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
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?
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
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
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.
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
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.
Next Video: ua-cam.com/video/12K_aP1KPFk/v-deo.html
Final Exams and Video Playlists: www.video-tutor.net/
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
I am so glad i found your channel !!! I cant describe my happiness :) !!!!
Me 2...
It's like the happiness during sex.. It's just too much😂
The most helpful channel in youtube.
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.
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
🕴️🧔
saving students academics, one video at a time ❤
I'm happy to see that you've hit a million subscribers. All the best!
:)
Love from India.
It's 2024 and I'm here, since grade 11 and I'm now a first year in uni. From SA 🇿🇦
The best channel on UA-cam
Really
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.
yo andrew I hope it is not to late but I will link a mega link that has a worksheet that solves this question hope this helps bro
I sent it hope it helps
@@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
Hello JG, can you kindly assit us by adding more content of fluid mechanics especially the introductory part...you are awesome!😊
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
Great video!!!
Best explanation.
This channel is very helpful to students such as myself.
Keep it coming brother
It is a well-explained tutorial. I now understand the basics. Thank you.
veryyy helpful!!! greetings from Croatia
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 god I found you 😩😭💕💕💕💕
Best explanation
Your videos are awesome and helping me pass my power engineering exams!
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
best explanation ever
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
Iam so happy to find you 🥺🥺💕
This channel is awesome but I recommend that you call rho “rho” and not pe :)
oh gosh I finally understood the concept thanks to you!
This lecture is amazing! I am easy to understand! Thanks!
THAT"S WHY HE"S THE GOAT
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 so much!! this is very helpful 🥺
Good explain❤thankyou ❤
It's so simple with a good teacher ha
still using it as my reference thank you sir:)
why do we use aboluste pressure and not gauge pressure when the container is closed?
On example 5 where did the 101.3 come from?
Thank you so much
Sir can't we add both of the densities of water and oil...for bottom Guage press....in last question
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!
7:26 why didn't he minus the 101.3 kPa?
Great video!!
Very clear and useful. Thanks
When I get my diploma I'm just gonna mail it to this guy lol
THANKS A BUNCH
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.
why do we use absolute pressure and not gauge pressure when the container is closed?
Actual life saver
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)
i got a question.To calculate the gauge pressure why dont we consider too the weight of the water beneath that of oil?
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
thanks for the useful materials. very helpful
Great video, thanks ^^
great explaination ! thanh you
wouldnt the height for question b be 15+8, not just 15? or that a depth of 15
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.
So Helpful. THANKS
Thank u man
thank you
Thanks for your effort.....
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
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
THIS IS WAY BETTER THAN MY $9000 UNIVERSITY EDUCATION!
Thank you sir
why don't you put all fluid mechanics related videos in one playlist , please
thank u so much
يسعد امك ياشيخ على ذا الشرح الفنان
I love you Dude..I love youuu
Thank u
Why is the gauge pressure called the pressure "above" the atm pressure if it is the pressure "inside" the tank?
idk either :/ someone know?
ua-cam.com/video/MJ_zjWyRHX8/v-deo.html thiss video gives a good description as to why
maan, I love you
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.
a 3 hour Uni lecture in 13 minutes bro
dude you are a god
Coming here for physics course 😅
thanks!
where did u get the value of atm in roblem number 4
Constant value. 1 atm. = 101.325 kPA
Beast mode on
At the start I was waiting to hear "in this video"
Hey. This is easy!
So the pressure gauge formula is pgh?
yes
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
Show actual data of actual aneroid barometer in energy terms
You're a beast
It's best
How to get the unit of pascal out of (pgh)??! Anyone here now?!
عاشت ايدك
I wanna talk to the 27 people who disliked this video. LOL!!
Does 4.2 ATM mean 4.2 times the standard 1 ATM? Anyone know?
mike jackson yes
nice
Wait this just solved the whole unit im in
Ang galing mo
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
I love you
i love you
WTH is atm?
Atmospheric pressure
pa shout out po sa LU
can u solve our problems number 4 7 8 9 10
thanks :DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD =3333
God why is this concept so hard to understand
Density is ρ (Greek letter, rho), not "p".
Don't make extra layers of confusion.