Question

You apply 0.8 V to a patient's upper leg and record a current of 0.0016 A. The upper leg has a 12 cm diameter and 40 cm length. Estimate a patient's percent body fat based on this measurement. The resistivity of muscle is ρ m u s c l e = 13 Ω m and ρ f a t = 25 Ω m. Answer with the percent of the leg that is fat, but don't enter the percent sign.

Answer 1

0.8 0,0016 00 R A= To (0.12² To 0.0 113097 m² (x)0.0113097) m Af = (1-x) (0.0113097) mit SA Am AF soo SmL + SEL AF Am Sm of (LW² 500 SmL Aft ff L Am (13) (25) Co.uoj? 6(13)(0.40) Apt @ 5X6.40 RM Joo 13 52 + 52 - - 2کی = 2600 MF 5000 Am (2600) (0.0113097) (1-x) + 800)(u)(0.0113097) 29.4053 29.4053 at 5%.948668 x z (27.14 336) = 22.59489 0.8324 $ 83.24 °C fat = 100-83.24 = 16276 % -16.76%