course Phy 231 6/14 4:30 006. `query 6*********************************************
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Given Solution: The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions: • the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities justification in terms of definitions: if acceleration is uniform, then the v vs t graph is linear so that the average velocity is equal to the average of initial and final velocities • multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m justification in terms of definitions: ave velocity is ave rate of change of position with respect to clock time, so vAve = `ds / `dt, from which it follows that `ds = vAve * `dt • the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2 justification in terms of definitions: ave acceleration is ave rate of change of velocity with respect to clock time, so aAve = `dv / `dt. in this situation acceleration is uniform, so we can if we wish use just plain a instead of aAve The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters. A solution to the problem: Using the fourth equation of motion with the given information (`ds, a and vf) we have vf^2 = v0^2 + 2 a `ds , which we solve for v0 to get v0 = +- sqrt( vf^2 - 2 a `ds) = +- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) = +- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) = +- sqrt( 400 m^2 / s^2) = +- 20 m/s. If vf = 20 m/s then we could directly reason out the rest (vAve would be 25 m/s, so it would take 5 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - 20 m/s) / (2 m/s^2) = 10 m/s / (2 m/s^2) = 5 s. If vf = -20 m/s then we could directly reason out the rest (vAve would be (-20 m/s + 30 m/s) / 2 = 5 m/s, so it would take 25 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - (-20 m/s) ) / (2 m/s^2) = 50 m/s / (2 m/s^2) = 25 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok ********************************************* Question: An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance. If the positive direction is down the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the air resistance positive or negative? If the positive direction is up the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the automobile's acceleration positive or negative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the positive direction: velocity is positive Air resistance is negative In the negative direction: velocity is negative Acceleration is positive confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you are riding in the car your perception is that the forward direction is the one in which you are moving. You can stick your hand out the window and feel that the air resistance is in the 'backward' direction. Since the automobile is speeding up its acceleration is in its direction of motion. The velocity is down the hill. Thus the direction of the velocity and acceleration are both down the hill, and the direction of the air resistance is up the hill. Therefore • If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. • If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* Ok Question: (gen and univ phy) At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40,000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 10,000 families in the town (assuming 4 people/family) 1 m^3 is 1000 L; Convert the liters to m^3: 1200L / 1000L = 1.2 m^3 10,000 families* 1.2m^3 / day *365 days = 4.38*10^6 m^3/family Surface area of the lake is 50km^2 50km^2 (10^6m^2 / km^2) = 5*10^7 m^2 Depth of the lake *area = volume Depth = volume / area = 4.38*10^6 m^3/ 5*10^7 m^2 = .0876 meters confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should know how the milliliter, liter and cubic meter, three common measure of volume, are related: • a milliliter is the volume of a cube 1 cm on a side • a liter is the volume of a cube 10 cm on a side • a cubic meter is the volume of a cube 1 meter on a side so that • 1 liter = (10 cm)^3 = 1000 cm^3 or 1000 milliliters • 1 cubic meter = (100 cm)^2 = 1 000 000 milliliters • 1 cubic meter = 1 000 000 milliliters / (1000 milliliters / liter) = 1000 liters You should also understand the following images, which will allow you to visualize and thereby reason out these and similar relationships • It takes 10 sides of length 10 cm to make a 1 meter side, so to fill a cubic meter with 10 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A cubic meter is therefore 10 * 10 * 10 liters, or 1000 liters. • It takes 10 sides of length 1 cm to make a 10 cm side, so to fill a liter (a 10 cm cube) with 1 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A liter is therefore 10 * 10 * 10 milliliters, or 1000 milliliters. It is also helpful to visualize the relationship between a cubic meter and a cubic kilometer: • A kilometer is 1000 meters. • A cubic kilometer is therefore (1000 meters) ^ 3 = 1 000 000 000 m^3, or 10^9 m^3, or a billion m^3. • A cubic kilometer is visualized as 1000 layers each with 1000 rows each made up of 1000 one-meter cubes, for a total of 1000 * 1000 * 1000 = 1 000 000 000 one-meter cubes. Finally you should understand what a cylinder is and how we find its area. • The volume of a cylinder is equal to the area of its cross-section multiplied its altitude, as you saw in the q_a_initial_problems. • The links below explain prisms and cylinders, and their volumes, in elementary terms: The solution: 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. Visualize the surface of the lake as the base of a large irregular cylinder. The volume of a cylinder is the product of the area of its base and its altitude. • The volume of water corresponding to a depth change `dy is therefore `dy * A, where A is the area of the lake. • It might be helpful to think of a layer of ice several centimeters thick on top of the lake. The cross-sections of this layer all have very nearly the same size and shape, so it can be viewed as a cylinder with cross-sectional area equal to the area of the lake, and a thickness of several centimeters. 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It helps to draw out a bunch of pictures for these problems, because I kept confusing myself with the relationships between the liters and the meters and everything. Looking at a drawing of what you want to find helps a lot. ------------------------------------------------ Self-critique rating #$&* 3 Question: univ 1.74 (11th edition 1.70) univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 km, 0 degrees 3.5 km, 320 degrees Total magnitude = 5.8 km direction = 0 degrees Ax = 2cos0 = 2 ay = 2sin0 = 0 Bx = 3.5cos315 = 2.47 by = 3.5sin315 = -2.47 Cx = unknown cy = unknown For the direction to be 0, ry needs to be 0, because arctan 0 = 0 degrees 0 = 0-2.47+cy, therefore cy = 2.47 Then, for the magnitude to be 5.8 km, 5.8 = sqrt(rx^2 + 0^2), so rx = 5.8. Plug 5.8 in with the rest of the components to find cx 5.8 = 2+2.47+cx and cx = 1.33 So, you have the components of C: Magnitude of the vector = sqrt(1.33^2 +2.47^2) = 2.81 km Direction of the vector = arctan (2.47 / 1.33) = 61.7 degrees confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok Self-critique rating #$&* OK Question: `q**** query univ 1.86 (11th edition 1.82) (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product Your solution: Ax = 3.6cos70 = 1.23 Ay = 3.6sin70 = 3.38 Bx = 2.4cos210 = -2.08 By = 2.4sin210 = -1.2 a) Scalar product: 1.23(-2.08) + (3.38(-1.2) = -2.5584-4.056 = -6.61 b) Vector Product: Add Az = 0 and Bz = 0 Cx = Ay(bz)-by(az) = 0 Cy = az(bx)-ax(bz) = 0 Cz = ax(by)-bx(ay) = 1.23(-1.2)- (-2.08)(3.38) = -1.476 +7.0304 = 5.55 km = magnitude Direction = straight up the z axis confidence rating #$&*: Ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ** Self-critique (if necessary): To do the scalar product without components, you do magnitude of A *magnitude of B *cos(theta). To do the vector product without the components you do magnitude of A * magnitude of B *sin(theta). Self-critique rating #$&* 3 "