An 81.5-kg man stands on a horizontal surface.

(a) What is the volume of the man's body if his average density is 985 kg/m^3?
(b) What average pressure from his weight is exerted on the horizontal surface if the man's two feet have a combined area of 4.50 × 10-2 m^2.

Respuesta :

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10²  N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

 F= m g= 81.5 x 10 = 815 N      ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

[tex]P=\dfrac{F}{A}[/tex]

[tex]P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2[/tex]

P=181.11 x 10²  N/m²

(a) The volume of the man's body is 0.0827 m³

(b) The average pressure exerted on the ground is 177.48 x 10²  N/m²

Volume and pressure:

(a) Given information:

mass of the man m = 81.5 kg

Density, ρ = 985 kg/m³

No

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

(b) The weight of the man is given by:

F = mg

F = 81.5 × 9.8

F = 798.7 N      

Given that the area of the feet is:

A= 4.5 x 10⁻² m²

The Pressure P is given by:

P = F/A

P = (798.7) / (4.5 x 10⁻²) N/m²

P = 177.48 N/m²

Learn more about pressure:

https://brainly.com/question/8033367?referrer=searchResults