anonymous
  • anonymous
An opening of area 0.252 cm2 in an otherwise closed beverage keg is 40.1 cm below the level of the liquid (of density 1.01 g/cm3) in the keg. What is the speed of the liquid flowing through the opening if the gauge pressure in the air space above the liquid is (a) zero and (b)0.532 atm?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@IrishBoy123 can you help me?
IrishBoy123
  • IrishBoy123
toricelli
IrishBoy123
  • IrishBoy123
start here https://en.wikipedia.org/wiki/Torricelli%27s_law

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More answers

anonymous
  • anonymous
i know how to do in letter a however, in a letter b i can't seem to get the answer :(
IrishBoy123
  • IrishBoy123
ok so we may have to go back to Bernoulli, again from that Wiki page: \[gz+{p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\] but here we seem to have different pressures so \[gz+{p_{\color{red}{inside}}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\] sove for v from there with a) \(p_{inside} = 0\) and b) \(p_{inside} = ????\) where ??? is (0.5* atmospheric pressure) in SI units
IrishBoy123
  • IrishBoy123
1 thing to be careful about it's gauge pressure so inside for a \(P = P_{atm}\) for b), i think you add a further 0.5 atm been a while
ganeshie8
  • ganeshie8
i think gauge pressure is the deviation from atmospheric pressure..
anonymous
  • anonymous
This question is kinda of confusing
anonymous
  • anonymous
I have the same problem also and seems can't get the answer
anonymous
  • anonymous
thank you guys for helping me :)
IrishBoy123
  • IrishBoy123
yes, ganesh, my recollection too, and i have had a check as well so for internal gauge = 0 you get \(gz+{p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\) \(\implies gz={v^2 \over 2}\) and hence Toricelli for the second where internal gauge = 0.532 you get \(gz+{1.532p_{atm}\over\rho}={v^2 \over 2}+{p_{atm}\over\rho}\) \(\implies gz+{0.532p_{atm}\over\rho}={v^2 \over 2}\) so you plug the numbers [SI uits] into a fuller Bernoulli equation to solve
anonymous
  • anonymous
oh i't not equal to zero :)) but the answer is the same as my a
ganeshie8
  • ganeshie8
Hey, still here ?

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