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A gold ingot weighs 5.50 ibs. If the density of gold is 19.31 g/cm^3, and the length and width of the ingot are 12.0 cm and 3.00 cm respectively, what is the height of the ingot? A) 3.59 cm B) 1.34 times 10^3 cm C) 10.2 cm D) 6.50 times 10^-3 cm

Respuesta :

Answer:

A) 3.59 cm

Explanation:

Given that :-

The density of the gold ingot = [tex]19.31\ g/cm^3[/tex]

Given that:- Mass = 5.50 lbs

Also, considering the conversion of lbs to g as shown below:-

1 lb = 453.592 g

Thus,

Mass = [tex]5.50\times 453.592\ g[/tex] = 2494.756 g

The volume = Length*Breadth*Height

Given that:- Length = 12.0 cm , Breadth = 3.00 cm

Considering the expression for density as:-

[tex]Density=\frac{Mass}{Volume}[/tex]

[tex]19.31=\frac{2494.756}{12.0\times 3.00\times Height}[/tex]

Solving for height, we get that:-

Height=3.59 cm

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