Assignment3_RESUBMIT_query

course Phy 241

?????}?T?|???{?assignment #003

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

??··?????}·?d

Physics I

06-26-2006

??\???????????

assignment #003

??··?????}·?d

Physics I

06-26-2006

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

09:52:19

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)

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

RESPONSE -->

na

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

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

09:52:23

** 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. **

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

RESPONSE -->

na

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

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

10:08:40

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 -->

Problem 1.34 in my Physics text book reads as follows: Use a scale drawing to find the x and y components of the following vectors. For each vector the numbers given are i) the magnitude of the vector and ii) the angle, measured in the sense from the positive x axis toward the positive y axis, that it makes with the positive x axis, Find a) magnitude 9.30 m, angle 60.0 degrees, b) magnitude 22.0 km angle 135 degrees, c) magnitude 6.35 cm, angle 307 degrees.

My solution is as follows:

A) cos 60=Ax/9.30 ; Ax=4.65 m

sin 60=Ay/9.30 ;Ay=8.05 m

B) sin 135= Bx/22.0 ; Bx=15.56 km

cos 135=By/22.0; By=-15.56 km

C) sin 307=Cx/6.35; Cx=-5.07

cos 307=Cy/6.35 Cy=3.82

What does the information in parenthesis mean? Do I have the wrong textbook?

The problem in your text is actually #36; it was mislabeled here. As far as I know there aren't many errors of this nature in the assignments and queries, but there was one here.

In any case you appear to understand the necessary trigonometry.

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

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

10:11:02

** 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 -->

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

"

This looks good. Let me know if you have questions.