ARCHIVED POSTS
< Earlier Kibitzing · PAGE 5 OF 277 ·
Later Kibitzing> |
| Jun-02-06 | | themadhair: Only if you require IRS on your back. |
|
| Jun-02-06 | | whatthefat: <WillC21>
It's a nice problem. What I'd find most interesting is to watch the reading on the scale as the last link of the chain is landing. As it's landing, the reading on the scale will be 3 times the weight of the chain. Just after it's landed and the chain's at rest, it must of course return to 1 times the weight of the chain. I think it'd be interesting to look at the time over which this transition occurs - the dependence on the length of chain links would be nice. |
|
| Jun-02-06 | | WillC21: Here's a non-physics brain teaser that I was working on the other day. I did not make this up, and it took me about 15 minutes to solve it: <Problem> John owns a computer on which he has access to a random number generator. What makes this random generator unique is that it can only generate numbers from 1-10, inclusive. He simply hits "space bar" to generate a random number. If John hits the space bar key three times, and thereby generates three random numbers, what is the probability of John getting 2 DIFFERENT numbers in the range 1-5(inclusive)? |
|
| Jun-02-06 | | WillC21: Assume the random generator is not defective, in so far as each number it generates is "perfectly" random. |
|
| Jun-02-06 | | whatthefat: <WillC21>
For x not y, and both in range 1-5, we have 3 possible arrangements:
xxy, xyx, yxx. All 3 have equal probability:
Prob. = 5/10*1/10*4/10 = 1/50
Thus, Total Prob. = 3/50 |
|
| Jun-02-06 | | WillC21: <whatthefat> I'll look over that solution tomorrow, as I think there's a slight mistake there, but for now let me leave you with a problem that I have a verifiable solution to(I'll check your solution to both problems tomorrow): <Problem> A car slides without friction down a ramp described by a height function "h(x)", which is smooth and monotonically decreasing as x increases from 0->L. "L" is the distance along the ground from where the ramps vertical leg intersects the ground to a point where a loop begins at the end of the ramp. The loop is circular(with radius "R"). Gravitational acceleration is a constant "g" in the negative "h" direction, obviously. If the velocity is zero when x=0, what is the minimum height h0 = h(0) such that the car goes around the loop NEVER leaving the track? |
|
| Jun-03-06 | | AniamL: <WillC21> Yeah, wouldn't the solution be: 5/10 (chance to get 1-5)
times
4/10 (chance to get 1-5, but not the same as the first) times
5/10 (chance to get 6-10)
times
3 (3 possible arrangements, as whatthefat noted)
= 3/10 = 30%? |
|
| Jun-03-06 | | whatthefat: <WillC21>
I solved the problem interpreting it as: all 3 numbers generated were in the 1-5 range, and there were 2 distinct numbers generated.<AniamL>'s way is also correct - under his interpretation. As for the new problem, you need to find the case where the centripetal force required to go through the loop is exactly provided by the gravitational force, at the top of the loop. By cons. of energy, the KE at the top of the loop will be mg*(h0-2R) = mv^2/2
i.e., v^2 = 2(h0-2R)
We require that F_cent=F_grav
So, mv^2/r=mg
=> 2mg(h0-2R)=mgR
So h0=2.5R, which is independent of g, as we might expect. |
|
| Jun-03-06 | | zarra: <AniamL> Here's another trick proof for you: How to proof that a flea is as heavy as an elephant? Let the elephant's mass be M, the flea's mass m, and the sum of the masses S. Therefore we have (1) M + m = S
(2) M = S - m
(3) M - S = -m
Now multiply equations (2) and (3):
M^2 - MS = m^2 - mS
Add (S/2)^2 to both sides:
M^2 - MS + (S/2)^2 = m^2 - mS + (S/2)^s
This is the same as
M^2 - 2M(S/2) + (S/2)^2 = m^2 - 2m(S/2) + (S/2)^2
Now apply the square of a binomial -formula:
(M - S/2)^2 = (m - S/2)^2
Thus we must have
M - S/2 = m - S/2
M = m
So the elephant's mass is equal to the flea's mass. |
|
| Jun-03-06 | | Catfriend: <zarra> Ah, yet another pretty trick people often use to <befuddle>, as <AniamL> calls it:) y^2=x^2 doesn't mean that x=y..
<WillC21> Yeah, I'm quite interested in physical and mathematical riddles! I visit <chessgames.com> much, much less than, say, a year ago, so I won't be a fast solver, but I might add some tough stuff myself. |
|
| Jun-03-06 | | ganstaman: How about this: Can every even integer greater than 2 be written as the sum of 2 (not necessarily unique) prime numbers? If so, prove it. If not, find a counter-example. For example: 4=2+2, 6=3+3, 8=5+3, 10=5+5, 16=3+13 or 5+11, etc. If you figure this one out, I'll give you a million dollars. But you realize that this is actually an unproven conjecture. If we could actually figure this out on an internet chess forum while mathematicians can't figure it out, I'm pretty sure we can earn more than a million dollars to cover the costs of me paying you. But our chances are slim. Oh well. |
|
| Jun-03-06 | | themadhair: <All> If you have heard of/seen this before then keep stum (quiet). Don't spoil it for those who haven't. <The Puzzle>There are three mathematicians, Erdõs, Einstein and Euclid. Now being the @#$%-stirrer that he is Erdõs decides to have some fun in the way only Paul Erdõs can. He chooses two numbers and tells the product to Einstein and the sum to Euclid. Now Einstein knows that Euclid knows the sum and Euclid knows that Einstein knows the product. They are told that both the product and sum are less than 100, and that both numbers are greater than 1. They meet to discuss the problem. The conversation is as follows: Euclid:I don't know the two numbers.
Einstein:I knew you wouldn't know.
Euclid:Now I know the two numbers.
Einstein:Now I know the two numbers.
So what are the two numbers? (both whole and natural) |
|
| Jun-03-06 | | themadhair: Addendum - <ganstaman>'s statement of Goldbach's conjecture above is useful for solving. |
|
| Jun-03-06 | | AniamL: Oohhh, very nice, <zarra>! Thankfully I've been brainwashed into saying "plus or minus square root of" whenever I take a root. :) |
|
| Jun-03-06 | | WillC21: <whatthefat> Good job on the physics problem. Correct! <whatthefat><AniamL> On the number generator problem, all it says is that 2 DIFFERENT numbers of the 3 are in the (1-5) range, inclusive. This does not mean that the 3rd number has to also be in the range(although it could be). For example, if it churned out numbers 4,5, and 10 this satisfies the criteria. If it churned out 4,5, and 1 this satisfies the criteria. If it churned out 4,5,5 this satisfies the criteria. If it churned out 4,4,6 this does NOT satisfy the criteria. Having clarified things, I'll let you guys work on it once more before posting the solution or assuming you can't solve it. |
|
| Jun-03-06 | | whatthefat: <WillC21: This does not mean that the 3rd number has to also be in the range>
Okay, well that was <AniamL>'s interpretation, so his solution is probably what you're looking for! |
|
| Jun-03-06 | | AniamL: Actually, my solution assumed that the third number could NOT be between 1 and 5... so the real answer should be: 5/10 (chance for 1-5)
times
4/10 (chance for 1-5 but not the first)
times
1 (last number can be anything)
times
3 (different ways)
= <60%>! Pretty good odds. |
|
| Jun-03-06 | | WillC21: I'm sorry, you are both incorrect still. Would you guys like the solution or to have more solving time? |
|
| Jun-03-06 | | ganstaman: I haven't fully read the other solutions, so sorry if this is a repeat. 3 numbers from 1-10 makes 1000 combinations (10*10*10). Now let's find the combinations that satisfy our conditions. The first number can be from 6-10, the second from 1-5, and the third from 1-5 but different from the second. This is a total of 5*5*4=100 combinations. We also have similar situation where the 6-10 number is in the second or third position. This means that 300 total combinations look like this. Also, all 3 numbers can be from 1-5. All 3 numbers can be different. This possibility contains 5*4*3=60 combinations. Alternatively, two numbers may be the same. This occurs in 3*(5*1*4)=60 combos. All in all, we have 300+60+60=420 combinations that meet our conditions. 420/1000 = 42%. I have to be correct because 42 is the answer to everything, right? |
|
| Jun-03-06 | | WillC21: <ganstaman> Bravo! Correct. |
|
| Jun-03-06 | | WillC21: Nicely explained as well. Clear and concise. |
|
| Jun-04-06 | | whatthefat: <AniamL: times
1 (last number can be anything) >
This is the problem under this interpretation of the problem, since the number of orderings depends on whether the third number is the same as one of the first 2. <ganstaman>'s case by case solution is necessary then. |
|
| Jun-04-06 | | whatthefat: Here's a problem that's a little trickier than you might think: A projectile is fired at angle theta to the horizontal, from a vertical height h>0, under a constant downward gravitational force. What angle theta(h) maximizes the horizontal distance covered before the projectile impacts the ground? N.B. For h=0, the solution is trivially theta=pi/4. It's also simple to see that for h>0 we require 0<theta<pi/4. |
|
| Jun-04-06 | | AniamL: <whatthefat> Ahhh, I see. Good reference, at least! I just finished reading the "increasingly misnamed five book trilogy" a couple weeks ago. |
|
| Jun-04-06 | | themadhair: <whatthefat>
Is it given by
sin (theta)= sqrt[u^2/(2u^2+2hg)]?
where u is the initial velocity. |
|
 |
|
ARCHIVED POSTS
< Earlier Kibitzing · PAGE 5 OF 277 ·
Later Kibitzing> |
Kibitzing
