Chapter Questions and Problems 1. A 200-kg load is hung on a wire having a length of 4.00 m, cross-sectional area 0.2times 10^-4m^2 and Young's modulus6 8.0times (10^-10N)/(m^2) What is its increase in length? 2. A steel wire of diameter 1 mm can support a tension of 0.2 kN. A cable to support a tension of 20 kN should have diameter of what order of magnitude? 3. A 300-kg hammer strikes a steel spike 230 cm in diameter while moving with speed 20.0m/s The hammer rebounds with speed 10.0m/s after0.110 s. What is the average strain in the spike during the impact? 4. If the shear stress in steel exceeds 4.0times 10^8N/m^2 the steel ruptures Determine the shearing force necessary to (a) shear a steel bolt 1.0 cm in diameter and (b) punch a 1.00-cm -diameter hole in a steel plate 0.500 cm thick. 5. Lead has a greater density than iron and both are denser than water. Is the buoyant force on a lead object greater than, less than, or equal to the buoyant force on an iron object of the same volume? 6. When an object is immersed in a liquid at rest,why is the net force on the object in the horizontal direction equal to zero? VII 7. When water freezes it expands by about 9.0% What pressure increase would occur inside your automobile engine block if the water in it froze?(The bulk modulus of ice is 2.0times 10^9N/m^2) 2.xcorig 8. A40 -cm tall glass is filled with water to a depth of 30 cm. A. What is the gauge pressure at the bottom of the glass? B. What is the absolute pressure at the bottom of the glass? 9. Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0 -cm diameter pipe in the basement under a pressure of 3.03times 10^5 Pa, what will be the velocity and pressure in a 2.6 -cm diameter pipe on the second floor 5.0 m above? 10. Calculate the absolute pressure at an ocean depth of 1.0times 10^3m Assume that the density of the water is 1.025times 10^3kg/m^3 and that P_(atm)=1.01times 10^5Pa

1. The increase in length can be calculated using the formula for strain: strain = (stress / Young's modulus). The stress is the force divided by the cross-sectional area. So, stress = (200 kg * 9.8 m/s^2) / (0.2 * 10^-4 m^2) = 9.8 * 10^7 N/m^2. Therefore, strain = (9.8 * 10^7 N/m^2) / (8.0 * 10^-10 N/m^2) = 1.225 * 10^17. The increase in length is then strain * original length = 1.225 * 10^17 * 4.00 m = 4.9 * 10^-14 m.<br /><br />2. The tension a wire can support is proportional to the cross-sectional area of the wire. Since the tension is 100 times greater, the cross-sectional area must be 100 times greater. The diameter of the wire is proportional to the square root of the cross-sectional area, so the diameter must be the square root of 100, or 10 times greater. Therefore, the diameter of the cable should be 10 mm.<br /><br />3. The average strain in the spike during the impact can be calculated using the formula for strain: strain = (stress / Young's modulus). The stress is the force divided by the cross-sectional area. The force is the mass of the hammer times its velocity, or 300 kg * 20.0 m/s = 6000 kg*m/s. The cross-sectional area of the spike is π * (0.115 m)^2 = 0.0419 m^2. Therefore, stress = 6000 kg*m/s / 0.0419 m^2 = 1.43 * 10^8 N/m^2. The Young's modulus of steel is approximately 2.0 * 10^11 N/m^2. Therefore, strain = (1.43 * 10^8 N/m^2) / (2.0 * 10^11 N/m^2) = 0.00715.<br /><br />4. (a) The shearing force necessary to shear a steel bolt 1.0 cm in diameter can be calculated using the formula: shearing force = shear stress * cross-sectional area. The shear stress is given as 4.0 * 10^8 N/m^2. The cross-sectional area of the bolt is π * (0.5 * 10^-2 m)^2 = 7.85 * 10^-4 m^2. Therefore, shearing force = 4.0 * 10^8 N/m^2 * 7.85 * 10^-4 m^2 = 3.14 * 10^8 N.<br />(b) The shearing force necessary to punch a 1.00-cm-diameter hole in a steel plate 0.500 cm thick can be calculated using the formula: shearing force = shear stress * area of the hole. The shear stress is given as 4.0 * 10^8 N/m^2. The area of the hole is π * (0.5 * 10^-2 m)^2 = 7.85 * 10^-4 m^2. Therefore, shearing force = 4.0 * 10^8 N/m^2 * 7.85 * 10^-4 m^2 = 3.14 * 10^8 N.<br /><br />5. The buoyant force on an object is equal to the weight of the fluid displaced by the object. Since lead has a greater density than iron, a lead object will displace more water than an iron object of the same volume. Therefore, the buoyant force on a lead object will be greater than the buoyant force on an iron object of the same volume.<br /><br />6. When an object is immersed in a liquid at rest, the net force on the object in the horizontal direction is equal to zero because the pressure exerted by the liquid on the top and bottom surfaces of the object is the same. This is due to the fact that the pressure in a fluid increases with depth.<br /><br />7. The pressure increase inside your automobile engine block if the water in it froze can be calculated using the formula: pressure increase = bulk modulus * strain. The bulk modulus of ice is given as 2.0 * 10^9 N/m^2. The strain is the change in volume divided by the original volume. The change in volume is 9.0% of the original volume. Therefore, pressure increase = 2.0 * 10^9 N/m^2 * 0.09 = 1.8 * 10^8 Pa.<br /><br />8. A. The gauge pressure at the bottom of the glass can be calculated using the formula: gauge pressure = density * gravity * depth. The density of water is 1000