The answer is option (A) 1.0x10 5 lb Given, Density of Iron = 7.87 g/cm 3 The re

Question

The answer is option (A) 1.0x105 lb

Given,

Density of Iron = 7.87 g/cm3

The required conversions are:

1 inch =2.54cm

1lb = 453.6g

1 yard = 0.9144m = 91.44cm

Convert the given yard and depth measurement into cm, then you get

120 yards = 120*91.44 = 10972.8 cm

60 yards = 60*91.44 = 5486.4 cm

1mm = 0.1cm

Volume of the football = (10972.8)*(5486.4)*(0.1) cm = 6,020,116.992 cm3 (because the formula for voulume=length*breadth*depth)

We need to find mass of Iron.

We know that Density = Mass / Volume

So formula for Mass = Density * Volume

So now substitue the values of volume and density in the above formula. Then we get

Mass = (7.87 g/cm3 ) * (6,020,116.992 cm3)

Mass = 47,378,320.73g = 47 Mg

Now convert grams into lb, 1 lb = 453.6g

So Mass = 47,378,320.73g = 47 Mg = 1.0x 105 lb

So the answer is option (A) 1.0x105 lb

Explanation / Answer

20. Iron has a density of 7.87 g/cm3. What mass of iron would be required to cover a football playing surface of 120 yds x 60. yds to a depth of 1.0 mm? (1 inch -2.54 cm; 1 lb 453.6 g; 1 yard 0.9144 m) A) 1.0×105 lb B) 6.4×104 lb C) 6.8 x 103 Ib D) 4.7 x 10 lb E) 3.7×108 lb

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