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Alex17521 [72]
3 years ago
7

If you had 8 balls and 7 of them were a certain weight, and 1 of them was heavier, how could you find the heaviest ball. All the

balls look the same. How can you find the heaviest ball?
Physics
1 answer:
dybincka [34]3 years ago
7 0

Answer:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Explanation:

There are two method of comparing the balls 1) using a balance  2) by  only 2 weighings.

Make the following groups - --- (1,2,3),(4,5,6),(7,8)

Step 1. compare the Weigh (1,2,3) and (4,5,6)

there are 2 possible outcomes:

1---both the group are of same weight. and named as (Case A)

2--- one of the group is heavier than other and named as  (Case B)

Step 2. Let examine both case

In Case A --in this case, now compare the weight of 7th and 8th ball. By this you have recognize the heavier ball by 2 weighing method.

In Case B -- considered the heaviest group (assume group (1,2,3) is heavy), from this group take randomly two ball and compare the their weight. out of these two ball, one  is heavy else the third ball is.

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3 years ago
Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

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Explanation:

The volume of the alcohol is ...

  (33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

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The total volume is ...

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So, the resulting density is ...

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The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

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The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

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