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Finger [1]
3 years ago
6

Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance

+/- 0.000005 m^3/kg)?
Engineering
1 answer:
Svetradugi [14.3K]3 years ago
6 0

Answer:

specific\ volume=0.00097\ m^3/kg

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 kg/m^3

Density of sea water = 1025  kg/m^3

We know that

Density=\dfrac{mass}{Volume}   ---1

Specific volume

specific\ volume=\dfrac{Volume}{mass}    ---2

From equation 1 and 2

We can say that

specific\ volume=\dfrac{1}{density}\ m^3/kg

specific\ volume=\dfrac{1}{1025}\ m^3/kg

specific\ volume=0.00097\ m^3/kg

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Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
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Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

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Given:

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Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

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40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

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