28

The physics answer is, an accelerometer will detect all accelerations relative to an inertial frame. If you're in free fall being accelerated by a gravitational field, the answer is actually no, because a frame in free fall is inertial, even though for most purposes it's more useful to treat it as an accelerating frame. So to your questions in order, 1a Yes, ...


10

No. Since every particle in your body accelerates at the same rate, there are no forces between them and no compression or tension. This is identical to being in freefall in a gravitational field. An astronaut in orbit, for example, is affected by the force of gravity at all times, but doesn't "feel" anything because they are in free fall. Half an ...


6

Newton's second law $$F = \frac{dp}{dt}$$ only holds in inertial frames. While $v_r$ is defined as the velocity of the released gas relative to the rocket, it is also equal to the change in velocity $\Delta v$ of the fuel as observed from any inertial frames. This is a consequence of Galilean relativity, which holds at low velocities. Thus the formula $$F = \...


6

Yes. Via the equivalence principle if you accelerate with respect to a locally inertial reference frame, then you will not be able to distinguish the effects of acceleration from a gravitational field. So you will effectively feel a gravitational force acting in the direction of acceleration, with a magnitude proportional to the magnitude of the acceleration....


5

For a motion with constant $||\dot x||$ in $n$ dimensions the velocity vector travels on an $(n-1)$-sphere and its "velocity", $\ddot x\,,$ (what we call acceleration) has a constant magnitude (not a constant direction). In 3D I can imagine many trajectories on the $2$-sphere that have constant velocity and acceleration and are not helical. Helical ...


4

For a particle executing SHM, its displacement $x$ from its mean position at some time $t$ is given as $$ x(t) = A\cos (\omega t + \phi), $$ where $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is the initial phase offset. Its velocity is then given as (by differentiating the above with respect to time) $$ v(t) = -A\omega\sin (\omega t + \...


4

You need to know over what period $t$ in seconds that the speed went from $0$ to Mach 10. You also need to convert the speed equivalent of Mach 10 into $ms^{-1}$, and call this speed $v$. From here you would calculate the acceleration$^1$ $$a=\frac{v}{t}$$ and noting that $9.8ms^{-2}=1g$, then $$\frac{a}{9.8}=Xg$$ where $X$ is the g-value you are looking ...


3

According to my teacher it is possible to keep the block at rest or even accelerate it in the upward direction along the the inclined plane of wedge. I think 🤔 you are confusing the words of your teacher. What your teacher is actually trying to say is that it is possible to keep the block at rest or even accelerate it in the upward direction with respect ...


3

For an object to appear "flashed", I am guessing it would move at the same speed as say an airplane propeller, since they also appear "flashed". Such propellers have on average a rotational speed of $\approx 2500rpm \approx 43rps$ and the length of their blades are on average $1.5m$. This would give each blade (at the tip of it) a speed ...


3

This derivation actually uses $v_{rel}$ because the author found more convenient to write the equations this way. But this does not mean that the system is being described from the rocket reference frame.


2

If you think of force in terms of potential: $$F=-dU/ds$$ Where U is the potential, and s the displacement. If the potentials are linear, that is, if we can calculate the total potential by adding the various partial potentials that our object is experiencing, then the overall force will also be the linear sum of partial forces. With that context out of the ...


2

Shouldn't the elevator experience the same force but downwards? Yes. On the other hand, if you want to know the effect of the equal and opposite forces on the person and elevator individually, you need to apply Newton's second law $F_{net}=ma$ to the person and elevator individually. Hope this helps.


2

Yes, you said it. The accelerated observer knows that he/she is in presence of gravity or, at least, moving with some acceleration. Furthermore, another important point to take into account is that the non accelerated observer does not see any curvature of the light path, in this case the trajectory of light is a straight line going downwards (see picture ...


2

If $\vec F$ is the only force acting on the object then $\vec a$ must be parallel to $\vec F$, so $\theta =0$, $\cos \theta=1$, and $\vec F=m\vec a$. On the other hand, if $\theta$ is not zero then there must be one or more other forces acting on the object so that the net force $\vec F_{net}$ is parallel to $\vec a$, and $\vec F_{net}=m \vec a$. In this ...


2

As physicists we usually don't get lost into existential questions about it. If some case would arise where this thing is not differentiable, it would likely be so bizarre that it would be obvious that the classical approach has some problems. But I can guess you are more a mathematician at heart, and I will try to bring some elements along those lines In ...


2

Acceleration and speed a fundamentally different things. Speed is the change in position over time. Acceleration is the change in speed over time. Relativity limits speed not acceleration. One hint that this was based on confusion were the units. Speed would have been $m/s$. You correctly showed units of acceleration in $m/s^2$.


2

And yet, an accelerometer made up of only particles with the same mass-to-charge ratio wouldn't read anything, because all its constituents would accelerate at the same rate. Does this mean that proper acceleration isn't always reflective of what acceleration is "felt" by a system? In Newtonian mechanics this indeed would be true: it is ...


2

Consider a particle on a table. Let it have an initial velocity towards the edge of the table. At some point, the particle leaves the table. At this point, the acceleration is undefined. The vertical component of acceleration experiences a jump from 0 to -1 g. This is because the normal force from the table suddenly vanishes. Discontinuity in normal force ...


2

The force F, which you have drawn horizontally, could actually have been drawn in a number of different directions, including the direction opposing the starting acceleration. The tension in the string changes and the required size of F changes as the direction of F is adjusted. If F points in the direction opposing the initial acceleration then the size of ...


2

The fly is flying in the air in your car. The air in your car is moving at the same speed as your car, more or less. When you brake or accelerate the change in the speed of your car is small compared with the mean speed of the molecules of the atmosphere in your car, so there is no gross tidal movement of the air as a consequence. The fly therefore moves ...


2

Q1. The statement is not true. In 3D, if you impose the restrictions $$\|\dot{x}\| = 1 \,\,\, \text{ and } \,\,\, \|\ddot{x}\| = \kappa_0$$ then there are infinitely many curves with this property that are very far from being helixes or circles and having differing geometric properties. This follows for example from the canonical Frenet-Serret frame of a ...


2

No. Let's consider the case where you're out floating in free space, and suddenly a large moon pops into existence in your vicinity, which you start accelerating toward due to gravity. Before the moon appears, you're obviously in an inertial frame. After the moon appears, you're in free-fall toward it, which is also an inertial frame. See the equivalence ...


2

Remember that Pascal's law tells us that a fluid exerts its pressure equally in all directions, so if we draw all the forces that the mercury exerts on the container is in they will look something like this: So while you are quite correct that the mercury is exerting a force to the right on the cork, it is also exerting a force to the left on the opposite ...


2

First imagine what would happen if the system were in outer space without gravity. The bob would be floating around, and when the car accelerates to the right, the top of the rope would be pulled to the right, drawing more rope after it until the rope were pulled taut horizontally with the bob trailing along at its left end. That's what you're seeing here, ...


1

In Brownian motion particles bounce around randomly because of jostling from the surrounding medium. Mathematically this can be modelled as a random process that does not have any derivatives. Statistical mechanics generalizes Newtonian mechanics for this kind of motion with random processes. In practice particles move in a straight line until they get ...


1

In the laboratory frame, The equation of motion given by $$N\cos\theta-mg=0$$ $$N\sin\theta =ma$$ In Accelerating frame, The equation of motion given by $$N\cos\theta-mg=0$$ $$N\sin\theta-ma=0$$ Both are equivalent.


1

The sign of acceleration is the direction of the acceleration. If positive the acceleration is in the positive direction. If negative then the acceleration is in the negative direction. There are 4 cases: If the object is moving in the positive direction has a negative acceleration it is slowing down. Likewise if an object is moving in the negative ...


1

No, that's not correct. $\vec a$ is not directed horizontally, the prefactor $\cos\theta$ doesn't change this fact as it only can change the magnitude, not the direction.


1

Newton's second law states that the direction of acceleration is also the same as the direction of the net force. The term $macos(\theta)$ is therefore not to be considered the direction of acceleration due to the force $F$. So $a$ is the correct acceleration.


1

What you are looking for is the magnitude, the value, of the force in a particular direction. So skip the vector arrow symbols. You are completely right that $ma\cos(\theta)$ will give you the correct horisontal component you are looking for. In this particular case, the original force is already pointing in the direction of this component, though, so $\cos(\...


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