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2 years ago

7

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

/m^2. By how much is the column shortened if it is 7.9 m high? Express your answer to two significant figures and include the appropriate units.

1 answer:

vodka [1.7K]2 years ago

3 0

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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