Fluids (part 12)
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Great tutorials! Very clear!
what if we assume that area of surface of fluid (sof) is larger than area of the hole by a factor of k? the velocity of the fluid coming out of the hole will then vary with time. I tried figuring out the velocity as a function of time and i got v = k sqrt[(2k^2gh(t))/(k - 1)] where h is the height of the sof wrt the hole... and then i tried solving for h(t) by setting up (dh/dt)^2 = (2gh(t))/(k^2 - 1) (I hope thats right)... Can someone pls help me solve this?
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thankyou very much ,concepts got cleared as easily as fluids flow :P
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2 hours not wasted thank you my friend
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Thank you so much for these videos! They are much easier to understand compared to lecture notes and you make it really simplified.
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thnk you sooooooooooo much
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These videos are so helpful, and could you please continue on with hydraulic grade line and energy grade line concept? Thanks
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the 1 dislike is someone who is sad that its ending.
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this video can make me understand more my doctor. so i rely thank you
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Thank you!
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The one person who dislikes this video is CRAZY!!! Really useful video. Thankk you for uploading these videos. God bless you!!
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Thanks for the Fluids Videos!
DYNALINKGHANIMA 2 years ago 17
Thank you for the fluid videos! I watched them all!
gruelin1 2 years ago 14