u/Repulsive-Peak4442

Suppose we have 2 molecules and one moves on the other with 10m/s. The pressure wave travelled from one molecule to other with a speed of 10m/s so this should be the speed of sound. Same principle applies when there's billions of them no? Why? Does my example hold but in reality since there's billions of them the system gets more chaotic and so the speeds are not the same due to energy losses, not perfectly elastic collisions, different speeds​ of each individual molecule etc?

So from what I've understood choked flow occurs if a fluid flows in a pipe with a velocity of mach 1 and no matter how much more we decrease the cross-sectional area of the pipe it won't go past mach 1. Why is that? I've heard explanations saying that mach 1 is the speed that information travels from the right to the left (assuming that the fluid flows from left to right) and so if we decrease the area from the right part of the pipe the information will not be able to travel back to say​ ​"​hey the pressure dropped here, I want​ you to push more". But I don't want that kind of explanations, I want to really understand what happens there. Even if we have to look at the macroscopic level. I want to understand the mechanism. Also I am asking this because I'm trying to understand how the De Laval Nozzle works so if anyone can help me there feel free to drop an answer to this post:​

https://www.reddit.com/r/AerospaceEngineering/comments/1vq87hu/how_does_a_de_lavar_nozzle_workand_what_is/

Many explanations from my perspective seem incomplete because they do never actually explain the why but only the what happens. Why does adding a diverging section on a throat which gases reach velocities of magnitude Mach 1, accelerate them even more reaching supersonic speeds instead of slowing them down? From what I've read this phenomenon occurs due to the gases being compressible instead of incompressible but still that does not explain the why. Why a compressible's fluid speed increases when the cross-sectional area gets bigger? Is it the opposite of what would happen with an incompressible fluid? It seems to. Also what if we keep just the converging and throat sections only and keep making the throat smaller and smaller? I've watched a video saying that Mach1 is the max speed the gases can have at the throat and if we keep making it smaller it doesn't make them faster. How is that? Can someone give me an in-depth analytical explanation?

Many explanations from my perspective seem incomplete because they do never actually explain the why but only the what happens. Why does adding a diverging section on a throat which gases reach velocities of magnitude Mach 1, accelerate them even more reaching supersonic speeds instead of slowing them down? From what I've read this phenomenon occurs due to the gases being compressible instead of incompressible but still that does not explain the why. Why a compressible's fluid speed increases when the cross-sectional area gets bigger? Is it the opposite of what would happen with an incompressible fluid? It seems to. Also what if we keep just the converging and throat sections only and keep making the throat smaller and smaller? I've watched a video saying that Mach1 is the max speed the gases can have at the throat and if we keep making it smaller it doesn't make them faster. How is that? Can someone give me an in-depth analytical explanation?

Let's use the BMP180 Barometer as an example. The bmp180 measures pressure due to the piezoresistive effect. The piezoresistive effect tells us that if you apply mechanical strain and therefore pressure on the bmp180, its resistance changes. But in a model rocket then how can it give accurate results if it is not only the atmospheric pressure applied on it? First of all when the rocket accelerates there is that force from the engine ehich is applied on the bmp and therefore a pressure. Secondly, when the rocket flies the faster it goes the less the pressure around it too due to the Bernoulli's Principle. The bmp might not get affected from that since it is mounted inside the rocket which has an atmospheric pressure of 1 atmoshoere but then how would it even understand the hight if the pressure applied on it is always the one from sea level​

If say I connect a wire to a lamp and only to the negative pole of the battery, why doesn't the lamp turn on? If the electrons from the battery are what pushes the electrons of the wire then they will all start moving towards the lamp right? Then why does the circuit needs to be closed? I just thought of something that I don't know if it is the reason. Is it because the electrons from the wire push back the electrons from the battery and so the circuit has to be closed so they all will move towards the same direction?

I am currently making a Rocket called "Stellar" which I want to launch when it's ready but I do not know the laws yet. I am not in any Rocketry Club as as far as I know there is none in Greece. At which areas am I allowed to launch a Model Rocket, with what class of Rocket motors, what is the maximum velocity, acceleration, hight my rocket can accomplish etc?

Hello everyone 👋! How are you? So suppose there is a car and the driver presses the gas. The engine will apply torque on the axles of the wheels and therefore there will be a force on the wheels from the axle. Say that force is 10N, it gets translated to the contact patch area of the tire and therefore the static friction the tire applies on the road is the same as the static friction the ground applies on the tire. But then the net force on the tire is 10N from the axle minus 10N from the static friction that the ground responds with to the 10N of static friction that the tire applies on the ground which means =0 so the Fnet of the tire is 0. That sounds logical at first because there is no slipping but then this should mean that the tire must not rotate? What is happening here? Some may say that the friction force from the ground is the only external force applied to the car (neglecting all the others) and so this is what accelerates it. But the car is a composite of many different bodies, it is a body system. If we study the tire and as a body alone then it should not rotate.

Hello everyone 👋! How are you? So suppose there is a car and the driver presses the gas. The engine will apply torque on the axles of the wheels and therefore there will be a force on the wheels from the axle. Say that force is 10N, it gets translated to the contact patch area of the tire and therefore the static friction the tire applies on the road is the same as the static friction the ground applies on the tire. But then the net force on the tire is 10N from the axle minus 10N from the static friction that the ground responds with to the 10N of static friction that the tire applies on the ground which means =0 so the Fnet of the tire is 0. That sounds logical at first because there is no slipping but then this should mean that the tire must not rotate? What is happening here? Some may say that the friction force from the ground is the only external force applied to the car (neglecting all the others) and so this is what accelerates it. But the car is a composite of many different bodies, it is a body system. If we study the tire and as a body alone then it should not rotate.

How are you? I previously asked a question about how a rolling tire stops after a while if I let it go, if static friction applied on it from the ground is what moves it. I came to a conclusion that is because of rolling resistance, neglecting drag and all the other internal frictions like the axle-wheel one. (btw if my conclusion is wrong, feel free to correct me) What is exactly rolling resistance in depth? I have read that it has to do just with two things. Hysteresis of the tyre, and displacement of the normal force that the ground applies on the tyre. But I do not know if it has to do with these, or with more or less than these. (I don't even know how they work but that is what I am trying to understand :) If anyone will help me to make me understand that in depth I would really appreciate it and I would be really grateful Thank You.

How are you? I previously asked a question about how a rolling tire stops after a while if I let it go, if static friction applied on it from the ground is what moves it. I came to a conclusion that is because of rolling resistance, neglecting drag and all the other internal frictions like the axle-wheel one. (btw if my conclusion is wrong, feel free to correct me) What is exactly rolling resistance in depth? I have read that it has to do just with two things. Hysteresis of the tyre, and displacement of the normal force that the ground applies on the tyre. But I do not know if it has to do with these, or with more or less than these. (I don't even know how they work but that is what I am trying to understand :) If anyone will help me to make me understand that in depth I would really appreciate it and I would be really grateful Thank You.

Hello everyone 👋! How are you? Suppose we have a rear wheel drive car and we press the gas. From what I have understood, the rear tyres will apply a force on the road opposite to the direction the car will travel (in the same direction they are rotating). If they don't slip, then this force is static friction from the tyres on the road and therefore the road will respond with an opposite of that static friction again, but with the same magnitude. The front wheels on the other hand, will experience a static frictional force from the ground in the opposite direction that the car is travelling because they get pushed from the rear wheels' static frictional force maybe (im not so sure about that) can someone clarify? And also, my question is, if say we have an isolated tyre and we apply a force on it to keep it rolling, since the static friction is not resisting its (not rotational) motion then why when we let it go it stops? Also if static friction from the ground and the force the engine applies on the wheel then the resultant force on the wheels should be zero right? So then, what pushes the car forwards? If you want me to rephrase something, tell me. I would really appreciate it if you gave a whole explanation that covers my questions

Hello everyone 👋! How are you? Suppose we have a rear wheel drive car and we press the gas. From what I have understood, the rear tyres will apply a force on the road opposite to the direction the car will travel (in the same direction they are rotating). If they don't slip, then this force is static friction from the tyres on the road and therefore the road will respond with an opposite of that static friction again, but with the same magnitude. The front wheels on the other hand, will experience a static frictional force from the ground in the opposite direction that the car is travelling because they get pushed from the rear wheels' static frictional force maybe (im not so sure about that) can someone clarify? And also, my question is, if say we have an isolated tyre and we apply a force on it to keep it rolling, since the static friction is not resisting its (not rotational) motion then why when we let it go it stops? Also if static friction from the ground and the force the engine applies on the wheel then the resultant force on the wheels should be zero right? So then, what pushes the car forwards? If you want me to rephrase something, tell me. I would really appreciate it if you gave a whole explanation that covers my questions

Hello everyone 👋! How are you?

According to Newton's third law, when a force is applied on a body, that body applies a force on the body which applies force on it, equal in magnitude and negative in direction. So that means that, when I am walking my foot applies a force on the ground in the opposite direction of my movement, and the ground applies an equal force back on me but in the direction I am moving. Which means that the reaction force from the ground is what causes my movement. But friction is what let's us walk right? It is that reaction force. But how can friction be that reaction force if it has some limit? If say I apply 5N on the ground, by newton's third law it should apply 5N back on me. If I apply 20N and the maximum static friction is 10N then I start slipping because not static but kinetic friction acts on me with say 8N. How is that? I apply 20N on the hround and it responds with 8N?

Hello everyone 👋! How are you? I am getting better and better at physics but there is a thing that pops up on my way suddenly that makes me so confused I wonder if I lack previous knowledge or has to do with momentum which I haven't studied yet. That thing is Action-Reaction. Basically, when a force $\vec{F_{BA}}$ is acted on a body, the body acts a force $\vec{F_{BA}}$ which is equal in magnitude and negative in direction. Ok, this sounds reasonable, if I push on a wall with 10N, it pushes me with 10N. But here comes the paradox in my mind. If say, I apply a force of 10N on my table to move it, it should apply 10N on me. I get it moving and accelerating, but if it is applying 10N on me, then shouldn't I accelerate backwards? Or even, stay stationary? Hear me out. Say I apply 10N on the table, if so then the table should apply 10N on me too acording to newton's third law so $F_{net}=10N-10N=0N$ so my hand should have zero acceleration, that is, its velocity should not change

Hello everyone 👋! How are you? It is true, and I can agree that $K=\frac{1}{2}mv^{2}$ because it is derived from the Work equation. Ok. But. We can say that $K=K_{1}+K_{2}+...$ right? (Check my whole derivation image) But Ais which I've asked tell me thay it is wrong even though I told Knowunity to give me an example with numbers, and it turned out that my equations worked. Here is what it said: ### Parameters $$\[ m = 2 \text{ kg} \]$$ $$\[ F_1 = 10 \text{ N} \]$$ $$\[ F_g = 20 \text{ N} \]$$

Work-Energy Theorem (ΘΜΚΕ)

$$\[ F_{net} = \sqrt{F_1^2 + F_g^2} = \sqrt{10^2 + 20^2} = 22,36 \text{ N} \] \[ a = \frac{F_{net}}{m} = \frac{22,36}{2} = 11,18 \text{ m/s}^2 \] \[ t = 2 \text{ s} \] \[ v = a \cdot t = 11,18 \cdot 2 = 22,36 \text{ m/s} \] \[ K = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 2 \cdot (22,36)^2 = 500 \text{ J} \]$$

Conversion to custom equations

$$\[ v_1 = \frac{F_1}{m} \cdot t = \frac{10}{2} \cdot 2 = 10 \text{ m/s} \] \[ v_g = \frac{F_g}{m} \cdot t = \frac{20}{2} \cdot 2 = 20 \text{ m/s} \] \[ \cos(\theta) = \frac{F_1}{F_{net}} = \frac{10}{22,36} \approx 0,447 \] \[ \cos(\phi) = \frac{F_g}{F_{net}} = \frac{20}{22,36} \approx 0,894 \] \[ K_1 = \frac{1}{2} m v_1 v \cos(\theta) = \frac{1}{2} \cdot 2 \cdot 10 \cdot 22,36 \cdot 0,447 = 100 \text{ J} \] \[ K_g = \frac{1}{2} m v_g v \cos(\phi) = \frac{1}{2} \cdot 2 \cdot 20 \cdot 22,36 \cdot 0,894 = 400 \text{ J} \] \[ K = K_1 + K_g = 100 + 400 = 500 \text{ J} \]$$

Hello everyone 👋! How are you? Some days ago I came up to the Drag Equation and I searched how it's derived. I show that Bernoulli's Equation is used in order to do so, and that meant I searched how that one is derived. I found a derivation using Work and change in kinetic energy (W=∆K). But that work should be zero right? Let's say we habe a pipe that has an area A1 at its left end and a smaller area A2 at its right end. Since W=Fnet•∆x and when say a particle is at the left end (far away from the stenosis) there are equal pressures on it from any directions and so there are equal forces acting on it from every direction so they cancel out. Which means Fnet=F1-F1=0 so doesn't that mean W=0? In the derivation they are using not Fnet but F but why only one component? Fnet is what changes the particle's kinetic energy (where for example, when the particle is at the left end or at the right end there is no net force acting on it and so no acceleration but at the stenosis the neft force acting on it is what accelerating it not the force itself but the net force)