Timeline for How does Newton's third law apply to an object interacting with air?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 22, 2015 at 9:02 | comment | added | SchrodingersCat | @JanDvorak I admit I didn't know that a space shuttle can slow down from 8 km/h to less than one percent of that speed without spending any fuel of its own. So what is your opinion regarding this? I'll be happy to counteract your logic. | |
| Oct 21, 2015 at 16:02 | comment | added | John Dvorak | " you cannot touch an air molecule as such and air molecules have very low or no rigidity" - so, how does a space shuttle slow down from 8 km/h to less than one percent of that speed without spending any fuel of its own, again? | |
| Oct 21, 2015 at 3:04 | comment | added | jpmc26 | Strictly speaking, it's $ F_{avg} t $ or $ \int F dt $, but whatever. The oversimplification is understandable for your audience. =) +1 | |
| Oct 20, 2015 at 16:22 | comment | added | Peter - Reinstate Monica | @Toughquestions You apply force to your hand while you accelerate the fist, but you are the one decelerating it as well (and if you throw violently, you can hurt yourself in the process). The air is not applying a significant force to your fist. With water you can have a resistance at normal speeds. With air, you must drive maybe 70 mph (my Mach 0.1) to feel your 100N when you stick the hand out of the passenger window, and then you do feel the air's force. | |
| Oct 20, 2015 at 16:21 | comment | added | SchrodingersCat | Birds have a very light body and an aerodynamic shape to use air drag to its advantage. So they can use reaction of air by flapping their wings. $\mathbf{But \,\ your \, fist \,\ is \,\ much \,\ heavier}$. | |
| Oct 20, 2015 at 16:18 | comment | added | Tough questions | Then how do birds fly? They use reaction of air by flapping their wings.air must give birds enough reaction. | |
| Oct 20, 2015 at 16:05 | comment | added | SchrodingersCat | Your force is not the same, your $\mathbf{impulse}$ is the same. Impulse = $F \times t$ = $mv - 0$. When you say you are punching both violently, you mean you are experiencing the same impulse, not same force. | |
| Oct 20, 2015 at 15:57 | comment | added | Tough questions | Am I not punching air with 100 Newton? If not then why not? Cause I am doing the same thing...punching the air as violently as punching the wall... – Tough questions just now edit | |
| Oct 20, 2015 at 15:43 | history | answered | SchrodingersCat | CC BY-SA 3.0 |