How To Find The Value of Absolute Zero Temperature

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  • Опубликовано: 26 окт 2024

Комментарии • 28

  • @sarafathossain6212
    @sarafathossain6212 3 года назад +20

    why this channel is so underrated?

    • @foodplanet120
      @foodplanet120 3 года назад +4

      Only three videos are there

    • @NaN_000
      @NaN_000 Год назад

      @@foodplanet120 🗿

    • @lit1041
      @lit1041 Год назад +1

      I'm sure if there were more videos on it, it would become popular.

  • @resonatedvirtue8746
    @resonatedvirtue8746 2 года назад +8

    Thank you so much sir! Your videos solve all the doubts my teachers couldn't solve. Just Great job!

  • @jialixx
    @jialixx 2 года назад +3

    Awesome video. I subscribed to your channel even thought there are only three videos. Keep up with the good stuff!

  • @patboily300
    @patboily300 Год назад +2

    Very good, and most likely 100% correct, but would you need many more measurements, especially at colder (sub-zero Celcius) temperatures to make sure that the relationship is not asymptotic?

  • @mumbaiverve2307
    @mumbaiverve2307 Год назад +1

    Beautifully explained !

  • @TaufikAngga-xk6ej
    @TaufikAngga-xk6ej Год назад

    wow your video is very clear all 3 video is good video thanks for share this, Im from indonesia. Just good job sir.....

  • @ShivanahuMishra-c6e
    @ShivanahuMishra-c6e 2 месяца назад

    Keep it up ❤ excellent content very well explained

  • @omsingharjit
    @omsingharjit Год назад

    Keep continuing your work on physics experimental setup.

  • @superbaddctv
    @superbaddctv 2 года назад

    Only 3 vids and I'm impressed. New Subscription. We need more

  • @dwilley8
    @dwilley8 Год назад

    Great job.

  • @vivekthumu8992
    @vivekthumu8992 3 года назад +1

    Super videos brother......
    Very understandable.....

  • @NoCantsAllowed
    @NoCantsAllowed Год назад

    Don't you think the portion of the beaker exposed to ambient air will affect that of the submerged portion containing the helium?

  • @rajeshmondal7134
    @rajeshmondal7134 3 года назад

    Good work,

  • @shrawankhaling8892
    @shrawankhaling8892 3 года назад +2

    Absolutely enlightening!

  • @TakuTJuly
    @TakuTJuly 4 года назад

    We love your videos. Please upload more frequently😖😊

  • @ravindrabhamre4254
    @ravindrabhamre4254 4 года назад +1

    Good video

  • @thomasolson7447
    @thomasolson7447 Год назад

    Why not use this with arctan instead of a ratio with 2pi in e? I eyeballed the tangents. (760/7463+1240/7463)/(1-1240/7463*(760/7463))=14926000/54753969≈0.2726012428432. He is approximately 0.280048237974... He might be 7/25, which is actually a rational triangle 7^2+24^2=25^2. MORE NOBLE GASSES PLEASE! Err, the slope is [L]^3/E or [L]*[T]^2*[M]^(-1)?

  • @Arun_Kumar_x86
    @Arun_Kumar_x86 3 года назад

    wow awsome dude , upload more videos please

  • @zapzed8079
    @zapzed8079 4 года назад

    Awesome

  • @MahmudulHasanss
    @MahmudulHasanss 2 года назад

    Bro give mor video

  • @siddharthannandhakumar6187
    @siddharthannandhakumar6187 2 года назад

    ❤️👏

  • @saroosalah9762
    @saroosalah9762 3 года назад

    Arabic too plz🙏🏻🙏🏻🙏🏻

  • @saroosalah9762
    @saroosalah9762 3 года назад

    Plz put Arabic subs tile

  • @saroosalah9762
    @saroosalah9762 3 года назад

    Arabbbiccc subtitles