course PHY232 I am not sure, am I supposed to submit the homework assignment here? I finished most of the homework assingment, of couse the quesitons you asked were the ones I did not get. ?v?????c???[?assignment #001001.
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16:26:58 Most queries in this course will ask you questions about class notes, readings, text problems and experiments. Since the first two assignments have been experiments, the first two queries are related to the experiments. While the remaining queries in this course are in question-answer format, the first two will be in the form of open-ended questions. Interpret these questions and answer them as best you can.
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RESPONSE -->
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16:28:30 Suppose you measure the length of a pencil. You use both a triply-reduced ruler and the original ruler itself, and you make your measurements accurate to the smallest mark on each. You then multiply the reading on the triply-reduced ruler by the appropriate scale factor. Which result is likely to be closer to the actual length of the pencil? What factors do you have to consider in order to answer this question and how do they weigh into your final answer?
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RESPONSE --> The orignal ruller itself would give the most accurate measurment. The triply-reduced ruler scale factor may not be accurate. confidence assessment: 2
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16:29:57 Answer the same question as before, except assume that the triply-reduced ruler has no optical distortion and you know the scale factor accurate to 4 significant figures.
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RESPONSE --> With no optical distortion and a scale facotor accurate to 4 significant fiures would depend on how accurate my measurment need to be. If the measurment need to be accurate to within 5 signifcant figures, then the triply-reduce ruler would be problematic. confidence assessment: 2
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16:31:15 Suppose you are to measure the length of a rubber band whose original length is around 10 cm, measuring once while the rubber band supports the weight of a small apple and again when it supports the weight of two small apples. You are asked to report as accurately as possible the difference in the two lengths, which is somewhere between 1 cm and 2 cm. You have available the singly-reduced copy and the triply-reduced copy, and your data from the optical distortion experiment. Which ruler will be likely to give you the more accurate difference in the lengths? Explain what factors you considered and how they influence your final answer.
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RESPONSE --> The singly -reduced copy will give the most accurat measurment since the error is not propogated. confidence assessment: 2
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16:35:11 Later in the course you will observe how the depth of water in a uniform cylinder changes as a function of time, when water flows from a hole near the bottom of the cylinder. Suppose these measurements are made by taping a triply-reduced ruler to the side of a transparent cylinder, and observing the depth of the water at regular 3-second intervals. {}{}The resulting data would consist of a table of water depth vs. clock times, with clock times 0, 3, 6, 9, 12, ... seconds. As depth decreases the water flows from the hole more and more slowly, so the depth changes less and less quickly with respect to clock time. {}{}Experimental uncertainties would occur due to the optical distortion of the copied rulers, due to the the spacing between marks on the rulers, due to limitations on your ability to read the ruler (your eyes are only so good), due to timing errors, and due to other possible factors. {}{}Suppose that depth changes vary from 5 cm to 2 cm over the first six 3-second intervals. {}{}Assume also that the timing was very precise, so that there were no significant uncertainties due to timing. Based on what you have learned in experiments done in Assignments 0 and 1, without doing extensive mathematical analysis, estimate how much uncertainty would be expected in the observed depths, and briefly explain the basis for your estimates. Speculate also on how much uncertainty would result in first-difference calculations done with the depth vs. clock time data, and how much in second-difference calculations. {}{}How would these uncertainties affect a graph of first difference vs. midpoint clock time, and on a graph of second difference vs. midpoint clock time? {}How reliably do you think the first-difference graph would predict the actual behavior of the first difference? {}Answer the same for the second-difference graph. {}{}What do you think the first difference tells you about the system? What about the second difference?
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RESPONSE --> confidence assessment: 0
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????????????x?assignment #002 002. `query 1 Physics II 02-01-2009
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22:09:48 query introset 5 # 12: Finding the conductivity given rate of energy flow, area, temperatures, thickness of wall. Describe how we find the conductivigy given the rate of energy flow, area, temperatures, and thickness of the wall?
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RESPONSE --> H= ka * (tf - ti)/L So, if we know the flow (H), tempertures (tf-ti), and the length, we can then solve for (a) the thermal conductivity
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22:10:45 ** The rate at which thermal energy is conducted across for a object made of a given substance is proportional to the temperature gradient (the rate at which temperature changes across the object), and to the cross-sectional area of the object. The conductivity is the constant of proportionality for the given substance. So we have Rate of thermal energy condction = conductivity * temperature gradient * area, or R = k * `dT/`dx * A. For an object of uniform cross-section `dT is the temperature difference across the object and `dx is the distance L between the faces of the object. In this case the equation is R = k * `dT / L * A and we can solve to get k = R * L / (`dT * A). **
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RESPONSE --> Ok
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22:12:17 Explain how energy flow, for a given material, is affected by area (e.g., is it proportional to area, inversely proportional, etc.), thickness and temperature gradient.
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RESPONSE --> The energy flow is directly proportional to the area and inversly proportional to the length. H = ka *(tf-ti)/L
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22:13:21 ** CORRECT STUDENT ANSWER WITHOUT EXPLANATION: Energy flow is directly proportional to area inversely propportional to thickness and directly proportional to temperature gradient GOOD STUDENT ANSWER WITH EXPLANATIONS, PLUS INSTRUCTOR COMMENTARY: The energy flow for a given material increases if the area increases. This is because the more area you have the wider a path something has to go through so more of it can move through it. Just like a 4 lane highway will carry more cars in a given time interval than a two lane highway will. So the relationship of energy flow to area is proportional. Energy flow, however is inversely proportional to thickness. This is because although the thermal energy flows through the material, the material impedes it. So if the thickness increases the thermal energy will have to travel farther through the resistance and be impeded more. ** Also for given temperature difference, greater thickness `dx implies smaller temperature gradient `dT / `dx. Small temperature gradient doesn't 'drive' the energy flow as much. Energy flow is also proportional to the temperature gradient. Meaning if the difference in the two temperatures is greater then the energy will move faster from one side to the other. Temperature gradient is not difference in temperatures, it's difference in temperature per unit of distance across the material. Temperature gradient is `dT / `dx, not just `dT. Greater temperature gradient means greater difference in temperature over any given small distance increment. The greater the temperature difference across this increment the more energy will flow. **
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RESPONSE --> I need to give a more consice answer
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22:21:10 principles of physics and general college physics 13.8: coeff of expansion .2 * 10^-6 C^-1, length 2.0 m. What is expansion along length if temp increases by 5.0 C?
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RESPONSE --> Linear expansion Change in the length is equal to the coefficent of linerar expansion for the material times the initial length times the change in temperature. The change would be equal to 2 x 10^-6
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22:21:37 The amount of the expansion is the product of the coefficient of expansion, the original length and the temperature difference: `dL = alpha * L0 * `dT = .2 * 10^-6 C^(-1) * 2.0 m * 5.0 C = 2 * 10^-6 m. This is 2 microns, two-onethousandths of a millimeter. By contrast the coefficient of expansion of steel is 12 * 10^-6 C^(-1);using this for the coefficient of expansion yields a change in length of 24 * 10^-6 m, or 24 microns, which is 240 times as much as for the given alloy.
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RESPONSE --> I understand
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22:23:13 query general phy 13.12: what is the coefficient of volume expansion for quartz, and by how much does the volume change? (Note that Principles of Physics and University Physics students do not do General Physics problems)
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RESPONSE --> The coeffient of volume expansion for quartz is equal to 0.12 x 10^-5 Do not need to do this problem
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22:23:21 ** The coefficient of volume expansion for quartz is 1 x 10^(-6) C^(-1). The sphere has diameter 8.75 cm, so its volume is 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3, approx.. We therefore have dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 **
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RESPONSE -->
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22:24:09 query univ 17.103 (15.93 10th edition) (Note that Principles of Physics and General College Physics students don't do University Physics problems).
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RESPONSE --> I understand
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22:25:03 ** The ice doesn't change temperature until it's melted, at which time it is in the form of water with the specific heat of water. Also the steam will come to temperature Tf so its change in thermal energy after being condensted will be 4186 J / (kg K) * .035 kg * (Tf - 100 C). I prefer to say that the sum of all the thermal energy changes is zero, so that we don't have to worry about taking a negative of a negative (which you should have done on your right-hand side, and which would have avoided the negative result). I would write the equation as follows: [.446 kg * 390 J/kg*K * (Tf - 0 C)] + [0.0950 kg * 4186 J/kg*K *(Tf - 0 C)] - .0350 kg * 2.256 x 10^6 J/kg + 4186 J / (kg K) * .035 kg * (Tf - 100 C) = 0. Noting that change in temperature of a Kelvin degree is identical to a change of a Celsius degree this gives you 170 J/C * Tf + 390 J/C * Tf - 79000 J - 14000 J + 140 J / C * Tf = 0 or 700 J / C * Tf = 93000 J, approx. or Tf = 130 C. This isn't possible--we can't end up warmer than the original temperature of the steam. We conclude that not all the steam condenses and that the system therefore reaches equilibrium at 100 C, with a mixture of water and steam. Our energy conservation equation will therefore be [.446 kg * 390 J/kg*C * (100 C - 0 C)] + [0.0950 kg * 4186 J/kg*C *(100 C - 0 C)] - mCondensed * 2.256 x 10^6 J/kg = 0 where mCondensed is the mass of the condensed steam. This gives us 17000 J + 39000 J - mCondensed * 2.3 * 10^6 J/kg = 0 or mCondensed = 56000 J / (2.3 * 10^6 J/kg) = .023 kg. We end up with .095 kg * .023 kg = .118 kg of water and .035 kg - .023 kg = .012 kg of steam. **
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RESPONSE --> OK
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22:28:32 query univ phy 17.100 (90 in 10th edition): C = 29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T . Give your solution to this problem.
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RESPONSE --> I did not finish this problem in the homework
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22:32:54 ** Specific heat is not constant but varies with temperature. The energy required to raise the temperature of 3 moles by dT degrees while at temperature T is 3 mol * C * dT = 3 mol * (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) * dT. You have to integrate this expression from T= 27 C to T = 227 C, which is from 300 K to 500 K. Antiderivative of (29.5 J/mol K + (8.2 + 10^-3 J/mol K^2) T) is F(T) = 29.5 J / (mol K) * T + (8.2 + 10^-3 J/mol K^2) * T^2 / 2. Simplify and apply Fundamental Theorem of Calculus (find F(500) - F(300) if you think the temperature T is in Kelvin or F(227) - F(27) if you think it's in Celsius; this isn't specified in the problem and while the units tend to imply Kelvin temperature the resulting specific heats would be unrealistic for most real substances), then multiply by the constant 3 moles. The result for Kelvin temperatures is about 20,000 Joules. **
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RESPONSE --> I will rework the problem
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22:33:18 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> I did not finish the assignment
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22:39:01 **The final mass of the system is .525 kg, meaning that .525 kg - (.340 kg + .150 kg) = .035 kg of steam condensed then cooled to 71 C. The thermal energy change of the calorimeter plus the water is .150 kg * 420 J/(kg C) * 56 C + .34 kg * 4187 J / (kg C) * 56 C = 83,250 J, approx. The thermal energy change of the condensed water is -Hf * .035 kg + .035 kg * 4187 J / (kg C) * (-29 C) = -Hf * .035 kg - 2930 J, approx. Net thermal energy change is zero, so we have 83,250 J - Hf * .035 kg - 4930 J = 0 so that Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. **
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RESPONSE --> I will rework the problem
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