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anonymous
 4 years ago
What is the magnitude of the force on the 10 charge in the figure ?
anonymous
 4 years ago
What is the magnitude of the force on the 10 charge in the figure ?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327758982728:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the resultant force should be...dw:1327759214456:dw something like that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0broke down x and y components for force on + on  and  on  and then added...not getting right answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's not the right answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is not easy. it is a vector quantity. therefore you must have to consider directions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you have to wait until i solve it for you. ok elicia85?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure and can you please show steps??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes i will show steps to you. wait. i am here.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is diagram first look at this and understand this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is 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
 4 years ago
Best ResponseYou've already chosen the best response.0wait i have to go and just come in 2min

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hello. 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0F15 can be resolve into factors. along xaxis and along yaxis.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0forces along yaxis can be calculated as, F15y=(kq1q2)/r² sin45 and F5y=(kq1q2)/r² forces along xaxis can be calculated as, F15x=(kq1q2)/r² cos45

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0finally. sum forces and you will get your answer.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you got it? it is done.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's exactly what i did but i will try again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you show your solution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well you still have to add the F5 on F10 don't you? which will only consist of y direction force

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my answer was 4.07*10^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well you still have to add the F5 on F10 don't you? which will only consist of y direction force on yaxis: F10y=F15sin45  F5 on xaxis: F10x= F15x finally, \[F _{10}=\sqrt{(F _{10y})^{2}+(F _{10x})^{2}}\] This is complete.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0shayan can you help me?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0of course fryad. although we are brothers. arn't we?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you i'll try again!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are ever welcome. Regards, Read my Profile. :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes ,sure thank you , please prove this for me A.B=cos& A.B

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok w8. let me analyze.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The three vectors above form the triangle AOB. using Law of Cosines , \[ab^{2}=a^{2}+b^{2}2ab\cos \theta\] using properties of dot product. \[ab^{2}=(ab).(ab)\] \[=a.aa.bb.a+b.b\] \[=a^{2}+b^{2}2a.b\] original becomes, \[ab^{2}=a^{2}+b ^{2}2ab\cos \theta\] and, \[a^{2}2a.b+b^{2}=a^{2}+b ^{2}2ab\cos \theta\] now cancel terms which are same and you will got \[a.b=ab\cos \theta\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that is all. Regards, Read my Profile. :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0zor zor supas aziyatm dait

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is this language?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what is your language?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok. now good luck. take care everyone.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am beginner in physic,thus I want help me for any time if you can my brother?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure sure. i am often available here. you can post your question.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is all kind of you for every thing.can I recognize you in facebook?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Just 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
 4 years ago
Best ResponseYou've already chosen the best response.0good point, i wasn't using 45 though, just used the measurements given. i still need to resolve, haven't got to it yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0elica85 are you here?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0elica I assume you are adding the components separately and then using the Pythagorean Theorem to calculate the magnitude of the resultant, right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/4f245a39e4b0a2a9c2667f2f We need you in mathematics section. kindly help us.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank 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 ^^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are ever welcome. we are made for help. pleasure.
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