hello sir! thank you for this simple yet understandable tutorial :) i have a question though, when an object is just at sea level (say, an automobile tire with given 24.2 psi) do i still need to convert the given gauge pressure to absolute?
For Question 5. How did you know Gauge pressure was using the oil and not the water? Is it because the oil is in contact with the atmosphere and the water isn't?
I think #4 is wrong, you swapped Pg and Pt, so the #'s are correct but the total pressure is 307.1 KPA and the gauge is 305.8 (Pg = P - Po, where Po is 1 atm)
So helpful 😇 may please do for me this : A wooden cylinder with cross-section area of 50 cm^2 floats in water. The visible part of the cylinder is 8 cm. 1. Using Archimedes' principle derive the following equation for a partially submerged cylinder (height of the cylinder / height of the cylinder below fluid surface = density of fluid /density of object) 2. Determine the height of the cylinder below the water surface using equation in 1. 3. Determine the weight of the cylinder.
If you compute the absolute pressure of oil and water separately, you would arrive with different answer. But in this case, gage pressure were computed first before adding the atmospheric pressure.
Im confused by this video because at the beginning you describe the guage pressure as being the difference between a total pressure and the atmospheric pressure, but then further on you calculate the guage pressure of the water and oil but there is no mention of finding the difference from the atmospheric pressure. What am I missing, cheers
hello mr just wanted to report a potential mistake in your calculations (please correct me if im wrong) at 12:18 the addition of the both gauge pressures [58800+14700] adds up to 695800 which you have mistakenly written as 205800 pa . PLEASE tell me if im going wrong somewhere
Can I ask a question, what if the given in the last example is the specific gravity? How can I compute that, for example the Poil=.5 and the H20=50kg/m^3
Thanks for this great explanation. However, in example #4, shouldn't the atmospheric pressure at 50 meters below sea level be equal to 101.3 x 6 atm considering that the atmospheric pressure until a depth of 10 meters will remain constant at 1 atm then it will increase every 10 meters deep by 1 atm.
at 10:16, how did you know that the weight of the atmosphere was 101.3 kpa? I thought this was for sea level. Did you just make that assumption? thanks
