EPT![[image loading]](http://i63.photobucket.com/albums/h156/gglawl69/stuck.jpg)
any help really appreciated, answer is: 7.17*10^-8 C
Physics help
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Ftrunkz
Australia2474 Posts
any help really appreciated, answer is: 7.17*10^-8 C | ||
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StorrZerg
United States13919 Posts
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Ftrunkz
Australia2474 Posts
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Hot_Bid
Braavos36399 Posts
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Cloud
Sexico5880 Posts
Then you say that force is equal to the magnetic force calculated with the Lorentz equation or something. To be honest, I saw this shit like 5 years ago so I can't point to you exactly what to do. | ||
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Freyr
United States500 Posts
Use this information in conjunction with Coulomb's law (electrostatics, not magnetism) and the wonders of trigonometry to solve for the charge (which is equal in each balloon). Essentially what Cloud said, minus the magnetism bit. | ||
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GOB
50 Posts
![[image loading]](http://img413.imageshack.us/img413/5259/stuckf.jpg) | ||
micronesia
United States24771 Posts
edit: The steps I used to do it symbolically (not using a calculator right now) 1) Split the triangle into two right triangles 2) Call the hypotenuse L (which is 100) and one of the legs d/2 (since the entire base of the triangle is 30, d/2 is 15... it's probably smarter to define the entire base of the triangle as 2d but oh well) 3) Find the third leg by using the pythagorean theorum 4) Set up free body diagrams for all three objects... just to check: The mass has a downward force of gravity m*g, and two tension forces each going at an angle that you can calculate by using the triangle from above. 5) Solve for T using this free body diagram. 6) The free body diagram for a balloon should be an upward buoyant force (no downward gravity force since we just assume the buoyant force is that much weaker to compensate), a lateral electrostatic force repelling the opposite balloon, and tension going down at an angle and away from the electrostatic force. 7) Add up the forces in the x direction using Newton's second law. Plug in what you know for T, cos of the angle you can find by using your triangle from earlier, and the other force is simply Coulomb's law kQ^2/d^2. 8) Solve for Q. | ||
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Ftrunkz
Australia2474 Posts
wall of possibly incomprehensible working coming, if anyone can decipher and manage to see where i've gone wrong or where my understanding is completely messed up it'd be greatly appreciated =]: the force due to the charges of the balloons on each other is: F= (Q^2 * K) / r^2 = (Q^2 * 8.99 * 10^9) / .36 = 2.50*(10^10) * Q^2 so that force must equal the horozontal force from the tension in the strings. the verticle componant of the strings combined is mg of the little 5g triangle thing. therefor the verticle componant on one of the strings is F=(.005 * 9.8) / 2 = 0.0245 then i need to work out the triangles, doing so ended up getting me a 1:0.8 ratio (whole string vs verticle componant). So the tension force in the string is 0.0245 * (1/.08) = 0.03 thennnn i need to work out the horozontal componant, which from the diagram in the question is a ratio of 100:60, so 0.03 * .6 = 0.018375 (i left out a lot of figures earlier, adding them now) so, that 0.018375 = electrostatic force = 2.5 * 10^10 Q^2 solving for Q gave me 8.57*10^-7, unfortunatly not the answer in the... answers... =[. | ||
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gyth
657 Posts
Are you sure your answer is right? | ||
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paper
13196 Posts
On November 06 2009 11:59 GOB wrote: two circles and your first reaction is mouse eyes? | ||
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NukezaFlyin
United States10 Posts
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randombum
United States2378 Posts
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Lemonwalrus
United States5465 Posts
Seriously. | ||
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Ftrunkz
Australia2474 Posts
On November 06 2009 12:08 gyth wrote: I get 5.6e-7C Are you sure your answer is right? I'd THINK they would be, since they are probably the ones used to mark the exam in 2007. But it's not outside the realm of possibility =o. If someone else confirms your answer i'll gladly assume its right over the ones givin, haha and to the guy that said my ratios are wrong: That's probably likely, since i dont know where the hell else i'm going wrong =[ | ||
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micronesia
United States24771 Posts
On November 06 2009 12:14 randombum wrote: Lol Micronesia always finds these physics threads so quickly. Do you do a search for them whenever you log on or something? I would have been the first response except I was driving home from parent conferences when this thread was made :p btw I updated my earlier post to show some suggested steps | ||
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gyth
657 Posts
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Ftrunkz
Australia2474 Posts
Thanks heaps for the speedy replies everyone  much appreciated :Dedit: wait, now im getting gyth's answer, 5.6*10^-7C... lol. I guess the answers were wrong? Either way the questions killed me enough over the past hour or two that i'll assume that answer is right and move on, haha... and yay dragoon. | ||
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micronesia
United States24771 Posts
On November 06 2009 12:08 gyth wrote: I get 5.6e-7C Are you sure your answer is right? Hm I got the same thing using google to calculate everything (which sucks so I have no confidence in this). I'm trying to decide if there is a mistake in this method... sqrt((2*.005*9.8*.3^3)/(9*10^9*sqrt(1-.3^2))) | ||
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AcrossFiveJulys
United States3612 Posts
On November 06 2009 11:59 GOB wrote: Rofl. So I guess the whole face is edible? | ||
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