A 75 kg hiker climbs a mountain 1076 m high in 3 hours. Assuming the average O2 consumption during the climb?
99832
i believe it is around 9.5%.\n\nBecause Efficiency = O/p Power / Input power * 100.\n\nOutput is the given one \n= 4 * 104 J/min * 3 * 60 min = 74880 J \nInput is  = 75 * 1076 * 9.8 J ( mgh - mg * 0 = Potential energey) \nthefore, eff. is 9.5%\n\nplease reply at autoprash90@yahoo.co.in
262916
4 OH-(aq) --->  O2(g) + 2H2O(l) + 4e-
57426
That would be Mt. Everest in the Himalayas (Nepal/Tibet) with 29,035 ft. or 8,850 m.
889917
80 km/hr * 10^5cm/km * 1hr/3600sec = 2222 cm/sec
501508
3853 feet (1175 meters)?
102121
Drew, this is another conservation of energy calculation.  Just like the other one I responded to.  So I'm sure you can get it now but I'll go over this one with you.  so the inital speed after dimensional anaylsis is 27.78 m/s (remember there is 3600s in an hr and 1000m in a km).  So using 1/2mv^2 = mgh (by the way the masses cancel).  you will get a total height of 39 meters (2 sig figs!).. Hope this helps
410398
The partial pressure of O2 at 10 000 feet altitude.   Well, the air pressure at sea level is about 100 kilopascals.   At 10 000 feet it's about 70 kilopascals.   About a fifth of the air is oxygen so the partial pressure of O2 up there is about 14 kilopascals.
888459
v(t)= -gt - ve(ln((m - rt)/m))\n\nx(t) = ∫v(t) dt\n\n= ∫[-gt - ve(ln((m - rt)/m))] dt (Note ∫lnu du = ulnu - u + k)\n\nLet u = (m - rt)/m Thus du = -r/m dt ie dt = -m/r du\n\nSo x(t) = -½gt² - ve*-m/r (u(lnu - 1) + c\n= mve/r [ (m - rt)/m * (ln((m - rt)/m)  - 1] - ½gt²\n= ve(m - rt)/r * (ln((m - rt)/m) - 1) - ½gt²\n\nIf g = 9.8 m/s^2, m = 30000 kg, r = 155 kg/s, and ve = 3000 m/s and t = 1 minute = 60 s\n\nx(60) = 3000 * (30000 - 155 * 60)/155 * (ln (30000 - 155 * 60)/155 - 1) - ½ * 9.8 * 60²\n\n≈ 138390 m\n\n= 138.39km
672620
2. An oxygen molecule moving at 560 meters/second.
946074
avg. speed = distance/time = 75/11 = 6.82 m/s
1113728
The airplane is taking off from a plateau 3856[ft] in elevation, and is climbing at a rate of 1000[ft] per minute. \n(Rate of climb X # of minutes) + elevation from which airplane took off
330600
It depends on a many things.  \n\nThe most popular starting point for a trip to the summit of Everest is Kathmandu and from here it takes several days to reach Everest Base Camp.  At this point climbers will stop and spend several days acclimatizing.  It's from this point onwards that the true ascent of Everest begins, although reaching Base Camp is an undertaking in itself as it's at 19,000 feet (5,800 metres).\n\nThe journey from Base Camp to Summit involves establishing a series of camps higher and higher up the mountain and these are usually given the imaginative title of Camp 1, Camp 2 etc.  It's not unusual for a climber at a high camp to retreat to a lower one for a day or more - usually because the rarified air at altitude causes problems (primarily pulmonary and cerebral oedema).\n\nAnother major factor is the weather and it's not uncommon for climbers to become storm-bound and have to sit out the storm where they are, in some cases this can delay them by days.\n\nPeople tend to climb Everest and other similarly high mountains in one of two ways.  They either use 'seige' tactics with huge numbers of people, porters, back up teams, assistants etc or they adopt the 'Alpine Style' which is basically to be self-sufficient.\n\nTo answer your question, a small team adopting the Alpine Style and not encountering serious problems would take 4 to 6 days from Base Camp to summit and usually a day or two less to descent.\n\nIf you took all expeditions into account then the average time for ascent is probably in the order of 2 weeks and the descent about 7 to 10 days.
528462
the blood flow at speed of 75-82 kms per hour.
1109710
28.38 degrees N, 77.12 degrees E
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
193517
You don't provide enough data to determine this.  The gas law PV=nRT has variables you need to define.
1031939
temperature decreases with increase in altitude