Question

Calculate the buoyant force due to the surrounding air on a man weighing 900 N . Assume his average density is the same as that of water. Suppose that the density of air is 1.20 kg/m3

Answer 1

Answer:

1.1 N

Explanation:

Given:

ρa = 1.2 kg/m^3 [air density]

ρw = 1000 kg/m^3 [water density]

F = ρa V g

V = m / ρw = W / (ρw g)

F = (ρa / ρw) W

F = (1.2 / 1000) 900 = 1.08 N

Answer 2

The buoyant force due to the surrounding air on a man weighing 900 N is 1.08 Newton

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Further explanation

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

$\large {\boxed {P = F \div A} }$

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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Given:

Density of Air = ρ_air = 1.20 kg/m³

Densiy of the man = ρ_water = 1000 kg/m³

Weight of the man = w = 900 N

Asked:

Buoyant Force = F = ?

Solution:

We could use Archimedes' principle to solve the problem as follows:

$F = \rho_{air} g V$

$F = \rho_{air} g \frac{m}{\rho_{water}}$

$F = \rho_{air} \frac{mg}{\rho_{water}}$

$F = \frac{\rho_{air}}{\rho_{water}}w$

$F = \frac{1.20}{1000} \times 900$

$F = 1.08 \texttt{ Newton}$

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Conclusion:

The buoyant force due to the surrounding air on a man weighing 900 N is 1.08 Newton

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Learn more

• Minimum Coefficient of Static Friction : https://brainly.com/question/5884009
• The Pressure In A Sealed Plastic Container : https://brainly.com/question/10209135
• Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure