Problem 19

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. **

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

16:00:05

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).

......!!!!!!!!...................................

RESPONSE -->

ok

.................................................

......!!!!!!!!...................................

16:00:10

** 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. **

......!!!!!!!!...................................

RESPONSE -->

ok