What velocity does the milk exit the nipple with to be able to have that trajectory?
What velocity does the milk exit the nipple with to be able to have that trajectory?
I second this. It is the option in the three dots when viewing a stream of posts, right? I recall it had the icon of a finger and and arrow and was called “Swipe mode”.
Imagine sitting in a bathroom stall, with a gorilla pounding full force on the door, you just waiting for it to break, knowing full well that you will be crushed soon.
Just like Tolkien wrote it.
To make room for the pennies
I will give you an upvote for maintaining critical thinking, but I will say that the book is not that bad. And it is really well written and interesting, but might be favoring SBF more than other books. The truth is probably somewhere in the middle.
Read “Going Infinite” which describes the rise and tall of FTX and SBF. “isn’t smart” is not true, but he is definitely not “human smart”, but rather “math smart”.
This also hides post if searching for it or showing specific NSFW communities.
What kind of “guess the billionaire” game is this?
Or import this
No, it is because it is not a gun.
Make more money elsewhere and hold your own pizza party with the extra money.
Like pihole? I recently set it up and it is not blocking YouTube ads (as other commenters have also pointed out).
Is it possible to learn power?
But what has feelings?
Came here to say all of this, so thank you for saving me keystrokes :)
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.
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} ]
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.