Map The traditional method of measuring a person\'s body-fat ratio is to weigh t

Question

Map The traditional method of measuring a person's body-fat ratio is to weigh them on dry land and then weigh them while submerged in water. The body-fat ratio can be calculated using the difference in these weight measurements. Unfortunately, this method requires that the person become submerged in water, which is a problem for those suffering from hydrophobia (fear of water) As an angel investor, you give money to companies that you think have highly marketable products or services in exchange for partial ownership of the company. A company has mailed a proposal to you seeking money to market a new device for measuring a person's body-fat ratio with no need for the person to get wet. Here's how: The person in the tigure is standing on a weight scale in a sealed chamber. Initially the chamber is maintained at standard atmospheric pressure. However, the pressure is then lowered and the new weight is noted Derive an expression for the density of the person in terms of his weight before (w1) and after (w2) the chamber is evacuated. Use P1 and P2 for the density of the air before and after the chamber is evacuated nsider a typical adult weighing 189.0 lb and possessing a density of 987.0 kg/m9. How low will the chamber need to drop in pressure at minimum to measure the person's density if the weight scale can only measure to the nearest 0.20 lb? (Assume air is an ideal gas and the chamber is maintained at standard temperature as the air is evacuated. The density of air at sea level is roughly 1.22 kg/m3.) (Scroll down for more questions) Number atm

Explanation / Answer

let volume of the person be V,

initial density of air = rho1

initial buoyant force = B1 = rho1*V*g

initial scale reading = W1 = rho*V*g - rho1*V*g ( where rho is density of the person)

final density of air = rho2

final buoyant force = B2 = rho2*V*g

final scale reading = W2 = rho*V*g - rho2*V*g

hence

V = W1/(rho - rho1)g

W2 = W1(rho - rho2)/(rho - rho1)

W2*rho - W2*rho1 = W1*rho - W1*rho2

rho = (W2*rho1 - W1*rho2)/(W2 - W1)

rho*V = 189 lb = 85.729kg

rho = 987 kg/m^3

rho1 = 1.22 kg/m^3

least count of weight scale = 0.2 lb = 0.0907185 kg = 0.889948485 N

then, V = 0.08685 m^3

W1 = 85.729*9.81 - 1.22*0.08685*9.81 = 839.96205183 N

W2 = W1 + 0.889948485 = 840.852000315

840.852000315 = 85.729*9.81 - RHO2*0.08685*9.81

rho2 = 0.17545 kg/m^3

hence the air pressure has to be dropped till the density becomes this

p = rho*R*T

for iso thermal case

p1/rho1*R = p2/rho2*R

1 atm/1.225 = p2/0.17545

p2 = 0.143 atm

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.