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A solid sphere of brass (bulk modulus of 14.0 āœ• 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.)

Respuesta :

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus = Ā 14.0 Ɨ 10Ā¹ā° N/mĀ²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = Ļ g h = 1030 Ɨ 9.81 Ɨ 2.4 Ɨ 1000

Ā  Ā  Ā  Ā  Ā  Ā  Ā  Ā = 24.25 Ɨ 10ā¶ Ā N/mĀ²

Volume = [tex]\dfrac{4}{3} \pi r^3[/tex]

Ā  Ā  Ā  Ā  Ā [tex]\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}[/tex]

Bulk modulus is equal to

[tex]B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }[/tex]

now

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

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

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

Ī” r = -0.0156 m

change in diameter

Ī” d = -2 Ɨ 0.0156

Ī” d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

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