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22

You are right. From continuity of the incompressible fluid you have $$A_1 v_1 = A_2 v_2.$$ So obviously the velocity is changing. Thus the fluid is accelerated, and therefore there must be a force causing this acceleration. In this case the force comes from the pressure difference between the wide and the narrow part of the pipe. (image from ResearchGate - ...

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There is more mass per area behind than ahead of the constriction so since ppressure is force divided by area there develops a pressure difference

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The simple answer is pressure. As you state, from the continuity equation you can see that the velocity $v_2>v_1$. The next step is a momentum balance (like any balance in fluid dynamics: $\frac{d}{dt}=in-out+production$). The momentum flowing into the system is smaller than the momentum flowing out of the system ($\rho A_1 v_1^2 < \rho A_2 v_2^2 = \... 5 Standing on the surface of the Earth, the reference frame at rest relative to yourself is certainly not a frame with Minkowski metric. Here is the proof: release an object, so that it is in free fall. There is relative acceleration between the object and the chosen frame. Hence the frame is not inertial and its metric is not Minkowskian. To define a tangent ... 5 seems to me that this should imply that a local observer standing on the earth (so not free falling at all) should be considered as an accelerating, non inertial frame. Yes, an observer standing on the earth is not inertial in relativity. The definitive test is to have the observer carry a good accelerometer. In this case it will indicate an acceleration of ... 5 Instead of thinking of velocity and acceleration as two different things, simply think of acceleration as the name we've given to any change in velocity. Then it suddenly is obvious that what we call acceleration is present in the very instance the velocity changes; not a moment before or a moment after. Because that change is acceleration, just in other ... 4 Only if the mass of the accelerating body is changing; or in the theory of relativity, depending on how you define acceleration. Newton's second law reads$\vec{F}=m\vec{a}$, so in Newtonian mechanics, if the force is constant, mass must change if acceleration is to change. As an example, consider a rocket propelled by a constant force in space, expending ... 4 Warning: Non-rigorous math up ahead. Usually we handle instantaneous things in physics with the Dirac Delta function, which is qualitatively an infinite spike at some given point. We can use this here to express the instantaneous force in terms of an impulse per time: $$F(t)=J_0\delta(t)$$ so that the total change in momentum of the object is given by $$\... 3 Using g\approx10\,\mathrm{m/s^2} for the standard acceleration of gravity on Earth, so a=10^7\,\mathrm{m/s^2} and$$\Delta v=a\Delta t$$we get \Delta t=1\times10^{-9}\,\mathrm{s}, that is, the extremely short period is about 1 nanosecond. Plugging that into$$\Delta s=\frac12a\Delta t^2$$we get that the object moves 5 picometres, 0.1 the Bohr radius ... 2 Re. "at any instant the velocity is determined by the acceleration at that instant." No. At a given point, the velocity and acceleration at an instant are independent of each other. Given acceleration at an instant, t, velocity at that instant can take on any value. However if acceleration a' is given for an infinitesimal interval of time, ... 2 "at any instant the velocity is determined by the acceleration at that instant." No. The velocity at an instant, t_1, is determined by the accelerations at all previous times going back to a time t_0 at which we know the 'initial' velocity. That's because we can use the accelerations to calculate the change in velocity, leading up to time t_1.... 2 For simple near-Earth calculations, the value of g changes very little. Using your derivitive, the value of g changes at a rate of -3.08\times10^{-6} m/s^2 per meter of height. So, even at 10 km of elevation (where airliners fly), the value of g has only changed by about 0.01 m/s^2. Compared to the value of 9.8 m/s^2, this is entirely ... 2 The point of:$$ F = m\ddot r = mg $$is that it is tractable for introductory physics classes. It is also sufficient to solve many problems to reasonable accuracy. If you want to get exact, you're equation is fine. For a perfectly spherical non-rotating Earth of uniform density that does not have a moon, or an atmosphere. For a realistic Earth: you have ... 2 You are forgetting the gist of relativity: the man falling is in motion with respect to you. According to him, he experiences no forces so he would call himself at rest, as the man floating in space would. It has nothing to do with geodesics. For him, he would just hover around. So, what makes him touch he ground then? He can argue that the ground moves up ... 2 Note: The following answer has been given from a classical mechanics point of view. It does not deal with any quantum mechanical phenomenon arising due to the following scenarios. Theoretical analysis An infinitely powerful impulse can, theoretically, exist. Because, mathematically$$J=\Delta(mv)=\Delta p=p_{\rm final}-p_{\rm initial}=p_{\rm final} \qquad \... 2 This is because of greater shear strain you will experience if you run on the train while its accelerating. But us humans do not deform when subjected to this stress but they fall down. There are two factors involved here Shear force(assuming uniform acceleration) Without using complicated terms, let's assume that you're a passenger(A) watching another ... 1 Velocity is the integral of acceleration with time, and if your acceleration curve has a finite value at$t=0$and zero otherwise, the area under the curve is zero. This means, your first interpretation is correct, and the body will not move. The rule to remember is: $$\Delta \text{(momentum)} = \text{(impulse)}$$ The change in momentum equals the impulse. ... 1 I think you are essentially talking about impulse. You are right that$\int_0^t F(u)du$would be zero if the force is applied for an instant, but any real force is applied for at least some time and so the integral cannot be zero. However, it may be very small as in your case where the force is not of a very large magnitude. The same goes with acceleration, ... 1 The fact that the equation doesn't tell you the acceleration is 0, doesn't mean it implies the opposite. If A doesn't imply B, that doesn't mean that not B is true. It might be that B is still true but you just can't infer it from A. In this case A is "the equation is satisfied" and B is "the acceleration is 0". The correct way of ... 1 You have graphed position s as a function of time:$s(t) $in Physics notation. But you call time$t$your independent variable$x$, and your$f$refers to the mathematical shape of the function rather than the resulting Physical quantity position$s$, which is the dependent variable, a.k.a.$ y$in maths. The derivative with respect to t would be the ... 1 An inertial frame near the Earth's surface is heading towards the center of the earth with an acceleration 9.81 ms$^{-1}$. A frame stationary wrt to the surface is not inertial in GR. Instead in this non-inertial surface frame we experience the "fictitious" force that we call gravity. The curvature of spacetime is responsible for the tidal ... 1$g$is normally considered constant in cases like this, because over small scales that's a good approximation: it's easy to see that at a height$h$above the surface, then $$\frac{\Delta g}{g} \approx \frac{R^2}{(R + h)^2}\quad\text{and = for a spherical Earth with no Moon etc}$$ And you can plug in values for this: for$h = 100\,\mathrm{m}$it's about$\...

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Yes. The tangential and normal components of the acceleration are orthogonal, so by the Pythagorean theorem, the magnitude of the resultant acceleration is $$\sqrt{(g \cos \theta)^2 + (g \sin \theta)^2} = g\sqrt{\cos^2 \theta + \sin^2\theta}=g.$$

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You do not see a difference in forces. You see a difference in pseudoforces. In both cases the observe has exactly 0 forces on it. However, in a non-inertial frame, we find that there is an acceleration on the object, which is basically the inverse of the non-inertial behavior of the frame in the first place. The key to understanding these things is that ...

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