Question:
If a stack of 40 nickels fits snugly in the coin wrapper shown, how thick is 1 nickel? Round to the nearest hundredth.
Coin wrapper volume = 27,480 mm^3
Coin diameter = 21.18 mm
Answer:
The thickness of a coin is: 1.95mm
Step-by-step explanation:
Given
-- volume
--- the diameter
Required
The thickness of each coin
This implies that, we calculate the height of each coin.
First, calculate the radius (r) of the coin
![r = \frac{1}{2}d](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B1%7D%7B2%7Dd)
![r = \frac{1}{2} * 21.18mm](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%2021.18mm)
![r = 10.59mm](https://tex.z-dn.net/?f=r%20%3D%2010.59mm)
The volume of the coin is calculated as:
![V =\pi r^2H](https://tex.z-dn.net/?f=V%20%3D%5Cpi%20r%5E2H)
Where H represents the total heights of the 40 coins
![27480 = 3.14 * 10.58^2 * H](https://tex.z-dn.net/?f=27480%20%3D%203.14%20%2A%2010.58%5E2%20%2A%20H)
Solve for H
![H = \frac{27480 }{ 3.14 * 10.58^2}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B27480%20%7D%7B%203.14%20%2A%2010.58%5E2%7D)
![H = \frac{27480 }{ 351.480296}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B27480%20%7D%7B%20351.480296%7D)
The height of a coin is calculated by dividing H by 40
![h = \frac{78.1836145944}{40}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B78.1836145944%7D%7B40%7D)
![h = 1.95459036486](https://tex.z-dn.net/?f=h%20%3D%201.95459036486)
![h \approx 1.95mm](https://tex.z-dn.net/?f=h%20%5Capprox%201.95mm)
<em>Hence, the thickness of a coin is: 1.95mm</em>