Viewer's Request: Physics: Fluid Statics #1: Helium Balloon

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Күн бұрын

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We will solve: A balloon with a volume of 600m^3 and mass of 500kg is filled with helium. When the balloon is allowed to ascend, how high=? will it rise in 10 seconds?
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• Viewer's Request: Phys...

Пікірлер: 40

@ahmadazeem8279 4 жыл бұрын

Prof,Ur video really help me a lot. i learnt almost everything from u for my physics and maths course in uni life. Ur explanation really good 100%

@fizixx 4 жыл бұрын

This is a very enjoyable video, thank you :)

@rudrashiva 4 жыл бұрын

My Favourite is Physics, combine it with python. You'll rule the world.

@M.Ashraf.Bhatti 4 жыл бұрын

Excellent sir

@mehmetali4626 4 жыл бұрын

Thank you. Could you solve another problem about change density air depend altitude. So what is accelerator function depend altitude.

@eliudaunda2745 Жыл бұрын

d) A helium party balloon, assumed to be a perfect sphere, has a radius of 18.0 cm. At room temperature, (20o C), its internal pressure is 1.05 atm. Find the number of moles of helium in the balloon and the mass of helium needed to inflate the balloon to these values. [5mks]

@MichelvanBiezen Жыл бұрын

Use the equation PV = nRT n = PV/RT = (1.05 ATM) ( 101,300 Pa/ATM) (4/3) (pi) (0.18m)^3 / [ (8.315 J/mol K) ( 20 + 273) K ]

@tsoojbaterdene7793 4 жыл бұрын

Thank you so much.😍😘🤗😄😃😀🙏

@TALKmd Жыл бұрын

Viewer's Request Videos are great stuff (:

@MichelvanBiezen Жыл бұрын

Yes, our viewers come up with some very interesting problems. 🙂

@anilsharma-ev2my 4 жыл бұрын

What is workdone by an aneroid barometer in joules ?

@zuhail339 4 жыл бұрын

I just discovered your channel and I want to thank you so much sir

@MichelvanBiezen 4 жыл бұрын

Welcome to Ilectureonline!

@zuhail339 4 жыл бұрын

@@MichelvanBiezen one question sir how much physics I'll need to study your astronomy course?

@historyisthebest5831 4 жыл бұрын

@@zuhail339 It somewhat depends on what level of astronomy are you taking. If you're majoring in astronomy, you will need to study at least most of the aspects of Physics. The two subjects are pretty much interlocked with each other.

@zuhail339 4 жыл бұрын

@@historyisthebest5831 but as beginning path your playlist is enough right? I mean it's really alot content !

@historyisthebest5831 4 жыл бұрын

@@zuhail339 My playlist?? Do you mean videos on my channel? LOL.

@wowi1802 Жыл бұрын

thanks great video, it openend my understanding in way's, i cant discribe here :)🤯

@MichelvanBiezen Жыл бұрын

Glad it was helpful! 🙂

@wowi1802 9 ай бұрын

@@MichelvanBiezen but how about this.. kzfaq.info/get/bejne/mMh4ltue35OdZ30.html

@wowi1802 9 ай бұрын

you dont need g

@nathanielkuri 3 жыл бұрын

I need more tutorials examples

@wingkwongst19 4 жыл бұрын

Professor: Please calculate at what height will the helium balloon be blasted in the atmoshere? As the balloon ascend there must be a height in the atmosphere that the density of air in the atmoshere is equal or lesser than the density of helium inside the balloon.

@MichelvanBiezen 4 жыл бұрын

Balloons, cannot make it to space. The higher they go the less atmosphere there is to push them up. Eventually the buoyancy force becomes so small that the balloon stops going up.

@gentlyschannel4193 4 жыл бұрын

@@MichelvanBiezen I think he might be asking at what point the balloon will burst due to pressure differential.. pretty sure.

@wingkwongst19 4 жыл бұрын

@@MichelvanBiezen , There must be an altitude that the buoyancy force is so strong that outside atmosheric no longer able to suppress it and the balloon bursted. pls refer to this video: kzfaq.info/get/bejne/p5l0eKSZndjdYYk.html

@MichelvanBiezen 4 жыл бұрын

Weather balloons are made to go to very high altitudes where the outside pressure is very small. The balloon material is strong enough to withstand the pressure difference. Ultimately when they reach a height of 18 km to 30 km, the balloons will burst. The reason why thy go so high, is because they are made of very stretchy material (latex) and so they expand a lot.

@ameerunbegum7525 4 жыл бұрын

Hi sir, Omar present..................

@sailokesh4423 4 жыл бұрын

Hello Michel sir congratulations on reaching 6 lakh subscribers 🙂🥳 Sir by the way few days ago I have requested you to explain a question regarding thermodynamics and also sir you have said that it is an interesting question. So sir please tell me when will you explain that problem and I will be excitingly waiting for that day with all my friends who are preparing for JEE exam (a very big competitive exam). Sir no problem how many days it may take but please please please tell me when you are explaining that question so that I can inform my friends sir.🙏🙏🙂🙂 And here is that question sir:- prntscr.com/tizvsv

@sailokesh4423 4 жыл бұрын

Sir it will be very helpful to us if you explain that thermodynamics problem because we are not only asking that problem for learning subject from you but also to improve our problem solving skills from you sir ....... Thank you 🙂🙂🙏

@MichelvanBiezen 4 жыл бұрын

Yes, let me give you some hints. (Was that the problem with the Temperature vs Entropy graph?). Also by the way, we are starting a series of videos on how to prepare and take the JEE advanced exam. Here is what you need to do for the thermo problem. Delta S = Q/T so Q = T x delta S Q cold = ToSo (area under the lower curve). Q hot = ToSo + (1/2) (2To-To)So W = Qh = Qc = (1.5)ToSo - ToSo = 0.5 ToSo Efficiency = W/Qh = 0.5 ToSo / 1.5 ToSo = 1/3 (33.3 %) Q3 in your diagram = 0 because delta S = 0

@sailokesh4423 4 жыл бұрын

@@MichelvanBiezen sir thank you for giving hint ,i almost understood but I can't understand that Qh and Qc and it will be more clear if you explained it on white board and i am keenly waiting for that white board explanation. ..... Also sir i am very very glad that you are starting new series for JEE advanced exam , it will very helpful to so many students and also sir start those series as soon as possible because the time left for the jee advanced exam is shrinking..

@eliudaunda2745 Жыл бұрын

Help me that one please.

@MichelvanBiezen Жыл бұрын

Use the equation PV = nRT n = PV/RT = (1.05 ATM) ( 101,300 Pa/ATM) (4/3) (pi) (0.18m)^3 / [ (8.315 J/mol K) ( 20 + 273) K ]