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1

Assuming that the pressure change across the cone is small (e.g., no significant density changes for the flowing gas), use the continuity equation. With constant density, this simplifies to $A_1 \cdot v_1 = A_2 \cdot v_2$, where $A$ is the cross sectional area of the flow stream and $v$ is the velocity of the flow stream. If you additionally need the ...


0

I'm pretty sure that we can't do any better with trig identities; it looks like any such expression would have a term vaguely like $\sqrt{1 + r \cos \theta}$, and then the square root ruins the nice intuition. In lieu of that I'll offer a derivation of the group velocity that uses no fancy math at all! Hopefully that's what you wanted, even if you didn't ...


0

Mathematically, the friction is described by a couple of Heaviside functions $\Theta$, asuming your body is of length $d$ and the transition from non-friction to friction occurs at $x_0$ with $x_1 = x_0 - d/2$ and $x_2 = x_0 + d/2$. $~~~~~~~~~F_\mu (x,v) = -\mu \cdot v \cdot \left[ \Theta(x-x_1) \cdot \Theta(x_2-x) \cdot \cfrac{x-x_1}{x_2-x_1} + ...


0

That's why when you want to integrate numerically the vertical acceleration you have to subtract 1g from the vertical acceleration you measure from the accelerometer


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I would use the traditional drag equation: F = 0.5 * p * v^2 * Cd * A Reasoning: Stokes' law is for very small Reynolds number flows only (much less that 1), and I don't believe the more common drag model has any such restriction. The problem is that you will need to know the drag coefficient of the ball in question to use this force model. Sources: I ...


1

Depends on the Reynolds number. Stokes can be used for purely laminar flow. Complete answer to be found here.


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This is a subtle problem. Assuming that the normal force that is involved in friction is strictly dependent on how much of the block is over that surface, it is seen that the friction force increases steadily as more and more of the block moves over that surface. This means that the amount of friction force and the acceleration are dependent on the ...


0

Assuming the body comes to rest before it completely enters the region, the body will trace out a partial sine curve in the space of $(t,x)$. Suppose the body is of length $L$ has uniform mass $M$. We let $x$ correspond to the positive horizontal distance from the interface. Then the force of kinetic friction (with coefficient $\mu_k$) is \begin{cases} ...


1

There is another aspect somehow overlooked by the other answers. Consider a pile of iron filings accelerated towards a magnet. If you were to arrange so that they all have the same magnetic force per unit mass they would appear to experience no force relative to each other while being accelerated towards the magnet, and if you had weak bonds holding them ...


1

We don't need to appeal to relativity to explain why you don't feel any force in free fall. Plain old Newtonian mechanics predicts that too. What you actually feel when you feel a force being applied to you is that the external force applies only to a small part of your body (the soles of your feet if you're standing up and feel the normal force from the ...


0

A hydrogen atom (I.e. a bound electron-proton pair) can travel a distance, during which time the particles must necessarily have the same average velocity; if the electron is oscillating about the proton then its instantaneous velocity will cross over from less than the proton's to greater than the proton's at times, such that there are instants where the ...


0

It depends on the nature of the particles. If we are bosons there is no limitation at all (example is a photon). If we are talking about two fermions (electrons, protons...), then we need to make sure that they do not occupy the same state. Now, what does define the state of a particle? In classical physics the answer would be velocity and position. In ...


0

Why does a free-falling body experience no force despite accelerating? Because there is no force acting upon it. If you look at some pictures of the principle of equivalence, you will find that they typically depict a guy in a rocket accelerating through space. There's a force on his feet, he can feel it. They also depict a guy standing on the surface ...


39

Before telling you why an observer in free fall does not feel any force acting on him, there are a couple of results that should be introduced to you. Newton's second law is only valid in inertial frames of reference: To measure quantities like the position, velocity, and acceleration of an object, you need a coordinate system $(x,y,z,t)$. Now the ...


5

Well, everything depends on what you mean by "to experience a force". I suspect that you are thinking of some psycho-physical idea. Indeed both floating in space and freely falling we perceive similar sensations. The reason is simply due to the fact that, in both situations, all particles of our body moves with the same speed (due to a spatially uniform ...


1

You need a coordinate system to decide a body’s position, velocity, acceleration, momentum or force on it. Assume the body is in free fall near the Earth. 1) First consider a coordinate frame (3 perpendicular rods and a clock) with its origin in free fall near the free falling body. By the equivalence principle we know the rods are falling in unison with ...


23

It is incorrect to link the feeling of being accelerated to being accelerated itself. You can be under constant velocity or be continuously accelerated, yet you need not feel anything at all. Let me explain. The reason you feel compressed or stretched when you are accelerated in a lift is because of the presence of the normal force from the ground on you. ...


3

falling in a gravitational field is physically indistinguishable from floating in interstellar space Yes. Indeed, this is one of the founding principles of general relativity and is (one of the forms of) the equivalence principle. Your argument is that we can feel acceleration, and gravity makes you accelerate, so shouldn't you feel acceleration while ...


0

Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference: $$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$ where $s$ denotes the space fixed frame and $r$ denotes the rotating frame. ...


0

For a friction less drop we simply use Newtons law: $$\frac{dv}{dt} = g$$ and solving for $v$ gives $v(t) = gt$. Introducing the friction component into the differential equation, I make the approximation that the drag is laminar: $$ \frac{dv}{dt} = g - \frac{\gamma}{m} v$$ The solution of this is: $v(t) = \frac{mg}{\gamma} - \frac{mg}{\gamma} ...


1

Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do. It might help to start with the speed an object attains after falling a short distance, ignoring air resistance and assuming the effect of gravity is constant. If we fall from Mount Everest (about 8000 meters) to sea level, we can treat gravity ...


1

Okay, so here's the basic physics: it comes down to differential equations. Big objects in air tend to see $v^2$ drag. Plugging this into $F = m a$, this means that something which is falling straight downward with speed $v$ will obey the nonlinear differential equation: $$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length $\lambda$. The terminal ...


-3

Is it true to say that all matter in the universe is travelling with velocity c through spacetime? Actually, no. Because there is no motion through spacetime. Spacetime models motion through space over time, but because it includes the time dimension, it's totally static. And whilst this model works very well, we live in a world of space and motion, ...


0

You're correct. The speed is only the magnitude of the velocity. However, since the speed doesn't change it cannot change direction, as that would require acceleration. Of course, that assumes the speed is only in one dimension, because it could plot the speed of something with a circular motion, or any other motion for that matter. A spaceship in a circular ...


1

The simple answer is that displacement, in this context, is distance. There are other uses, such as the weight of a ship, but that is not germane. Consider that the area under the line is x times y. In this case the x axis has the units of seconds, and the y axis has the units of meters per second. So when multiplying them out, $$sec\times\frac{meters}{sec} ...


1

The graph shows a constant speed, which is positive, so we know that the velocity is not changing sign. Depending on the context of the graph (is it dealing with 1D motion or curvilinear motion) we could tell a lot. Technically, on a traditional velocity vs time graph, one is plotting a component of velocity, complete with signs. I don't think the plot is ...


0

Terminal velocity is the velocity at which the force air resistance is equal and opposite to gravity in the direction of motion (ignoring friction for the moment). In free fall, the force of gravity is $F_g=mg$ where $m$ is the mass of the object in question, and $g$ is the gravitational acceleration at Earth's surface. On an incline, however, the force of ...


0

Going down a track at an angle reduces the acceleration of gravity by the cosine of the angle. If you then ignore all friction except air resistance (not a good assumption), the terminal velocity goes down as the square root of the acceleration of gravity. If you want to get it right it won't be easy because there is a lot going on. If you want it about ...


3

Suppose your object is a sphere with a radius $r$ and mass $m$. The aerodynamic drag on a sphere is given by: $$ F_{drag} = \tfrac{1}{2}C_d \rho \,\pi r^2 \,v^2 \tag{1} $$ where $\rho$ is the density of the air and $C_d$ is the drag coefficient. The drag coefficient varies with speed (the NASA article I linked shows how $C_d$ changes with speed) but over a ...


0

Imagine 3 objects. One is a flat piece of paper. The second is an identical piece of paper rolled into a ball. The third object is the exact same shape and size as the rolled up piece of paper but it's made of iron. If you drop the flat paper and the rolled up paper at the same time, the rolled up paper hits first because it has less air resistance due to ...


0

If u want to find exit pressure of control volume you have consider, you can use Bernoulli equation since you know exit velocity profile. Since boundary layer is thin and no flow separation, this flow is more or less irrotaional flow and I'm assuming the flow is steady. $$ p_1 +\rho \frac{v_1^2}{2} =p_2 +\rho \frac{v_2^2}{2}$$ Here $p_1$, $v_1$ are inlet ...


1

This is one of those questions that can drive you crazy, since there is a great deal assumed and not stated. Let me try an alternate possibility. Conceivably, the problem wants you to assume that, when moving, the overall speed of the raindrop remains fixed at 5 m/sec, but it travels in a straight line at an angle due to horizontal wind forces, and this may ...


2

If the raindrop's vertical velocity is constant as the train is both stationary and moving, the time taken for the raindrop to travel down the window would be: $$t = \frac{1\ \text{m}}{5\ \text{m}/\text{s}} = 0.2\ \text{s}$$ Remember, the time $t$ would not depend on the speed of the train. The exercise also specifically states that the raindrop's vertical ...


0

Another way of looking at this: The vertical component of the raindrop's velocity vector is 5 m/sec downward, and the horizontal component is 30 m/sec across. By the Pythagorean method, the resultant velocity vector is 30.41 m/sec diagonal. The vertical component of the raindrop's displacement vector is 1 meter, and as the raindrop is pushed horizontally ...


0

As others have said, escape velocity is a speed, not a velocity. As to why, see the etymology of the word velocity: early 15c., from Latin velocitatem (nominative velocitas) "swiftness, speed," from velox (genitive velocis) "swift, speedy, rapid, quick," of uncertain origin. Velocity used to mean speed, and we still say things like "high velocity ...


0

After some investigation it turns out that my question is a bit of a convoluted way to ask for the momentum, which contains the information as follows. The momentum-energy relation $$ E(t)^2/c^2 - \vec{p}(t)^2 = m_0^2$$ with $\vec{p}$ the momentum, $E$ the engery and $m_0$ the invariant rest mass, simplifies to $$ \vec{p}(t)^2 = E(t)^2/c^2 $$ for ...


2

Yes, escape velocity should really be escape speed. The Wikipedia article on escape velocity states this explicitly. I doubt there is any logical reason for using the term escape velocity and I suspect it is an accident of history. You might want to ask on the History of Science SE how the term originated - a quick Google failed to retrieve any information ...


3

I know this is an old thread, but I had to figure this out for a problem on my physics homework. What helped me to understand this is to think about 2 objects on a spinning disk, one being close to the center of the disk and one being close to the outside of the disk. Angular (rotation) speed deals strictly with the angle. How long does each object take to ...



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