Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Spehss _]

New question. Haven't solved the other one, thought I solved my error but wound up getting the same final result after trying to solve the integral of py-y^₂ using partial fractions.

Spoiler: New question. (click to show/hide)

Find the solution of the initial value problem dy/dx=e^_x+y where y(1)=0

e^_x+y = e^_x*e^_y

so now I can separate y and x

1/e^_ydy=e^_xdx

e^_-ydy=e^_xdx

Integral of the y function is -e^_-y

Integral of x function is e^_x+c

-e^_-y=e^_x+c

Not really sure what to do after that. Pretty sure I'm supposed to remove the e bases. To do that I'd use natural logs. But the natural log of a negative number isn't a real answer. Soooo I seem to be stuck.

[Spehss _]

Consider multiply everything by -1. Remember, c is a constant. It can be positive, negative, or whatever else you wish it to be. So, 2c=c, -c=c, e^c=c, etc.

But then e^x would be -e^x. Then if I use natural log to get e^y down to y, and (I'm pretty sure) I'd have to also use natural log on the other components, and ln(-e^x) doesn't exist just like ln(-e^y) doesn't exist.

[Spehss _]

Physics question involving static equilibrium.

Spoiler (click to show/hide)

"A rope of negligible mass is stretched horizontally between two supports that are 3.44 meters apart. When an object weighing 3160 Newtons is hung at the center of he rope, the rope sags by 35.0 centimeters. What is the tension in the rope?"

The answer key in the back of the book gives an answer of 7.92 kN, or ~7920 N.

I don't see how I'm supposed to work out this problem using the information provided to get that answer. In a state of static equilibrium, the linear momentum p, angular momentum l, and net forces and net torque are all zero. So aside from the 3160N applied on the center of mass of the rope-weight system, the other forces must apply a total equivalent force in the opposite direction. I think. I guess? I'm having a hard time taking the (seemingly sparse) material the textbook talks about in the chapter and actually applying it to the word problems given by the textbook.

Also having to find the tension in the rope throws me off. Rope problems seem to always give me trouble.

[TheBiggerFish]

Triangles?

[Mostali]

Triangles?

Yes.  It is a vector/right triangle problem.  7920 is approx. (3160)(sqrt(35^2+172^2))(1/2)(1/35).  The only numbers that shouldn't be obvious are the 1/2 and the 172, which are both from the rope splitting the weight.

[Powder Miner]

well, chemical math worksheet that has me up at 12:30 going insane: BUFFERS. I've already made a shot at this question and failed (multiple choice, I didn't get any of the answers):
"The amount (in grams) of sodium acetate (Molecular Weight=82.0) to be added to 500.0 mL of 0.200 molar acetic acid (Ka=1.8*10^-5) in order to make a buffer with pH=5.000 is what?"

[crazysheep]

1. Use Ka and the dissociation equilibrium equation to find the concentration of acetate ions, [Ac] in terms of the other variables.
2. Use the known variables to solve for [Ac].
3. Assume that all the [Ac] is contributed by the complete dissociation of sodium acetate in the solution, and then you can solve for the mass of sodium acetate you need.

You should find that you need roughly 14.76g of sodium acetate to make up this buffer solution.

[Powder Miner]

Huh. I somehow managed to get exactly half of that on my first try around. But that confirms my base idea (solving for acetate with the magic of algebra) was a good idea! Thank you for responding! I should be able to figure out exactly how I managed to half my answer, I think.

Well, I had the calculations down to needing .180 molarity of sodium acetate to get the desired pH, and then I multiplied that by .5L (the original acetic acid volume) to get .090 mol, which translated to 7.38 g of sodium acetate.

[crazysheep]

No worries, dissociation constants can be a pain in the backside :p

edit: Did you remember to account for the starting concentration of the acetic acid?

[Powder Miner]

Yes. I figure I should put out my full process:
pH=5 log[H+]=-5 10^-5=1*10^-5 (insert cheeky tautology note here) [H+]=1e-5
1.8e-5=(1e-5)[C2H3O2-]/[HC2H3O2]
.5L*.2M acetic acid=.1 M HC2H3O2
R HC2H3O2 = H+ + C2H3O2-
I  0.1 M         0       0
C -x             +x      (whatever gets added for buffer stuff)   
E .1-x          1e-5    y

.1-1e-5=9.999e-2
1.8e-5=(1e-5)(y)/(9.999e-2) (all in molarity)

y=(1.83-5M*9.999e-2M)/1e-5M
y=.180 M
.180 M*.5 L=.090 mol (not sure what other volume I'd use but the preexisting solution volume...)
.090mol*82g/mol=7.38g

[crazysheep]

Yup, I think I see what you did wrong: the molarity of the acetic acid solution only needs to be incorporated once, so either you do that at the end, or do that at the start. Here you did it twice (both at the start and at the end), so that's why the mass you calculated is half of the one I calculated.

[Powder Miner]

Yeah, I figured that was the case, I'm just not too sure how to avoid it since my calculations are in molarity and m...
oh my god
i've made a very very simple mistake
the .1 wouldn't be in molarity, that's a mole value i calculated, I shouldn't incorporate that yet
i see what you mean

wow, thank you
finally, the problem is done

(I mean, it's number 2 out of 15 but actually getting what I'm doing will make the rest of the sheet go by MUCH more quickly)

[crazysheep]

Good luck with the rest of the problem sheet then

[Shadowlord]

What the fuck kind of insane troll logic is this?

From https://www.maa.org/sites/default/files/pdf/upload_library/22/Polya/07468342.di020786.02p0470a.pdf which was linked from https://golang.org/src/math/big/int.go?s=16674:16711#L703 which is the source code to the big integer stuff in Google's Go language.

[TheBiggerFish]

Something something consequent.

I don't know if there's a formal term, but that's not a proper proof by negation or whatever.

E:I cannot into logic apparently.