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38

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 ...


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. ...


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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


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

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 ...


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 ...


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 ...


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 ...


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

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.


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 ...


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

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) ...


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 ...


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 ...


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 ...


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

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, ...


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

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 ...



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