50.) a 26 m tall statue of buddha in tibet is covered with 279 kg of gold. if the gold
was applied to a thickness of 0.0015 mm, what surface area (in square units) was
covered? [gold's density is 19,320 kg/m?]

Respuesta :

The surface area of the statute covered with the gold is 9,627.32 m².

What is Volume?

Volume is a scalar quantity expressing the amount of three-dimensional space enclosed by a closed surface.

Volume of the statute covered with gold

The volume of the statute covered with gold is calculated as follows;

Volume = mass/density

Volume = (279 kg) / (19,320 kg/m³)

Volume = 0.0144 m³

Surface area of the statute covered with gold

The surface area of a solid object is a measure of the total area that the surface of the object occupies.

V = S.A x h

where;

  • S.A is surface area
  • h is thickness

S.A = V/h

S.A = (0.0144) / (0.0015 x 10⁻³)

S.A = 9,627.32 m²

Thus, the surface area of the statute covered with the gold is 9,627.32 m².

Learn more about surface area here: https://brainly.com/question/76387

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