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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 ...


1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.


1

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2} $ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2} $ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...


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


1

There are 4 standard kinematic equations from Newtonian mechanics, and you need what is usually considered to be the fourth equation. $\mathrm{(final\ velocity)^2 = (initial\ velocity)^2 + 2 \times acceleration \times distance}$ You know the initial velocity, final velocity, and distance. Solve for acceleration.


0

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

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

So indeed, the torque $\vec \tau = \vec r \times \vec F$ slows this thing down, and assuming that the force is never radial but only tangential, this is just $\tau = r ~ F$ where r is the radius of the circle and $F$ is the magnitude of the braking force. Newton's law $\frac{d\vec p}{dt} = \sum_i \vec F_i$ becomes for angular problems $\frac{d\vec L}{dt} = ...


1

I find that sometimes intuition works better with more extreme examples. Let's change the problem up a bit. Instead of some itty-bitty difference in acceleration (a vs 2a), lets choose a big acceleration for a short period of time. In this modified example, both A and B are going to be fired out of a cannon. The firing is going to take just 0.1s, but ...


0

I'm assuming you used the following kinematic equation in your solution: $$d = x_0 + v_{0}t + \frac{1}{2}a_{0}t^{2}$$ Remember that after the initial acceleration, the particles will have covered some distance and gained some velocity. Hence, in the second step where you have a new acceleration, you no longer have a zero initial position and velocity. I got ...


14

Let's draw a graph of velocity against time for the two particles $A$ and $B$. For convenience I've made the total time $2t$: The red line shows the velocity for particle $A$ while the green line shows the velocity for particle $B$. When we draw a velocity:time graph the distance travelled is the area under the line. More precisely it is the integral of ...


0

I think your answer is incorrect. First the intuitive argument: Assume you had stopped both particles after half of the time. It is obvious that in this case they had covered the same distance. At first B is first, but in the second half A would catch up. If you don't stop the particles before they continue with a different acceleration also the velocity ...


0

The same reasoning that goes for racing cars: Acceleration is important if you are slow, because then the relative change in velocity over time will be high edit: (also the velocity is the accumulated acceleration, meaning that the earlier you have high accelaration the better because you will "keep" that to the end)


0

It's true that a 600 hp engine does not necessarily have to be generating that amount of power in order to get from point A to point B. But even though you manage the throttles so that both cars accelerate at the same rate, the larger engine has greater internal cylinder displacement, and therefore consumes more air and fuel per engine revolution than the ...


0

Unfortunately, I see a premise here that is difficult to prove true or false. Mainly because it is almost impossible to use a 600hp engine in a way similar to an 80hp engine and compare them directly. Any car that has a 600hp engine will have other components to support it. Compared with an 80hp car, that might be a larger, more robust transmission, ...


0

I had to go back to this paradox again because something came up that proved to be even more critical in it's resolution. Forget the explanation thus far given - there is no paradox here - because there is no experiment - the scenario for the experiment makes one very flawed assumption - that is there are three separate systems, when in fact there's only ...


0

It is not as sophisticated as it looks like first take the speed of the car at the end of the slope $v_1$ so the overall energy of the object would be $E = \frac{1}{2}mv_1^2$. Assuming that at the start the car has no kinetic energy. Now we simply remove the gravitational potential from it, so: $W = \frac{1}{2}mv^2 - mgh$ Then to find the velocity the ...


0

This gets a little complicated, at least for high precision. As a really simple first version, you can assume no frictional losses. No aerodynamic drag, either. Then, starting at the top of the hill, measure the car's speed and calculate its kinetic energy. At the bottom of the hill (with a height h), measure the speed again and calculate the kinetic ...


3

The technical term for this force is the radiation reaction force and it is electromagnetic in nature. Maxwell's equations do not describe this phenomenon simply because they're not meant to; it's like asking the heat equation to describe the chemical reactions that happen in a fire. Maxwell's equations describe the electric and magnetic fields generated ...


1

Let's say their acceleration and velocities are not equal. Then we can define a relative acceleration $a_r$ and velocity $v_r$ between the two. After a time $t$, the distance between them would be $$x= \frac {1}{2} a_r t^2 = v_r t$$ Since $x$ might not be equal to zero as $t$ tends to greater values, the bodies will stop being in contact.


0

You asked two simple questions - I will give two simple answers. I don't know if I can use uniform circular motion equation since v is not constant At the very instant that the curve starts, the velocity is given by $\sqrt{2gh}$ - and for that first instance it is constant. So yes, you can use uniform circular motion Where is the g-force directed ...


0

I don't know if I can use uniform circular motion equation since v is not constant The equation for centripetal force is independent of whether the motion is uniformly circular or not. However, irrespective of the radius of the track, the velocity at that point, and the weight of the roller-coaster, or whether the equation for centripetal force is ...


0

You're good. Yeah, you can pretty much assume that it's a constant velocity, as long as $h \gg r.$ As I recall, the expressions involved are extremely simple as long as you don't try to figure out exactly what's happening in time: actually solving the Euler-Lagrange equations gives you some sort of $\int d\theta / \sqrt{a + b \sin\theta} = t$ equation for ...


2

There are two accelerations involved: The gravitational acceleration $g$ that points down, and the centripetal acceleration $a_r = \frac{v^2}{r}$ that points along the radius vector of the curve. The component of the gravitational acceleration that is tangential to the curve does not contribute to the g-force as it accelerates the cart and us in this ...


2

SECTION A : Free fall of roller coaster into circular motion (kinetics) Suppose that the roller coaster, called from now on "particle", is at rest at point A ($\:\upsilon_{A}=0\:$) and starts free falling till point B where it starts its circular motion. Well-known is that at B the speed is $\:\upsilon_{B}=\sqrt{2gh}\:$ under the assumption of no energy ...


1

\begin{equation} \mathbf{a}=\dfrac{\Delta \mathbf{v}}{\Delta t} \tag{01} \end{equation} \begin{equation} \vert\mathbf{a}\vert=\dfrac{\vert \Delta \mathbf{v} \vert}{\vert\Delta t\vert}=\dfrac{\upsilon \vert\Delta \phi\vert}{\vert\Delta t\vert}=\dfrac{\upsilon \vert\Delta s\vert}{r \vert\Delta t\vert}=\dfrac{\upsilon \cdot \upsilon \vert\Delta t\vert}{r ...


1

In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant. So, when the object has to transverse a curve trajectory, the main ...


4

Imagine a object steadily traversing a circle of radius $r$ centered on the origin. It's position can be represented by a vector of constant length that changes angle. The total distance covered in one cycle is $2\pi r$. This is also the accumulated amount by which position has changed.. Now consider the velocity vector of this object: it can also be ...


2

The roller coaster falls a height $h$ before entering the loop (the path of the RC will look like a J. The straight bit of the J is of length $h$), and I assume it started with zero velocity. Within the loop it's total kinetic energy will equal the total potential energy lost, which is $$ E_\text{kin}=mg(h+r\sin\theta). $$ Here $\theta$ is defined such that ...


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 ...


0

You ask: If this is the case, then can we say that "acceleration" or changed frames of reference are not required to resolve the twin paradox? and you say: they can be viewed as they are both in inertial frames of reference from each other's perspective The answer is that acceleration is required to resolve the twin paradox and the two observers ...


2

This is a good (and notoriously difficult) question. I'm going to follow the explanation given by Crispino, Higuchi, and Matsas in their review 0710.5373, but you should be aware there are different answers out there and also there is no (uncontroversial) experimental test of this effect. Having said all of that, the basic picture I have (and is given in ...


1

So, just to recap the Twin Paradox, it is a variation of the paradoxes of relative motion of reference frames Alice and Bob, created by the statement "Alice sees Bob's clocks moving slowly, but Bob also sees Alice's clocks moving slowly." The simplest such paradox, in my opinion, is "what if Alice calls Bob up and they talk on the phone? One of them surely ...


1

So you can use the uniform circular motion equations, but you must use different values for $v$ since the v is not constant due to acceleration from gravity. I will show the g-force experienced at the beginning of the loop, the side of the loop (so a height $r$ above the ground), and the top of the loop (a height $2r$ above the ground). For the cart ...


0

The distance between successive lines is 1,3,5... making their absolute position 1,4 (1+3), 9 (1+3+5)... This follows from double integration of the equation of motion with a constant force: $$v = \int \frac{F}{m} dt = a\; t\\ x = \int v\; dt = \int a\; t\; dt = \frac12 a t^2$$ So the position increases quadratically with time for constant acceleration ...


0

What is the acceleration of the space ship, if you turn off the earth's gravitational force? The acceleration of the ship is $$ a = \frac{F + G}{m} $$ where $F$ is thrust of the engine (force), $m$ is mass of the rocket and $G=-mg$ is gravity force. If thrust is such as to exactly cancel gravity, it is $F=mg$ so the acceleration with no gravity would ...


0

As @Floris said, rocket motors provide thrust (reaction force), not power. The thrust is just the mass times velocity of exhaust, per second. The power is the mass times velocity squared, per second (over two). Example, shooting a rifle bullet has low reaction force, but high energy. Shooting a bowling ball with the same reaction force takes a lot less ...


2

If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy. And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from $$P\cdot t = \frac12 m v^2\\ v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$ If you ...


1

You conclude that constant acceleration is appropriate for one of a few reasons: You read in the question text that you should treat it that way. If (1) does not apply, you read in the question text that some physical situation obtains and you know or suspect that this situation is usually well represented by a constant acceleration. If neither (1) nor (2) ...



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