A 8.1 pound bucket about 3/4 full of water sits on a scale that reads 85.3 pound

Question

A 8.1 pound bucket about 3/4 full of water sits on a scale that reads 85.3 pounds. A standard concrete block that measures 7-5/8 x 7-5/8 x 15-5/8 inch and weighs 33.5 pounds is lowered into the bucket and suspended by a rope that is attached to a spring scale. The block is completely in the water and no water spills out of the bucket. The spring scale reads 17.0 pounds after the concrete block is suspended in the water. Determine the following.

A) The volume of the water. (gallons)

B) The density of the block. (lb/ft3)

C) The percent of coring, or voids, in the block. (%)

D) The reading of the scale the bucket rests on after the block is suspended. (lb)

E) The concrete block is now suspended in a new liquid. The spring scale reading is 18.8 pounds. Determine the density of the new liquid (lb/ft3)

Will rate Lifesaver for good, clear, concise steps.

Explanation / Answer

A) 85.3-8.1=weight of water

volume=mass/density, density should be around 1000kg/m^3 in metric.

B) you could use the density equation again:

density=mass/volume, so 33.5 lb/(7-5/8 x 7-5/8 x 15-5/8 inch) should yield pounds per cubic INCH. Remember to convert.

C) For this one, I believe that you need to already know the density of concrete. Google says it is 149.8 lb/cft, so you can calculate the discrepancy between the expected and observed weights of the block with these dimensions.

((observed in (B)-expected)/expected)*100 yields the percent of the weight missing.

D) The block SHOULD weigh 33.5 lb, but the spring scale reads as 17.

33.5-17=weight of the block supported by the water, so add that to 85.3: the reading on the scale BEFORE the block.

E) The water displaced by the block originally is equal to 33.5 lb -17 lb.

The water displaced by the block in the new medium is 33.5-18.8

16.5=(water)Vg

14.7=(medium)Vg

16.5/(water)=14.7/(medium)

So...

medium=(14.7*water)/16.5

Hope I helped!

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