Young's Modulus & Vernier Scales - A-level Physics

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  • Опубликовано: 25 авг 2024

Комментарии • 66

  • @aaronbrooker8098
    @aaronbrooker8098 4 года назад +205

    Think I've found my 'free science lessons' of A Level

  • @retooluvyuhx5569
    @retooluvyuhx5569 6 лет назад +190

    WE TRULY APPRECIATE YOUR EFFORTS AND TIME SIR

  • @benb758
    @benb758 7 лет назад +72

    Is this 1 video every day? That's kind of insane. (And extremely helpful)

    • @ScienceShorts
      @ScienceShorts  7 лет назад +19

      +Ben B Not anymore! I'm done for now :D

    • @Zravi1234
      @Zravi1234 7 лет назад +7

      All your videos are absoulutely amazing and saved my A-Level, I've started rewatching all of them. Just wondering what % of the A-Level would you say you have covered and have you gone over all the required practicals?

    • @ScienceShorts
      @ScienceShorts  7 лет назад +21

      Thank you! There's not a lot I teach on top of what's covered in the videos (I don't do any of the other AQA optional topics), apart from the odd little bits of knowledge missing, like metastable states, for example - but this is stuff that's easy to check on with the spec. So I'd say I've covered ~95% of the course. The important thing to do is answer questions to improve your exam technique and pick up on the subtleties the exam board look for in your answer.

    • @Zravi1234
      @Zravi1234 7 лет назад +11

      I have been doing some practice questions my teacher has set me, however due to the new reformed A levels there is only 1 past paper and 1 specimen available. Any advice on how to aid my revision even further?

    • @retooluvyuhx5569
      @retooluvyuhx5569 6 лет назад +13

      you can do papers of other exam boards then regarding the topics you choose, if you have only one specimen and 1 past paper then that wouldn't be enough for the revision as u can get a bunch of differently asked questions that can be really tricky, Idk but I would do that if I was in your shoes. idk for how long this comment has been here I just thought that my reply might help in a sort of way

  • @mohamedrashed6643
    @mohamedrashed6643 5 лет назад +18

    watching this vid 30 mins before exam its a great help thx

    • @vahe057
      @vahe057 5 лет назад +1

      Lol same

  • @musfirakhurshid
    @musfirakhurshid 5 лет назад +50

    Chris Hemsworth teaches Physics. Wow! 👌🏼

  • @elliewhiter
    @elliewhiter 6 лет назад +6

    so so helpful. I have spent so long trying to read a vernier scale and now I finally understand how to

  • @chenulgomes8234
    @chenulgomes8234 4 года назад +3

    Hey, I don’t know who you are but you have been pulling me through my exam. Thanx

  • @muhammadanas7791
    @muhammadanas7791 3 месяца назад

    Got my A-level physics Paper 1 in 3 days. I'm killing it

  • @elliotverry730
    @elliotverry730 6 лет назад +4

    A little bit of me died every time I saw I had to draw out another graph haha, great video though thanks for the help :)

  • @yikesQuakes
    @yikesQuakes 3 года назад

    Your explanation is truly beautiful sir as clear as crystal

  • @stanislavshelemekh2162
    @stanislavshelemekh2162 4 года назад +1

    Very very very well explained!

  • @lethbophilip6365
    @lethbophilip6365 4 месяца назад

    Apprghe knowledge fam...
    Be blessed 🙌 ❤

  • @emilythompson3678
    @emilythompson3678 3 года назад

    man you are going to save my life

  • @dyxtopia21
    @dyxtopia21 4 года назад +1

    This was very helpful thank you

  • @rachelallsop6524
    @rachelallsop6524 5 лет назад

    This is so helpful thank you !!! really helped explaining the graphs and why it means what it does ^-^

  • @lethbophilip6365
    @lethbophilip6365 4 месяца назад

    Appreciate 😢

  • @invictusgaming3622
    @invictusgaming3622 6 лет назад +7

    hi sir! what about the features of a force-extension (or stress-strain) graph for rubber? how will the gradient look?

    • @ScienceShorts
      @ScienceShorts  6 лет назад +13

      Curvy. Have a look at OCR 2017 paper 2, there's a question on it.

  • @selleryy
    @selleryy 2 года назад

    Great video, thank you Sir.

  • @kanyakinanti
    @kanyakinanti 3 года назад

    Very helpful. Thank you, Sir

  • @Swaglander
    @Swaglander 2 года назад

    thanks for the help

  • @GamingGuyGus
    @GamingGuyGus 6 лет назад +2

    legend

  • @sakifnewaz7212
    @sakifnewaz7212 5 лет назад

    Thank you very much

  • @s3lim77
    @s3lim77 3 года назад +2

    Ahh it was Mr Rhys from Malmesbury Science all along...

    • @ScienceShorts
      @ScienceShorts  3 года назад +3

      Nope. Rather, Mr Rees was Science Shorts all along.

    • @s3lim77
      @s3lim77 3 года назад

      @@ScienceShorts had to guess which Rees it was going to be :/

    • @ScienceShorts
      @ScienceShorts  3 года назад +1

      Don't worry, even my pupils spell it wrong.

  • @wafa2075
    @wafa2075 4 года назад +2

    How does extension increase with length?

  • @d.k.6015
    @d.k.6015 5 лет назад +2

    Ohmygod bless you

  • @aitchisoniantutor1625
    @aitchisoniantutor1625 3 года назад

    Very helpful

  • @SonLyAu
    @SonLyAu 7 лет назад

    Thank you

  • @pranjalbanerjee6986
    @pranjalbanerjee6986 2 года назад +1

    Is l the strengthen length or unstressed lagers?

  • @ibrahimabulwifa4296
    @ibrahimabulwifa4296 5 лет назад

    brilliant

  • @evolvinggenius198
    @evolvinggenius198 3 года назад

    very cool

  • @seemadevi-mk8pz
    @seemadevi-mk8pz 6 лет назад

    Very good explanation. Thanku

  • @Koltary
    @Koltary 5 лет назад

    I need definitions for hard and malleable and the other characteristics

  • @williamwelmans3026
    @williamwelmans3026 4 года назад +1

    What can I say sir! "Excellent" comes to mind! " Wish I had you as a Teacher!"

  • @xgenarater
    @xgenarater 5 лет назад

    Would u not do a force extention graph instead of a mass extention graph?

  • @moodymonstrosity1423
    @moodymonstrosity1423 4 года назад

    I have a question:
    When you made the equation for the youngs modulus :- E =FL/A x extension, why didn't you take a constant k because in hooke's law we had a constant while forming an equation F = k(x)
    Please clear this for me as I am really confused. Thankyou.

    • @radowanahmedbaized3597
      @radowanahmedbaized3597 4 года назад +1

      Youngs modulus is the constant for stress-strain graphs. "K" is the constant for Force-Extension graphs for Hooke's law.

    • @moodymonstrosity1423
      @moodymonstrosity1423 4 года назад

      @@radowanahmedbaized3597 huh?

    • @sangodan3031
      @sangodan3031 3 года назад +1

      @@moodymonstrosity1423 you only need k when trying to calculate an extension by a force, in this you already have the force and the extension so k is irrelevant

  • @luqmanhakim-kq7so
    @luqmanhakim-kq7so 6 лет назад

    CAN YOU SHOW ME THE YIELD POINT SIR?

  • @miesht430
    @miesht430 5 лет назад

    hi sir this is not for edexcel right

    • @maazraza4895
      @maazraza4895 4 года назад

      this is on the Edexcel syllabus

  • @samirhussain458
    @samirhussain458 7 лет назад

    you god

  • @kieran7610
    @kieran7610 6 лет назад

    What i've never understood is why the yield point decreases in stress and then increases again.

    • @alymaghraby6923
      @alymaghraby6923 6 лет назад +2

      You dont need to know this, but in simple terms: The graph doesnt take into account the changes in surface area when the strain increases. So that little dip in stress is just due to the area increasing while the wire is extending.

  • @courtneyallen5685
    @courtneyallen5685 5 лет назад

    100

  • @yuki002x
    @yuki002x Год назад

    that vernier caliper tho 😭

  • @musfirakhurshid
    @musfirakhurshid 5 лет назад +3

    Chris Hemsworth teaches Physics. Wow! 👌🏼