Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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 ;/

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

you there?
yeah, im here.. trying to understand
|dw:1359605406061:dw| these need direction due to vectors .. it says A direction is 16.6 degrees sooooo
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?
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 \[B_x=3.30\]
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 \[R_x=A_x+B_x\] \[R_y=A_y+B_y\]
R_x=4.25 R_y= 3.16
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
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 \

Not the answer you are looking for?

Search for more explanations.

Ask your own question