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 cnm diameter ball that has a specific gravity of 1.08. Assume the viscosity m/s of water is 1.0x10-3 Pa-s. What do you calculate as the terminal velocity of this ball in water? include units with answer) A: 84.4m/s C: 84.41 A. # tries: 1 | Show Details 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 236 17.97 pts. 10er B. full. The initial height of the water in # tries: 1 | Show Details the tub is 2.6 ft and the water goesCheok out a 2.4 inch diameter drain in the bottom of the tub include units with answer) The celebrated Lazydog has recently added 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 e change in gage pressure of 0.68x10-3 atm. The viscosity of air is 0.018x10 Pa S (include units with answer) 18.33 pts,4.0% C. # tries: 0 Show Details xert a Fomat Cheok Hints: 21

Answer 1

b.

Radius of the drain, r = 1.2 inch = (1.2 / 12) ft = 0.1 ft

Cross section of the bathtub, A = 2.0 * 4.4 = 8.8 ft2

Velocity of exit from the drain, v = (2gh)1/2

where 'h' is the height of the water in the tub at a given time 't'

So,

πr2 (v dt) = A dh

=> πr2 (2gh)^1/2 dt = A dh

=> (1/h1/2) dh = (πr2/A)(2g)^1/2 dt

Integrating both sides with limits, we get,

2.6∫1.3 (1/h^1/2)dh = (πr2/A)(2g)^1/20∫t dt

=> 2[h^1/2]2.6,1.3 = (πr2/A)(2g)^1/2[t]0t

=> 2(2.6^1/2 - 1.3^1/2) = (πr2/A)(2g)^1/2t

=> t = 2A(2.6^1/2 - 1.3^1/2) / πr2(2g)^1/2

=> t = 2 * 8.8 * (2.6^1/2 - 1.3^1/2) / [π * 0.1^2 * (2 * 32.2)^1/2]

=> t = 32.97 s

c.

Using Poisuille's equation,

Q = πR^4 (P2-P1)/8nl
R = (8nlQ/π (P2-P1))^.25
R = (8*0.018*10^-3*18*(90/16*60)/π*0.68*10^-3*101325)^.25
R = 0.03255 m
Diameter = 2R = 0.0651 m