In the figure is a picture of a device similar to the one demonstrated in the le

Question

In the figure is a picture of a device similar to the one demonstrated in the lecture. The air passes through a large-diameter glass tube, then through a small-diameter tube, and finally through a large-diameter tube again. Suppose the diameters of the two sizes of tubes are 9.2 mm and 5.3 mm, and suppose that the velocity of the air in the large-diameter tube is 4.68 m/s.

(a)

Find the velocity of the air in the small-diameter tube.

m/s ( ± 0.2 m/s)

(b)

Find the difference in pressure of the air in the large- and small-diameter tubes. The density of air is 1.29 kg/m33.

Pa ( ± 2 Pa)

(c)

The vertical tubes contain colored water. Find the difference in height of the water in the middle tube and the two outer tubes. Neglect viscosity of air. This means that in this problem, the water in the two outer tubes are at the same height, contrary to what really happens, as shown in the figure.

cm ( ± 0.02 cm)

Explanation / Answer

a )

using equation

A1 V1= A2 V2

V2 = A1 V1 / A2

V2 = 3.14 X ( 9.2 X 10-3 / 2 )2 X 4.68 / 3.14 X ( 5.3 X 10-3 / 2 )2

V2 = ( 9.2 X 10-3 / 2 )2 X 4.68 / ( 5.3 X 10-3 / 2 )2

V2 = 14.1 m/sec

b )

the difference in pressure of the air in the large- and small-diameter tubes is

= ( V2 - V1 )2 / 2 g

= 4.52

c )

the difference in height of the water in the middle tube and the two outer tubes is

= denisty X g X ( V12 - V22 ) / 2 g + 4.52

= 1.29 X 9.8 X ( 4.682 - 14.12 ) / ( 2 X 9.8 + 4.52 )

= 92.72 N/m2

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