Example 12.6 squeezing a Brass Sphere A solid brass sphere is initially surround

Question

Example 12.6 squeezing a Brass Sphere A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.00 x 105 NVm2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth where the pressure is 2.00 x 107 N/m2. The volume of the sphere in air is 0.35 m3. By how much does this volume change once the sphere is submerged? SOLVE IT Conceptualize Think about movies or television shows you have seen in which divers go to great depths in the water in submersible vessels. These vessels must be very strong to withstand the large pressure under water. This pressure squeezes the vessel and reduces its volume. Categorize We perform a simple calculation involving the following equation, so we categorize this example as a substitution problem. AP R volume stress AF A volume strain AV Vi AV Vi VAP Solve the equation above for the volume AV change of the sphere: Substitute the numerical values: 0.35 m 2.00 x 10' N/m 1.00 x 10 N/m AV 6.10 x 10 100 N/m2 0.000114. m3 AV 3.9963*10 The negative sign indicates that the volume of the sphere decreases. Typical values for Elastic Moduli Young's Modulus Shear Modulus Bulk Modulus Substance (NM?) (N/m (N/m 11 11 11 Tungsten 3.5 x 10 1.4 x 10 2.0 x 10 11 10 100 Steel 2.0 x 10 8.4 x 10 6. x 10 11 10 11 Copper 1 x 10 4.2 x 10 1.4 x 10 10 10 10 Brass 9.1 x 10 3.5 x 10 x 10 10 10 10 Aluminum 7.0 x 10 2.5 x 10 7.0 x 10 6.5-7.8 x 10 2.6-3.2 x 10 5.0 5.5 x 10 100 Glass 10 10 10 Quartz 5.6 x 10 2.6 x 10 2.7 x 10 Water 2.1 x 10 10 Mercury 2.8 x 10 MASTER IT HINTS GETTING STARTED I M STUCK! The volume of the steel sphere is measured to be 0.76 m n air at standard atmospheric pressure. Now we put it in a vacuum jar (pressure assumed to be zero). What is the change in volume? AV

Explanation / Answer

formula for change in volume is

del V = -Vi del P/B

= - 0.76 m^3 ( 0- 1.013 * 10^5)/6 * 10^10

=0.128 * 10^-5 m^3

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