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6

Is it because the acceleration is too weak? It is too weak with respect to the four forces we measure. The fact that the four known forces are so much stronger means that agglomerates of particles, up to the scale of galaxies are not internally affected, they keep their structure intact, like the famous raisins in the rising bread. It is only at the ...


5

Your observations are spot on. I usually write Newton's second law this way: $\vec{a} = \vec{F}/m$. This form makes it clear that the law is a relationship between the dynamic variables force and mass, and the kinematic variable, acceleration. $F$ and $m$ describe the situation, $a$ is the result. Cause and effect, if you will. In fact, that's one ...


4

In a completely ideal world, where air resistance was not present, the bomb would continue to move forward and the same horizontal speed as the airplane it was dropped from. Gravity only acts in the perpendicular direction, thus has no effect on the horizontal component of the bomb's velocity. In reality though, air friction reduced the speed of the bomb as ...


3

The accelerometer measures the negative of gravity plus any upwards acceleration see NOTE#1 $$ acc = -(g+\ddot{x}) $$ and you want to integrate $\ddot{x}$ to get speed $v=\dot{x}$ and position $x$. So your expressions should be $$v(t)=-\int_0^t ( acc+g)\,{\rm d}t \\ x(t)=-\int_0^t \int_0^t ( acc+g)\,{\rm d}t\,{\rm d}t$$ You also know that the final ...


2

The issue here is that the initial momentum needs to come from somewhere. I suppose, if we could track the momentum of various things in space, we could "hitch a ride" from it. Like netting a comet or taking advantage of space debris striking the craft. That's a properly analogous situation to your basketball/skateboard system. However, you seem to be ...


2

There is a sense in which your suggestion is correct. The gravitational field of the Earth (above its surface) is described by the Schwarzschild metric. There are several different ways of writing this down, and the one relevant to this question is called the river model - the link is to a scientific paper, but it's quite readable even if you ignore all the ...


2

Like Ross said, you need to try to remove the overall constant part of your acceleration data. Most accelerators report gravity all of the time. So even if your phone was sitting still on a desk, it would report and acceleration of -9.8 m/s^2 "down", however down lined up with the axes of your phone. Looking at your data a bit, for the first 1/3 of your ...


2

If your accelerations are in $g$, you should multiply the second term in column C by $9.8$. The velocities will then be in m/sec and your integration to make D will be in meters. You may then want to apply any constraints you know of. If you know the velocity starts and ends at zero, you might want to add or subtract a constant acceleration to the data to ...


2

There are many devices that can measure rate of a quantity without using approximate derivative and integral. I will give some example here: Pitot tube Pitot tube is a device used to measure velocity of a body with respect to flow. This device uses Bernoulli's equation. For computing the velocity using Pitot tube, the total pressure and static pressure ...


2

Years ago, the speedometer in a car moved the needle by spinning a magnet. The physical rotation of the driveshaft turned the cable inside the assembly. The spinning cable is attached to a magnet. The needle is mounted on a disk attached to a spring which provides rotational counter-force. The spinning magnet attempts to spin the disk, but the spring ...


2

The problem with supercavitation is that it occurs at a pressure drop, not an increase. The thing that kills you on impact is the pressure in front of you - the water that cannot get out of the way fast enough. Skin drag on the body happens later - when more of the body is already submerged. Most likely you are dead by then... Let me try (with my rusty ...


2

While aliasing is a real concern for anything where sampling is involved1, it is really unlikely for this to cause a big zero-peak in frequency space, for an accelerometer on an animal (where it's unlikely that there's an oscillation precisely in sync with the sampling clock). Of course it's possible that a lot of the measured signal had nothing to do with ...


1

Without knowing more about the specific situation, the first likely culprit is aliasing. This occurs when you are sampling at a frequency $\nu_s$ that is too low compared to the largest frequency that is sizeably present in your timeseries. More specifically, Nyquist's theorem guarantees you that if the highest frequency present is $\nu_M$, then a sampling ...


1

The question is phrased in terms of dynamical concepts like force and mass, but there's a more fundamental kinematical answer that trumps these issues. If an object is moving with speed $u$, and you then apply a boost $v$, the object's new speed is not $u+v$ but rather $(u+v)/(1+uv/c^2)$. This is always less than $c$. Therefore it's not possible to ...


1

In the context of linear motion (as BMS correctly points out in a comment of a different answer), the magnitude of acceleration is a measure of how much speed you are gaining per second. The difference with the acceleration vector is that the vector form also encapsulates the direction in which this gain in speed is happening. So as an example an ...


1

Does it need to "slow down" to the new speed of light? Or does a new photon get generated? One has to keep clear in this case the difference between a photon and an electromagnetic wave. An individual photon is an elementary particle, its wavefunction is given by the quantum mechanical form of Maxwell's equations and the square of this wave function ...


1

According to the shell theorem, there's no gravitational attraction inside a spherical shell of matter. At the center of the earth, its entire mass is arranged around you in spherical shells, so the gravitational acceleration due to the earth is zero. You'd still feel the gravity of the sun and moon and other external objects, as well as a (small) ...


1

It'll be exactly the acceleration due to gravity, because no other force is acting on it. Edit answering comment: You're thinking about velocity. Acceleration isn't something an object has. Objects accelerate because there's an external force acting on them. When B is towing A, it exerts a force on A which causes A to accelerate. When B lets go, that ...


1

A better electrical analogy to Newton's second law might be inductance: $$ V = L \frac{\mathrm d i}{\mathrm d t} $$ The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not ...


1

We should think a bit more carefully what force, mass and acceleration really are: For simplicity, we consider a classical point particle. The force $\vec F$ is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, $\vec F = - \nabla V$, but it need not be. In general, it is some ...


1

In my view, an expression like V=RI, practically speaking, isn't much different. You have to be concrete with a real-world example. I can connect a resistor across a voltage source, and then I can only alter the voltage or change the resistance value to change the current; I cannot change the current directly in this configuration. Your statement that ...


1

Acceleration is dv/dt v is limited by the velocity of light. delta(t) is limited by Planck time, in the sense that physics in smaller time intervals is not known ~5.39106(32) × 10−44 s One Planck time is the time it would take a photon traveling at the speed of light to cross a distance equal to one Planck length. Theoretically, this is the smallest ...


1

There is actually such a mechanism that you're describing. The means to achieve that is through radiation pressure from light. This is called solar sail, and would achieve space exploration without having to expend proppelents for the main movement production.


1

Here the average acceleration can be understood as follows: The particle going from A to B along a half circle with speed 1m/s can be here viewed as the particle going from A to C and again to A with an initial velocity 1m/s as shown in the figure: The particle in the half circle, will move under a centripetal acceleration which is always directed towards ...


1

Is there a difference between "average acceleration" and centripetal acceleration? Yes, in fact they're almost completely unrelated. The average acceleration is defined as $$\vec a_\text{avg} = \frac{\Delta\vec v}{\Delta t}$$ It is one quantity that partially describes the motion of a particle over an extended time. In other words, average ...



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