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kompoz [17]
3 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.
Physics
1 answer:
vodka [1.7K]3 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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Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba
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Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

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Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

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As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

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T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

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3 years ago
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