# Tag Info

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A circle is defined in the following way: The ratio of the perimeter $C=2\pi r$ and the diameter $d=2r$ is constant. This constant ratio is $\pi$. Note that this definition holds only when Euclidean space i.e. flat space-time is considered. When we consider the curved space-time of general relativity, this definition doesn't hold as the space-time is no ...

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The question you quote appears to suggest that you are so far from other objects that their gravitational effects can be overlooked. If that is the case then the answers are: a. You cannot determine whether the ship has any specific speed relative to anything else. Indeed, the implicit assumption that the ship can have an absolute speed of 0.8c without any ...

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Say you have a body pivoted about a point. The velocity vector of a point located at $\boldsymbol{r}$ from the pivot is $$\boldsymbol{v} = \boldsymbol{\omega} \times \boldsymbol{r}$$ What this means is the velocity is vector field (a vector that depends on position only) and that for each location there is only one velocity vector associated with that ...

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Speed-Up / Sped-Up From Google's definition: noun an increase in speed, especially in a person's or machine's rate of working. Some examples: I sped-up (accelerated) to catch my friends in front of me. Bobby will need to speed-up (accelerate) to overtake the race leader. The rocket fired its booster engines and sped-up (accelerated) to 10 km/s

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If you need a word, coin it, define it in your writing and use it consistently. I would use hastening or quickening. For instance, an object going around in a circle at constant speed (thus angular velocity) isn't quickening, though it is constantly accelerating.

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Impulse required to start or stop something moving. Of course, there's mass involved, so you might want to take Impulse/mass

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g-force is typically used to express the magnitude only, but the words are generally used interchangeably; laymen typically referring to the magnitude only.

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Consider a variation on the Atwood machine where instead of hanging straight down, the two masses $m_1$ and $m_2$ each sit on a separate frictionless plane at an angle of $\theta_1$ and $\theta_2$ to the vertical. The planes are joined along a ridge and the string connecting the two masses runs over the ridge, again without friction. The simple Atwood ...

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I think you are on the right track. Instead of looking at a net force on the string, which would not be possible as @Marco Ocram points out, consider that the net force is acting on a system consisting of the string and masses connected to each end. Then the system, including the string, can be considered accelerating with no net force on the string itself. ...

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The arrangement you describe is impossible. The tension of the string will be 70N. Whatever was trying to restrain the end of the string with a force of 60N will be subject to a force of 70N by the string. As a result it will accelerate subject to a net force of 10N. The reaction on the string will be 70N.

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In English, it seems that: Position is a vector. Distance/length is a name of its magnitude. Velocity is a vector. Speed is a name of its magnitude. Acceleration is a name of a vector and its magnitude. Force is a name of a vector and its magnitude. Momentum is a name of a vector and its magnitude. ... Velocity/speed as well as position/length seem to be ...

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Starting from Newton's 2nd law: \begin{align} F_\text{net} &= ma\\ \therefore\quad F_\text{net} &= ma\,\frac{g}{g}\\ \therefore\quad \frac{F_\text{net}}{mg} &= \frac{a}{g} \end{align} where $g\approx 9.8$ ms$^{-2}$ is the acceleration due to (Earth's) gravity.

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The graph looks like the velocity of a ball rolling on the floor and bouncing sideways between two walls. The line is straight (I presume that is what you meant to draw), which means there is no acceleration between two bounces. The corners are sharp, which means the reversal is sudden. A perfectly sharp corner on the graph would mean that an infinitely ...

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It's hard to say for sure, since it's not clear what assumptions have gone into the creation of the graph. I would expect the upward slopes to be quite steep as the ball very rapidly changes from travelling towards the floor (negative velocity) to travelling away from the floor (positive velocity). If we assume a fairly idealized system (no energy lost on ...

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It may be a velocity-time graph, but if so, it doesn't represent anything like a bouncing ball. So describing what is happening at some point may be difficult. Under the influence of gravity alone, all objects near the earth's surface have a near-constant acceleration. On a velocity-time graph, a constant acceleration would be a constant slope. So a ...

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If the question says in next part that electron accelerate. It mean the initial velocity was zero, but acceleration was non zero. If it doesn't, then both are zero.

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When pondering relativistic effects it can help to consider them from a reciprocal perspective. Suppose I am on the ship, and my speed has reached 0.9c, say, relative to Earth. At that point I cut my engines and coast past you on Earth. In my frame I am at rest and it is you who are racing by at 0.9c. You get into your car and accelerate down a road away ...

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Being at rest refers to zero velocity. The moment the electric field is turned on, at t=0, say, the particle will feel a force and hence an acceleration

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In the rocket, it seems that the amount of energy per unit time stays constant. When the astronauts look outside they see al the other objects in the universe move faster and faster (it seems these objects are in free fall in a homogeneous gravitational field). This means that the astronauts see the time on these objects move slower and slower. So say, for ...

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If you know how much fuel (mass) you need to travel from the earth to the next object , you can answer the question how much energy you need , remember that energy is equivalent to mass. I found the answer in this dokument http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html $$\frac{m_F}{m_L}=\exp\left({\frac{a\,T}{c}}\right)-1\tag ... 1 But from the astronaut's reference frame: the spaceship is simply accelerating at 10m/s2 and so the force on the spaceship is constant. Then why would we need huge amounts of energy to accelerate the spaceship? I think the astronaut interprets the rocket's need for a huge amount of energy in another way. He observes that as he recedes from the earth, more ... 0 From the rocket's frame of reference, the rocket is at rest and the Earth is travelling faster and faster, approaching c. In both frames of reference, the relative velocity approaches c so the energy needed diverges. I'm not sure whether this answers your question. 1 Supposing that the propellant for the bullet is not affected by the different environment, the bullet will accelerate slightly faster due to the (much) thinner atmosphere. There is no other difference up to the gun's muzzle. Beyond that, the bullet will decelerate slower, due to both lower gravitational acceleration and thin atmosphere. The maximum height ... 2 The dot product \mathbf{v}\cdot\mathbf{a}, using \mathbf{F}=m\mathbf{a}, is the power$$P = \mathbf{F}\cdot\mathbf{v}$$divided by the mass. Power is the rate of change of kinetic energy mv^2/2, so your dot product is the rate of change of the kinetic energy per unit mass:$$\mathbf{v}\cdot\mathbf{a} = \frac{d}{dt}\left(\frac12 v^2\right).$$That ... 2 When you are in free fall you are subjected to the force of gravity. But you do not "feel" or "sense" the force. In other words you are not subject to contact forces. In practice g forces are surface-contact forces between objects expressed in multiples or fractions of the force of gravity on your mass. When you stand on the ground you experience a contact ... 1 It's definitely measuring force as opposed to acceleration. For example, in freefall, we say there are 0 g's acting on the body; but it's accelerating. Conversely, a person standing on Earth's surface has no vertical acceleration, only a vertical force equal to 1g; yet we say it experiences 1 g in the vertical direction. G-force is a measure of ... 2 The third law asserts that whenever two objects are exerting a force upon each other, such as gravity, the two objects are both accelerating towards the common center of mass. (And this of course generalizes to any number of mutually force exerting objects.) In other words, the common center of mass of the objects involved will remain in inertial motion. ... 0 One way to think of it is that Earth and Apple are accelerated in reference to their common center of gravity. 3 Apple's and Earth's accelerations are with respect to which observer (or which reference frame)? In any inertial frame, you will measure the same acceleration of the Earth and the same acceleration of the apple. One example is the one in which the Earth is initially at rest (neglecting the Earth's motion about the sun, etc). Another could be the frame in ... 1 Now when the system is accelerating at a rate greater than 𝑔, what would it feel like? Would he feel heavier? Or less heavy? I am sure he would not feel weightless. This is my question. Thanks It seems to me that the answers to your questions all depend on how one defines the "feeling of weight". There seems to be general agreement that feeling "... 2 If you were being accelerated downward at 2g via a rope tied around your legs, you would feel the same as if you were hanging upside down with no acceleration. When in free fall, you would accelerate downward at g. The rope would be applying a force to your legs to accelerate you at an extra g. This force would be the same as if you were hanging upside ... 1 Not sure I completely understand the scenario you're describing or maybe some "upwards"s and "downwards"s are swapped. But the key determiner of what things feel like is relative acceleration of different parts of your body. If you're in freefall your arms, legs, kidneys, everything, are all being accelerated together at g so you feel weightless. If you were ... 5 The shape of the track is identical to the graph of the marble's potential energy vs. distance. Flipping the graph upside down shows you a graph of the marble's kinetic energy by distance, because a loss of PE means a gain in KE. If you put the two kinetic energy graphs on top of one another, you can see that graph Y has higher kinetic energy than graph X at ... -1 You are right in remarking that they will have the same velocity at the end of the track. But Y will start losing potential energy, and converting it into kinetic energy, much earlier than X. So it will soon start covering a lot of ground, while X moves much more slowly. So it will reach the end of the track much earlier. Sure, towards the end, it will not ... 1 You are correct, and this is the equivalence principle. As per GR, the inertial and gravitational mass are equivalent. In the theory of general relativity, the equivalence principle is the equivalence of gravitational and inertial mass, and Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body ... 1 No, you cannot without further assumptions and information. You can't even calculate the speed. You can only calculate the average speed and by the first theorem of calculus that the body would have this speed at least at one time frame, but can have different speeds otherwise. 1 To clarify from the comments: You can’t. You require information about how the journey progressed in terms of its velocity Whilst this is a technicality, it is entirely correct - you would need to know the initial velocity or the final velocity in order to truly answer this question. However, it seems implicit to take the initial velocity to be zero ... 0 I don't see it any more paradoxical then hibernating yourself until the calculation is complete. If you have an event A in spacetime with coordinates (t,r) and another event B with coordinates (t',r'), then the amount of time passed for an observer going from A to B would depend, in general, on his trajectory. You don't even need gravity for this, special ... 0 I do not believe your statement to be true, as what you're measuring now is not the absolute acceleration of the train but rather the acceleration of the train relative to the ball. 0 Would it be possible for you to draw a free body diagram of your system? Realistically, the best way to solve this problem would be to integrate the motor force output as a function of distance. This will give you the work done on the mass by the motor. Does your servo motor control the speed with proportional control? Or does it take a short cut and control ... -1 In my personal opinion, second law is more or less enough, as other two laws can be derived from it. First law:$$ \sum\vec{F} = m\frac{\textrm{d}\vec{v}}{\textrm{d}t} = 0 ,\\ \Rightarrow \vec{v}=\textrm{const} $$Third law$$ \vec{F}_{A\to B} \neq 0, \vec{v_{_B}}=\textrm{const},\\ \Rightarrow \vec{F}_{A\to B} + \sum_i\vec{F_i} = 0 , \\ \Rightarrow \...

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The first law is a statement about the "natural" state of an object. It does not make reference to the specific form of the second law. But keep in mind that the word "force" does appear in the statement (..., unless...). So you question is more about linguistics and mapping English into pure logic than it is about physics. The second law is not needed ...

2

If you examine the equation $a = v \frac{dv}{ds}$ when $v \rightarrow 0$, then it turns out that, $\frac {dv}{ds} \rightarrow -\infty$. So when $v$ is exactly $0$, the acceleration takes the indeterminate form of $a=(0)(-\infty)$. Let me explain why. As the velocity tends to $0$, the small change in the displacement i.e. $ds$ also tends to zero (it gets ...

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If you have a parabolic trajectory $$s = \frac{1}{2} g t^2$$ with velocity $$v = \frac{ds}{dt} = gt$$ then note $$v = \sqrt{2gs}$$ and $$\frac{dv}{ds} = \frac{\sqrt{g}}{\sqrt{2s}}.$$ Note that the function $1/\sqrt{s}$ is infinite at $s = 0$. So the equation $$a = v \frac{dv}{ds}$$ is true at $t = 0$, A.K.A. $s = 0$, when $v = 0$ because $\frac{dv}... 2 Since you don't mention the medium that the object is falling in, one could assume the medium is air, but it also could be falling in a liquid. In either case, since air can be (is) treated as a fluid, the object is affected by drag, which is a type of friction or force acting opposite to the relative motion of the object and is proportional to the square of ... 2 Let's assume that the drag force is given as $$F_{drag}=-bv$$ then the net acceleration is given as $$a=g-\frac{bv}{m}$$ now differentiate this equation with respect to time then,$$\frac{da}{dt}=-\frac{ba}{m}$$ Now it is clear from this equation that the slope is negative and the curve falls according to the given equation:- $$a=exp(\frac{-bt}{m})+constant$$... 2 In an ideal scenario, yes it would be straight, but in this case, you have to take into account drag forces present. Drag force$F_D=C_DA\frac{\rho v^2}{2}=kv^2$, where$k$is some constant. As such, for an object vertical falling motion, velocity increases, and that means drag force increases at an increasing rate. (Consider a$y=x^2$graph). Since ... 2 Suppose that the drag force is proportional to the speed of the falling object then the equation of motion of the falling object is$ma = mg -kv \Rightarrow \dfrac {da}{dt} = - \dfrac k m a$. So the slope of an acceleration against time graph is negative and gets less negative as the acceleration decreases. A similar result can be obtained if the ... 0 In Special Relativity, it is assumed that only relative velocity has an effect on the ideal-clock rate, as read from another inertial observer. It is implicitely postulated that the clock's acceleration has no effect, or else you would need some universal acceleration to compare, like there's an universal constant$c\$ to compare velocities. On dimensional ...

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