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The displacement vectors A and B both have magnitudes of 3.30 m. The direction of vector A is 16.6 degrees. Find the magnitude of |A+B|, |A-B|, |B-A|, and |A-2B| graphically. PLEASE HELP ;/

Physics
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each vector has a y and x component
i and j :) ?
well you're given a magnitude

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Other answers:

|dw:1359604762762:dw|
you there?
|dw:1359605259789:dw|
yeah, im here.. trying to understand
|dw:1359605406061:dw| these need direction due to vectors .. it says A direction is 16.6 degrees sooooo
|dw:1359605476238:dw|
so in order to make sense of each vector you need to break them down into right triangles that show components of x and y i'll start with A|dw:1359605592468:dw|
now using trig functions you can find the components correct ? can you do that?
using sohcahtoa right?
A_x=sin(16.6)*A?
yes
i got 3.16 as side A_y..
so now we have \[A_x=Acos(16.6)\] \[A_y=Asin(16.6)\]
you had them switched around =]
and 0.95 as a_y.. A=3.30 right?
yep sounds correct
atleast according to my diagram lol
alright so the B is easy since there is only one component to it... which is the \[B_y\]
B=3.30 but theres no theta?
yes but it's a simple line on the y axis so it only has one component an x is what i meant =/
so you have \[B_x=3.30\]
|dw:1359606759899:dw|
now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components
"like components" ? 8.1?
so B is 3.30 and for A do I add a_x and a_y and A?
well lets do it like my book says \[R_x=A_x+B_x\] \[R_y=A_y+B_y\]
\[|A+B|=\sqrt{R_x^2+R_y^2}\]
R_x=4.25 R_y= 3.16
|dw:1359607102751:dw|
5.3 m?
\[B_x=3.30\] \[A_x=3.30cos(16.6)\] \[A_y=3.30sin(16.6)\]
6.6 m?
\[R_x=6.46\] \[R_y=.943\] \[|R|=\sqrt{6.46^2+.943^2}\]
i'm getting 6.5
6.53 to be more exact
yeah i got that too now. i rounded wrong earlier.
so for A-B would i just subtract?
yes to do it easily i'm trying to remember what it looks like lol|dw:1359607683207:dw|
a-b is the diagonal from left to right if i remember right so you have to take A and move it infront of b like this
|dw:1359607791579:dw|
oh my lordy... i give up... ;/ my grade is good enough.. 96.3% on this assignment. I'm done with this question.. i still have to do a lab report and study for calc 2... thanks for the help though ;/
I'll go back later when i have more imte tomorrow and try to figure it out lol.
i looked at my book you have to face the vectors in certain ways so a-b says to face the vecotrs inward to eachother \

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