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.
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?
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
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.
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)
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
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
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So, if you use a gauge to measure the air pressure in your tires is that absolute pressure or gauge pressure?
This channel is saving my life in school right now. Thank you so much for all these physics videos and examples
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Me 2...
It's like the happiness during sex.. It's just too much😂
you're the best dude. helping me solve questions in an instant, what the teacher couldn't do in a week lol, thank you so much.
The most helpful channel in youtube.
Thanks for allowing me to achieve the best results in science during my high school
and now again in university.
yo, that's not a diver. That's definitely a corpse.
(Amazingly helpful video as always)
You’re the best 👍 great way of explaining. Much better then my professor 👩🏫
Lol
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I'm happy to see that you've hit a million subscribers. All the best!
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Love from India.
saving students academics, one video at a time ❤
The best channel on TH-cam
Really
Great video!!!
Best explanation.
This channel is very helpful to students such as myself.
Keep it coming brother
Hello JG, can you kindly assit us by adding more content of fluid mechanics especially the introductory part...you are awesome!😊
Your videos are awesome and helping me pass my power engineering exams!
It is a well-explained tutorial. I now understand the basics. Thank you.
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.
@Kurosaka it isn't here :(
@Kurosaka youtube deletes links
So paste the link and add a letter at the end so that we can copy paste it and remove the letter
veryyy helpful!!! greetings from Croatia
Best explanation
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oh gosh I finally understood the concept thanks to you!
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It's 2024 and I'm here, since grade 11 and I'm now a first year in uni. From SA 🇿🇦
This lecture is amazing! I am easy to understand! Thanks!
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It's so simple with a good teacher ha
best explanation ever
This channel is awesome but I recommend that you call rho “rho” and not pe :)
Great video!!
Very clear and useful. Thanks
Thank you so much
lmao who's here just because they can't solve their homework?
I'm here because my teacher didn't teach us lol.
Preparing for a test
thanks for the useful materials. very helpful
Good explain❤thankyou ❤
Great video, thanks ^^
still using it as my reference thank you sir:)
THAT"S WHY HE"S THE GOAT
When I get my diploma I'm just gonna mail it to this guy lol
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?
THANKS A BUNCH
Actual life saver
great explaination ! thanh you
So Helpful. THANKS
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
THIS IS WAY BETTER THAN MY $9000 UNIVERSITY EDUCATION!
Thanks for your effort.....
Thank u man
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.
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?
im wondering abt tht as well
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)
thank you
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
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
i already found out how, thnks for the video.
Thank you sir
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.
maan, I love you
Sir can't we add both of the densities of water and oil...for bottom Guage press....in last question
I love you Dude..I love youuu
why do we use aboluste pressure and not gauge pressure when the container is closed?
why don't you put all fluid mechanics related videos in one playlist , please
thank u so much
i got a question.To calculate the gauge pressure why dont we consider too the weight of the water beneath that of oil?
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
When the question doesn't tell you anything about the atmosphere level you must assume it to be the sea level atmosphere
@@avafaghihi9763 so I always use that number if it isn't given?
@@ניין-י9ש yeah but it's usually given in exams.
@@avafaghihi9763 ohh thanks!
why do we use absolute pressure and not gauge pressure when the container is closed?
يسعد امك ياشيخ على ذا الشرح الفنان
On example 5 where did the 101.3 come from?
Hey. This is easy!
Thank u
a 3 hour Uni lecture in 13 minutes bro
Why is the gauge pressure called the pressure "above" the atm pressure if it is the pressure "inside" the tank?
idk either :/ someone know?
th-cam.com/video/MJ_zjWyRHX8/w-d-xo.html thiss video gives a good description as to why
dude you are a god
Coming here for physics course 😅
Beast mode on
Our teacher made us memorize a formula. PSI = Height in Inches x Specific Gravity / 27.7" H2O. I'm not sure why though. Can you explain this formula??
no
7:26 why didn't he minus the 101.3 kPa?
wouldnt the height for question b be 15+8, not just 15? or that a depth of 15
I wanna talk to the 27 people who disliked this video. LOL!!
At the start I was waiting to hear "in this video"
thanks!
i think there is a mistake cause you are adding kpa to Pa without considering conversion before so basically the Patm is used as kpa
where did u get the value of atm in roblem number 4
Constant value. 1 atm. = 101.325 kPA
Why did u calculate volume of cylinder in btw 4th example to find gauge pressure formulae..pls say...
Vasu Ram he just use it as reference
He just tell that to let you know how pgh was derived.
why is Patm @8:28 101.3?
P=pgd
= (13.6)(9.8)(0.760)
= 101.2928Pa if significant figures that will be 1.01x10^5Pa
Pressure of the atmosphere
1 atm = 101.3kPa
So the pressure gauge formula is pgh?
yes
Wait this just solved the whole unit im in
You're a beast
It's best
Show actual data of actual aneroid barometer in energy terms
for question number 4 i think he miscalculated the absolute pressure. it should be 502.25k + 101.3 = 502.35k.
anyone else ?
No. It's 101.3 kP, not 101.3 P. So the calculation is 502.25 kP + 101.325 kP
Ang galing mo
عاشت ايدك
nice
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How to get the unit of pascal out of (pgh)??! Anyone here now?!
i love you
WTH is atm?
Atmospheric pressure
Does 4.2 ATM mean 4.2 times the standard 1 ATM? Anyone know?
mike jackson yes
God why is this concept so hard to understand
pa shout out po sa LU
Any Chem Eng students?
I am here
thanks :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD =3333
can u solve our problems number 4 7 8 9 10