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There is a jar full of nails, i have 50 nails at my disposal, there is no other jar that i can weigh or use. The jar is oddly-shaped, not perfectly symmetrical, it is smaller at the top of the jar and it gets bigger going down... is there any way of using graphs maybe?

2007-08-22 13:10:49 · 6 answers · asked by Alex T 2 in Science & Mathematics Other - Science

There is a jar full of nails, i have 50 nails at my disposal, there is no other jar that i can weigh or use. The jar is oddly-shaped, not perfectly symmetrical, it is smaller at the top of the jar and it gets bigger going down... is there any way of using graphs maybe? Oh and you can't open the jar, you can only look (wish it was that easy).

2007-08-22 13:38:17 · update #1

6 answers

A couple of approaches:
1) Weigh the jar full of nails. Measure the jar as best you can and estimate its weight. Subtract the estimated weight from the total. Divide by the weight of the 50 nails, times 50.

2) Measure the jar and estimate its volume. Put the 50 nails in another container and estimate their volume. Volume of the jar divided by volume of 50 nails times 50 is your estimated count.

2007-08-22 19:22:09 · answer #1 · answered by injanier 7 · 0 0

You could melt the jar and the nails, seperate the two phases, form new nails with an exact count, and form a new jar. Then, tell someone how many nails you made and put in the new jar.

There is no way to determine the number of nails inside the container with the information you've been provided.

You cannot do it by volume, as the nails can group together differently and cause inconsistencies in the measurement. Fill the jar ten times, and based on how the nails land in the jar, there will be more or less of the space consumed by empty air each time you filled it.

The nails will each have a different weight, and this precludes using weight as an accurate measurement. The difference is small, but significant. If you knew the maximum weight variation, you would be able to accurately judge the weight of the nails up to the quantity where the accumulated maximum error of each nail equaled the minimum weight of one nail. At this point, your measurement is now uncertain and it is impossible to measure accurately based on weight.

So, if this is a question in school, tell your teacher it's a bad question and they need to go take some refresher courses on physical metrology. Asking questions that hurt true critical thinking doesn't help to train better professionals.

2007-08-26 03:33:35 · answer #2 · answered by AJ R 3 · 0 0

Find an identical jar and weigh it empty. Now find a nail of the size that is in the jar.

Now weigh the jar with the nails in it. Subtract the weight of the empty jar from the weight of the jar with the nails. Divide the difference by the weight of a single nail.

Get a good night's sleep.

2007-08-22 22:37:14 · answer #3 · answered by aviophage 7 · 0 0

Weigh the 50 nails, then empty the nails in the jar and weigh them, then divide the 50 nail weight into the weight of the jar nails. Now multiply.

2007-08-22 20:31:25 · answer #4 · answered by billy brite 6 · 1 0

If you can model the geometry of the jar, you can write a numerical random packing program to determine the statistically likely number.

You could do a similar thing by determining the mean depth of the jar, computing the statistical number of nails along any perpendicular. Find the tranmitivity of light integrated over that crosssection. The density comparison should help arrive at a reasonable number.

2007-08-23 13:32:29 · answer #5 · answered by jcsuperstar714 4 · 0 0

Come on now, do your own homework!

2007-08-22 20:18:42 · answer #6 · answered by Marc G 6 · 1 0

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