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Chapter 10: Problem 24

An anchor, made of cast iron of bulk modulus \(60.0 \times 10^{9} \mathrm{Pa}\) and of volume \(0.230 \mathrm{m}^{3},\) is lowered over the side of the ship to the bottom of the harbor where the pressure is greater than sea level pressure by \(1.75 \times 10^{6} \mathrm{Pa} .\) Find the change in the volume of the anchor.

Short Answer

Expert verified
Answer: The change in volume of the anchor is \(6.77 \times 10^{-6}\:m^3\).

Step by step solution

01

Write down the given values

We have the values for initial volume, bulk modulus, and change in pressure. Initial Volume: \(V = 0.230\:m^3\) Bulk Modulus: \(B = 60.0 \times 10^9\:Pa\) Change in Pressure: \(ΔP = 1.75 \times 10^6\:Pa\)
02

Plug in values into the formula

We will now plug in the values into the formula: ΔV = \(V\frac{ΔP}{B}\) ΔV = \((0.230\:m^3)\frac{1.75 \times 10^6\:Pa}{60.0 \times 10^9\:Pa}\)
03

Calculate the change in volume

Now we will simplify the expression and calculate the change in volume: ΔV = \((0.230\:m^3)\frac{1.75 \times 10^6\:Pa}{60.0 \times 10^9\:Pa} = 6.77 \times 10^{-6}\:m^3\) So the change in volume of the anchor is \(6.77 \times 10^{-6}\:m^3\).

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