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15 November, 17:03

You drop an irregular piece of metal into a container partially filled with water and measure that the water level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

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Answers (2)
  1. 15 November, 18:26
    0
    Density = 1464.8kg/m3

    the density of the metal is 1464.8kg/m^3

    Explanation:

    Given;

    Mass m = 450g

    Density = Mass/Volume = m/V

    Volume V = change in height * base area = ∆h * A

    ∆h = 4.8cm

    A = 8*8 = 64cm^2

    V = 4.8*64 = 307.2cm^3

    Density = 450g/307.2cm^3

    Density = 1.4648g/cm^3

    Density = 1.4648 * 1000kg/m^3

    Density = 1464.8kg/m3

    the density of the metal is 1464.8kg/m^3
  2. 15 November, 20:57
    0
    1464.84 kg/m³

    Explanation:

    Density = mass/volume.

    D = m/v ... Equation 1

    But from Archimedes principle,

    every object immersed in water will displaced an amount of water equal to its own volume

    Therefore,

    v = v' ... Equation 2

    Where v = volume of the irregular object, v' = volume of water displaced.

    Since the base of the container is a square,

    Then,

    v' = L² (d) ... Equation 3

    Where L = length of the square base of the container, d = rise in water level.

    Substitute equation 3 into equation 1

    D = m/L²d ... Equation 4

    Given: m = 450 g = 0.45 kg, L = 8 cm = 0.08 m, d = 4.8 cm = 0.048 m

    Substitute into equation 4

    D = 0.45 / (0.08²*0.048)

    D = 0.45/0.0003072

    D = 1464.84 kg/m³
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