go to http://phet.colorado.edu/simulations/sims.php?sim=Ladybug_Motion_2D and click on Run Now. Directions: 1. A Labybug was crawling in a circle around a flower like in the picture below. a. Sketch what you think the velocity and acceleration vectors would look like. b. If the flower is the “zero” position‚ what would the position vector look like? c. Use Ladybug Motion 2D to check your ideas. Make corrections if necessary 2. Suppose the bug
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Variables used in this lab were “x” for position of the object‚ “v” for velocity of the object‚ and “a” for acceleration of the object. Understanding the graphical representation of motion was important in helping students understand how position‚ velocity‚ and acceleration are affected with a moving object over a certain period of time. Using a motion detector and an Xplorer GLX‚ a calculator that graphed our distance velocity‚ and acceleration‚ students were able to create graphs for the information
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Investigating downstream changes on the Afon Caerfanell INTRODUCTION AIM: To investigate how selected parameters change downstream on the Afon Caerfanell. Hypothesis: 1. The velocity of Afon Caerfanell increases downstream 2. The velocity of the river increases down the stream as the angle of the slope increases. RIVER DEFINITION A river is the natural course of the water‚ which goes from a higher point‚ to the lowest point
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the end of this two-experiment laboratory‚ students ideally will know how to analyze displacement‚ velocity‚ and acceleration in terms of time for objects in motion with a constant acceleration in a straight line. In addition‚ students will master how to calculate the slope of a displacement-time graph to determine the velocity of an object in motion at a constant velocity and the slope of a velocity-time graph to determine the acceleration of an object. Materials In experiment 1‚ students prepare
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------------------------------------------------- Velocity Velocity is defined as a measure of the distance an object travels in a stated direction in a given length of time. Thus velocity is speed in a stated direction. Velocity is referred to as a vector quantity because it possesses both size and direction‚ the size being speed. Where speed only tells us how fast or slow an object is moving it gives no reference of direction velocity is used as a more complete measure as it not only gives
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stop? 3. where is car #1 when car #2 changes direction? 4. what is the average velocity of car #1 in km/hr? 5. estimate the instantaneous velocity (in km/hr) of car #2 at t = 43 min ? x / km car #1 20 10 car #2 0 20 10 30 40 t / min [1.2] Here is a plot of velocity versus time for an object that travels along a straight line (positive direction to the right) with a varying velocity. v/ m/s 1. at what time(s) is the object at rest? 20 2. what is the average
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position and velocity are displayed under the time. Click Start again and write down at least 2 observations about this simulation below. Activity 2: Click the Show Grid box and make sure that System Centered and Show Traces are checked too. Drag the slider bar all the way to the left for the most accuracy. Click Reset‚ then change Body 1’s mass to 500 and its x and y position and velocity to 0. Change Body 2’s mass to 30‚ its x position to 200‚ and y position and x and y velocity to zero. Reset
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Merrily We Roll Along! Purpose: To investigate the relationship between distance and time for a ball rolling down an incline. Data: Table A | Time (s) | Incline 25° | Distance (cm) | Trial 1 | Trial 2 | Trial 3 | Average | 20.5 | 0.31 | 0.32 | 0.29 | 0.31 | 41 | 0.47 | 0.27 | 0.38 | 0.37 | 61.5 | 0.51 | 0.52 | 0.31 | 0.45 | 82 | 0.67 | 0.54 | 0.45 | 0.55 | 102.5 | 0.69 | 0.90 | 0.58 | 0.72 | 123 | 0.88 | 0.67 | 0.58
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Science 10. P 393 Investigation 13A Time (sec) Displacement (cm down) Velocity ( down) 0 0 0 0.1 0.6 = Df – Di = 0.6 – 0 = 0.6 = 0.2 1 = Df – Di = 1 – 0.6 = 0.4 = 0.3 3.3 = Df – Di = 3.3 – 1 = 2.3 = 0.4 5.9 = Df – Di = 5.9 – 3.3 = 2.6 = 0.5 7.4 = Df – Di = 7.4 – 5.9 = 1.5 = 0.6 8.7 = Df – Di = 8.7 – 7.4 = 1.3 = 0.7 10.1 = Df – Di = 10.1 – 8.7 = 1.4 = 0.8 11.4 = Df – Di = 11.4 – 10.1 = 1.3 = 0.9 12.9 = Df – Di = 12.9 – 11.4 = 1.5 = 1.0 11.8 = Df – Di = 11.8 – 12.9 = -1.1 = 1
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AP Physics Slinky Velocity Lab Group: Asaf Yankilevich‚ Lily Greenwald‚ Yaeli Eijkenaar‚ Michal Antonov 2/23/15 Materials ● Slinky ● Spring weight ● Force measurer ● Measuring Tape ● Timer Procedure 1. The first slinky’s mass was weighed‚ using a scale‚ and its tension was measured using a force measurer 2. The slinky was stretched to 4m. 3. The linear mass density was solved for‚ by dividing the mass by the length. 4. The theoretical velocity was solved for‚ using the equation
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