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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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Most popular questions from this chapter

Spider silk has a Young's modulus of $4.0 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}\( and can withstand stresses up to \)1.4 \times 10^{9} \mathrm{N} / \mathrm{m}^{2} . \mathrm{A}$ single webstrand has across-sectional area of \(1.0 \times 10^{-11} \mathrm{m}^{2}\) and a web is made up of 50 radial strands. A bug lands in the center of a horizontal web so that the web stretches downward. (a) If the maximum stress is exerted on each strand, what angle \(\theta\) does the web make with the horizontal? (b) What does the mass of a bug have to be in order to exert this maximum stress on the web? (c) If the web is \(0.10 \mathrm{m}\) in radius, how far down does the web extend? (IMAGE NOT COPY)
(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 $\sin ^{2} \omega t+\cos ^{2} \omega t=1.1$.]
A brass wire with Young's modulus of \(9.2 \times 10^{10} \mathrm{Pa}\) is $2.0 \mathrm{m}\( long and has a cross-sectional area of \)5.0 \mathrm{mm}^{2} .$ If a weight of \(5.0 \mathrm{kN}\) is hung from the wire, by how much does it stretch?
It takes a flea \(1.0 \times 10^{-3}\) s to reach a peak speed of $0.74 \mathrm{m} / \mathrm{s}$ (a) If the mass of the flea is \(0.45 \times 10^{-6} \mathrm{kg},\) what is the average power required? (b) Insect muscle has a maximum output of 60 W/kg. If \(20 \%\) of the flea's weight is muscle, can the muscle provide the power needed? (c) The flea has a resilin pad at the base of the hind leg that compresses when the flea bends its leg to jump. If we assume the pad is a cube with a side of \(6.0 \times 10^{-5} \mathrm{m},\) and the pad compresses fully, what is the energy stored in the compression of the pads of the two hind legs? The Young's modulus for resilin is $1.7 \times 10^{6} \mathrm{N} / \mathrm{m}^{2} .$ (d) Does this provide enough power for the jump?
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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