course PHY201 w|lfߑ൹Tassignment #006
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09:13:31 General Physics 1.42. At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40000
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RESPONSE --> Assuming no input to the lake, no leaching and no evaporation: (40 000 * 1200 l * 1000cc)/50km^2 4.8e10cc/50e10cc = 0.096cm/day 0.96 cm/day * 365 day/yr = 35.040 = 35 cm/yr confidence assessment: 3
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09:14:24 ** A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube. A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2. 1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day. Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year. The volume of the lake corresponding to a depth change `dy is `dy * A, where A is the area of the lake. The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2. `dy * A = Volume so `dy = Volume / A = 4.3 * 10^6 m^3 / (5 * 10^7 m&2) = .086 m or 8.6 cm. This estimate is based on 4 people per family. A different assumption would change this estimate. STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract. INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth. Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side. You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares. If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation. ANOTHER INSTRUCTOR COMMENT: The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model. COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2 INSTRUCTOR COMMENT: Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares. 1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. **
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RESPONSE --> ok self critique assessment: 3
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09:14:32 univ 1.70 univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement
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RESPONSE --> ok confidence assessment: 3
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09:14:37 ** Letting the three vectors be A, B and C (C unknown) and the x axis point east we have A at 0 degrees, B at 315 degrees and the resultant at 0 degrees. Ax = 2, Ay = 0 (A is toward the East, along the x axis). Bx = B cos(315 deg) = 2.47, By = B sin(315 deg) = -2.47. Rx = 5.8, Ry = 0. Since Ax + Bx + Cx = Rx we have Cx = Rx - Ax - Bx = 5.8 - 2 - 2.47 = 1.33. Since Ay + By + Cy = Ry we have Cy = Ry - Ay - By = 0 - 0 - (-2.47) = 2.47. C has magnitude sqrt( Cx^2 + Cy^2) = sqrt ( 1.33^2 + 2.47^2) = sqrt(7.9) = 2.8, representing 2.8 km. C has angle arctan(Cy / Cx) = arctan(2.47 / 1.33) = 61 deg. **
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RESPONSE --> ok self critique assessment: 3
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09:14:45 **** query univ 1.82 (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product
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RESPONSE --> ok confidence assessment: 3
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09:14:52 ** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT: A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz 3.6 * 2.4 * cos (140 deg) = -6.62 To check for consistency we can calculate the components of A and B: Ax = 3.6 * cos(70 deg) = 1.23 Ay = 3.6 * sin(70 deg) = 3.38 Bx = 2.4 * cos (210 deg) = -2.08 By = 2.4 * sin(210 deg) = -1.2 dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough. ---------------------- Cross product: | A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554. Finding the components we have (Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k = ((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k = 0 i + 0 j + 5.55 k, or just 5.55 k, along the positive z axis ('upward' from the plane). INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward. The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. **
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RESPONSE --> ok self critique assessment: 3
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course Phy201 ςPa٨뢜|͊kwassignment #007
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09:26:51 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
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RESPONSE --> vAve = 'dv/'dt = 50/5 =10cm/s if v0 = 0 then vf = 20cm/s aAve = 'dv/'dt = 20/5 = 4cm/s/s vAve = 'dv/'dt = 50/3 =16.67cm/s if v0 = 0 then vf = 33.33cm/s aAve = 'dv/'dt = 33.33/3 = 11.11cm/s/s vAve = 'dv/'dt = 50/2 =25cm/s if v0 = 0 then vf = 50cm/s aAve = 'dv/'dt = 50/2 = 25cm/s/s confidence assessment: 3 confidence assessment: 3
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09:27:01 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
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RESPONSE --> ok self critique assessment: 3
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09:27:40 `q002. What are the ramp slopes associated with these accelerations?
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RESPONSE --> .5/50 = 0.010 1/50 = 0.020 1.5/50 = 0.030 confidence assessment: 3
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09:27:48 For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03.
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RESPONSE --> ok self critique assessment: 3
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09:31:22 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
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RESPONSE --> If a curve trendline were added it would be increasing at a decreasing rate. A linear trandline has an r^2 value of .966 (Excel generated) confidence assessment: 3
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09:32:51 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
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RESPONSE --> My graph was slope vs acceleration - if it were acceleration vs slope then it would have been increasing at an increasing rate. self critique assessment: 3
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09:36:58 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
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RESPONSE --> 0,.007 .05,45 .001 confidence assessment: 2
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09:37:37 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
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RESPONSE --> ok self critique assessment: 3
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09:41:38 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
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RESPONSE --> 110.686 s / 100 cycles = 1.11 s/cycle confidence assessment: 3
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09:41:48 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
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RESPONSE --> ok self critique assessment:
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09:44:35 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
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RESPONSE --> T = 2'pi'sqrt(g)*'sqrt(L) 110.686 = 2*3.1415'sqrt(g)*'sqrt(30) g = 10.345 confidence assessment: 2
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09:45:41 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
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RESPONSE --> ok self critique assessment: 3
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