Author Topic: Lets see if you can figure out this..  (Read 5502 times)

Offline Seifer

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Lets see if you can figure out this..
« on: August 25, 2004, 11:33:50 PM »
Lets see if you guys can figure out this.. I already know the answer and state why you think its that answer.

This problem should be done without looking up or using any known referenced values.It can of course be solved by looking up the value for the diameter of the earth, but do not solve it this way.

The earth is approximately a sphere. Image that a string is wrapped snugly around the equator of a perfectly smooth earth. Suppose that we now add 15 feet to the length of the string and shape the longer string into a smooth circle with its center still at the center of the earth. Can a two-year old walk under the string? Defend your answer.

Offline Sunlyte

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Re:Lets see if you can figure out this..
« Reply #1 on: August 26, 2004, 04:18:34 AM »
looks like nobody knows the answer but i'll say no because its just a STRING!!!

Offline amievil

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Re:Lets see if you can figure out this..
« Reply #2 on: August 26, 2004, 07:35:32 AM »
Semantics > you, sorry :(
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Offline HAPPYGOLUCKY

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Re:Lets see if you can figure out this..
« Reply #3 on: August 26, 2004, 08:28:49 AM »
I'm no math-whiz, but here's my shot at it:

Assume we're dealing with perfect circles, both of which have the same center point (the center of the earth). We'll call the equator circumference c1, and the larger circle c2. The circumference of any cirlce is:
c=2#r     (# is pi, equal to 3.14 for this problem  ;))

so, c2 = c1 + 15 (the additional 15 ft)

Now, we'll assume the minimum height of the child is 4 ft. So, the difference in the radius of circle one and the radius of circle 2 must equal 4 ft minimum.

adjusting our original formula for circumference by plugging in what we know/assume, we get:
c1+15  = 2#(R+4)

(c1+15)/(R+4) = 2#

Now, we have a ratio. we know that as C increases, R must also increase. The larger the values, the smaller the effect of the added string length. So, the less distance between the strings. The smaller the value of c, the greater the distance between the strings. we will not worry about the earth's circumference for a bit. We will theoretically reduce the circumference (and hence length of the string) to approximately zero. because c/r = 2# for circles, this will reduce the R to an even smaller number. Essentially, we have reduced the equation to this:
15/4 = 2# (or approx 6.28)

This is not true. Thus, for all values of our new circle's circumference, the distance between the strings would be less than 4 ft (our proposed minimum). It would start around 2.4 ft and get rapidly smaller as the circumference of the original circle increased.

At least, that's my argument.  
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cuteflyz89

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Re:Lets see if you can figure out this..
« Reply #4 on: August 26, 2004, 09:42:48 AM »
holy crap! how can u ppl be so smart like that? That looks so confusing I hate math and stuff like that....yea lol I dunno the answere and I dont feel like figuring it out, I did my math for the day.

Offline weredragon

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Re:Lets see if you can figure out this..
« Reply #5 on: August 26, 2004, 10:12:50 AM »
I'll just say that the Earth is so freaking huge that just adding 15 ft. to the string would not be nearly enough at all.

Also, assuming that the child would walk under the string at just a random place on Earth, chances are, that random place would be the ocean. You know, since there's a lot of it and all. So then I guess the kid could swim under it, or at least drown under it, but no walking.

Offline Cabbit

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Re:Lets see if you can figure out this..
« Reply #6 on: August 26, 2004, 10:18:18 AM »
Lets see if you guys can figure out this.. I already know the answer and state why you think its that answer.

This problem should be done without looking up or using any known referenced values.It can of course be solved by looking up the value for the diameter of the earth, but do not solve it this way.

The earth is approximately a sphere. Image that a string is wrapped snugly around the equator of a perfectly smooth earth. Suppose that we now add 15 feet to the length of the string and shape the longer string into a smooth circle with its center still at the center of the earth. Can a two-year old walk under the string? Defend your answer.

No, 15 feet of length = almost nothing in relationship to the radius. This is of course assuming the two-year old can walk.

I'm no math-whiz, but here's my shot at it:

Assume we're dealing with perfect circles, both of which have the same center point (the center of the earth). We'll call the equator circumference c1, and the larger circle c2. The circumference of any cirlce is:
c=2#r     (# is pi, equal to 3.14 for this problem  ;))

so, c2 = c1 + 15 (the additional 15 ft)

Now, we'll assume the minimum height of the child is 4 ft. So, the difference in the radius of circle one and the radius of circle 2 must equal 4 ft minimum.

adjusting our original formula for circumference by plugging in what we know/assume, we get:
c1+15  = 2#(R+4)

(c1+15)/(R+4) = 2#

Now, we have a ratio. we know that as C increases, R must also increase. The larger the values, the smaller the effect of the added string length. So, the less distance between the strings. The smaller the value of c, the greater the distance between the strings. we will not worry about the earth's circumference for a bit. We will theoretically reduce the circumference (and hence length of the string) to approximately zero. because c/r = 2# for circles, this will reduce the R to an even smaller number. Essentially, we have reduced the equation to this:
15/4 = 2# (or approx 6.28)

This is not true. Thus, for all values of our new circle's circumference, the distance between the strings would be less than 4 ft (our proposed minimum). It would start around 2.4 ft and get rapidly smaller as the circumference of the original circle increased.

At least, that's my argument.  

That's what I said + numbers and equations. >>

So then I guess the kid could swim under it, or at least drown under it, but no walking.

I would guess that most mammals would be hard pressed to do anything at all under it.
« Last Edit: August 26, 2004, 10:22:11 AM by Cabbit »

Offline RandomDan

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Re:Lets see if you can figure out this..
« Reply #7 on: August 26, 2004, 12:54:27 PM »
This problem should be done without looking up or using any known referenced values.It can of course be solved by looking up the value for the diameter of the earth, but do not solve it this way.

The earth is approximately a sphere. Image that a string is wrapped snugly around the equator of a perfectly smooth earth. Suppose that we now add 15 feet to the length of the string and shape the longer string into a smooth circle with its center still at the center of the earth. Can a two-year old walk under the string? Defend your answer.

No.

This BSS is shite as it does not support unicode so § is pi.

Assuming concentric circles, qua Fonzy:
(1) c0 = 2§r0
Increase circumference by A.
(2) c1 = 2§r0 + A = 2§r1
The difference between the radii is the minimum height of the child, h.
(3) r1 = r0 + h
Thus,
(4) 2§r1 = 2§(r0 + h) = 2§r0 + A
The extra height gained for an addition of A to the circumference
(5) h = A/2§

For any 15ft increase in circumference we have a 15/2§ ft increase in radius, or 2.38732415 ft

An average two year old is about 3 feet high, so assuming that they do not duck, no, they cannot.  It is worth noting that the initial radius drops out the mathematics, the height gained is independent of the initial circumference, which is counter-intuitive.

You people really out to use the metric system.
(Edit: corrected a mistake)
« Last Edit: August 26, 2004, 12:57:15 PM by RandomDan »
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Offline ZeroRyoko1974

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Re:Lets see if you can figure out this..
« Reply #8 on: August 26, 2004, 01:30:10 PM »
Sure she can, she just has to walk south of a string in a parallel direction to the string.  She is now walking under the string.
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Offline HAPPYGOLUCKY

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Re:Lets see if you can figure out this..
« Reply #9 on: August 26, 2004, 01:36:22 PM »
Yeah. Concur on 2.38 ft. Of course, I rounded it rather than go out 8 decimal places, but what do ya want from a Physicist.  ;D

ps-Metric system is for squares. Once America rules the world, we'll adopt our system, and metric weenies will have to go underground.

I think it's intuitive when you consider the fixed relationship between C and R of a circle, then just imagine an infinitely small circle and add 15 ft. I thought it was sorta like finding the area under a curve by filling the space with an infinite number of infinitely-narrow rectangles. But, those techniques are probably not available to most people at this site (I didn't learn it until college anyways  :-\), so I tried to explain myself in other terms. But, I'll default to the resident expert (you) to make it clear. If anyone wants to know about the evolution of avian tRNA (or something equally gay), I 'll step in.  ::)
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Offline RandomDan

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Re:Lets see if you can figure out this..
« Reply #10 on: August 26, 2004, 03:03:07 PM »
Yeah. Concur on 2.38 ft. Of course, I rounded it rather than go out 8 decimal places, but what do ya want from a Physicist.  ;D

I no longer have a calculator, I just typed "15/2pi" into Google and copy-pasted the answer.

If anyone wants to know about the evolution of avian tRNA (or something equally gay), I 'll step in.  ::)

For the record, I'm in the middle of trying to collect enough information to work out how long a virtual e+e- pair can exists for in a vacuum.  >_>
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Offline HAPPYGOLUCKY

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Re:Lets see if you can figure out this..
« Reply #11 on: August 26, 2004, 04:54:42 PM »
For the record, I'm in the middle of trying to collect enough information to work out how long a virtual e+e- pair can exists for in a vacuum.  >_>

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Offline RandomDan

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Re:Lets see if you can figure out this..
« Reply #12 on: August 27, 2004, 01:16:15 AM »
For the record, I'm in the middle of trying to collect enough information to work out how long a virtual e+e- pair can exists for in a vacuum.  >_>
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I know nothing about transfer ribonucleic acid >_>;

It's not that tricky, I was trying to derive the (Delta E)(Delta t) < h/4pi formula; I gave up, it was far too tricky the way I was going about it.  Briefly, you are allowed to "borrow" energy, Delta E, from nothing so long as you "return" it within some time, Delta t.  So, if you wanted to create an electron-positron pair (each has a rest mass of 0.511 MeV), you need to "borrow" about 1.022 MeV or 1.6352x10-13 J.  So that means that so long as they are not moving, the longest that they can 'exist' for is about 1.36x10-45 seconds; not very long.
.. not really that interesting


Or, in English, the entire world is subject to particles popping in and out of existence all the time, but this is okay so long as they disappear within the time allowed by quantum uncertainty.

Though I have a question:

You have been given two ropes of varying thickness (which means that the rate of burning is random, and not linear) and a box of matches.  All you have been told is that each rope takes an hour, in total, to burn if lit from an end.  How can you time 45 minutes using what you have been given?
« Last Edit: August 27, 2004, 01:47:54 AM by RandomDan »
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Offline HAPPYGOLUCKY

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Re:Lets see if you can figure out this..
« Reply #13 on: August 27, 2004, 09:03:00 AM »
Damn you, Dan. You made use my friggin' brain at work. I hate that.
I dunno if this is right, but I think it'll work. I would've tested it, but my boss has a "no lighting fires in your cubicle" policy.

I assumed the two ropes are identical (hey, you didn't say I couldn't).
Lay the two ropes next to each other. Light one rope at both ends simultaneously. Both ends will burn towards each other. When the flames meet, mark that point on the 2nd rope. Cut the 2nd rope in two at that point. Light both ends of either rope.
The first rope burn time is 30 min. The second rope burn time is 15 min. That makes 45 min.
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Offline RandomDan

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Re:Lets see if you can figure out this..
« Reply #14 on: August 27, 2004, 09:33:56 AM »
I assumed the two ropes are identical (hey, you didn't say I couldn't).
Lay the two ropes next to each other. Light one rope at both ends simultaneously. Both ends will burn towards each other. When the flames meet, mark that point on the 2nd rope. Cut the 2nd rope in two at that point. Light both ends of either rope.
The first rope burn time is 30 min. The second rope burn time is 15 min. That makes 45 min.

Close, but I'm going to have to say no.  The ropes are not identical.  Anyway, I didn't give you anything to cut the rope with  :P
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