Uncertainties in Graph Gradients and Y-Intercepts Using Lines of Worst Acceptable Fit

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  • Опубликовано: 4 ноя 2024

Комментарии • 29

  • @aleezajehangir178
    @aleezajehangir178 4 года назад +6

    Great video, thanks!

    • @PhysicsHQ
      @PhysicsHQ  4 года назад +2

      You're welcome!

    • @aleezajehangir178
      @aleezajehangir178 4 года назад +2

      @@PhysicsHQ Just keep uploading more videos like these 👍

    • @PhysicsHQ
      @PhysicsHQ  4 года назад +3

      I’ve got the follow up video for analysing log graphs coming soon.

  • @annarae834
    @annarae834 2 года назад +1

    Wow this explained everything so well!! Thank you so much for this video!

    • @PhysicsHQ
      @PhysicsHQ  2 года назад +1

      You’re welcome ☺️
      All the best with your studies!

  • @SGayanFernando
    @SGayanFernando 5 месяцев назад +1

    This video helped me a lot! Thank you very much!

    • @PhysicsHQ
      @PhysicsHQ  4 месяца назад

      You're welcome! Glad it helped.

  • @masked_musings
    @masked_musings 2 года назад +2

    really love this, very clear explanation. Is it possible to ask for the ppt you used? Would really appreciate it

    • @PhysicsHQ
      @PhysicsHQ  2 года назад +2

      Thanks very much!
      It’s a Keynote presentation not PowerPoint. But yes I can share that via my website & TES. I’ll ping a link.

    • @PhysicsHQ
      @PhysicsHQ  2 года назад +3

      Hi again. Just uploaded it here as free download: wonkylogic.co/a-level-physics-hq
      Will put it on TES too but on my website no login needed for download.

  • @kidjuke6262
    @kidjuke6262 2 года назад +1

    clean presentation👌👌👌

    • @PhysicsHQ
      @PhysicsHQ  2 года назад

      Thanks. All the best with your studies.

  • @taejgaul5991
    @taejgaul5991 2 года назад +1

    Thank you so much! 🥰 love the hair btw 😅

    • @PhysicsHQ
      @PhysicsHQ  2 года назад

      😊 thanks. I am at the mercy of what my hair does day to day!
      All the best - especially if you have exams coming up.

  • @ihenwoko38
    @ihenwoko38 2 месяца назад

    Don’t still get how you got your error bars though
    Did you just take a random pick at the +/- values?

  • @saieshengovender7109
    @saieshengovender7109 3 года назад +3

    thanks a lot sir

    • @PhysicsHQ
      @PhysicsHQ  3 года назад +1

      Thanks! You’re welcome.

  • @naweyeeminn23
    @naweyeeminn23 Год назад

    May you please drop the link for the tips for not starting from the zero origin .

  • @jamieashworth_
    @jamieashworth_ 3 года назад

    Surely you should have a modulus sign on the bottom as if the gradient is negative you'll get a negative uncertainty

    • @PhysicsHQ
      @PhysicsHQ  3 года назад +2

      That’s right. A negative uncertainty wouldn’t make sense.

    • @jamieashworth_
      @jamieashworth_ 3 года назад

      @@PhysicsHQ okay was just saying because you didn't include a modulus on the bottom of your formula

    • @PhysicsHQ
      @PhysicsHQ  3 года назад +1

      Yes I see that now. Had forgotten at the time that the standard percentage difference equation needed modification.

  • @insertusername4822
    @insertusername4822 4 года назад +1

    Thx alot, great video

  • @eatinsomtin9984
    @eatinsomtin9984 3 года назад +1

    Does %c mean the the percentage uncertainty?

    • @PhysicsHQ
      @PhysicsHQ  3 года назад +1

      %c is the percentage uncertainty in the y-intercept.

  • @Proph77
    @Proph77 2 года назад

    How can I change the percentage to a definite value

    • @PhysicsHQ
      @PhysicsHQ  2 года назад +2

      The percentages are percentage uncertainties in gradient or intercept. They are related to a quantity of interest - e.g. resistivity (ρ) - so apply uncertainty calculation rules to determine percentage uncertainty in that quantity. Then multiply percentage uncertainty by the value of the quantity for the absolute uncertainty of the quantity.
      E.G. gradient = ρ/A ⇒ %ρ = %gradient + %A
      If ρ = 2.3 × 10⁻⁸ Ω m
      And if %ρ was determined to be 5.8% then absolute uncertainty is 1 × 10⁻⁹ Ω m.
      So the absolute uncertainty of the gradient or intercept is not of so much interest but the absolute uncertainty of the final value is of interest.