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We jump ahead to modern times and find Lazydog having lunch at Shoney's. While pounding down...

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.

1. 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 4.3 cm diameter ball that has a specific gravity of 1.22. Assume the viscosity 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)

2. The distinguished Lazydog now decides to calculate how long it would take to drain a 2.8 ft wide by 5.2 ft long bathtub from full to half full. The initial height of the water in the tub is 2.2 ft and the water goes out a 2.3 inch diameter drain in the bottom of the tub.
(include units with answer)

3. The celebrated Lazydog has recently added a 87 m3 workshop on to his house. What diameter must a 16 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 a change in gage pressure of
0.63x10-3 atm. The viscosity of air is 0.018x10-3 Pa⋅s.
(include units with answer)

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Answer #1

Part-1

Given is:-

diameter 4.3cm 4.3X ГО 772

now,

we know that

by stoke's law

mg - F_b = 6 pi eta r v

where Fb buoyant force

or

v_t = rac{2}{9}r^2 (rac{ ho _s - ho_l}{eta }) g

where  ho _s is the density of solid(ball) = 1.22 = 1220 and  ho_l = density of liquid(water =1000)

by plugging all the values we get

2,4.3 x 101220 1000 9 vt =-( 10

which gives us

Uy 221.469772/s

Part-2

Using flow equation  A_1v_1 = A_2 v_2

T2312)2 2,8 52

which gives us

ひ1 0.0021/

Now,

0.002 2gy dt ;H = 2.2 ft or 0.67056m

A |H (V2 -1) al eq-1

and

0.002

by plugging all the values in eq-1 we get

0.67056m-1 (V2 - 1) 0.002

which gives us

t54.2s

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