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November 26, 2014

November 26, 2014

Posted by **JOAN** on Saturday, November 24, 2012 at 4:57am.

- UNIVERSITY PHYSICS 1 -
**JOAN**, Saturday, November 24, 2012 at 5:00amPLEASE ANSWERS MY ASSIGNMENT

- UNIVERSITY PHYSICS 1 -
**Steve**, Saturday, November 24, 2012 at 5:11amIf I read your somewhat confusing vectors correctly, they are (in units of b)

-6.30i

-4.00cos40i - 4.00sin40j

3.00cos50i - 3.00sin50j

-5.00j

the distances traveled in each case is

6.30

4.00

3.00

5.00

adding up to 18.30

the directions are all given. just read 'em off

- UNIVERSITY PHYSICS 1 -
**JOAN**, Saturday, November 24, 2012 at 5:41amwhat the answer to letter c. how can i find the answer please illustrate to me

- UNIVERSITY PHYSICS 1 -
**Steve**, Saturday, November 24, 2012 at 10:46amadding up the i and j values, the total displacement is

-12.4i - 4.87j = 13.32 @ 158.6°

|-12.4i - 4.87j| = √(12.4^2 + 4.87^2)

tan158.6° = -4.87/-12.4

- UNIVERSITY PHYSICS 1 -
**JOAN**, Sunday, November 25, 2012 at 5:28amhow did u get -12.4i and 4.87j

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