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kozerog [31]
3 years ago
5

A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does

the diameter of the sphere decrease as it sinks to a depth of 2.40 km? (Take the density of ocean water to be 1,030 kg/m3.)
Physics
1 answer:
astra-53 [7]3 years ago
7 0

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus =  14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

               = 24.25 × 10⁶  N/m²

Volume = \dfrac{4}{3} \pi r^3

         \dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}

Bulk modulus is equal to

B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }

now

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }

(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

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Density <em>ρ</em> is mass <em>m</em> per unit volume <em>v</em>, or

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