• 0 Posts
  • 21 Comments
Joined 1 year ago
cake
Cake day: July 21st, 2023

help-circle

  • I think you need to take the same approach as the British during WW2 with the Enigma. They could decrypt the messages and know when attacks would happen, but if they stopped every attack, the Nazis would know and change encryption device. So you need to accept that some people will die and only mitigate the disaster in small but impactful ways.



  • petersr@lemmy.worldtoGreentext@sh.itjust.worksAnon defends Michael Jackson
    link
    fedilink
    arrow-up
    52
    arrow-down
    3
    ·
    edit-2
    29 days ago

    His father was a complete piece of shit and treated him horribly and also he was never really allowed a proper childhood.

    So I have heard some people argue that perhaps he was not a pedophile, but rather just really mentally ill, partly believing to be a kid that just wanted child playmates. Still wrong for an adult to act like this, but perhaps there was no sexual misconduct.

    Not sure what to believe and perhaps I just don’t want him to be a villain, but I would like to hear if there is some concrete evidence.











  • From ChatGPT:

    Me: Water exits a nozzle at coordinate (0, 1.7m) and hits the ground at (2m, 0). Assume the water exits the nozzle at 90 degree (i.e. straight to the right). How can we determine the speed of the water?

    (here I made some rough assumptions based on the image, and assumed that her feet were at (0, 0)

    Answer: The speed of the water exiting the nozzle can be determined using the principles of projectile motion.

    1. Determine the time of flight:

      The water is projected horizontally from a height of 1.7 meters and hits the ground at 2 meters horizontally. The vertical motion is governed by the equation: [ y = y_0 + v_{y0} t - \frac{1}{2} g t^2 ] Here, ( y_0 = 1.7 ) meters, ( v_{y0} = 0 ) m/s (since there is no initial vertical velocity), and ( g = 9.8 ) m/s² (acceleration due to gravity).

      The water hits the ground when ( y = 0 ): [ 0 = 1.7 - \frac{1}{2} \times 9.8 \times t^2 ] Solving for ( t ): [ 1.7 = 4.9 t^2 ] [ t^2 = \frac{1.7}{4.9} ] [ t = \sqrt{\frac{1.7}{4.9}} \approx 0.59 \text{ seconds} ]

    2. Determine the horizontal velocity:

      The horizontal distance covered is 2 meters. Since horizontal velocity ( v_x ) is constant (there is no horizontal acceleration), the distance can be given by: [ x = v_x t ] Here, ( x = 2 ) meters and ( t \approx 0.59 ) seconds: [ v_x = \frac{2}{0.59} \approx 3.39 \text{ m/s} ]

    Therefore, the speed of the water exiting the nozzle is approximately 3.39 m/s.