To Smart people (and to future Chemists & Mathematicians)

[Kazaxat]

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?

[Judge]

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

[NuclearSilo]

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

[NuclearSilo]

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

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?

[Deadsolid]

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

I 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.

[/Pi]

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

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

[Vindicator]

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

I 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.

[NuclearSilo]

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)

[ThiefzV2]

Nuclear, do u know the answer to this one:

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.

[No reply]

Let f(x) = 0.

[ThiefzV2]

Let f(x) = 0.

OMG why did I not think of that... big help there... thanks.

[No reply]

np

[Vindicator]

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

[NuclearSilo]

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

[ThiefzV2]

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

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?

[No reply]

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

[Vindicator]

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.

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

[No reply]

@ 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).

[NuclearSilo]

Nuclear, do u know the answer to this one:

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.

[/Pi]

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

[TOloseGT]

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

[ThiefzV2]

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:

[TOloseGT]

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

[Vindicator]

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...

[NuclearSilo]

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

[No reply]

x is the horizontal distance traveled in meters

Solve for x, gives us 80 m for Y(0)= 160. (you have written 150 instead)

[TOloseGT]

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.

[ThiefzV2]

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.

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.

[Vindicator]

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

[ThiefzV2]

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.