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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360530853167:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\vec{F}_{net}=(\vec F_{41}\vec F_{31}\cos(45))\hat i +(\vec F_{21}\vec F_{31}\sin(45)) \hat j\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360531227825:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[F_{net}=\left( \frac{kq_4q_1}{r_{41}^2}\frac{kq_3q_1}{r_{31}^2}\sin(45)\right) \hat i \left(\frac{kq_2q_1}{r_{21}^2}\frac{kq_3q_1}{r_{31}^2}\cos(45)\right) \hat j\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[q_1=3\times10^{9}\] \[q_2=9\times10^{9}\] \[q_3=9\times10^{9}\] \[q_4=9\times10^{9}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360531777377:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I get \[(72974.5 N)\hat i\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1which is wrong...

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1.000392 should be the answer

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360532089439:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.18 orders of magnitude is quite an error

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1it sure is ...haha

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1the charges were given in nano coulombs. I converted them to coulombs. \[q_1=+3nC=3\times 10^{9}\]

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Newtons \hat{i}

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1maybe I read the question wrong

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

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1dw:1360532909843:dw oh my gosh ...yes I think soooo

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I'll do it on wolfram... one sec

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1I 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1and 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1silly question... \[F_{31}=F_{31x}+F_{31y}\] is this true?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1or is it \[F_{31}=\sqrt{F_{31x}^2+F_{31y}^2}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1ignoring 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

phi
 one year ago
Best ResponseYou've already chosen the best response.1as 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1why is "the force from the bottom left is < 9*3/8 * 1/sqrt(2) i" Why did you divide by 8

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

phi
 one year ago
Best ResponseYou've already chosen the best response.1I 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1Yep that's correct...I'm still trying to make sense of it...one second...

JenniferSmart1
 one year ago
Best 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1what happened to k

phi
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1shoot me....I continue to get .0000082137 \[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)}{0.04m^2}\]\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)(\frac 1{\sqrt{2}})}{0.08m^2}\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)}{0.04m^2}\]\[\frac{(8.99\times10^9N\cdot m^2/C^2)(9\times10^{9}C)(3\times10^{9}C)(\frac 1{\sqrt{2}})}{\sqrt0.08m^2}\]

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