Physics 115A Autumn 2000 - Chapter 10 Homework Solutions: Pressure and Buoyancy, Assignments of Physics

Download Physics 115A Autumn 2000 - Chapter 10 Homework Solutions: Pressure and Buoyancy and more Assignments Physics in PDF only on Docsity! Physics 115A Autumn 2000 Chapter 10 Homework Problem Solutions 5. From the masses we have mwater = 98.44 g – 35.00 g = 63.44 g; mfluid = 88.78 g – 35.00 g = 53.78 g. Because the water and the fluid occupy the same volume, we have SGfluid = fluid/ water = mfluid/mwater = (53.78 g)/(63.44 g) = 0.8477. 9. (a) The force of the air on the table top is F = PA = (1.013 × 105 N/m2)(1.6 m)(1.9 m) = 3.1 × 105 N (down). (b) Because the pressure is the same on the underside of the table, the upward force has the same magnitude: 3.1 × 105 N. This is why the table does not move! 11. There is atmospheric pressure outside the tire, so we find the net force from the gauge pressure. Because the reaction to the force from the pressure on the four footprints of the tires supports the automobile, we have 4PA = mg; 4(240 × 103 N/m2 )(200 × 10–4 m2) = m(9.80 m/s2), which gives m = 1.96 × 103 kg. 13. The pressure from the height of alcohol must balance the atmospheric pressure: P = gh; 1.013 × 105 N/m2 = (0.79 × 103 kg/m3)(9.80 m/s2)h, which gives h = 13 m. 18. The minimum gauge pressure would cause the water to come out of the faucet with very little speed. This means the gauge pressure must be enough to hold the water at this elevation: Pgauge = gh; = (1.00 × 103 kg/m3)(9.80 m/s2)(41 m) = 4.0 × 105 N/m2. 24. When the balloon and cargo float, the net force is zero, so we have Fnet = 0 = Fbuoy – mHeg – mballoong – mcargog; 0 = airgVballoon – HegVballoon – mballoong – mcargog; 0 = (1.29 kg/m3 – 0.179 kg/m3))π(9.5 m)3 – (1000 kg + mcargo), which gives mcargo = 3.0 × 10 3 kg. 26. Because the mass of the displaced water is the apparent change in mass of the sample, we have ∆m = waterV. For the density of the sample we have = m/V = (m/∆m) water = [(63.5 g)/(63.5 g – 56.4 g)](1.00 × 103 kg/m3) = 8.94 × 103 kg/m3. From the table of densities, the most likely metal is copper. 30. When the ice floats, the net force is zero. If the fraction of the ice that is above water is f, we have (1.025) wg(1 – f )V = (0.917) wgV, which gives f = 0.105 (10.5%). Fnet = 0 = Fbuoy – miceg = swg(1 – f )V – icegV, or mballoong Fbuoy mcargog mHeg