Urgent Help Needed

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November 18th 2007, 05:10 PM
cheyanne

Urgent Help Needed

Hello all:

I could really use some help with two problems:

1. A steel band is stretched tightly around the center of the Earth (the radius of the Earth is 6400 km). Then supposed the wire is cut and its length is increased by 20 m. It is then placed around the planet so that it is the same distance from the Earth at every point. Could you walk under the band? Why or why not?

2. Two hats contain red and blue marbles with the following distributions: Hat #1 has 2 red and 8 blue marbles, and hat #2 has 8 red and 2 blue marbles. Kathy selects a hat at random and randomly pulls out 2 marbles (without replacement). Both are blue. If she switches hats, what is the probability that she will now select a red marble?
November 18th 2007, 05:29 PM
Jhevon

Quote:

Originally Posted by cheyanne

Hello all:

I could really use some help with two problems:

1. A steel band is stretched tightly around the center of the Earth (the radius of the Earth is 6400 km). Then supposed the wire is cut and its length is increased by 20 m. It is then placed around the planet so that it is the same distance from the Earth at every point. Could you walk under the band? Why or why not?

Let $R$ be the radius of the Earth
Let $C$ be the circumference of the Earth, that is, $C = 2 \pi R$

Now, originally, the length of the band is $C$ . however, it is increased by 20m. now, this increase will lead to a corresponding increase in the radius of the center of the Earth to the band. Let this increase be $x$ meters.

Thus we have,
$C + 20 = 2 \pi (R + x)$

$\Rightarrow C + 20 = \underbrace {2 \pi R}_{\mbox {which is C}} + 2 \pi x$

$\Rightarrow C + 20 = C + 2 \pi x$

$\Rightarrow 2 \pi x = 20$

$\Rightarrow x = \frac {10}{\pi}$ meters

$\approx 3.183$ meters.

this is how high the band will be above te ground.

since i am 15 meters tall, i cannot walk under it, can you?

this ingenious solution is due to Soroban, see here

using this method, we didn't even need to know the radius or circumference of the Earth. the method i suggested in the post i directed you to requires it though. that's why Soroban is the genius
November 18th 2007, 05:38 PM
Soroban

Hello, cheyanne!

The first one is a classic problem . . .

Quote:

1. A steel band is stretched tightly around the Equator of the Earth.
(The radius of the Earth is 6400 km).
Then suppose the wire is cut and its length is increased by 20 m.
It is then placed around the planet so that it is the same distance
from the Earth at every point.
Could you walk under the band? Why or why not?

Let $R$ = the radius of the Earth.
The length of the band is: . $2\pi R$

Then the band is extended to: . $2\pi R + 20$

Let $R + h$ = the radius of the larger circle.
Then we have: . $2\pi(R + h) \;=\;2\pi R + 20\quad\Rightarrow\quad2\pi R + 2\pi h \:=\:2\pi R + 20$
. . $2\pi h \:=\:20\quad\Rightarrow\quad h \:=\:\frac{10}{\pi} \:\approx\:3.18\;m$

This is over 10 feet . . . Yes, we can walk under it.
November 18th 2007, 05:41 PM
Jhevon

Quote:

Originally Posted by Soroban

Hello, cheyanne!

The first one is a classic problem . . .

Let $R$ = the radius of the Earth.
The length of the band is: . $2\pi R$

Then the band is extended to: . $2\pi R + 20$

Let $R + h$ = the radius of the larger circle.
Then we have: . $2\pi(R + h) \;=\;2\pi R + 20\quad\Rightarrow\quad2\pi R + 2\pi h \:=\:2\pi R + 20$
. . $2\pi h \:=\:20\quad\Rightarrow\quad h \:=\:\frac{10}{\pi} \:\approx\:3.18\;m$

This is over 10 feet . . . Yes, we can walk under it.

I beat you with your own solution :D...sorry about that:(

i love this solution though, you can get by with an amazingly small amount of information