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Kazaxat
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Post subject: Re: Math HW Help Posted: Sat Apr 12, 2008 8:52 pm |
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NuclearSilo wrote: Kazaxat wrote: Quote? Im not going to waste my time just to prove a real fact for u. Use search button then search for all posts by D2U, u'll get your answer. no one is going to try prove what you say for u  u made a statement it's your job to prove it. and if you are so intelligent like you claim to be, then perhaps solve the problems posed by Vindicator and Thiefz? or u cant?
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Judge
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Post subject: Re: Math HW Help Posted: Sat Apr 12, 2008 9:01 pm |
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Kazaxat wrote: NuclearSilo wrote: Kazaxat wrote: Quote? Im not going to waste my time just to prove a real fact for u. Use search button then search for all posts by D2U, u'll get your answer. no one is going to try prove what you say for u  u made a statement it's your job to prove it. and if you are so intelligent like you claim to be, then perhaps solve the problems posed by Vindicator and Thiefz? or u cant? He doesn't claim to be intelligent, he is. You on the other hand appear to be threatened by his apparent intelligence. If you want to prove you're better at math than silo, beat him at his own game, rather than spewing verbal fodder such as; "blah,blah". He made a statement, that same statement was backed earlier by charts and past post. Rather than begging others to do it for you, how about you take some initiative and search for the post. He provided a key-word and a author, unless that is ~ your afraid of the truth. As much as it is his job to back his statement it is yours, in a two person discussion, to disprove it. Silo has solved countless problems in the past, he doesn't have to prove a thing. You, however, at the fear of losing whatever credibility you still have must solve said problems. Do that, then smack talk 
_________________ “Experience hath shewn, that even under the best forms (of government) those entrusted with power have, in time, and by slow operations, perverted it into tyranny” - Thomas Jefferson
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NuclearSilo
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 9:04 pm |
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Vindicator wrote: e^(2x)-e^x + 5 = 11 e^(2x) - e^x = 6 2x - x = ln6 x = ln6
false because ln(a-b) =/= lna - lnb Here is the solution: let t=e^x the equation above become t^2 - t + 5 = 11 => t^2 - t - 6 = 0 delta = (-1)^2 - 4*1*(-6) = 25 2 solutions: t1 = -2 t2 = 3 since e^x > 0 for all x => the only solution is 3 t = e^x = 3 => x = ln3
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NuclearSilo
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Post subject: Re: Math HW Help Posted: Sat Apr 12, 2008 9:08 pm |
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Kazaxat wrote: NuclearSilo wrote: Kazaxat wrote: Quote? Im not going to waste my time just to prove a real fact for u. Use search button then search for all posts by D2U, u'll get your answer. no one is going to try prove what you say for u  u made a statement it's your job to prove it. and if you are so intelligent like you claim to be, then perhaps solve the problems posed by Vindicator and Thiefz? or u cant? I think u need to stfu. Believe it or not is not my business. Searching is not my job either. At least what u said doesnt based on anything too Quote: so you are saying PicoMon got the HP formula and dmg formula by reversing the server code? how would that be possible? Proove it, that it is not possible?
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Deadsolid
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 9:11 pm |
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NuclearSilo wrote: Vindicator wrote: e^(2x)-e^x + 5 = 11 e^(2x) - e^x = 6 2x - x = ln6 x = ln6
false because ln(a-b) =/= lna - lnb Here is the solution: let t=e^x the equation above become t^2 - t + 5 = 11 => t^2 - t - 6 = 0 delta = (-1)^2 - 4*1*(-6) = 25 2 solutions: t1 = -2 t2 = 3 since e^x > 0 for all x => the only solution is 3 t = e^x = 3 => x = ln3I just learned this stuff the other day and my knowledge is limited but if that problem is as simple as it first looked than I believe Vindicator is right. I may be wrong though.
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/Pi
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 9:46 pm |
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ThiefzV2 wrote: C) What is the molarity (M) of a solution of 50% NaOH?
Weight-volume percentage: X% = (Xg/100mL) * 100 Or put it simply, the percentage is the amount of the solute in grams. So we have 50g of NaOH from a solution of 50% NaOH. Molarity: M = mol/L Mole: mol = g/m.w. (m.w. = molecular weight) With 50g of NaOH: mol. of NaOH = 50g NaOH/39.9971g NaOH = 1.25009 mol. of NaOH With 1.25509 mol. NaOH: M = 1.25509/(100mL/1000mL) We need the volume in liters so 100mL = 1L/1000mL M = 12.5509 ThiefzV2 wrote: D) What is the concentration of a strong acid with a pH of 1.2?
When it comes to pH, the molarity is based on the amount of protons you have: M = 1/(1 * 10^xpH) With a pH of 1.2, you'll have: M = 1/(1 * 10^1.2) M = 0.0631
Last edited by /Pi on Sat Apr 12, 2008 9:51 pm, edited 1 time in total.
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Vindicator
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 9:49 pm |
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Joined: Jul 2007 Posts: 1734 Location: L-A-B
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Deadsolid wrote: NuclearSilo wrote: Vindicator wrote: e^(2x)-e^x + 5 = 11 e^(2x) - e^x = 6 2x - x = ln6 x = ln6
false because ln(a-b) =/= lna - lnb Here is the solution: let t=e^x the equation above become t^2 - t + 5 = 11 => t^2 - t - 6 = 0 delta = (-1)^2 - 4*1*(-6) = 25 2 solutions: t1 = -2 t2 = 3 since e^x > 0 for all x => the only solution is 3 t = e^x = 3 => x = ln3I just learned this stuff the other day and my knowledge is limited but if that problem is as simple as it first looked than I believe Vindicator is right. I may be wrong though. unfortunately, she is right. I was going too fast to pay attention to the fact it is subtraction and therefore lna - lnb = ln(a/b). Id still like to see someone solve my begginer integrals and derivative questions, and if anyone can figure out the physics spring question.
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NuclearSilo
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 9:56 pm |
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Joined: Aug 2006 Posts: 8834 Location: Age of Wushu
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It's been a long time i didnt touch derivative and integral. Im a programmer though. Question 1: derivative of LOG(5)(x^2 + 3x - 1) LOG(5)x = lnx / ln5 (LOG(5)x)' = x'/xln5
=> (2x + 3)/((x^2 + 3x - 1)ln5)
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ThiefzV2
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:00 pm |
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Nuclear, do u know the answer to this one: Quote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground.
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Last edited by ThiefzV2 on Sat Apr 12, 2008 10:27 pm, edited 1 time in total.
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No reply
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:08 pm |
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np 
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Vindicator
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:09 pm |
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with a trajectory of y = 160-(x^2 / 40) (im assuming x squared over 40 right?) Then y = 0 when x = 80. Therefore it takes 80 seconds to hit the ground. Using X = V(initial)t + .5at2 where: x= distance traveled in the x direction v(initial) = initial velocity in the x direction t = time a = acceleration in the x direction. Neglecting air friction and wind, acceleration is 0 m/s^2 therefore it is x = (24m/s)(80s) = 1920m edit: whoops, its 24m/s not 25 
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Last edited by Vindicator on Sat Apr 12, 2008 10:24 pm, edited 1 time in total.
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NuclearSilo
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:10 pm |
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Integral of 6^(5x) dx: Formula: a^x = e^(ln(a^x)) = e^(xlna) integral of e^(xlna) = e^(xlna)/lna Replace a with 6 and x with 5x
integral = e^(5xln6)/5ln6
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ThiefzV2
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:15 pm |
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Prophet Izaach wrote: ThiefzV2 wrote: C) What is the molarity (M) of a solution of 50% NaOH?
Weight-volume percentage: X% = (Xg/100mL) * 100 Or put it simply, the percentage is the amount of the solute in grams. So we have 50g of NaOH from a solution of 50% NaOH. Molarity: M = mol/L Mole: mol = g/m.w. (m.w. = molecular weight) With 50g of NaOH: mol. of NaOH = 50g NaOH/39.9971g NaOH = 1.25009 mol. of NaOH With 1.25509 mol. NaOH: M = 1.25509/(100mL/1000mL) We need the volume in liters so 100mL = 1L/1000mL M = 12.5509 ThiefzV2 wrote: D) What is the concentration of a strong acid with a pH of 1.2?
When it comes to pH, the molarity is based on the amount of protons you have: M = 1/(1 * 10^xpH) With a pH of 1.2, you'll have: M = 1/(1 * 10^1.2) M = 0.0631 You got the 2nd one right but not the first one (not even close) ... hmmmm what could be the reason? Nuclear, did u read my previous post?
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No reply
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:16 pm |
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Quote: with a trajectory of y = 160-(x^2 / 40) (im assuming x squared over 40 right?) Then y = 0 when x = 80. Therefore it takes 80 seconds to hit the ground ... I'm sure you mean 24 m/s, not 25 m/s
Last edited by No reply on Sat Apr 12, 2008 10:44 pm, edited 1 time in total.
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Vindicator
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:23 pm |
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NuclearSilo wrote: Integral of 6^(5x) dx: Formula: a^x = e^(ln(a^x)) = e^(xlna) integral of e^(xlna) = e^(xlna)/lna Replace a with 6 and x with 5x
integral = e^(5xln6)/5ln6 never seen it quite like that before...just follow the rule of a^u where: a is a constant u is a function of x integral of a^u = (a^u)/(lna) + c where c is any constant during an undefined integral. Not sure what yours is reduced, but theres no need to use e's, its just longer. No reply wrote: Quote: with a trajectory of y = 160-(x^2 / 40) (im assuming x squared over 40 right?) Then y = 0 when x = 80. Therefore it takes 80 seconds to hit the ground ... I'm sure you mean 24 m/s, not 25 m/s, other than that you're correct. lol thanks, thought it was 25 
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No reply
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:30 pm |
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@ NaOH task: NaOH (aq) + H2O (l) --> Na(+) + OH(-), so really, it's all about the solubility of NaOH (some of it will not react and form Na+ + OH- , and there you got the concentration).
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NuclearSilo
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:34 pm |
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ThiefzV2 wrote: Nuclear, do u know the answer to this one: Quote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. I dont see the point of giving the speed 24m/s here, since u have given the equation of hte trajectory. And are u sure it's 160? Because at x=0, the height is 160 and not 150. Therefore, u have chosen the origin of your graph 10m above the bird. So when ever bird hits the ground it's at altitude y=10 in the graph. So the equation needs to resolve is : 10=160-(x^2/40) And the solution is x=root(6000)=20*root(15) m Since u said x is the distance, there is not need that the speed of flying involves.
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Last edited by NuclearSilo on Sat Apr 12, 2008 10:39 pm, edited 2 times in total.
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/Pi
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:35 pm |
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ThiefzV2 wrote: You got the 2nd one right but not the first one (not even close) ... hmmmm what could be the reason?
Oh snaps, you got the answer? I think I got it right, unless my mistake has something to do with sig figs and units. T.T
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TOloseGT
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:38 pm |
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ThiefzV2 wrote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. i'm not sure about vindicator's answer. the trajectory doesn't matter here. this is a freefall question. first, figure out time (t): use this eqn--> change in Y = velocity initial - (0.5)(acceleration of gravity)(t^2) -150 = 0 - (0.5)(9.8 )(t^2) t ~= 5.53s then distance of x = velocity of x * time x = 24 m/s * 5.53 s x = 132.72 m
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Last edited by TOloseGT on Sat Apr 12, 2008 10:44 pm, edited 1 time in total.
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ThiefzV2
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:41 pm |
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Vindicator wrote: with a trajectory of y = 160-(x^2 / 40) (im assuming x squared over 40 right?) Then y = 0 when x = 80. Therefore it takes 80 seconds to hit the ground. Using X = V(initial)t + .5at2 where: x= distance traveled in the x direction v(initial) = initial velocity in the x direction t = time a = acceleration in the x direction. Neglecting air friction and wind, acceleration is 0 m/s^2 therefore it is x = (24m/s)(80s) = 1920m edit: whoops, its 24m/s not 25  lol what did you just do? the answer greatly differs from your answer  here is a pic i draw in paint of what u are tyring to find: 
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TOloseGT
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:43 pm |
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thiefz, is the question asking what the x distance the prey traveled is? if it is, then look at my answer.
if u'r looking for the arc distance, then that's a whole other business =s
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Vindicator
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:43 pm |
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TOloseGT wrote: ThiefzV2 wrote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. i'm not sure about nuklear's answer. the trajectory doesn't matter here. this is a freefall question. first, figure out time (t): use this eqn--> change in Y = velocity initial - (0.5)(acceleration of gravity)(t^2) -150 = 0 - (0.5)(9.8 )(t^2) t ~= 5.53s then distance of x = velocity of x * time x = 24 m/s * 5.53 s x = 132.72 m The question itself is flawed. The given values above the ground, speed, and trajectoral equation do not match...at all. There are many answers depending on what information you choose to use. Dude, if your making questions, you need to specify what you want found... horiz, vert, and arc distance are all different...
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NuclearSilo
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:45 pm |
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TOloseGT wrote: ThiefzV2 wrote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. i'm not sure about nuklear's answer. the trajectory doesn't matter here. this is a freefall question. first, figure out time (t): use this eqn--> change in Y = velocity initial - (0.5)(acceleration of gravity)(t^2) -150 = 0 - (0.5)(9.8 )(t^2) t ~= 5.53s then distance of x = velocity of x * time x = 24 m/s * 5.53 s x = 132.72 m Read the question again. He talked about the trajectory equation. Not the velocity equation y=height x=distance
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No reply
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:47 pm |
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Quote: x is the horizontal distance traveled in meters Solve for x, gives us 80 m for Y(0)= 160. (you have written 150 instead)
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TOloseGT
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:48 pm |
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to me, that question is a simple free fall/figure out x distance question. if he wants the arc distance, the given values aren't enough.
i think he made a typo, it was supposed to be 160m above ground, not 150. however, my formulas still stand, just switch 150 to 160 and figure out from there.
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ThiefzV2
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:49 pm |
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Vindicator wrote: TOloseGT wrote: ThiefzV2 wrote: B) An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground. i'm not sure about nuklear's answer. the trajectory doesn't matter here. this is a freefall question. first, figure out time (t): use this eqn--> change in Y = velocity initial - (0.5)(acceleration of gravity)(t^2) -150 = 0 - (0.5)(9.8 )(t^2) t ~= 5.53s then distance of x = velocity of x * time x = 24 m/s * 5.53 s x = 132.72 m The question itself is flawed. The given values above the ground, speed, and trajectoral equation do not match...at all. There are many answers depending on what information you choose to use. Dude, if your making questions, you need to specify what you want found... horiz, vert, and arc distance are all different... I copied the question directly from the problem booklet... heres exactly what i said. Quote: An eagle is flying at 24 m/s at an altitude of 160m when it accidentially drops its prey. The parabolic trajectory of the falling prey is described by the equation y=160-(x^2/40) until it hits the ground, where y is its height above the ground and x is the horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until it hits the ground.
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Vindicator
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:50 pm |
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TOloseGT wrote: to me, that question is a simple free fall/figure out x distance question. if he wants the arc distance, the given values aren't enough.
i think he made a typo, it was supposed to be 160m above ground, not 150. however, my formulas still stand, just switch 150 to 160 and figure out from there. yep, although if you do simple free fall with the given values of 160m and 24m/s, then the equation he provided for the trajectory doesnt even come close to the actual situation 
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ThiefzV2
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Post subject: Re: To Smart people.Help (and to future Chemist & Mathematician) Posted: Sat Apr 12, 2008 10:50 pm |
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TOloseGT wrote: to me, that question is a simple free fall/figure out x distance question. if he wants the arc distance, the given values aren't enough.
i think he made a typo, it was supposed to be 160m above ground, not 150. however, my formulas still stand, just switch 150 to 160 and figure out from there. Yeah I did make a small typo but I corrected it a while ago though. :/ And yes the distance traveled is the length of that arc as in my pic of paint I posted. It should be longer than the distance of the pythagorean theorem since it's not straight if u know what i mean.
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