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Post by Mykal Watsrobe on Sept 29, 2008 19:39:49 GMT -5
Super magnets. Dart guns. "Cat"-a-paults (lolz Kittie launchers). Rocket launchers. Impossible homework. Best class ever.
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Post by anthony on Sept 29, 2008 22:45:16 GMT -5
I WANT TO TAKE AP PHYSICS TOO BAD MY AP CALC CLASS IS 6TH PERIOD AND SO IS AP PHYSICS LOL
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Post by Mykal Watsrobe on Oct 2, 2008 21:34:25 GMT -5
ROFLWAFL
My dad just taught me how to derive the range equation.
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Post by Mykal Watsrobe on Oct 2, 2008 21:57:21 GMT -5
R=Vo2sin(2Ø)/g
x=xo+Vot+(1/2)at2
x=xo=0 cancel out the x's
0=Vot+(1/2)at2 factor 0=t(Vo+(1/2)at) solve for zeros t=0,2Vo/g
sinØ=Vy/Vo
Vy=VosinØ=vertical velocity
cosØ=Vx/Vo
Vx=VocosØ=horizontal velocity
plug into time equation above t=2VosinØ/g
Range=horizontal velocity x time it takes for the object to reach its vertical peak and fall to the ground again (t=2VosinØ/g) R=VocosØ x 2VosinØ/g R=VocosØ x Vo2sinØ/g R=V22sinØcosØ/g
2sinØcosØ=sin2Ø so.....
R=V2sin2Ø/g
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Post by Mykal Watsrobe on Oct 2, 2008 21:59:46 GMT -5
It looks a lot more complicated than it really is. It's much easier to explain it than it is to type it.
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Post by Mykal Watsrobe on Oct 7, 2008 22:25:21 GMT -5
This has nothing to do with Physics, but I didn't feel like making a new thread for Chemistry. So the plan for The Lord of the Constants 2 has been set in motion. Jenny Ogelsby's group in Mrs. Wagner's class needs help with the video. They're making a sequel to our Mole Day video from last year. www.veoh.com/videos/v1360980Ss9dsjs5We have to make this episode of the great trilogy the best we can. They're (supposedly) meeting at Conner's house this Sunday.
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Post by Mykal Watsrobe on Oct 21, 2008 22:49:50 GMT -5
So on Facebook, Birju was asking around for people from AP Physics for help with the really long homework. Apparently, he was stuck on one of the easier questions (in my opinion). It's just a bunch of simple equations.
A skier with a mass of 75kg is holding two ropes, forming 45º angles north and south of east, attached to two boats, both exerting a force of 600N on the skier. (a)What is the force of friction if the skier is moving at a constant velocity? (b)Assuming friction remains the same, what would the skier's acceleration be if the two boats exerted a force of 700N on the skier?
a.) Simple. If you draw a free-body diagram (which I can't draw here, but I wish I could), it makes it a lot easier because you can see all the work you're doing. Anyway, use the equation cosØ=adj/hyp. You get cos45º=600N/Ff. So Ff=600N/cos45º. You end up with a force of 850N west.
b.) Friction remains the same, but forward force changes. So redo the above equation for 700N. Fx=700N/cos45º=990N Then find the sum of the forces. Since friction is in a negative direction, subtract friction from the forward force. Fx=990N-850N=140N To get acceleration, plug that answer into the equation F=ma. 140N=75a a=1.87m/s2
Why was that so confusing?
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Post by anthony on Oct 22, 2008 9:40:48 GMT -5
Even I knew how to do that. Tell Birju that he should trade his AP Physics spot with my CP Physics spot. I'll be keeping the seat warm.
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Post by epicfailguy on Oct 22, 2008 15:29:11 GMT -5
i just saw LOTC. perfect movie
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Post by Mizagium on Oct 22, 2008 16:52:11 GMT -5
R=V o2sin(2Ø)/g x=x o+V ot+(1/2)at 2x=x o=0 cancel out the x's 0=V ot+(1/2)at 2factor 0=t(V o+(1/2)at) solve for zeros t=0,2V o/g sinØ=V y/V oV y=V osinØ=vertical velocity cosØ=V x/V oV x=V ocosØ=horizontal velocity plug into time equation above t=2V osinØ/g Range=horizontal velocity x time it takes for the object to reach its vertical peak and fall to the ground again (t=2V osinØ/g) R=V ocosØ x 2V osinØ/g R=V ocosØ x V o2sinØ/g R=V 22sinØcosØ/g 2sinØcosØ=sin2Ø so..... R=V 2sin2Ø/g *head asplaods*
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