Pre-lab Density of an Unknown Metal A 10cm x 10 cm x 10cm block of steel ( steel

Question

Pre-lab Density of an Unknown Metal

A 10cm x 10 cm x 10cm block of steel (steel = 7900 kg/m3) is suspended from a spring scale. The scale is in Newtons.

What is the scale reading if the block is in air?

What is the scale reading after the block has been lowered into a beaker of

water and is completely submerged?

When analyzing a sample of ore, a geologist finds that it weighs 2.00 N in air and 1.13 N when immersed in water. What is the density of the ore? What assumptions did you make to answer this question?

Explanation / Answer

Volume = 10*10*10 = 1000cm^3 => 0.001m^3.

a.Mass = Volume*Density.
            = (0.001)*7900

            = 7.9Kg
W = 7.9*9.8

    = 77.42N

2..M =Volume(D-D ' )

         = 0.001*(7900-900)

         = 0.001*7000

         = 7Kg

W = 7 *9.8

    = 68.6N.

Bouyant force = 2 - 1.13 N = 0.87 N

now bouyant force = weight of water displaced = volume*density

density of water = 1000 kg/m^3

volume*1000 = 0.87

volume of water displaced = 8.7*10^(-4) .

Now this should also be volume of the ore ....(Note that here I assumed that ore was completely immersed in water)

weight of ore = 2N

so its mass = 2/9.8 kg = 1/4.9 kg

density of ore = mass/volume = 1/(4.9*8.7*^(-4)) = 234.57 kg/m^3

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