How many pieces of paper will fit under the string?

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AI Thread Summary
The discussion centers around a mathematical problem involving the circumference of the Earth and the height of a string wrapped around it after adding 10 feet. The initial calculations suggested an unrealistic number of sheets of paper that could fit under the string, leading to confusion. Participants clarified the correct approach by using the formula for circumference and solving for the height of the string above the Earth's surface. The final consensus indicates that approximately 1909.8 sheets of paper, each 0.01 inches thick, would fit under the adjusted string. The importance of proper unit conversion and understanding the relationship between circumference and height was emphasized.
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Homework Statement



The Earth is a sphere with a radius of 24, 843 miles. You wrap a piece of string around
the Earth so that it fits snuggly. THEN, you cut it, add 10 feet to the string, and adjust
it so that it has equal height all around the world. Question: How many pieces of
paper will fit under the string (thickness = 0.01 inches).

Homework Equations



C=2pi(R)

The Attempt at a Solution



C1=2pi(24,843)
C2=2pi(24,843.00189)

C2-C1=0.1189997 miles

0.1189997 miles/ 2p i= R= 0.001893939miles

0.001893939mile x 5280/1mi = 9.999998174 feet

9.999998174 feet x 12in/1 ft =119.9999781 in

119.9999781 in/0.01 in = 11999.99781 sheets of paper

My answer seems unrealistic. I thought to solve this problem I would subtract the two circumferences and then divide the answer by 2pi to find the length of the gap and then i would use conversion factors to figure out the amount of papers that would stack up.
 
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It's several time too big.
 
Did I do it wrong?
 
C = 2π(R)

And C + 10 = 2π(R+h), where h is the height of the string above the surface of the sphere. Expand the right side out. Then substitute C for the 2πR on the right side & solve for h.
 
2∏(24848) +10 = 2∏(24848) + 2∏h

10=2∏h

10/2∏=h

h=1.5915

h/0.01 = 159.154 sheets of paper

Is this correct?
 
Did you convert inches to feet? I make it fewer than 2000 sheets.
 
SammyS said:
C = 2π(R)

And C + 10 = 2π(R+h), where h is the height of the string above the surface of the sphere. Expand the right side out. Then substitute C for the 2πR on the right side & solve for h.

This gives C + 10 = 2π(R) + 2π(h)
      = C + 2π(h)
Solve for h.

This gives h in feet. multiply by 12 to get inches.

I agree with NascentOxygen's calculation.
 
I think I got it this time. Thanks for the help.

2πr +10 = 2πr+ 2πh
2π(24848)+10 = 2π(24848)+ 2πh

10π=2h

10/2π=h

h=1.5915 ft
h=1.5915ft x (12 in)/(1 ft)
h/0.01 = 1909.8 sheets of paper
 
Looks good.

Notice that you really didn't need to plug in 24848 ft for r.
 

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