anonymous
  • anonymous
What is the magnitude of the force on the -10 charge in the figure ?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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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:1327758982728:dw|
anonymous
  • anonymous
the resultant force should be...|dw:1327759214456:dw| something like that
anonymous
  • anonymous
broke down x and y components for force on + on - and - on - and then added...not getting right answer

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anonymous
  • anonymous
450*10^-9 N
anonymous
  • anonymous
can you explain?
anonymous
  • anonymous
it's not the right answer
anonymous
  • anonymous
why?
Shayaan_Mustafa
  • Shayaan_Mustafa
it is not easy. it is a vector quantity. therefore you must have to consider directions.
Shayaan_Mustafa
  • Shayaan_Mustafa
you have to wait until i solve it for you. ok elicia85?
anonymous
  • anonymous
sure and can you please show steps??
Shayaan_Mustafa
  • Shayaan_Mustafa
yes i will show steps to you. wait. i am here.
anonymous
  • anonymous
ok thank you
Shayaan_Mustafa
  • Shayaan_Mustafa
you are welcome.
Shayaan_Mustafa
  • Shayaan_Mustafa
here is diagram first look at this and understand this.
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Shayaan_Mustafa
  • Shayaan_Mustafa
is this pic making sense to you? if i do anything wrong here then kindly point out me. you too faryad, hei, elicia85 and an unknown person.
anonymous
  • anonymous
looks good
anonymous
  • anonymous
thank you Shayan
Shayaan_Mustafa
  • Shayaan_Mustafa
wait i have to go and just come in 2min
Shayaan_Mustafa
  • Shayaan_Mustafa
hello. it is taking too much time for me. lets solve step by step here. ok. force of +15nC will be represented by F15 force of -10nC will be represented by F10 force of -5nC will be represented by F5 ok.
Shayaan_Mustafa
  • Shayaan_Mustafa
now as you can see. on -10nC there are two forces acting on it. 1)F15 which is attractive and at an angle of 45 2)F5 which repulsive.
Shayaan_Mustafa
  • Shayaan_Mustafa
F15 can be resolve into factors. along x-axis and along y-axis.
Shayaan_Mustafa
  • Shayaan_Mustafa
forces along y-axis can be calculated as, F15y=(kq1q2)/r² sin45 and F5y=(kq1q2)/r² forces along x-axis can be calculated as, F15x=(kq1q2)/r² cos45
Shayaan_Mustafa
  • Shayaan_Mustafa
finally. sum forces and you will get your answer.
Shayaan_Mustafa
  • Shayaan_Mustafa
have you got it? it is done.
anonymous
  • anonymous
it's exactly what i did but i will try again
Shayaan_Mustafa
  • Shayaan_Mustafa
can you show your solution?
anonymous
  • anonymous
well you still have to add the F5 on F10 don't you? which will only consist of y direction force
anonymous
  • anonymous
my answer was 4.07*10^-3
Shayaan_Mustafa
  • Shayaan_Mustafa
well you still have to add the F5 on F10 don't you? which will only consist of y direction force on y-axis: F10y=F15sin45 - F5 on x-axis: F10x= -F15x finally, \[F _{10}=\sqrt{(F _{10y})^{2}+(F _{10x})^{2}}\] This is complete.
Shayaan_Mustafa
  • Shayaan_Mustafa
any question?
anonymous
  • anonymous
shayan can you help me?
Shayaan_Mustafa
  • Shayaan_Mustafa
of course fryad. although we are brothers. arn't we?
anonymous
  • anonymous
thank you i'll try again!
Shayaan_Mustafa
  • Shayaan_Mustafa
you are ever welcome. Regards, Read my Profile. :D
anonymous
  • anonymous
yes ,sure thank you , please prove this for me A.B=cos& |A|.|B|
Shayaan_Mustafa
  • Shayaan_Mustafa
ok w8. let me analyze.
Shayaan_Mustafa
  • Shayaan_Mustafa
The three vectors above form the triangle AOB. using Law of Cosines , \[|a-b|^{2}=|a|^{2}+|b|^{2}-2|a||b|\cos \theta\] using properties of dot product. \[|a-b|^{2}=(a-b).(a-b)\] \[=a.a-a.b-b.a+b.b\] \[=|a|^{2}+|b|^{2}-2a.b\] original becomes, \[|a-b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta\] and, \[|a|^{2}-2a.b+|b|^{2}=|a|^{2}+|b ^{2}|-2|a||b|\cos \theta\] now cancel terms which are same and you will got \[a.b=|a||b|\cos \theta\]
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Shayaan_Mustafa
  • Shayaan_Mustafa
that is all. Regards, Read my Profile. :D
anonymous
  • anonymous
zor zor supas aziyatm dait
Shayaan_Mustafa
  • Shayaan_Mustafa
kia?
Shayaan_Mustafa
  • Shayaan_Mustafa
what is this language?
anonymous
  • anonymous
what is your language?
Shayaan_Mustafa
  • Shayaan_Mustafa
urdu and english.
anonymous
  • anonymous
I am kurdish
Shayaan_Mustafa
  • Shayaan_Mustafa
ok. now good luck. take care everyone.
anonymous
  • anonymous
I am beginner in physic,thus I want help me for any time if you can my brother?
Shayaan_Mustafa
  • Shayaan_Mustafa
sure sure. i am often available here. you can post your question.
anonymous
  • anonymous
That is all kind of you for every thing.can I recognize you in facebook?
anonymous
  • anonymous
Just skimming through this.... You should recognize that the angle is not 45 degrees because it's a rectangle, not a square, so the values above would not yield the correct answer.
anonymous
  • anonymous
good point, i wasn't using 45 though, just used the measurements given. i still need to resolve, haven't got to it yet
Shayaan_Mustafa
  • Shayaan_Mustafa
elica85 are you here?
Shayaan_Mustafa
  • Shayaan_Mustafa
oh yes. so angle is not 45degree. it can be calculated as, \[\tan^{-1} (0.01/0.03)=18.43degrees.\] Kindly use this angle instead of 45.
Shayaan_Mustafa
  • Shayaan_Mustafa
thanks to Jemurray3.
anonymous
  • anonymous
elica I assume you are adding the components separately and then using the Pythagorean Theorem to calculate the magnitude of the resultant, right?
Shayaan_Mustafa
  • Shayaan_Mustafa
hi jemurray3
anonymous
  • anonymous
Hi there Shayaan
Shayaan_Mustafa
  • Shayaan_Mustafa
http://openstudy.com/study#/updates/4f245a39e4b0a2a9c2667f2f We need you in mathematics section. kindly help us.
anonymous
  • anonymous
thank you everyone, i did get the right answer finally. i was confused about +/-. i thought the - charge on negative would make it negative direction going down y, making it double negative...so i was just getting calculations wrong. it was 18 degrees ^^
Shayaan_Mustafa
  • Shayaan_Mustafa
you are ever welcome. we are made for help. pleasure.

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