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I'm adding my vectors incorrectly. Please give me a min to draw it.
 one year ago
 one year ago
I'm adding my vectors incorrectly. Please give me a min to draw it.
 one year ago
 one year ago

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JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1360530853167:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\vec{F}_{net}=(\vec F_{41}\vec F_{31}\cos(45))\hat i +(\vec F_{21}\vec F_{31}\sin(45)) \hat j\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1360531227825:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[F_{net}=\left( \frac{kq_4q_1}{r_{41}^2}\frac{kq_3q_1}{r_{31}^2}\sin(45)\right) \hat i \left(\frac{kq_2q_1}{r_{21}^2}\frac{kq_3q_1}{r_{31}^2}\cos(45)\right) \hat j\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[q_1=3\times10^{9}\] \[q_2=9\times10^{9}\] \[q_3=9\times10^{9}\] \[q_4=9\times10^{9}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1360531777377:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
I get \[(72974.5 N)\hat i\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
which is wrong...
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
.000392 should be the answer
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1360532089439:dw
 one year ago

phiBest ResponseYou've already chosen the best response.1
8 orders of magnitude is quite an error
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
it sure is ...haha
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
the charges were given in nano coulombs. I converted them to coulombs. \[q_1=+3nC=3\times 10^{9}\]
 one year ago

phiBest ResponseYou've already chosen the best response.1
First, what units is the answer given in ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
Newtons \hat{i}
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
maybe I read the question wrong
 one year ago

phiBest ResponseYou've already chosen the best response.1
your approach is ok. I would write down the components of each vector
 one year ago

phiBest ResponseYou've already chosen the best response.1
the k 9e9 times 9e9 gives 81. that still leaves a 3e9 are you double counting the k ?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
dw:1360532909843:dw oh my gosh ...yes I think soooo
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
I'll do it on wolfram... one sec
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
silly question... \[F_{31}=F_{31x}+F_{31y}\] is this true?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
or is it \[F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}\]
 one year ago

phiBest ResponseYou've already chosen the best response.1
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 1e18 C^2 * 9e9 NM^2 /C^2 * 1e4 M^2 (to fix the cm) that factor is 9e5 and our result is 4.3635*9e5 = 0.0003927 N for both the i and j components
 one year ago

phiBest ResponseYou've already chosen the best response.1
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.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
why is "the force from the bottom left is < 9*3/8 * 1/sqrt(2) i" Why did you divide by 8
 one year ago

phiBest ResponseYou've already chosen the best response.1
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)
 one year ago

phiBest ResponseYou've already chosen the best response.1
your 2.8 was just an approximation
 one year ago

phiBest ResponseYou've already chosen the best response.1
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
Yep that's correct...I'm still trying to make sense of it...one second...
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[(9\cdot 3 \cdot\frac 14+(9)3\cdot \frac 18 \cdot \frac 1{\sqrt2})\hat i\] and for \hat j as well Yep that makes sense
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
what happened to k
 one year ago

phiBest ResponseYou've already chosen the best response.1
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 1e9 (twice! to change nC to C) and by 1e4 cm^2 per meter^2 to get the units correct.
 one year ago

phiBest ResponseYou've already chosen the best response.1
if we use 8.99e9 for k, we match the book's answer 0.000392
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
shoot me....I continue to get .0000082137 \[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)}{0.04m^2}\]\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)(\frac 1{\sqrt{2}})}{0.08m^2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.1
\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)}{0.04m^2}\]\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)(\frac 1{\sqrt{2}})}{\sqrt0.08m^2}\]
 one year ago

phiBest ResponseYou've already chosen the best response.1
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 \( (2e{2}\ m)^2\)= \( 4e{4}\ m^2\)
 one year ago

phiBest ResponseYou've already chosen the best response.1
and \( 2 \sqrt{2} \) cm squared is written as 8e4 m^2
 one year ago
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