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.

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.

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

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.

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.

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. >_>

Quote from: RandomDan on August 26, 2004, 03:03:07 PMFor 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. >_>Fonz's head explodes.

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.