TIPERS PHYSICS PDF

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Actions Shares. Embeds 0 No embeds. No notes for slide. Tipers sensemaking tasks for introductory physics 1st edition hieggelke solutions manual 1. All masses are given in the diagram in terms of M, the mass of the smallest block. Note that the order of two items ranked as equal is not important, but some instructors encourage students to use alphabetical order. Masses are given in terms of M, the mass of the smallest block.

For stacks A and D the average mass of blocks in the stack is the total mass divided by three, which is 3M for both. The slope of a line at a given point is given by the change in the value of the quantity represented on the vertical axis divided by the change in the value of the quantity represented on the horizontal axis for a small interval near that point.

More simply, the line has the same steepness at all points along it. The slope of a graph is given by the rise divided by the run. Lines C and D rise 2 units over a run of 4 units, giving a slope of 0. Lines B and E rise 1 unit over a run of 4 units and have a slope of 0.

Line A has no rise, and its slope is zero. The rise of the graph divided by the run is largest at D the line segment at D rises 6 units for a run of 1 unit, giving a slope of 6 , smaller at B and C the slope for the line segment that B and C are on has a rise of 5 units over a run of 4 units, so the slope is 1. A B C D Rank the slopes of the diagonals between the points marked with dots for these rectangles.

For Case A the rise is 9, and the run is 6, and the difference between rise and run is 3. For Case B, the rise is 8 and the run is 12 and the difference is minus 4. If something is wrong, identify and explain how to correct all errors.

If this statement is correct, explain why. The student is determining slope by looking at the difference between rise and run, rather than the ratio. In addition, the student is determining the sign of the slope based on this difference, rather than on whether the horizontal change and the vertical change along the line have the same sign. Explain your reasoning. Answer: The same in both cases. Each graph rises 2 units in a run of 4 units, so each graph has a slope of 0.

For a curved graph like this one, the closer the points are that you use to calculate the rise over the run, the more acccurate the value you obtain for the slope at that point. In the limit that the points are extremely close together, the value of the slope is the same as the slope of the tangent to the curve.

In this case, the slope of the tangent to the curve is steepest at point C, less steep at point D, zero at point B, and negative at point A. Since the largest y-value is Answer: While the number of hits has increased, the growth is hardly dramatic.

The graph only makes the growth look good because the y-axis starts at , rather than at zero. There are sometimes good reasons for starting a graph with a nozero y value for the starting the left most value on the x- axis, but at times this is also done intentionally in hopes that people will misinterpret the results a trick known as a disappearing baseline. We appreciate our devoted readers. A graph of the distance of the cat from the dog as a function of time is shown.

What, if anything, is wrong with this calculation? If this is correct, explain why. Answer: This method of calculating a slope is only valid in the Distance, 20 20 16 16 12 12 8 8 4 4 1 1 2 3 4 5 6 7 8 Time, seconds special case that the object in this case, the cat is moving at a constant rate and if it started from a position of zero.

Here, the cat started 8 meters away from the dog, and changes speed at time 5 seconds. In general, the slope of a line is equal to the change in the y-axis quantity divided by the change in the x-axis quantity. A student graphs the height of the water in the glass as a function of time as shown: Height, h h Time, t Cross- section Perspective What, if anything, is wrong with this graph? Answer: The graph is correct. The volume of water that is added to the glass is constant per unit time.

Since the cross-sectional area of the glass is constant a circle of constant diameter and the volume is the area times the height, the height increases at a constant rate. The glass is filled using a tap with a constant flow rate of 4 ml per second.

Answer: According to the graph, it takes more time to fill the glass to the first mark than it does to the second, and that the time it takes to increase the height of water one unit of height decreases as the water level rises.

But since the glass is wider at the top than at the bottom, and the flow rate the volume of water added to the glass per unit time is constant, the time it takes to increase the height of water one unit should increase with height, not decrease: Height, h Time, t. The glass is filled using a hose with a constant flow rate of 4 ml per second. Answer: According to the graph, it takes the same amount of time to fill the glass to the first mark as it does to the second, and the time it takes to increase the height of water one unit of height is constant as the water level rises.

But since the glass is narrower at the top than at the bottom, and the flow rate is constant, the time it takes to increase the height of water one unit should decrease with height, not decrease: Height, h Time, t. Near the sink is where the water is flowing the fastest. Andre Bela Carl None of them Explain your reasoning. Answer: Bela is correct. As the water falls after it leaves the faucet, it is speeding up, and the stream is also narrower.

These two effects exactly compensate, and the flow rate is exactly the same. The amount of water you collect per unit time can be calculated by taking the product of the area of the stream and the speed of the water, so if the water leaving the tap has a speed of 20 centimeters per second and the stream of water there has an area of 4 cm 2 , then the flow rate is 80 cm 3 per second. Further away from the tap, the speed might be 40 centimeters per second: we can predict that the area of the stream of water at that point will be 2 cm 2 , so that we still have a flow rate of 80cm 3 per second.

The flask is filled using a hose with a constant flow rate of 4 ml per second. Then, when the water reaches the neck, the flask stays a constant width, so the height increases at a constant rate. Answer: The rate at which the flask is filling is constant, but the rate at which the height is changing depends on the diameter of the flask.

At points where the flask is widest, the rate at which the height changes will be slowest, and the slope of the height versus time graph will be smallest. From B to C, the height is changing at a constant rate, since the width of the flask is not changing, so the slope of the graph is constant.

From the bottom to height A, the width of the flask is increasing, so the height-time graph will have a decreasing slope as time increases, and from A to B, , the width of the flask is decreasing, so the height-time graph will have an increasing slope with time.

The student is correct that the height will increase at a constant rate at the neck. Answer: This statement is almost correct: x and y are variables in this equation, and as such, they represent the number of cars and the number of trucks.

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TIPERs: Sensemaking Tasks for Introductory Physics

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Short activities that help students apply concepts and address known difficulties. Activities are designed so that they cannot be solved using "plug-and-chug. Changing Representation Tasks — given one representation, e. Comparison Tasks — these ask the student to determine which of two situations has a greater value for a quantity, or if the two situations have the same value for quantity. Conflicting Contentions Tasks — these tasks present two or three usually natural language statements about a situation and the goal is to decide which, if any, of the statements is correct.

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All masses are given in the diagram in terms of M, the mass of the smallest block. Explain your reasoning. Using the ranking chart, this answer could be expressed either as. Note that the order of two items ranked as equal is not important, but some instructors encourage students to use alphabetical order. Masses are given in terms of M, the mass of the smallest block.

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TIPERs: Sensemaking Tasks for Introductory Physics gives introductory physics students the type of practice they need to promote a conceptual understanding of problem solving. This supplementary text helps students to connect the physical rules of the universe with the mathematical tools used to express them. The exercises in this workbook are intended to promote sensemaking. The various formats of the questions are difficult to solve just by using physics equations as formulas.

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