• Maalus@lemmy.world
    link
    fedilink
    arrow-up
    3
    ·
    5 months ago

    Pretty sure the answer is just “40 minutes” and it is a question to make someone think about what they are doing rather than automatically solve every task.

    • argh_another_username@lemmy.ca
      link
      fedilink
      arrow-up
      2
      ·
      5 months ago

      But it’s still wrong, though, as the 9th is about 70 minutes.

      There’s even a myth saying that the 9th was the determinant for the length of the original CD.

      • lugal@lemmy.ml
        link
        fedilink
        arrow-up
        1
        ·
        5 months ago

        That’s how long it usually takes since usually it’s played with about 200 players

    • Colonel Panic@lemm.ee
      link
      fedilink
      English
      arrow-up
      1
      ·
      5 months ago

      That doesn’t sound like giving it 110% and being a team player. We are a family here. We need go getters. We gotta make it happen.

  • General_Shenanigans@lemmy.world
    link
    fedilink
    arrow-up
    0
    ·
    5 months ago

    Let’s say you put like 1000 violinists all in a big, long row. Then, have the first violinist play a note, then the second plays the very same note, then the third, and so on. Let’s say you could also time it so that at the very moment the sound wave from one violinist hits the next is when that one plays the note. Brrrrrrump! All the way across. Let’s also say you could time it perfectly so that the waves don’t cancel each other out. What would happen?

    • BedInspector@lemmy.world
      link
      fedilink
      arrow-up
      0
      ·
      5 months ago

      I think eventually you reach a point where previously played notes would lose all of their energy, meaning there’s probably an upper limit on how loud it would get for an observer at the end. Something something Doppler effect.

      • Vigge93@lemmy.world
        link
        fedilink
        arrow-up
        0
        ·
        5 months ago

        Not the Doppler effect, as that only applies to moving objects, but instead the inverse square law, where the energy of the sound wave decreases by the square of the distance from the origin, since it spreads in a sphere with the energy being spread across the surface of the sphere, resulting in a very quick dropoff in the loudness.

        • BedInspector@lemmy.world
          link
          fedilink
          arrow-up
          1
          ·
          4 months ago

          The sound source is moving in the above scenario relative to a stationary object. I’m not saying you’re wrong but that was my thinking.