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Mathematics 8 Online
OpenStudy (anonymous):

Can you show me how to solve? Don't answer. Just show me the steps.

OpenStudy (anonymous):

the radio station has placed pennies in a cylindrical glass jar. Each penny is 0.75 inches in diameter and 0.061 inches thick. If the cylindrical glass jar containing the pennies has a diameter of 6 inches and a height of 11.5 inches, how many pennies can fit inside the jar?

OpenStudy (anonymous):

Do I solve for the cylinder and the pennies than subtract? I'm not sure, I just want to make sure I am doing it right. I don't understand geometry all that well but im trying

OpenStudy (anonymous):

This is as far as I got: V of sphere = 4/3 (3.14)r^3 4/3 (3.14)(.375)^3 1.3(3.14)(.375)^3 1.53075^3 3.586846606296875 Someone please help me.

jimthompson5910 (jim_thompson5910):

the jar is a cylinder, not a sphere

jimthompson5910 (jim_thompson5910):

if you stack the pennies perfectly, they form a cylinder

OpenStudy (anonymous):

so do I use v= (3.14)r^2(h) instead?

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

This is very difficult to do correctly, Stacking pennies will give you cylinders, and you can play with the footprint of the pennies on the floor of the jar, but you can see that there is a lot of air space that will remain. I do not know the answer. Alternatively, you can shake them up and hope to fill space better than with the cylinders.

OpenStudy (anonymous):

like this ? V of cylinder = (3.14)r^2(h) (3.14)(3)^2(11.5) 3.14(6)(11.5) 216.66

jimthompson5910 (jim_thompson5910):

3^2 is not 6

jimthompson5910 (jim_thompson5910):

3^2 means "3 squared" and NOT "3 times 2"

OpenStudy (anonymous):

oh thank you, then the final answer woul be 324.99 right?

jimthompson5910 (jim_thompson5910):

that's the volume of the entire jar

jimthompson5910 (jim_thompson5910):

what's the volume of a single penny?

jimthompson5910 (jim_thompson5910):

each penny is a cylinder on its own

OpenStudy (anonymous):

that's what I thought this was? V of sphere = 4/3 (3.14)r^3 4/3 (3.14)(.375)^3 1.3(3.14)(.375)^3 1.53075^3 3.586846606296875

jimthompson5910 (jim_thompson5910):

the pennies aren't spheres

jimthompson5910 (jim_thompson5910):

the jar isn't a sphere either

OpenStudy (anonymous):

Then what would the formula be fore the pennies???

jimthompson5910 (jim_thompson5910):

same cylinder formula (just different radius and height)

jimthompson5910 (jim_thompson5910):

both are cylinders

jimthompson5910 (jim_thompson5910):

|dw:1399861837541:dw|

OpenStudy (anonymous):

oh ok like this V of pennies = (3.14)r^2(h) (3.14)(0.375)^2(0.061) (3.14)(0.140625)(0.061) 0.0269353125

jimthompson5910 (jim_thompson5910):

I can't draw, but imagine looking straight down (from a birds eye view) into the jar this is what you'd see |dw:1399861957000:dw|

jimthompson5910 (jim_thompson5910):

Notice the empty spaces, I'll come back to this in a sec

jimthompson5910 (jim_thompson5910):

So the volume of the jar is roughly 324.99 the volume of a single penny is 0.0269353125

jimthompson5910 (jim_thompson5910):

Divide the two: 324.99/0.0269353125 = 12,065.5737704918 then round down to the nearest whole number to get 12,065

OpenStudy (anonymous):

Years ago I did some serious study of the math associated with packing of objects into confined volumes. This question is nol suitable for the students, I suspect. The closest packing of the circles on the bottom of the jar is a good approach but not demonstrably the closest packing obtainable.

jimthompson5910 (jim_thompson5910):

so if you were to melt down each penny into a liquid, then pour all of that liquid into the jar, you could fit in about 12,065 pennies

jimthompson5910 (jim_thompson5910):

however, in my drawing we have solid circles that touch each other and form empty space so you'll have room for less than 12,065 pennies (not sure how much more less)

OpenStudy (anonymous):

oh wow. Thank you so much. I think I understand a bit better. :)

jimthompson5910 (jim_thompson5910):

but 12,065 is definitely the highest you can go

jimthompson5910 (jim_thompson5910):

you're welcome, I'm glad it's clicking now

OpenStudy (anonymous):

Thank you. :)

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