old faithful geyser in Yellow stone Park eroupts at approximately 1-hour interva

Question

old faithful geyser in Yellow stone Park eroupts at approximately 1-hour intervals. and the height of the fountain reaches 40.0m. (a) consider the rising stream as a series of seperate drops. analize the free fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c)what is the pressure (above atmosphere pressure) in the heated underground chamber 175m below the vent? you may assume that the chamber is large compard with the geyser vent.

Explanation / Answer

a) let the initial speed be v. we have that v^2=2Hg so that v=28(m/s). b) we have that P0+Dv^2/2=P0+D*g*H. so that we also have v^2=2*g*H so v=28m/s c)the chamber is large so velocity of water when it leaves the chamber is approximately zero. P=P0+D*v^2/2+D*g*h=P0+21.1e5 (Pa) so P-P0=21.1atm

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