A Question of Molarity

Originally posted by smomofo
55.6, if you are talking about water. So would that be considered concentrated, or dilute?

I drop a ball off a 50m building onto your head. The initial velocity of it is 4.67 m/s, ignore air resistance. How long does it take before it nails you?

Originally posted by cmsMaster
I drop a ball off a 50m building onto your head. The initial velocity of it is 4.67 m/s, ignore air resistance. How long does it take before it nails you?

It depends on the direction that you "dropped" the ball and the height of my head.

Originally posted by smomofo
It depends on the direction that you through the ball and the height of my head.

Directly downward on top of you. Assume your head is 2.6 meters off the ground...You're probably not that tall, but you might be on a box or something....

Basically ignore the angle, sin, cos. etc. aren't needed. Just solve for time using what I gave.

I didn't think you'd actually solve it though...

Originally posted by cmsMaster
Directly downward on top of you. Assume your head is 2.6 meters off the ground...You're probably not that tall, but you might be on a box or something....

Basically ignore the angle, sin, cos. etc. aren't needed. Just solve for time using what I gave.

I didn't think you'd actually solve it though...

2.67 s

Originally posted by smomofo
2.67 s

IDK, I didn't do it.

I'll assume it's right.


Did you use

50-2.6 = 4.67(t) + .5(-9.8)(t)^2

Because if so I think you did it right.

Originally posted by cmsMaster
IDK, I didn't do it.

I'll assume it's right.


Did you use

50-2.6 = 4.67(t) + .5(-9.8)(t)^2

Because if so I think you did it right.

At first I was going to, but I didn't feel like solving a quadratic. So first I determined the final velocity using Vf^2 = Vi^2 + 2ad, then solved for time using Vf = Vi + at.

Originally posted by smomofo
At first I was going to, but I didn't feel like solving a quadratic. So first I determined the final velocity using Vf^2 = Vi^2 + 2ad, then solved for time using Vf = Vi + at.

Oh man, that confused the hell out of me for a second. We don't use the same variables in our work...let me see if I've got this right...

Final Velocity squared = Inital Velocity ^2 + 2 acceleration * displacement.

Hrm, that might work...I'm going to solve it, and see if they match. Give me a couple minutes, no calc. with me right now.

Ok nevermind, I was about to solve, but wasn't able to find a square root button on the computer calc.

Originally posted by cmsMaster
Oh man, that confused the hell out of me for a second. We don't use the same variables in our work...let me see if I've got this right...

Final Velocity squared = Inital Velocity ^2 + 2 acceleration * displacement.

Hrm, that might work...I'm going to solve it, and see if they match. Give me a couple minutes, no calc. with me right now.

Ok neverm ...[text shortened]... nd, I was about to solve, but wasn't able to find a square root button on the computer calc.

Just check the answer by plugging 2.66998 into the equation you asked if I used earlier, but leave the negative sign off of 9.8, or you will be changing your sign convention in the middle of your solution.

Originally posted by smomofo
Just check the answer by plugging 2.66998 into the equation you asked if I used earlier, but leave the negative sign off of 9.8, or you will be changing your sign convention in the middle of your solution.

It checked.

And leaving the negative probably would have been fine.

The ball is falling, so naturally, y-yo will be a negative number.

Originally posted by cmsMaster
It checked.

And leaving the negative probably would have been fine.

The ball is falling, so naturally, y-yo will be a negative number.

The way you wrote it in your equation it would have been positive. And then what about your initial velocity --- it should have been negative as well, if you are going to say that "y - yo" is negative "naturally".

Originally posted by cmsMaster
I drop a ball off a 50m building onto your head. The initial velocity of it is 4.67 m/s, ignore air resistance. How long does it take before it nails you?

you guys sound so young. I doubt there has been any balls dropped yet, no matter what direction.

Originally posted by cmsMaster
3.21X10^23

I seem to remember it being 6.02X10^23

Originally posted by cmsMaster
Ha, you looked it up, nice job.

Chemistry sucks.

Physics rules.

Originally posted by jimslyp69
Where does that leave biology?

I can't believe you all missed the obvious answer to this.

Biology drools.

Originally posted by RookRAK
I seem to remember it being 6.02X10^23

you mean 6.0221415 x 10^23 😀

but that is Avogadro's number, which is the number of atoms in a mole, or in this case the number of molecules in a mole, not how many moles per liter.

Originally posted by Suzianne
you mean 6.0221415 x 10^23 😀

but that is Avogadro's number, which is the number of atoms in a mole, or in this case the number of molecules in a mole, not how many moles per liter.

Damn! Here I thought I was brilliant for remembering something from freshmen chemistry (exactly 30 years ago this fall) ... and to think I got A's in that course.

Yes, of course you're correct. So, what is the actual answer?