How many pieces of paper will fit under the string?

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Homework Help Overview

The problem involves calculating how many pieces of paper can fit under a string wrapped around the Earth, which is treated as a sphere with a specified radius. The string is adjusted after being cut and extended by a certain length, leading to a question about the height of the string above the Earth's surface and the corresponding number of paper sheets that could fit in that space.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods to calculate the height of the string above the Earth's surface after adding length to it. Some explore the relationship between the circumferences before and after the adjustment, while others question the accuracy of their calculations and conversions.

Discussion Status

The discussion includes multiple attempts at solving the problem, with participants providing different calculations and questioning each other's methods. Some guidance has been offered regarding the mathematical relationships involved, and there is an ongoing exploration of the correct approach to find the height and the number of sheets of paper.

Contextual Notes

There are mentions of potential conversion errors and the need to clarify units, particularly regarding the conversion from feet to inches. Participants are also considering the implications of their assumptions about the radius and the adjustments made to the string.

november1992
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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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