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Alja [10]
2 years ago
8

PLEASE HELP THIS IS TIMED!

Physics
1 answer:
Monica [59]2 years ago
4 0
3.5 because yea periodt
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3 years ago
A marble column of cross-sectional area 1.6 m^2 supports a mass of 26600 kg. The elastic modulus for marble is 5.0 times 10^10 N
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Answer:

Δ L = 2.57 x 10⁻⁵ m

Explanation:

given,

cross sectional area = 1.6 m²

Mass of column = 26600 Kg

Elastic modulus, E = 5 x 10¹⁰ N/m²

height = 7.9 m

Weight of the column = 26600 x 9.8

                                    = 260680 N

we know,

Young's modulus=\dfrac{stress}{strain}

stress = \dfrac{P}{A}

           = \dfrac{260680}{1.6}

           = 162925

strain = \dfrac{\Delta L}{L}

now,

Y = \dfrac{stress}{strain}

\Delta L = \dfrac{162925}{Y}\times L

\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9

   Δ L = 2.57 x 10⁻⁵ m

The column is shortened by Δ L = 2.57 x 10⁻⁵ m

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