How many pieces of paper will fit under the string?

The radius cancels out when you subtract the two circumferences. So you could have used r for the radius of the Earth, and r+h for the radius of the string + 10 feet. So you could have just written:2πr +10 = 2πr+ 2πh10=2πh10/2π=hh=1.5915 fth/0.01 = 1909.8 sheets of paperIn summary, the Earth is a sphere with a radius of 24,843 miles. After wrapping a string around the Earth and adding 10 feet to it, the height of the string above the surface is 1.5915 feet. This equals
  • #1
november1992
120
0

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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  • #2
It's several time too big.
 
  • #3
Did I do it wrong?
 
  • #4
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.
 
  • #5
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?
 
  • #6
Did you convert inches to feet? I make it fewer than 2000 sheets.
 
  • #7
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.
 
  • #8
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
 
  • #9
Looks good.

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

1. How do you determine how many pieces of paper will fit under the string?

The number of pieces of paper that can fit under a string depends on several factors, such as the length and thickness of the string, the size and weight of the paper, and the tension of the string. To determine an accurate estimate, a series of experiments and calculations must be conducted.

2. Is there a specific formula or equation to calculate the number of paper pieces?

While there is no specific formula for calculating the exact number of paper pieces that can fit under a string, there are several mathematical principles that can be used to approximate the answer. These include the Pythagorean theorem, the area of a rectangle, and the circumference of a circle.

3. Does the type of paper affect how many pieces can fit under the string?

Yes, the type of paper can greatly affect the number of pieces that can fit under a string. Thicker or heavier paper will take up more space and may reduce the number of pieces that can fit. Additionally, the shape and size of the paper can also impact the results.

4. Can other materials besides paper be used in this experiment?

Yes, other thin and flat materials such as plastic sheets, fabric, or even playing cards can be used in place of paper. However, the number of pieces that can fit may vary depending on the material's weight and flexibility.

5. Are there any real-world applications for this experiment?

While this experiment may seem like a simple and fun math challenge, it actually has practical applications in fields such as architecture, engineering, and design. Calculating the number of materials that can fit in a specific space is important in determining building materials, storage capacity, and packaging design.

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