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DochEvi [55]
3 years ago
11

When a person inhales, air moves down the bronchus (windpipe) at 14 cm/s. the average flow speed of the air doubles through a co

nstriction in the bronchus. assuming incompressible flow, determine the pressure drop in the constriction. (enter the magnitude.) pa?
Physics
2 answers:
True [87]3 years ago
6 0

v₁ = speed of air in bronchus = 14 cm/s = 0.14 m/s

v₂ = speed of air in constriction = 2 v₁ = 2 x 14 = 28 cm/s = 0.28 m/s

ρ = density of air = 1.225 kg/m³

ΔP = pressure drop in the constriction = ?

pressure drop in the constriction using Bernoulli's equation is given as

ΔP = (0.5) ρ (v²₂ - v²₁ )

inserting the values

ΔP = (0.5) (1000) ((0.28)² - (0.14)² )

ΔP = 29.4 Pa

True [87]3 years ago
4 0

Answer:

ΔP = 0.0360 Pa

Explanation:

v₁ = speed of air in bronchus = 14 cm/s = 0.14 m/s

v₂ = speed of air in constriction = 2 v₁ = 2 x 14 = 28 cm/s = 0.28 m/s

ρ = density of air = 1.225 kg/m³

Using Bernoulli's equation is given as

ΔP = (0.5) ρ (v²₂ - v²₁ )

ΔP = (0.5) (1.225) ((0.28)² - (0.14)² )

ΔP = 0.0360 Pa

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4 0
4 years ago
A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal
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<h3>1.03684m</h3>

Explanation:

Using the formula for calculating range expressed as;

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Substitute the given parameters into the formula and calculate H as shown;

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