anonymous
  • anonymous
I'm adding my vectors incorrectly. Please give me a min to draw it.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1360530853167:dw|
anonymous
  • anonymous
\[\vec{F}_{net}=(\vec F_{41}-\vec F_{31}\cos(45))\hat i +(\vec F_{21}-\vec F_{31}\sin(45)) \hat j\]
anonymous
  • anonymous
|dw:1360531227825:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[F_{net}=\left( \frac{kq_4q_1}{r_{41}^2}-\frac{kq_3q_1}{r_{31}^2}\sin(45)\right) \hat i \left(\frac{kq_2q_1}{r_{21}^2}-\frac{kq_3q_1}{r_{31}^2}\cos(45)\right) \hat j\]
anonymous
  • anonymous
\[q_1=3\times10^{-9}\] \[q_2=9\times10^{-9}\] \[q_3=-9\times10^{-9}\] \[q_4=9\times10^{-9}\]
anonymous
  • anonymous
|dw:1360531777377:dw|
anonymous
  • anonymous
@phi
anonymous
  • anonymous
I get \[(72974.5 N)\hat i\]
anonymous
  • anonymous
which is wrong...
anonymous
  • anonymous
.000392 should be the answer
anonymous
  • anonymous
|dw:1360532089439:dw|
phi
  • phi
8 orders of magnitude is quite an error
anonymous
  • anonymous
it sure is ...haha
anonymous
  • anonymous
the charges were given in nano coulombs. I converted them to coulombs. \[q_1=+3nC=3\times 10^{-9}\]
anonymous
  • anonymous
k=8.99E9
phi
  • phi
First, what units is the answer given in ?
anonymous
  • anonymous
Newtons \hat{i}
anonymous
  • anonymous
maybe I read the question wrong
phi
  • phi
your approach is ok. I would write down the components of each vector
phi
  • phi
the k 9e9 times 9e-9 gives 81. that still leaves a 3e-9 are you double counting the k ?
anonymous
  • anonymous
|dw:1360532909843:dw| oh my gosh ...yes I think soooo
anonymous
  • anonymous
I'll do it on wolfram... one sec
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%28%28%288.99e9%29%289e%28-9%29%29%283e%28-9%29%29%29%2F%28.02^2%29%29-%28%288.99e9%29%28-9e%28-9%29%29%283e%28-9%29%29%29%2F%28.028^2%29%29cos%2845%29%29
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=%28coulombs+constant+*3nC*4nC%29%2F%282cm^2%29++-+++%28coulombs+constant+*-9nC*3nC%29%2F%282.8cm^2%29
anonymous
  • anonymous
I found another error...this looks better http://www.wolframalpha.com/input/?i=%28coulombs+constant+*9nC*3nC%29%2F%282cm^2%29++-+++%28coulombs+constant+*-9nC*3nC%29%2F%282.8cm^2%29
anonymous
  • anonymous
and another error.... http://www.wolframalpha.com/input/?i=%28coulombs+constant+*9nC*3nC%29%2F%28%282cm%29^2%29++-+++%28coulombs+constant+*-9nC*3nC%29%2F%28%282.8cm%29^2%29
anonymous
  • anonymous
silly question... \[F_{31}=F_{31x}+F_{31y}\] is this true?
anonymous
  • anonymous
or is it \[F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}\]
phi
  • phi
ignoring the units (we can figure those out later) the force from the top left corner is <9*3/4 i 0 j> the force from the bottom right is < 0 i 9*3/4 j> the force from the bottom left is < -9*3/8 * 1/sqrt(2) i - 9*3/8 *1/sqrt(2) j> add up the components to get 27/4( 1- 1/2sqrt(2)) for both i and j components that number is 4.3635 now to get the units 1e-18 C^2 * 9e-9 N-M^2 /C^2 * 1e4 M^2 (to fix the cm) that factor is 9e-5 and our result is 4.3635*9e-5 = 0.0003927 N for both the i and j components
phi
  • phi
as for your "silly" question. if by F31x you mean < F31x 0> and by F31y < 0 F31y> then yes, F31= F31x+F31y the magnitude of F31 is as you wrote it.
anonymous
  • anonymous
why is "the force from the bottom left is < -9*3/8 * 1/sqrt(2) i" Why did you divide by 8
phi
  • phi
the distance (diagonal) is 2 sqrt(2). squared you get 4*2= 8 the 1/sqrt(2) is cos(45) (often written as sqrt(2)/2 but we don't need to write it that way)
phi
  • phi
your 2.8 was just an approximation
phi
  • phi
I think the resultant force vector is < 0.0003927 i 0.0003927 j> with a magnitude of 0.000555 N pointing in the opposite direction of the diagonal
anonymous
  • anonymous
Yep that's correct...I'm still trying to make sense of it...one second...
anonymous
  • anonymous
\[(9\cdot 3 \cdot\frac 14+(-9)3\cdot \frac 18 \cdot \frac 1{\sqrt2})\hat i\] and for \hat j as well Yep that makes sense
anonymous
  • anonymous
what happened to k
phi
  • phi
I was ignoring the constants and units, just to get a rough idea of what is going on. I see that both the i and j components are positive and equal to each other. but obviously you have to multiply every term by k, and by 1e-9 (twice! to change nC to C) and by 1e4 cm^2 per meter^2 to get the units correct.
anonymous
  • anonymous
Oh I see
phi
  • phi
if we use 8.99e9 for k, we match the book's answer 0.000392
anonymous
  • anonymous
shoot me....I continue to get .0000082137 \[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}\]\[-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{0.08m^2}\]
anonymous
  • anonymous
\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{-9}C)(3\times10^{-9}C)}{0.04m^2}\]-\[-\frac{(8.99\times10^9N\cdot m^2/C^2)(-9\times10^{-9}C)(3\times10^{-9}C)(\frac 1{\sqrt{2}})}{\sqrt0.08m^2}\]
phi
  • phi
First, I would not write down that expression. I would write down the 3 individual vectors (with components i and j even if 0) Second, your distance squared should be written \( (2e{-2}\ m)^2\)= \( 4e{-4}\ m^2\)
phi
  • phi
and \( 2 \sqrt{2} \) cm squared is written as 8e-4 m^2
anonymous
  • anonymous
finally!!!1
phi
  • phi
Good.

Looking for something else?

Not the answer you are looking for? Search for more explanations.