course Phy 231 ????????????assignment #003
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18:07:56 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 --> Not in either of those courses. confidence assessment: 3
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18:08:02 ** 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 smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> OK self critique assessment: 3
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19:48:01 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 --> The problem you are referring to is problem 31 not 34. The solution to problem 31 is: to find the magnitude, you first find the distance from the starting point to the 45 degree turn. We use c^2 = a^2+b^2 to find 4.8km, then add 3.1km to get 7.9m. To find the direction of the resultant displacement, I used the formula cos(a) = (2.6)/(7.9), a=71degrees north of east. The difference in problem numbers scares me, I hope I have the right book. I have the 11th edition. confidence assessment: 2
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19:57:57 ** 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 --> Oh okay my answer was wrong. To solve the problem you first find the components of the upper triangle and add the components x and y displacemtn of the lower triangle.Ax=3.1kmcos(45deg)=2.19.Ay=3.1kmsin(45deg)=2.19. Adding the x components together:2.19+4.0=6.19. Adding the y components together: 2.19+2.6=4.79. To find the displacement: c^2=6.29^2 + 4.79^2, c=7.9km, which is a little different from their answer. To find the angle of displacement: theta=arctan(4.79/6.19) = 38deg. self critique assessment: 2
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