course Phy 202 please respond relative to goal of CI am resubmitting some of these that have not shown up yet.
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11:37:04 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 --> ok
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11:49:30 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 --> using the triply-reduced is not accurate because the optical size reducer is not accurate. When measuering a pencil with this the pencil dose not fit so you have to measure half at a time which also affects the accuracy. confidence assessment: 3
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11:51:22 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 --> This increases the accuracy but it still dose not help the fact that the pencile dose not fit on the ruler which is where you loose most of the accuracy. confidence assessment: 3
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12:13:47 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 slightly reduced copier would be easier to use because it is easier to read, but if you kow the distortion either one should be eaqually acurate. confidence assessment: 3
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12:43:46 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 --> It would be changing at different rates. confidence assessment: 1
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12:47:42 Suppose the actual second-difference behavior of the depth vs. clock time is in fact linear. How nearly do you think you could estimate the slope of that graph from data taken as indicated above (e.g., within 1% of the correct slope, within 10%, within 30%, or would no slope be apparent in the second-difference graph)? Again no extensive analysis is expected, but give a brief synopsis of how you considered various effects in arriving at your estimate.
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RESPONSE --> within 10% confidence assessment: 2
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]koѴҘoZ|c assignment #002 002. Physics II 07-19-2008
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13:01:38 In your own words explain how the introductory experience with scotch tape illustrates the existence of two types of charge.
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RESPONSE --> When you stick the two pieces of tape together and pull them apart, they attract. When you attach the pieces to a table and place another piece on the table and pull it off, it is attracted to one piece and it repels the othe piece. This shows that the third piece interacted with two opposite types of charge. confidence assessment: 3
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13:12:13 In your own words explain how the introductory experience with scotch tape supports the idea that the force between two charged particles acts along a straight line through those particles, either attracting the forces along this line or repelling along this line.
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RESPONSE --> When the two pieces of tape were attracted, they went towards each other in a straight line. The same results were achieved when the two pieces of tape were attached to the table and the third was introduced. confidence assessment: 2
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13:16:14 In your own words explain why this experience doesn't really prove anything about actual point charges, since neither piece of tape is confined to a point.
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RESPONSE --> The pieces of tape were not fixed to a exact point because they could accidentally or deiberately be moved, therefore they are not necessarily point charges. confidence assessment: 2
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13:29:56 If one piece of tape is centered at point A and the other at point B, then let AB_v stand for the vector whose initial point is A and whose terminal point is B, and let AB_u stand for a vector of magnitude 1 whose direction is the same as that of AB_u. Similarlylet BA_v and BA_u stand for analogous vectors from B to A. Vectors of length 1 are called unit vectors. If the pieces attract, then in the direction of which of the two unit vectors is the tape at point A pushed or pulled? If the pieces repel, then in the direction of which of the two unit vectors is the tape at B pushed or pulled? Explain.
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RESPONSE --> If the pieces attract, the tape at point A is pulled toward B, so the vector is toward B. If the pieces repel, the tape at point B is pushed away from A, so the vector is away from A. confidence assessment: 3
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13:38:24 Using the notation of the preceding question, which you should have noted on paper (keep brief running notes as you do qa's and queries so you can answer 'followup questions' like this), how does the magnitude of the vector AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector compare with the distance between A and B?
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RESPONSE --> The magnitude of vector AB_v should be the same amount but opposite in sign. As the distance becomes greater, the magnitude of the vectors will become greater, whether that means more positive or more negative. confidence assessment: 3
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13:51:00 Using the notation of the preceding question, how is the force experienced by the two pieces of tape influenced by the magnitude of AB_v or BA_v?`aThe expected answer is that the force exerted by two charges on one another is inversely proportional to the square of the distance between them. So as the magnitudes of the vectors, which are equal to the separation, increases the force decreases with the square of the distance; and/or if the magnitude decreases the force increases in the same proportinality. The two pieces of tape are not point charges, so this is not strictly so in this case, as some parts of the tape are closer than to the other tape than other parts.
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RESPONSE --> The force becomes less as the magnitude of the vectors becomes less. The force also becomes less as the distance becomes greater. confidence assessment: 2
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14:09:17 Query introductory set #1, 1-5 Explain how we calculate the magnitude and direction of the electrostatic force on a given charge at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> Use the distance equation to find the distance between the charges and then use the charge in Coulomb's Law to find the force. confidence assessment: 2
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14:24:07 Explain how we calculate the magnitude and direction of the electric field at a given point of the x-y plane point due to a given point charge at the origin.
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RESPONSE --> First, find the distance using the distance equation. Then, multiply the charges and then divide by the distance. This will give you the force. confidence assessment: 2
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14:25:14 ** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where q1 is the charge at the origin. The electric field is therefore F / Q = k q1 / r^2. The direction is the direction of the force experienced by a positive test charge. The electric field is therefore directly away from the origin (if q1 is positive) or directly toward the origin (if q1 is negative). The direction of the electric field is in the direction of the displacement vector from the origin to the point if q1 is positive, and opposite to this direction if q1 is negative. To find the direction of this displacement vector we find arctan(y / x), adding 180 deg if x is negative. If q1 is positive then this is the direction of the field. If q1 is negative then the direction of the field is opposite this direction, 180 degrees more or less than the calculated angle. **
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RESPONSE --> OK self critique assessment: 3
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ʜȞHŜ assignment #003 003. Physics II 07-19-2008
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15:18:43 In your own words explain the meaning of the electric field.
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RESPONSE --> An electric field is the spherical region around an electric charge that describes the force. confidence assessment: 3
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15:36:01 Query Principles of Physics and General Physics problem 16.15 charges 6 microC on diagonal corners, -6 microC on other diagonal corners of 1 m square; force on each. What is the magnitude and direction of the force on the positive charge at the lower left-hand corner?
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RESPONSE --> 14.9 N towards the center of the square confidence assessment: 3
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15:38:54 ** The charges which lie 1 meter apart are unlike and therefore exert attractive forces; these forces are each .324 Newtons. This is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6 * 10^-6 C) / ( 1 m)^2 = 324 * 10^-3 N = .324 N. Charges across a diagonal are like and separated by `sqrt(2) meters = 1.414 meters, approx, and exert repulsive forces of .162 Newtons. This repulsive force is calculated using Coulomb's Law: F = 9 * 10^9 N m^2/C^2 * ( 6 * 10^-6 C) * ( 6* 10^-6 C) / ( 1.414 m)^2 = 162 * 10^-3 N = .162 N. The charge at the lower left-hand corner therefore experiences a force of .324 Newtons to the right, a force of .324 Newtons straight upward and a force of .162 Newtons at 45 deg down and to the left (at angle 225 deg with respect to the standard positive x axis, which we take as directed toward the right). This latter force has components Fy = .162 N sin(225 deg) = -.115 N, approx, and Fx = .162 N cos(225 deg) = -.115 N. The total force in the x direction is therefore -.115 N + .324 N = .21 N, approx; the total force in the y direction is -.115 N + .324 N = .21 N, approx. Thus the net force has magnitude `sqrt( (.21 N)^2 + (.21 N)^2) = .29 N at an angle of tan^-1( .21 N / .21 N) = tan^-1(1) = 45 deg. The magnitude and direction of the force on the negative charge at the lower right-hand corner is obtained by a similar analysis, which would show that this charge experiences forces of .324 N to the left, .324 N straight up, and .162 N down and to the right. The net force is found by standard vector methods to be about .29 N up and to the left. **
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RESPONSE --> OK self critique assessment:
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16:37:34 query university physics 21.68 (22.52 10th edition) 5 nC at the origin, -2 nC at (4 cm, 0). If 6 nC are placed at (4cm, 3cm), what are the components of the resulting force?
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RESPONSE --> x component = -2.06x10^-6 N y component = -3.25x10^-6 N confidence assessment: 3
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16:58:58 Query univ phy 21.76 (10th edition 22.60) quadrupole (q at (0,a), (0, -a), -2q at origin). For y > a what is the magnitude and direction of the electric field at (0, y)?
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RESPONSE --> confidence assessment:
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17:03:50 ** The magnitude of the field due to the charge at a point is k q / r^2. For a point at coordinate y on the y axis, for y > a, we have distances r = y-a, y+a and y. The charges at these distances are respectively q, q and -2q. So the field is k*q/(y - a)^2 + k*q/(y + a)^2 - 2k*q/y^2 = 2*k*q*(y^2 + a^2)/((y + a)^2*(y - a)^2) - 2*k*q/y^2 = 2*k*q* [(y^2 + a^2)* y^2 - (y+a)^2 ( y-a)^2) ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - (y^2 - a^2)^2 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [y^4 + a^2 y^2 - y^4 + 2 y^2 a^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) = 2*k*q* [ 3 a^2 y^2 - a^4 ] / ( y^2 (y + a)^2*(y - a)^2) . For large y the denominator is close to y^6 and the a^4 in the numerator is insignifant compared to a^2 y^2 sothe expression becomes 6 k q a^2 / y^4, which is inversely proportional to y^4. **
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RESPONSE --> OK self critique assessment: 3
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17:05:32 query univ 22.102 annulus in yz plane inner radius R1 outer R2, charge density `sigma.What is a total electric charge on the annulus?
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RESPONSE --> OK confidence assessment: 3
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17:05:58 ** The total charge on the annulus is the product Q = sigma * A = sigma * (pi R2^2 - pi R1^2). To find the field at distance x along the x axis, due to the charge in the annulus, we first find the field due to a thin ring of charge: The charge in a thin ring of radius r and ring thickness `dr is the product `dQ = 2 pi r `dr * sigma of ring area and area density. From any small segment of this ring the electric field at a point of the x axis would be directed at angle arctan( r / x ) with respect to the x axis. By either formal trigonometry or similar triangles we see that the component parallel to the x axis will be in the proportion x / sqrt(x^2 + r^2) to the magnitude of the field from this small segment. By symmetry only the xcomponent of the field will remain when we sum over the entire ring. So the field due to the ring will be in the same proportion to the expression k `dQ / (x^2 + r^2). Thus the field due to this thin ring will be magnitude of field due to thin ring: k `dQ / (x^2 + r^2) * x / sqrt (x^2 + r^2) = 2 pi k r `dr * x / (x^2 + r^2)^(3/2). Summing over all such thin rings, which run from r = R1 to r = R2, we obtain the integral magnitude of field = integral ( 2 pi k r x /(x^2 + r^2)^(3/2) with respect to r, from R1 to R2). Evaluating the integral we find that magnitude of field = 2* pi k *x* | 1 /sqrt(x^2 + r1^2) - 1 / sqrt(x^2 + r2^2) | The direction of the field is along the x axis. If the point is close to the origin then x is close to 0 and x / sqrt(x^2 + r^2) is approximately equal to x / r, for any r much larger than x. This is because the derivative of x / sqrt(x^2 + r^2) with respect to x is r^2 / (x^2+r^2)^(3/2), which for x = 0 is just 1/r, having no x dependence. So at small displacement `dx from the origin the field strength will just be some constant multiple of `dx. **
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RESPONSE --> OK self critique assessment: 3
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14:38:30 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 --> the rate of conductivity is greater as the cross sectional area increases and is greater as the temperature dfference is greater. It is less as the thickness becomes greater.
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14:39:21 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 --> ok
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14:40:45 ** 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 --> ok
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14:44:50 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 --> 0.000002m
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14:45:17 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 --> ok
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14:50:39 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 --> 1*10^-6 /C is the coefficient of volume expansion of quartz
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14:51:14 ** 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 --> ok
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14:51:24 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 --> ok
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14:52:05 ** 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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14:52:11 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 --> ok
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14:52:18 ** 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 --> ok
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14:52:25 University Physics Problem 17.106 (10th edition 15.96): Give your solution.
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RESPONSE --> ok
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14:52:30 **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 --> ok
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Æ䑊J\陸ٙ assignment #012 012. `query 2 Physics II 07-20-2008
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14:56:56 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> the temperature change, the final temperature minus the inital temperature times the mass times the specific heat gives the energy change.
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14:57:57 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> ok
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14:58:08 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE -->
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14:58:34 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> ok
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15:05:14 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> 930C
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15:07:34 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE --> 7.4%
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15:08:01 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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RESPONSE --> ok
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15:08:07 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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RESPONSE --> ok
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15:08:12 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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RESPONSE --> ok
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15:08:25 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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RESPONSE --> ok
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15:08:31 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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RESPONSE --> ok
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15:08:45 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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RESPONSE --> ok
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Rv^MܺZ assignment #012 012. `query 2 Physics II 07-20-2008
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15:21:34 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> Energy lost or gained equalls mass times specific heat times change in temperature
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15:21:47 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> ok
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15:21:54 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE --> ok
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15:22:06 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> ok
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15:22:37 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> 930 C
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15:22:53 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE --> 7%
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15:22:58 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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RESPONSE --> ok
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15:23:04 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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RESPONSE --> ok
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15:23:09 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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RESPONSE --> ok
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15:23:17 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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RESPONSE --> ok
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15:23:26 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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RESPONSE --> ok
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15:23:33 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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RESPONSE --> ok
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|䄰BOѡDP assignment #014 014. `query 4 Physics II 07-20-2008
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15:30:45 query introset plug of water from cylinder given gauge pressure, c-s hole area, length of plug. Explain how we can determine the velocity of exiting water, given the pressure difference between inside and outside, by considering a plug of known cross-sectional area and length?
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RESPONSE --> we can use Bernoulli's equation
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15:31:06 ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. If L is the length of the plug then the net force P * A acts thru dist L doing work P * A * L. If init vel is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). **
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RESPONSE --> ok
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15:31:12 prin phy problem 10.3. Mass of air in room 4.8 m x 3.8 m x 2.8 m.
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RESPONSE --> ok
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15:31:35 The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass is close to the average mass of a medium-sized person.
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RESPONSE --> ok
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15:31:40 prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person.
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RESPONSE --> ok
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15:31:45 The density of blood is close to that of water, 1000 kg / m^3. So the pressure difference corresponding to a height difference of 1.6 m is pressure difference = rho g h = 1000 kg/m^3 * 9.80 m/s^2 * 1.60 m = 15600 kg / ( m s^2) = 15600 (kg m / s^2) / m^2 = 15600 N / m^2, or 15600 Pascals. 1 mm of mercury is 133 Pascals, so 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. Blood pressures are measured in mm of mercury; e.g. a blood pressure of 120 / 70 stands for a systolic pressure of 120 mm of mercury and a diastolic pressure of 70 mm of mercury.
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RESPONSE --> ok
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15:36:40 prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force
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RESPONSE --> boyant force up 21189 N Force of gravity down 9114 N net force up 12075 N
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15:37:16 ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the net force is about buoyant force - weight = 20,500 N - 9100 N = 11,400 N If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. **
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RESPONSE --> ok
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15:37:21 univ 14.51 (14.55 10th edition) U tube with Hg, 15 cm water added, pressure at interface, vert separation of top of water and top of Hg where exposed to atmosphere. Give your solution to this problem.
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RESPONSE --> ok
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15:37:26 ** At the interface the pressure is that of the atmosphere plus 15 cm of water, or 1 atm + 1000 kg/m^3 * 9.8 m/s^2 * .15 m = 1 atm + 1470 Pa. The 15 cm of water above the water-mercury interface must be balanced by the mercury above this level on the other side. Since mercury is 13.6 times denser than water there must therefore be 15 cm / (13.6) = 1.1 cm of mercury above this level on the other side. This leaves the top of the water column at 15 cm - 1.1 cm = 13.9 cm above the mercury-air interface. } Here's an alternative solution using Bernoulli's equation, somewhat more rigorous and giving a broader context to the solution: Comparing the interface between mercury and atmosphere with the interface between mercury and water we see that the pressure difference is 1470 Pa. Since velocity is zero we have P1 + rho g y1 = P2 + rho g y2, or rho g (y1 - y2) = P2 - P1 = 1470 Pa. Thus altitude difference between these two points is y1 - y2 = 1470 Pa / (rho g) = 1470 Pa / (13600 kg/m^2 * 9.8 m/s^2) = .011 m approx. or about 1.1 cm. The top of the mercury column exposed to air is thus 1.1 cm higher than the water-mercury interface. Since there are 15 cm of water above this interface the top of the water column is 15 cm - 1.1 cm = 13.9 cm higher than the top of the mercury column. NOTE BRIEF SOLN BY STUDENT: Using Bernoullis Equation we come to: 'rho*g*y1='rho*g*y2 1*10^3*9.8*.15 =13.6*10^3*9.8*y2 y2=.011 m h=y1-y2 h=.15-.011=.139m h=13.9cm. **
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RESPONSE --> ok
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