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MrRa [10]
4 years ago
14

What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a

nd the surface tension of honey is 0.052 N/m.
Give your answer to the nearest integer (do NOT use scientific notation).
Physics
1 answer:
Vikki [24]4 years ago
5 0

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure =  P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

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(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

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d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

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F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

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F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

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So, we have:

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F_{No} = 771.125N

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