#$&* course myh 164 10/31 7pm SOLUTIONS/COMMENTARY ON QUERY 9......!!!!!!!!...................................
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00:36:10 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4^2 + 1^2 - 2*4*1*cos(120deg) 16+1 - (-4) = 21 Sqrt (21) = 4.58 a = 4.58 4.58^2 + 1^2 - 4^2 / 2*4.58*1, 20.98 + 1 - 16 / 9.16 5.98 / 9.16 Arccos(5.98/9.16) = 49.2 B = 49.2deg. 120 + 49.2 = 169.2, 180 - 169.2 = 10.8 C = 10.8deg
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Given Solution: a= sqrt(4^2 + 1^2 -2(4)(1)cos120) a = sqrt(16+1-8cos120) a = sqrt(17-8cos120) a = 4.58 b^2 = a^2 + c^2 -2accos`beta 4^2 = 4.58^2 + 1^2 - 2(4.58)(1)cos`beta 9.16cos`beta = 4.58^2 + 1^2 - 4^2 9.16cos`beta = 5.98 beta = cos^-1(5.98/9.16) beta = 49.2deg c^2 = a^2 + b^2 -2abcos`gamma 1^2 = 4.58^2 + 4^2-2(4.58)(4)cos`gamma 36.64cos`gamma = 35.98 gamma = cos^-1(35.98/36.64) gamma = 10.89
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query problem 7.3.32 500 ft tower 15 deg slope 2 guy wires 100 ft on either side of base on slope
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00:45:52 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The triangle on the right side of the tower will have a 75deg angle between the tower and the ground because 90 - 15 = 75. a = 500ft. b = 100ft. In order to find the length of the guy wire we have to solve for c. a^2 + b^2 - 2abcosC 250000+10000 - 2*500*100*cos(75deg) 260000 - 25881.9 = 234118.1 sqrt(234118.1) = 483.86 c = 483.86 ft. The triangle on the left side of the tower will have a 105deg angle between the tower and the ground because 90 + 15 = 105. a = 250ft. b = 100ft. Solve for c to find the correct length. a^2 + b^2 - 2abcosC 62500+10000 - 2*250*100*cos(105deg) 72500 - (- 12940.95) = 85440.95 sqrt(85440.95) = 292.3 c = 292.3 ft. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** GOOD STUDENT SOLUTION: A triangle containing the tower side and the side of the slope up the hill will have angle 75 deg between tower and hill. The triangle with the side down the slope has angle 105 deg between tower and hill. The guy wire will form the hypotenuse c on each side. If c1 = length of guy wire on the right hand side of the tower and c2 = length of guy wire on the left hand side of the tower then the Law of Cosines gives us (c1)^2 = a^2 + b^2 - 2ab cos `gamma = 500ft^2 + 100ft^2 - 2(500)(100)(cos75 deg) = 260,000 - 25,881.90451 (c1)^2 = 234118.0955 c = approx. 483.8575 feet (c2)^2 = a^2 + b^2 - 2ab cos`gamma = 500ft^2 + 100ft^2 -2(500)(100)(cos105 deg) = 260,000 - (-25881.90451) (c2)^2 = 285,881.9045 c = approx. 534.68 feet ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think the 2nd part of the given solution isn’t correct, because the question says that the guy wire only goes to the middle of the tower on the left side so it would be 250 ft. instead of 500 ft. in this answer. ------------------------------------------------ Self-critique Rating: 3
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20:29:55 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10sin(2pi/3 t)
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Given Solution: ** Modeling with reference-circle point moving counterclockwise at angular velocity omega on a circle of radius A, and using the sine function, displacement will be y(t) = A sin (`omega * t + theta0), where theta0 is the initial angular position on the reference circle. We know that the spring is at equilibrium when t = 0, so theta0 is either 0 or pi. Since the spring is initially moving downward, we conclude that theta0 is pi. The amplitude of motion will be 10 cm so A = 10 cm. The period of motion is 2`pi/`omega = 3 sec, so `omega = 2`pi/(3 sec), so the equation of motion is y = 10 cm sin( 2 pi / (3 sec) * t + pi). Alternatively we could model using the cosine function, though this is less intuitive for an object moving up and down: x(t) = A cos(omega * t + theta0), in which case since the spring is at equilibrium when t = 0, theta0 is pi/2 (motion at t = 0 therefore being in the negative direction) and we get x(t) = 10 cm cos(2 pi / (3 sec) * t + pi/2). STUDENT COMMENT: I don’t understand why + pi is necessary on the end INSTRUCTOR RESPONSE: One stated condition is that the object is moving downward when t = 0. The equation you gave would have had it moving upward. In order for the object to be at equilibrium, with a sine model, the angle on the reference circle must be either 0 or pi. If the angle is 0, then since the positive direction is counterclockwise (e.g., from the positive x axis into the first quadrant), the sine function will immediately move into positive values, indicating upward motion. If the angle is pi, then continued positive motion around the reference circle will take you into the third quadrant, where the sine takes negative values; this indicates downward motion.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** query problem 7.5.12 d=5 cos(`pi /2)t Describe the motion of the object YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A.)simple harmonic B.) 5 C.) 4 D.) 1/4 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** This equation is of form d = A cos(omega * t) with A = 5 and omega = pi/2, corresponding to motion on a reference circle of radius 5 at angular velocity pi/2 rad/sec. The period of motion is 2 pi / omega = 2 pi / (pi/2) = 4. So frequency is 1 / period = 1/4. The amplitude of motion is 5 units, the coefficient of the cosine function. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** query problem 7.4.14 b = 4, c = 1, alpha = 120 deg What is the area of the triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: .5*4*1*sin(120deg) = 1.73
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Given Solution: 3 ½ b c sin alpha .5(4)(1)sin120 = 1.73 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** query problem 7.4.28 cone from circle 24 ft diameter, 100 deg sector removed Find the area of the cone obtained when you fold join the cut edges.
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00:56:16 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the area of the cone we subtract the area of the circle from the area of the sector. Circle: pi* 12^2 = 452.39 Sector: .5*12^2*1.745 = 125.64 Cone: 452.39 - 125.64 = 326.75 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 100deg = 1.745rad Area of Circle = pi*r^2 = pi*12^2 = 452.389 Area of Sector = 1/2(r^2)(`theta rad) = 1/2(12^2)(1.745) = 125.64 Area of Cone = 452.389 - 125.64 = 326.75
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** query problem 7.4.28 cone from circle 24 ft diameter, 100 deg sector removed Find the area of the cone obtained when you fold join the cut edges.
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00:56:16 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the area of the cone we subtract the area of the circle from the area of the sector. Circle: pi* 12^2 = 452.39 Sector: .5*12^2*1.745 = 125.64 Cone: 452.39 - 125.64 = 326.75 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 100deg = 1.745rad Area of Circle = pi*r^2 = pi*12^2 = 452.389 Area of Sector = 1/2(r^2)(`theta rad) = 1/2(12^2)(1.745) = 125.64 Area of Cone = 452.389 - 125.64 = 326.75
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #*&!