Reply With QuoteFor the first one I got a difference in diameter of 11.4 inches. That would put the string 5.7 inches above the ground.
How high above the ground would a string that is three feet longer than the circumference of the earth have to be to form a perfect ring around the earth? Assume the earth is a perfect sphere and is exactly 25000 miles in circumference and the string is 25000 miles + three feet long. How high would that street be above the ground at all points (equidistant) to have the two ends meet so the string makes a perfect circle (ring) around the earth?
If you could fold a piece of paper in half 50 times (you can't but if you could), how thick would it be? Assume the paper is 1/500th of an inch thick to start.
Back to the perfect sphere earth, if a man 6'2" tall was standing on this earth, how far would the horizon be from him? Assume his eyes are exactly six feet above the ground.
Finally, you have two , two gallon buckets. Bucket A has one gallon of liquid A in it, and Bucket B has one gallon of liquid B in it. You take one cup of liquid A from bucket A and pour it into the bucket of liquid B. You stir it up, then you take one cup of that mix from bucket B and pour it into bucket A. Is there more liquid A in the B bucket or more liquid B in the A bucket?
For the first one I got a difference in diameter of 11.4 inches. That would put the string 5.7 inches above the ground.
When will the world learn that a million men are of no importance compared with one man? [Henry David Thoreau]
You are correct...Amazing isn't it?Originally Posted by Freedom&Liberty
Here's something even more amazing. That delta r is independent of C. The ball could be 25K miles like in the riddle above and the delta r is 5.7 inches OR it could be 25 ^25^25^25 trillion light years in circumference - and the delta r would be 5.7 inches OR it could be one inch in circumference (the string 37 inches) and the delta r is 5.7 inches. It is independent of C. That is mind boggling.
2 * pi * r = C
differentiate both sides:
2 * pi * dr = dC
dr = dC/(2*pi)
dC = 36 inches so dr always = 5.7 inches not matter what C is.
That is amazing.
#2) 35 million miles
Scandal? The government dispatched more firepower to arrest Nakoula Basseley Nakoula in Los Angeles than it did to protect its mission in Benghazi.
Correct: About 1/3 the distance to the sun. (Folding a piece of paper in half 50 times.)
I asked my dad this and he got it correct in about three seconds. Caveat: He was an electrical engineer who worked with radio antenna design. There is a rule of thumb relationship between antenna height and line of sight transmission radius.
That's similar to a puzzle I presented here several years ago. It was on a test I had in a math class in college. All it said was, "You have a solid sphere. Through the center of the sphere you drill out a hole exactly three inches long. How much of the sphere remains?"
It looks like the question is incomplete, but it provides all the data needed to calculate the correct answer. I remember it because I was the only one who came up with the correct answer and had to present the solution to the class at the blackboard.
May I ask, to clarify, "through the center of the sphere you drill out a hole.." Does that mean one end of the hole is at the center of the sphere and it reaches the exterior surface of the sphere so therefore the radius is three inches? Or does it mean the hole goes completely through the sphere so the diameter of the sphere is three inches?
Is the hole circular in cross section? Meaning is shape of the hole a cylinder?
The volume of the sphere is 1.33* pi*r^3. I don't know the radius of the hole drilled; I only know the length is three inches so I don't know the volume of the hole - unless I get more clarification above.
Last edited by Charles; 06-02-2012 at 11:09 AM.
Yes, a cylinder bored completely through the sphere, centered. Think of placing a ball on a drill press and making a ring shape out of it, a ring that is 3" wide, if that helps. (By wide, I mean the width of the material, not the diameter of the circle the ring describes. If it were a ring on a finger, it would be the width of the ring between the two knuckles, not the diameter of the finger.)
What is the volume of that ring?
Oh, your formula is incorrect. The term is 4/3, not 4.
Last edited by Chachma v'Oz; 06-02-2012 at 11:09 AM.
OK so there was more to the problem. Originally you didn't mention the hole went completely through the ball. (The ball could have had a radius of ten inches for example and you wrote the length of the hole was three inches).
ok it's 1:30am and just had a crack at the buskets in my head so please don't be to savage on me cause I probably missed something.... it's looking in my mind as if they turn up with equal ratios... with 1:4 opposing substances.... but like I said seems too easy I don't think I'm right
OK. This was 40 years ago. I wasn't precise enough the way I expressed it.
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
Bookmarks