[lionsiy]

there are about 12 more followup questions but they had TeX in them and I didn't feel like figuring out how to get em to copy paste right

Also this was a question from the beginning of the year. Seeing as I am taking calc at the same time as physics, it was considerably harder when I was a beginner...

[carpartsho]

You're not a physicist, you piece of ****.

When I saw that, my first thought was to quote it specifically to make sure you would too.

[DoctorTDent]

If you would have said earlier that it was sub-optimal to exercise early since there would always be more value in selling the option (just another form of cashing in really!!)

/

This post is utterly ridiculous. I suggest you read a basic financial math book. Real investors, if they wish to take a market view, will resell the option for >= the intrinsic value in the market (strictly greater if implied volatility is anything other than 0). If the price was ever less than the intrinsic value then there exists a simple arbitrage. Exercising early is always suboptimal, modulo transaction costs and some minor difficulties like shorting at a retail level.

Like I said, please pay attention

[Stoottnoiciek]

And no, selling and exercising are not just different forms of cashing in.

The point is that EVEN IF YOU'RE PREVENTED FROM EXERCISING EARLY the option price is higher than or equal to the intrinsic.

[PlayboyAtWork]

HC, maybe you can answer a question that Kuci got wrong.

Betelgeuse is a star located 640 ly away from the earth.

RA is 5h 55m

Dec is +7 deg 24'

What are the coordinates of betelgeuse in cartesian coordinates, assuming the Sun is 0/0/0

[Hamucevasiop]

I don't know much physics, but... wouldn't momentum be conserved? So tripling the mass of the system causes the final velocity to be one third of the original.

maybe Ben could explain this stuff:

Vi = 0

Vf = sqrt(S^2/2m)

vf= vi + a*t

sqrt(S^2/2m) = g*t

t = sqrt(S^2/2)/g

???

[F1grandprix]

The force equals Mu*N = Mu*2M*g. Even if there were friction between the slab and the surface, the coefficient doesn't change. The force changes. Ok. I missed the part where it said the connection between the ground and the slab was frictionless. I thought they were asking for that, not for the friction between the slab and the block.

[larentont]

this is ****ing hilarious what are you talking about krazyhorse

[SzefciuCba]

That's how I solved for the velocity. Energy is conserved. You can use momentum too.

The rest is kinamatics equation. I know the final speed, and the initial speed, the acceleration and from that I can find the time.

ROFL
ROFLROFLROFLROFL
I AM RIIIIIIDING THE ROFLCOPTER
BAHAHAHA

It's kinetic friction you dumb ass, we aren't accounting for the waste heat so in this system we don't conserve energy. Jesus christ.

[DurryVony]

Ben, even your algebra is wrong.

[dremucha]

I can't believe he got it that wrong after a day to think about it.

[AngelBee]

I gave this less than five minutes effort, but I'm a lawyer, not a physicist. Somebody correct me if I'm wrong:

Spoiler: t = S/(2*g*Mu)
vf = S/2
Ff = 0 (no opposing force being applied)
The real dilemma is who I bill for that five minutes.

[fiettariaps]

You have the wrong coefficients on your answers.

Typed it in as I had it written down without looking to see what it actually said (that is, if what I corrected is what you're referring to).

[Licacivelip]

What was your original answer solomwi?

Take the parentheses out of the expression for t. As it read, but inadvertently, I had Mu*g on the other end of the fraction.

[MrsGoo]

Why is it (S/2)?

edit: crosspost

[hoarrimilsora]

Your answer would be right if the masses were the same, I believe