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course Phy 241
8/26 10:30pm
Explain your reasoning as you answer each question. Not all questions are easy, but do your best on each:
If a ball requires 2 seconds to accelerate from rest along 3 dominoLengths of track, what was its average speed during that time?
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Avg V = (delta S)/ (total time). Delta S = 3 domino lengths. Total time= 2 second. (3 domino lengths)/(2 seconds). 3/2 or 1.5 dominolengths (DL) per second.
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Answer the same question if the time interval for the 3 dominoLength distance is 1 second, and again if the time is 1.5 second.
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1 second: 3 DL/1 second. 3 DL per second.
1.5 second: 3 DL/ 1.5 seconds. 2 DL per second.
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Does the average speed change as much between the 1 second and 1.5 second time intervals as between the 1.5 and 2 second intervals?
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1 - 1.5 second interval: decreased by 1 DL.
1.5 - 2 second interval: decreased by .5 DL.
1.5 - 2 second interval did not experience as much of a change as the 1 - 1.5 s interval.
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For the case where the ball required 1.5 seconds to coast from rest through 3 dominoLengths, the initial speed of the ball was zero. You have calculated its average speed. What final speed, averaged with the initial speed, would result in the same average speed?
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(X DL/s + 0 DL/s)/2 = 2 DL/s. 4 DL/s = X
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Assuming the final speed is the one you just calculated, how quickly was the speed changing during the 1.5 seconds? Your final answer to this question should be one single quantity.
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Rate of change. Velocity and time. (0,0) and (1.5, 4). (4-0)/(1.5-0) = 8/3.
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For the 1 second and 2 second intervals, how quickly was the speed changing?
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1 second: 3 DL/1 second. 3 DL per second. (x + 0)/2 = 3. Final v = 6. 6/1 = 6.
2 seconds: 1.5 DL per second. (x+0)/2 = 1.5. FInal v = 3. 3/2 = 1.5.
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If you had four dominoes with thicknesses 10 mm, 11 mm, 12 mm and 13 mm, and repeatedly scrambled and stacked them into stacks of two, it would be possible to get stacks of equal height and stacks of unequal height.
How many different heights are possible?
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3 possibilities with 10. 10 and 11. 10 and 12. 10 and 13.
2 poss. with 11 because 10 is now excluded. 11 and 12. 11 and 13.
Final poss. of 12 and 13.
Six different height possibilities for one stack.
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What height differences are possible between the two stacks (0 would be regarded as a height difference)?
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10 +11 and 12+ 13. 21 and 25. Diff= 4.
10+12 and 13+11. 22 and 24. Diff= 2.
10+13 and 11+12. 23 and 23. Diff= 0.
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What would be the probability of each of the possible height differences?
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1/3 chance each.
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If you had six dominoes with heights 10, 10.5, 11, 11.5, 12 and 12.5 cm, what height differences are possible when comparing stacks of three, and what is the probability of each on a random stacking?
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10, 10.5, 11 and 11.5, 12, 12.5. 31.5 and 36. Diff 4.5.
10, 10.5, 11.5 and 11, 12, 12.5. 32 and 36.5. Diff 4.5.
10, 10.5, 12 and 11, 11.5, 12.5. Diff 2.5.
10, 10.5, 12.5 and 11, 11.5, 12. Diff 1.5.
10, 11, 11.5 and 10.5, 12, 12.5. Diff 2.5.
10, 11, 12 and 10.5, 11.5, 12.5. Diff 1.5.
10, 11, 12.5 and 10.5, 11.5, 12. Diff .5.
10, 11.5, 12 and 10.5, 11, 12.5. Diff .5.
10, 11.5, 12.5 and 10.5, 11, 12. Diff .5
10, 12, 12.5 and 10.5, 11, 11.5. Diff 1.5.
4 variations to choose from.
Possibility of choosing 4.5 = 2/10 = 1/5.
Possibility of choosing 2.5 = 2/10 = 1/5.
Poss of choosing 1.5 = 3/10.
Poss of choosing .5 = 3/10.
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If each domino pip is a hemisphere and the diameter of a pip is 1/4 of the domino's width and 1/2 of its thickness, then what percent of the domino's volume would need to be removed to make a single pip?
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Assuming the length of the domino is twice the width. Pip diameter = x. Domino width = 4x. Domino thickness = 2x. Domino length (twice the width) = 8x. Domino volume = (4x)(8x)(2x) = 64 x^3. Hemisphere volume = (1/2)(pi/6) d^3.
When diam = 1, D Volume = 64.
H Vol = (pi/12)(1)^3 = pi/12.
(pi/12) is what percent of 64? In other words, what decimal must 64 be multiplied by to get pi/12?
I'm getting .0041, which is .41%, but that doesn't seem to make a whole lot of sense... I'm not sure where I went wrong though (assuming the answer is incorrect).
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How much would the presence of pips add to the percent uncertainty in the volume of a domino?
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Plus or minus .41%
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If pip radius is r then domino thickness is 4 r, width is 8 r and length is 16 r so volume is 512 r^3.
Pip volume is half of 4/3 pi r^3, or 2/3 pi r^3.
So pip volume is 2/3 pi r^3 / (512 r^3) = pi / (256 * 3 ) = .004, roughly, in good agreement with your result.
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If the thicknesses of a set of dominoes varies by as much as 5%, then how much uncertainty would you expect in the actual weight of an object which was found to be equal to that of 4 randomly selected dominoes?
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The variance continues to be 5% throughout the domino volume and weight calculations. Thus, 5% at most variation for a single domino, applied to 4 dominoes, gives 20% uncertainty in the weight of the object comparable to that of 4 dominoes.
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If the weight of a single domino could be off the as much as 5% of the weight of a 'reference domino', then the weight of 4 dominoes could be as much as 20% of the weight of the 'reference domino'.
However that wouldn't be 20% of the weight of the object itself, which would presumably be close to that of 4 reference dominoes, not just one.
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Very good, but be sure to see my notes.
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