artemis96
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please help

a piece of wood of mass 40g and uniform cross-sectional area of 2cm^2 floats upright in water. calculate the length of the immersed part of the wood​

Respuesta :

The length of the immersed part of the wood​ is 0.2m.

What is density?

The density is the ratio of the mass and the volume of the object. It is denoted by ρ.

ρ = mass/Volume = m/V

Volume = Area of cross-section A x height h

Given is a piece of wood of mass 40g =0.04kg and uniform cross-sectional area A= 2cm² =2x10⁻⁴ m² floats upright in water.

The density of water ,ρ = 1000 kg/m³

The density related to length of part immersed is

1000 = 0.04/ (2x10⁻⁴ x h)

h = 0.2m

Thus, the length of the immersed part of the wood​ is 0.2m.

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