course Phy 201 æWòø±îûÅÄvº‡…ù}i”ÆÝáûµË»assignment #006
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02:09:33 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 --> 1200 l/ day/ fam * 365 days * 10,000 fams = 4.4*10^9 cm^3 = 4.4*10^4 km ^3 . (4.4*10^4 km ^3) / 50 km ^2 = 880 km confidence assessment: 0
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15:06:46 ** 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 --> I don't understand where you got the number .001 m^3. Doesn't 1000 cm^3 = 10.00 m^3? self critique assessment: 2
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15:06:59 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 --> I'm in Phy 201 confidence assessment: 3
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15:07:04 ** 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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15:07:16 **** 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 --> I'm in Phy 201 confidence assessment: 3
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15:07:24 ** 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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