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Chapter 10: Problem 19

A sphere of copper is subjected to 100 MPa of pressure. The copper has a bulk modulus of 130 GPa. By what fraction does the volume of the sphere change? By what fraction does the radius of the sphere change?

Short Answer

Expert verified
Answer: The change in volume fraction is approximately -0.000769, which means the volume decreases by about 0.0769%. The change in radius fraction is approximately -0.00059, which means the radius decreases by about 0.059%.

Step by step solution

01

Understand Bulk Modulus formula

Bulk modulus (B) is the ratio of the change in pressure (∆P) to the fractional change in volume (∆V/V). The formula for bulk modulus is as follows: B = -∆P / (∆V/V) We have been given the bulk modulus (B = 130 GPa) and the pressure (∆P = 100 MPa). We now need to find the change in volume fraction (∆V/V).
02

Convert units

Before moving further, we need to convert the given pressure and bulk modulus values to the same unit. Since the bulk modulus is given in GPa, let's convert the pressure to GPa as well: ∆P = 100 MPa = 100 × 10^{-3} GPa = 0.1 GPa
03

Calculate the change in volume fraction

Now, we can plug in the given values into the bulk modulus formula and solve for the change in volume fraction (∆V/V): 130 GPa = -(0.1 GPa)/(∆V/V) Rearranging the equation to find (∆V/V): ∆V/V = -(0.1 GPa)/(130 GPa) = -0.000769
04

Relate change in volume fraction to change in radius fraction

In order to relate the change in volume to the change in radius, we need to consider the formula for the volume of a sphere: V = (4/3)πr^3 The fractional change in volume is given by: ∆V/V = [(4/3)π(r+∆r)^3 - (4/3)πr^3] / (4/3)πr^3 Simplifying and canceling out common terms: 1 + (∆V/V) = (r+∆r)^3/r^3
05

Calculate the change in radius fraction

Now, we can plug in the change in volume fraction ∆V/V = -0.000769 and solve for the change in radius fraction (∆r/r): 1 - 0.000769 = (r+∆r)^3 / r^3 Taking the cube root of both sides: (1 - 0.000769)^{1/3} = r+∆r / r Let x = (1 - 0.000769)^{1/3} x = r+∆r / r Now calculating the change in radius fraction (∆r/r): ∆r/r = x - 1 Using a calculator, we get: x ≈ 0.99941 ∆r/r = 0.99941 - 1 = -0.00059
06

Interpret the results

The change in volume fraction is -0.000769. This means the volume of the sphere decreases by approximately 0.0769% when subjected to a pressure of 100 MPa. The change in radius fraction is -0.00059. This means the radius of the sphere decreases by approximately 0.059% under the same pressure.

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Most popular questions from this chapter

Martin caught a fish and wanted to know how much it weighed, but he didn't have a scale. He did, however, have a stopwatch, a spring, and a \(4.90-\mathrm{N}\) weight. He attached the weight to the spring and found that the spring would oscillate 20 times in 65 s. Next he hung the fish on the spring and found that it took 220 s for the spring to oscillate 20 times. (a) Before answering part (b), determine if the fish weighs more or less than \(4.90 \mathrm{N}\). (b) What is the weight of the fish?
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A \(0.50-\mathrm{kg}\) object, suspended from an ideal spring of spring constant \(25 \mathrm{N} / \mathrm{m},\) is oscillating vertically. How much change of kinetic energy occurs while the object moves from the equilibrium position to a point \(5.0 \mathrm{cm}\) lower?
(a) Given that the position of an object is \(x(t)=A\) cos \(\omega t\) show that \(v_{x}(t)=-\omega A\) sin \(\omega t .\) [Hint: Draw the velocity vector for point \(P\) in Fig. \(10.17 \mathrm{b}\) and then find its \(x\) component.] (b) Verify that the expressions for \(x(t)\) and \(v_{x}(t)\) are consistent with energy conservation. [Hint: Use the trigonometric identity $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]
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