Converting Units: Apple and Cranberry Barrels, Cloud Water Volume and Mass

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SUMMARY

The discussion focuses on converting units related to apple and cranberry barrels, specifically addressing discrepancies in shipment volume and cloud water calculations. The apple barrel has a volume of 7056 cubic inches, while the cranberry barrel is 5826 cubic inches. A merchant selling 50 cranberry barrels results in a volume discrepancy of 61500 cubic inches, which converts to approximately 1000 liters. Additionally, the volume of water in a cylindrical cumulus cloud is calculated using the formula v=πr²h, with the cloud's dimensions provided as 2.5 km in height and 1.5 km in radius.

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  • Familiarity with the formula for the volume of a cylinder (v=πr²h)
  • Knowledge of basic unit conversions (e.g., kilometers to meters)
  • Awareness of water density (1000 kg/m³) for mass calculations
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I realize these may be easy to some of you, but I'm have a few issues. For the first one:

Two type of "barrel" units were in use in the 1920s in the United States. The apple barrel had a legally set volume of 7056 cubic inches. The cranberry barrel, 5826 inches. If a merchant sells 50 cranberry barrels of goods to a customer who thinks he is receiving apple barrels, what is the discrepancy in the shipment volume in liters?

The first disrepancy(in cubic inches) I get 61500. I multiplied both volumes by 50 and took the difference, was this the right approach? I then converted this number 61500 to L and got 1.007L which is not correct. hmm

2. A cubic centimeter in a typical cumulus cloud contains 50 to 500 water drops, which have a typical radius of 10 µm. For that range, give the lower value and the higher value, respectively, for the following.

(a) How many cubic meters of water are in a cylindrical cumulus cloud of height 2.5 km and radius 1.5 km?
to m3
(b) How many 1 liter pop bottles would that water fill?
to bottles
(c) Water has a mass per unit volume (or density) of 1000 kg/m3. How much mass does the water in the cloud have?

Do I use the v=PIxr^2xh formula to compute the volume asked for in part A? I have a feeling I'll be plugging in 50 and 500 for max and min values somewhere bu thonestly I don't even know where to start in this problem. Any help would be hihgly appreciated!
 
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Hello Lozek. For the first one, I would just calculate the difference in cubic inches per barrel and then just multiply by 50. But we get the same thing: 61500 cubic inches. Now, there are 61.024 cubic inches in a liter. Just looking at the two numbers, can you see that 61500 cubic inches is about 1000 liters? You know, I got about "61" thousand and each liter has about 61 cubic inches. Can you figure out how much exactly?

For the second one, we worked a similar one but didn't go down to bottles although I did want to figure bucket fulls but I digress. Click this link:

Link to the cloud problem
 
The first disrepancy(in cubic inches) I get 61500. I multiplied both volumes by 50 and took the difference, was this the right approach? I then converted this number 61500 to L and got 1.007L which is not correct. hmm
61500 cu in is what I get. Show your work for the conversion so we can find where you went wrong.


Do I use the v=PIxr^2xh formula to compute the volume asked for in part A? I have a feeling I'll be plugging in 50 and 500 for max and min values somewhere bu thonestly I don't even know where to start in this problem. Any help would be hihgly appreciated!

[itex]v = \pi r^2h[/itex] is the volume of a cylinder with radius r and height h. You know one droplet of water has a radius of 10 micrometers, its asking you how many water droplets fit in a 2.5km x 1.5km cloud. a lot of conversions in this one.
 

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