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anonymous
 3 years ago
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, AB, BA, and A2B graphically.
PLEASE HELP ;/
anonymous
 3 years ago
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, AB, BA, and A2B graphically. PLEASE HELP ;/

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0each vector has a y and x component

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well you're given a magnitude

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359604762762:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359605259789:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, im here.. trying to understand

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359605406061:dw these need direction due to vectors .. it says A direction is 16.6 degrees sooooo

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359605476238:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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 Adw:1359605592468:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now using trig functions you can find the components correct ? can you do that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0using sohcahtoa right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i got 3.16 as side A_y..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so now we have \[A_x=Acos(16.6)\] \[A_y=Asin(16.6)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you had them switched around =]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 0.95 as a_y.. A=3.30 right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0atleast according to my diagram lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright so the B is easy since there is only one component to it... which is the \[B_y\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0B=3.30 but theres no theta?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes but it's a simple line on the y axis so it only has one component an x is what i meant =/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so you have \[B_x=3.30\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359606759899:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now you just add the like components to get 2 sides and the magnitude will be the square root of those squared total components

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"like components" ? 8.1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so B is 3.30 and for A do I add a_x and a_y and A?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well lets do it like my book says \[R_x=A_x+B_x\] \[R_y=A_y+B_y\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[A+B=\sqrt{R_x^2+R_y^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359607102751:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[B_x=3.30\] \[A_x=3.30cos(16.6)\] \[A_y=3.30sin(16.6)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[R_x=6.46\] \[R_y=.943\] \[R=\sqrt{6.46^2+.943^2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.06.53 to be more exact

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i got that too now. i rounded wrong earlier.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so for AB would i just subtract?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes to do it easily i'm trying to remember what it looks like loldw:1359607683207:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ab 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1359607791579:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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 ;/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll go back later when i have more imte tomorrow and try to figure it out lol.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i looked at my book you have to face the vectors in certain ways so ab says to face the vecotrs inward to eachother \
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