Question

We jump ahead to modern times and find Lazydog having lunch at Shoney's. While pounding down a BBQ bacon cheeseburger (only 1470 calories and 49 fat grams), the good professor worked on the following problems. Grab yourself an 875 calorie Ultimate Hot Fudge Cake from Shoney's and help the prestigious professor solve these problems. NOTES IMAGES DISCUSS UNITS STATS HELP Part Description Answer Chk History The honored Lazydog fondly recalls the many times he has sliced a golf ball into the water. This leads him to decide to calculate the terminal velocity in water of a 4.4 cT diameter ball that has a specific gravity of 1.08. Assume the viscosity of water is 1.0x10 Pa-s. What do you calculate as the terminal velocity of this ball in water? (include units with answer) 18.33 pts, 110% A. # tries: o show Detalls Format Gheck 29: try penalty Hints: 1.0 The distinguished Lazydog now decides to calculate how long it would take to drain a 2.0 ft wide by 4.4 ft long bathtub from full to half full. The initial height of the water in the tub is 2.6 ft and the water goes out a 2.4 inch diameter drain in the bottom of the tub (include units with answer) 18.33 pts,410% 2% try pengry Hints: 01 B. # tries: 01 Show Details Forma! Chak The celebrated Lazydog has recently added a 90 m3 workshop on to his house. What diameter must a 18 m long air duct have if the heating and air conditioning system is to replenish the air in the room every 16 minutes? The pump can exert aFrm Chack gage pressure of 0.6x10-3 atm. Assume atmospheric pressure is 101.3 kPa and the viscosity of air is 0.018x10-3 Pa-s. (include units with answer) 18.33 pts.110% | # tries: 0 | Show Details C. Hints: 1.1

Answer 1

Terminal velocity V_t = gd^2/18 Mu (rho_s - rho) g = 9.81 m/s^2 d = 4.4 cm = 4.4 times 10^-2 m Mu = 1.0 times 10^-3 Pa.S Rho_s = 1.08 times 1000 = 1080 kg/m^3 Rho = 1000kg/m^3 V_t = (9.81) times (4.4 times 10^-2)^2/18 times 1 times 10^-3 times (1080 - 1000) = 84.4 m/s For tub to drain, we have to apply Bernoulli's equation. P_1/Sigma + V-1^2/2g + Z_1 = P_2/Sigma + V_2^2/2g + Z_2 P_1 = P_2 & Z_1 = 0 Therefore V_1^2 = V_2^2 + 2gh(t) assume V_2 < < V_1 Therefore D_1 < < D_2 Therefore V_1 = squareroot 2gh & V_2 = -dh/dt \r\n continuity equation V_1 A_1 = V_2 A_2 A = Area V = velocity squareroot 2gh (pi r^2) = -dh/dt pi R^2 dh/dt = -squareroot 2g middot r^2 Continuity equation. V_1 A_1 = V_2 A_2 squareroot 2gh (pi r^2) = -dh/dt (l times b) dh/dt = -squareroot 2g middot pi r^2/l times