course Phy 231 ??????????????assignment #003003. `Query 3
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14:38:03 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> 323 cm, or 3.23 m 1.80 m = 180 cm 180 cm + 142.5 cm = 322.5 cm The micrometers are disregarded because they fall so far below the number of significant figures we are using. They would not have effect on rounding. If we were multiplying, we'd not disregard them. The answer must be rounded to the number of significant figures that are in the number with the least sig. figs. in the question, which is three for 1.80m
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14:41:18 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing below .01 m can be distinguished. 142.5 cm is .01425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, good to within .00000001 m. Then theses are added you get 1.81425534 m; however the 1.80 m is only good to within .01 m so the result is 1.81 m. The rest of the number is meaningless, since the first number itself could be off by as much as .01 m. **
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RESPONSE --> I didn't convert my numbers to meters. But shouldn't the answer still be greater than one meter? 142.5 cm is greater than one meter, and we are adding that length to 1.80 cm.
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15:22:48 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> This question looks like a portion of the 1.38 problem. Travelling 4.0 km along a line, then 3.1 along a line that is 45 degrees from the first line results in a displacement that can be described by a vector R with components Rx =6.19 km and Ry = 2.19 km. Considering the 4.0 km vector ( vector A) to be oriented along the x - axis, the 3.1km vector (vector B) will be 45 degrees measured in the sense from 4.0 km vector, counterclockwise. Ax = 4.0, Ay = 0 With an angle of 45 degrees, the x and y components of vector be will be equal and are 2.19 km: 3.1 cos (45) = 2.19 These are added to the components of vector A: Rx = 4.0 km + 2.19 km = 6.19 km Ry = 0.0 km + 2.19 km = 2.19 km The magnitude of the resultant vector is the square root of the sum of the squares of the components Rx and Ry. = sqrt (38.32 km^2+ 4.796 km^2) = 6.57 km The angle of the resultant vector is the angle in the first quadrant that corresponds to the inverse tangent of the ratio of the y component to the x component: arctan ( 2.19 km / 6.19 km ) = 19.48 degrees. Rounding to the correct number of significant figures (two), the vector that describes the displacement of the delivery truck is 6.6 km , 19 degrees
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15:24:52 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE --> I didn't add the 2.6 km vector from the text's description of the problem. Did you want us to rework the problem with the vectors you gave, or refer to the problem in the text?
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