29:006 Solutions Bloomfield 5e. Pages 168, 169. Ex. 2: It floats because it
displaces enough water to provide a buoyant force equal to its weight. Ex. 3: If
water ...
29:006 Solutions Bloomfield 5e
Pages 168, 169 Ex. 2: It floats because it displaces enough water to provide a buoyant force equal to its weight. Ex. 3: If water does not get into the car, a large portion of the car’s overall volume is just air so the average density is less than the density of water. Ex. 4: Cold air is denser that warm air so the cold air remains above the food. Ex. 7: Water is denser than gas, so 1 kg of water takes up less space than 1 kg of gas. Ex. 8: The density of oil is less than the density of vinegar, so the oil floats on top Ex. 13: The jar is packed under vacuum so that the pressure inside is less than the pressure of the atmosphere outside. Before the jar is opened the dimple is pushed in, and when it is opened air rushes in equalizing the pressure and causing the dimple to pop. Ex. 21: The pressure of the water increases with depth, so there is greater pressure on the bottom of the dam. The bottom of the dam must be thicker than the top to handle this increased pressure. Ex. 22: The pressure exerted by the water increases with depth. Below a certain depth the pressure may be so high that water may leak through the seal in the watch.
Pages 169, 170 Problem 2: The front surface of the textbook has dimensions of 8 in x 10 in or 80 sq. in. Atmospheric pressure is roughly 15 pounds per sq. in (psi), so the force on the book is
F = P A = 15 psi x 80 sq. in = 1200 lb. Alternately the books dimensions are 20 cm by 25 cm (0.2m x 0.2 m) for an area of 0.05 m2. Atmospheric pressure is roughly 1x105 N/m2 (Pa) so the force is F = P A = 1x105 N/m2 x 0.05 m2 = 5000N.
Problem 6: The buoyant force is the weight of displaced water or FB = 10 kg x 10m/s2 = 100 N. The net force on the log then is Fnet = FB – Wlog = 100 N – 8 kg x 10m/s2 = 100 N -80 N = 20 N upward, the log will float. Problem 7: The boat must displace enough water so that the buoyant force equals its weight. Thus the buoyant force must be 1200 N.
The buoyant force is the weight of displaced water = FB = mdw g, where mdw is the mass of the displaced water, So 1200 N = mdw g = mdw x 10 m/s2
mdw = 120 kg.
Problem 15: For each 10 m of depth of water, the pressure increases by one atmosphere (105 Pa). So at 300 m the pressure increase is 300/10 = 30 times 105 Pa = 3 x 106 Pa. The total pressure is then the pressure of the atmosphere on the water’s surface plus the increased pressure at 300 m or
