A 60.0 kg runner expends 335 W of power while running a marathon. Assuming that 11.0% of the energy is delivered to the muscle tissue and that the excess energy is primarily removed from the body by sweating, determine the volume of bodily fluid (assume it is water) lost per hour in cm^3. (At 37.0°C the latent heat of vaporization of water is 2.41 x 10^6 J/kg.)
2007-11-28 07:55:29 · 3 answers · asked by Sasha 1 in Science & Mathematics ➔ Physics
The runner works at a rate of 335 W. 11% is transformed into kinetic energy. The problem implies that the remaining 89% must be removed through the evaporation of water.
89% of 335 W = 298 W = 298 J / s
298 W over a one-hour (3600-second) interval is 1.07 x 10^6 Joules.
The problem tells us that the heat of vaporization of water at 37 degrees (about body temperature) is 2.41 x 10^6 J / kg
So
1.07 x 10^6 J = ( m ) ( 2.41 x 10^6 J / kg )
Solve for m; that's the mass of wtaer (in kg) that must be evaporated per hour.
2007-11-28 08:12:41 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
a) Use one million/2KX^2 - you purely might desire to put in the numbers (X being the amplitude) b) KE=one million/2mv^2 at equilibrium factor KE = Max Mech ability Rearrange to offer ... root(KX^2 x m) = v 2) this time, use KE=one million/2mv^2 utilising v=one million.4. now you may remodel the respond for KE into your one million/2KX^2 =one million/2mv^2 and you get... root((mv^2)/ok) ... purely you would be wanting to apply ... v= one million.4 m=one million ok= 113 and thats you're new amplitude. desire i kinda helped ...
2016-10-09 21:37:44 · answer #2 · answered by Erika 4 · 0⤊ 0⤋
what the.................................got no idea
2007-11-28 07:57:49 · answer #3 · answered by Anonymous · 0⤊ 1⤋