# Tag Info

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Is this a valid frame? I mean valid in the sense that in Newtonian mechanics acceleration frames are not valid because they experience ghost forces. You do experience a "ghost" force in Rindler coordinates; there will appear to be a force opposite the direction of acceleration (much like when a car accelerates, objects in the car appear to have an ...

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I liked the above answer where we distinguish between flat spacetimes and curved ones, I feel like that gives you a good answer of questions (1) and (2). To answer question (3) I would question this premise that special relativity doesn't “do” acceleration, because I believe that that's the only thing it does. So a remarkable thing to my mind is that the ...

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The solution with negative $t$ has a physical meaning. It means that as well as the time in the future when the vehicles meet, there is also another time in the past (assuming all conditions stay the same) when the vehicles met. In other words, if we run time backwards starting from $t=0$ then the vehicles will meet at some negative time. Suppose a ball is ...

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You are defining the acceleration of A to start at t=0. You then end up with a quadratic equation involving the acceleration of A. It therefore would be incorrect to use the negative t solution as before t=0 there was no acceleration so the equation couldn't apply to t<0. Using the solution in my comment above, using the t=-3.7 sec solution we get x~-60 m ...

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As Butane mentioned, the negative solution has a physical meaning. Notice that car B always had a speed of 30, and A is accelerating but started with a smaller speed, and only reaches 20 at t=0. So the two cars would have meet in the past, assuming their accelerations have not changed.

-1

Basically because the quadratic formula doesn't know the concept of time, we do and when we employ the formula to get solution it does its job and we get two solutions to the equation. Because we have defined time such that it can't be negative, we ignore the negative result. Say hypothetically time could be imagined to be negative, and try to retrace the ...

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Surely what you have done is correct and when it comes to the elaboration of the steps that you take, solution to quadratic equation always yields 2 solutions (not necessarily different) either real or complex and in your physical system to determine which of the result is correct you need to check the physical boundaries.In your case you want to find time ...

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This can only be answered by pointing out the difference between special and general relativity. There is the historically motivated definition, which is still in widespread use in popular and semi-popular accounts, according to which special relativity only deals with inertial frames and coordinates (1a), while general relativity deals with accelerated ...

1

Your reasoning is all correct. Gravity is not viewed as a force in general relativity. Gravity is instead caused by the curving of space and time. All objects naturally travel on straight lines through space and time (when no force is applied), but those "straight" lines become curved geodesics when space-time is curved.

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In a given time interval $T = [t_1,t_2]$, the distance travelled during that interval is given by $\Delta x_T = \int_{t_1}^{t_2} v(t) \mathrm{d}t = \bar v_T \cdot(t_2-t_1)$, where $\bar v$ is the average velocity over the time interval. If the acceleration is constant, then $\bar v_T = \frac{v(t_1) + v(t_2)}{2} = v(\frac{t_1 + t_2}{2})$. Thus, the distance ...

1

There's a sense in which mass does increase with speed. However, to increase something's speed, you have to put energy into it. That energy also has mass, it also figures into center-of-mass calculations, and it's exactly equal to the mass gained by the object when it speeds up. So no scheme of this sort can work.

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HINT: You need to collapse the 2 equation for the distance that is : $x = \frac{1}{2}at^{2}+x_{0}$ knowing that $x_{firstVehicule} = x_{secondVehicule}$ when they intersect.

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Could you solve the problem if one object was stationary and the second object was accelerating towards the first at $0.5$ m/s^2 ? Can you see how this might help you solve the original problem ?

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The simplest method is to solve this problem from the frame of one of one of the particles. Alternatively you can use the displacement equations of particles.

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How do we define $G$s? The misnamed g-force, or "$G$s", on an object is the component of acceleration that result from every external force except gravity, scaled by 9.80665 m/s^2. I wrote "misnamed" because $G$s have units of acceleration, not force. I wrote "component" because just as one can break down the net force acting ...

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Note: G and g are different. g and g are also different. How G and g are different G is the (universal) gravitational constant. g, however, is the gravitational acceleration. G is a constant. It's used in the formula for the Law of Universal Gravitation (look it up if you need), which gives the gravitational force between any two objects, even you and a star ...

1

When forces act on a body, there are three main effects: the body may accelerate as a whole the body may be squeezed or stretched or otherwise distorted as a whole the body may be caused to rotate internal parts of the body may be forced to rearrange themselves When we want to measure the level of discomfort experienced by a pilot in an aircraft, or ...

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From your question, I gather that you have found the velocity $V(t)$ of the box that guarantees that, in the frame of reference of the box, bob is immobile. You have found that the acceleration $\mathrm{d}V/\mathrm{d}t$ was a nonzero constant. Thus, in the frame of reference of ground and along $x$ direction, the bob accelerates also by $\mathrm{d}V/\mathrm{... 0 well we can think of this very easily if we switch our focus towards the gravitational field ($\vec{E_{g}}$) instead of the force ; let us also say that one of our particles is fixed ; then we can see that the field due to the fixed mass is not constant in space and increases as the radial distance decreases ; hence the acceleration of the moving body ... 0 Consider any sphere which is defined by a distance$R$from the centre of any of the two masses$m$and$m'$. At all points on these spheres, the magnitude of the acceleration will be uniform. However, if you consider$R$to be decreasing then the magnitude of the acceleration will increase for all$r$<$R'$. For any fixed$R$you will have constant ... 0 The easiest way to understand this is using relative motion and pseudo forces. Assume at an instant you have a velocity of$u$, and the lift has a velocity of$v$, analyzing it in the lift frame, your velocity will be$u-v$. If the lift is accelerating with acceleration$a$, you'll feel a pseudo force equal to$ma$(where$m$is your mass) in the direction ... 1 The straight line from the origin crossing all hyperbolas is the line of simultaneity of an inertial frame with a given velocity. If tangents are made at any hyperbola at the crossing point with this line, they are parallel. That is why it is not necessary a Lorentz transformation. All that points of spacetime (along the straight line) are in the same (... 2 Accelerated coordinates are just a spacetime analog of polar coordinates. The curves of constant$x$in accelerating coordinates have different curvatures (different accelerations) for the same reason the circles of constant$r$in polar coordinates do. Consider a bend in a road of constant width, as seen in this road sign: Note that the sides of the road ... 0 Because gravity and acceleration are the same thing, does that mean that the time dilation when near a gravitating mass is the same time dilation as when you are at a very high speed? Are these two things connected/the same? No. In fact concerning GPS satellites SR effects and gravitational effects can compete each other. During the fist launches of GPS ... 0 Because gravity and acceleration are the same thing, does that mean that the time dilation when near a gravitating mass is the same time dilation as when you are at a very high speed? Are these two things connected/the same? The principle of equivalence relates an accelerated frame and a gravitational field. Of course, as you said, it is only valid locally ... 0 An accelerometer works by measuring the movement of a test mass - or the force required to prevent a test mass from moving. The measured acceleration is in the opposite direction to the movement of the test mass. The raw data from an accelerometer at rest relative to the surface of the earth will show a$1$g acceleration upwards. However, accelerometers are ... 0 If you want to travel between two points separated by a distance$d$, starting and ending at rest, and there is no limit on the maximum speed, then the shortest time is achieved by accelerating at maximum acceleration for the first half of the distance, and then decelerating for the second half. The time taken to cover a distance$\frac d 2$starting from ... 1 Velocity$\vec v(\vec r,t)$is the integral of acceleration$\vec a(\vec r, t)$with respect to time, and position$\vec r(t)$is the integral of velocity with respect to time. So if the object starts at the origin with velocity zero then at time$t\vec v(\vec r(t) ,t) = \int_0^t \vec a(\vec r(t), t) dt \\ \vec r(t) = \int_0^t \vec v(\vec r(t), t) dt$... 2 Newton's Third Law says that $$\vec F_{ij} = -\vec F_{ji}$$ where$\vec F_{ij}$is the force on$j$due to$i$. Newton's Second Law says that, if this is the only force acting on that object $$\vec F_{ij} = m_j \vec a_{ij}$$ where$m_j$is the mass of$j$and$a_{ij}$is the acceleration of$j$resulting from interaction with$i$. This gives us $$\lvert \vec ... 0 Since the wheels of the engine are driving the engine forward, it might be reasonable to assume that the engine is not subject to a backward force of friction. The 5000 N of friction would be divided equally among the 5 wagons. You need the force to accelerate 4 wagons. (Including the 4000 N of friction.) 0 A non-tangential force which has a tangential component will cause a change in angular velocity. Also an inward radial force which is larger than that required to maintain centripetal acceleration will reduce the radius, and conservation of angular momentum will cause an increase in angular velocity. 0 An alternative approach (to avoid being confused by gravity) is to use the principle of equivalence of relativity, and suppose that just after the cut, everything is in the outer space without gravity. The bottom mass is under a upwards force of mg due to the deflection of the spring, and an acceleration \frac{mg}{m} = g. The same force is applied by the ... 1 If you take a free fall at the equator and with Newton gravitation law you obtain this equation:$$m\,g=\frac{G\,m\,M}{(R+h)^2}$$because the free fall height h is much smaller then the earth radius R we can take the Taylor expansion and obtain:$$g=\frac{G\,M}{R^2}-\frac{2\,M\,G}{R^3}\,h+(0\,h^3)=g_r-g_h$$thus as long as the mass falling (free fall) g ... 0 Before the top spring is cut, the tensions in the bottom spring and the string are Mg and 3Mg respectively. We can show this by considering the equilibrium of the lower two masses. When it comes to considering accelerations when the top spring is cut, I found it instructive to think about what would happen if the string were actually another spring. This ... 2 We know that the tension in spring 2 is determined by its extension, which does not change immediately after spring 1 is cut. So the tension in spring 2 immediately after spring 1 is cut is Mg - which also means that the instantaneous acceleration of the lower mass is zero. If we assume that the tension in the string is T then the net downwards ... 3 Why an electron bound to a nucleus does not emit photons when accelerated? Because bound electrons have no individual identity, no orbits, otherwise they would neutralize on the nucleus, but orbitals which are probability loci. The whole atom/molecule is the quantum mechanical entity, i.e. obeys the rules of quantum mechanics. If an atom molecule has an ... 0 When they fall, it need not be that the string is taut always, The string only becomes taut if both the bodies are to pull away from each other. In this free fall case that you have shown, the first case and second case are practically equivalent. 1 The two masses will have the same acceleration. There is a thought experiment(due to Gallileo, I guess) to show that objects of different masses fall with same acceleration. It goes as follows: Suppose you have a heavier object and a lighter object. You tie them up (like u have done in the picture), so then if the lighter objects falls slowly(less ... 0 I think mutual interaction means force on the particles in the system due to their internal properties like gravitational force due to mass , electromagnetic forces due to charge and not due to an external source. If considering only gravitational force then , acceleration is induced not due to push . They are actually pulling each other and hence their ... 0 Because gravity and acceleration are the same thing, does that mean that the time dilation when near a gravitating mass is the same time dilation as when you are at a very high speed? In some ways, yes. However, remember that what causes time dilation near a g-mass is the difference in g-potentials rather than g-acceleration. g-potential per unit ... 2 It is the principle of equivalence that states that a uniform gravitational field and a uniformly accelerated frame of reference are equivalent. It is incorrect to say that gravitational time dilation and relative motion time dilation are the same thing since they are both results of different phenomena. One is a result of relative motion and the other is a ... 0 Acceleration and Force are two different concepts. The first has S.I. Units m/s^2 and the second Newtons. Accelerometers measure acceleration. Force sensors measure forces. A static force (such as the force of gravity) is a force with constant magnitude. therefore a constant force induces a constant acceleration. 0 Velocity is defined as displacement per unit time. But when the motion is accelerated one then you can't directly apply displacement = (velocity )(time) The above formula is applicable only when the motion is uniform for the given time i.e. your velocity remains the same in that given time. But for an accelerated motion ( i.e. velocity is not uniform ) ... 0 Considering m_1 and m_2 as a system, there is a net horizontal force F acting on them so their centre of mass must accelerate with acceleration \displaystyle a_0 = \frac F {m_1+m_2} This is true regardless of whether or not m_1 moves relative to m_2. The frictional force which opposes m_1 moving relative to m_2 has a maximum value of \mu m_1 ... 1 Free-fall acceleration is assumed to be the constant number g whereas the field strength is the acceleration just defined for any radius. The closer you get to the surface the closer the approximation gets. 3 For a given force acting on an object, the acceleration of such object will be inversely proportional to its mass. But in the case of gravity, the force is also dependent on the mass, so if you change the mass you also change the force, in such a way that the acceleration of the object due to another mass is independent of its own mass. There is no ... 1 It does depend on the earth's mass and because of this dependence that you got such a small acceleration of the earth in spite of a sufficient force acting on the earth. And acceleration is still inversely proportional to mass since$$a = \frac{F}{m}$$And what you did above is that just canceled the mass in the$F\$ term with the one in the denominator. So ...

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The force exerted on the apple by the Earth is equal to the force exerted on the earth by the apple. Because F=ma, and because the apple's mass is much less than the Earth's mass, the apple's acceleration is much greater than the Earth's acceleration ( but of course in the opposite direction). Your mistake can be avoided if you ignore force and only ...

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So here is the basic question I need you to answer in your head: is orbit a special case of free-fall? I think that will cut right to the essence of the question! We grew up on this planet. We like to think about things happening on this planet. It's very easy to forget that space exists, that the Earth is round, that gravity eventually diminishes. If you do ...

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10 kilometers per hour is minutely less than 2.8 kilometers per second (2.777), therefore an acceleration of 10 kilometers per hour per second is equal to almost 2.8 kilometers per second per second acceleration.

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