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In the comments you have said that this question is part of a project regarding interplanetary travel; and the values you have given seem to indicate that what you're really asking is the velocity of an object that was dropped from $500 \space km$ height. If that's the case you can find it using the principle of conservation of energy. $$E = - G {M_Em \over ... 0 By definition the acceleration is first derivative of the velocity over time, i.e. a(t) = dv/dt Then obviously v = ∫a(t)dt. If acceleration a is a constant then v = at (we all know that). Otherwise you need to calculate the integral. 0 Assume the ball slides down the slope without friction and that it starts from stationary (v=0) at a height above the horizontal h. During the friction free slide the object's potential energy mgh is converted to kinetic energy \frac{mv^2}{2}, so that: mgh=\frac{mv^2}{2} and: v=\sqrt{2mgh}. This speed vector is of course oriented parallel to ... 0 Another way to do this is to consider the energy (as you do not seem to be losing energy anywhere, eg with friction). You have an initial velocity (perhaps 0) which corresponds to a kinetic energy, and a potential energy (which is mgh). Then the potential energy is converted to kinetic energy, and you can simply figure it out from there. One important ... 2 Yes. One way to see this: if S' and S have coordinates x' and x, then by the usual rule we know that S' observes a distance of \Delta x = x-x' between them. Differentiating on both sides, we get \Delta v = v - v', \Delta a = a - a', and so on. In other words, velocity, acceleration, and all higher derivatives behave like you think they ... 1 Yes, you're correct. Since S' is an non-inertial/accelerated frame of reference, all objects within this frame is acted upon by a pseudo force that is proportional to the mass of the object and whose direction is opposite to the direction of acceleration of S'. The fact that S is accelerating(with respect to a rest frame on ground) with a=5 m/s^2 ... 0 Yes, as viewed from S', S accelerates at 2 m/s2. One way to think of this is just imagining S' as a rest frame, meaning that the rest frame is travelling with an acceleration of 3m/s2 through 'absolute space'. Alternatively, just comparing the velocities of both frames after each second confirms that they are moving apart with an acceleration of 2m/s2. 0 You can look at the problem like this: Whether or not one person gets to see all the other persons as connected for some period of time depends on 1) the disk size, say radius R 2) the duration \tau of the connection in the disk frame 3) the speed v of individual components on the rim relative to the disk. Let's denote T the astronauts' period of ... 0 When the people see the light they fire thrusters to get up to speed and move in a circle and then after that they fire their thrusters just to maintain uniform circular motion... OK, we have a disk, and light, and a rotating ring of people. No problem. So there are frames where all of the one events happen before all of the other events. And ... 0 This sounds like a homework question (edit below with more info once you explained that it isn't a homework question), so I'm just going to give hints. First of all, notice that power can be written as P = Fv So if the engine power is fixed then at very low speed it can develop very high force. But the maximum force is dictated by static friction, ... 0 Choosing your system carefully and drawing free-body diagrams is crucial! It is often easiest to start with defining your system as one big block. You can do this because the engine and wagons are all connected by rigid steel locks that cannot compress or stretch, meaning that if one car moves, the others have to move along at the exact same rate. If we ... 1 Well, you can think of this as a car pulling a wagon with a = 0.2 \text{ m/s}^2. Naturally, the car's motor needs to generate enough force to accelerate both the car itself and the wagon it is pulling. This is true of the situation in this question. The steam engine needs to generate enough force to accelerate both itself and the two wagons at a = 0.2 ... 2 Do black holes have a puff pastry point? No. If a person falls into a certain sized black hole they accelerate very fast, which increases the g forces on them. They don't feel any g forces at all, because there aren't any. These g forces flatten the person out into a pancake. There aren't any, a falling person doesn't flatten into a ... 1 You can actually (in principle) do an experiment for one of those. If you had a huge, dense, thin sheet of matter in empty space you could cut a person sized hole through it then attach a cylinder to the hole and put a flat bottom on the far away end. So on one side it looks like a flat tower coming up from a big plane and on the other side it looks like ... 1 You would have to apply a force upwards to stop the body. Stopping the gravity would stop the acceleration but not the speed that it already has. A good place to start to check the effects of g forces in a human body is this wiki Changing the mass won't stop the fall either. You cannot make the mass zero, you can cut legs and arms but I think you'll be ... 0 Partly right. By definition, acceleration is the time derivative of velocity and is defined AT time instant t. So we take the time derivative of a velocity function then evaluate it AT t. However, experimentally (and even computationally) we usually compute acceleration as the change in velocity \Delta v over a small but finite time interval ... 4 How did you integrate acceleration to get velocity? Note that \Delta v = \int_{t_i}^{t_f} a(t) dt But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v ... 3 To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information. Since they give you the solution and you just have to ... 0 I think it's a fun question. The circular movement does put a spin on a classic problem. In a nutshell, the ruler, and anything traveling with you would appear normal. We're ignoring that in your giant plexiglass carousel, the centrifugal force would turn you into silly putty, and the entire carousel would stretch and fly apart itself long before it ... 6 SUMMARY This is the essence of Ehrenfest's "Paradox". Your ruler is of course not shortened relative to the measurements it makes in a locally co-moving inertial frame, the circumference of a circle centered on the center of rotation for the merry-go-round rider really is longer, from that observer's standpoint, than its length when the merry-go-round is ... 0 The torque needed at the pivot A to rotationally accelerate an object by \ddot{\theta}i s$$ \tau_A = I_A \ddot{\theta} + c_x W $$where I_A = I_{com} + m c^2 is the mass moment of inertia at the pivot c is the total distance between the pivot and the center of mass c_x the horizontal distance between the pivot and the center of mass W is the ... 0 The answer first: given your axis of rotation, which is the x-axis, you are looking for the moment of inertia associated with that axis, which is the first, smallest principal moment (I_1). It only becomes that simple because in your setup the coordinate axes and symmetry axes are identical. Now for some comments: As @BowlOfRed also pointed out in the ... 0 "a stationary rocket ship in close proximity to ours, stationary relative to our reference frame" If the two ships are stationary with respect to one another, then their clocks will run identically (in fact, all physics will be identical in both ships) unless there are actual sources of gravity with non-negligible tidal forces across the distance separating ... 1 By your logic, there is no gravitational force on the ball while you are holding it, because it is not accelerating. And if there is no gravitational force you should be able to throw the ball as high as you want, right? The truth, though, is that there are two forces on the ball - one from your hand and one from gravity - which are in balance. Keeping that ... 0 The force that you "feel" is not just the current force of gravity but the sum effect of lots of gravitational force. There are a couple equivalent ways to sum up the forces involved. One is called "impulse" and "momentum". This says that when a (constant) force acts on an object for a unit of time, it gets an "impulse" equal to the product of the two. So ... 0 Let's say that you have a 1 kg object that you want to throw directly upward and have it reach a height of 1 meter after you let go of it. Using$$2a(y -y_o)=v^2-v_o^2$$with a constant acceleration of a=-g=-9.80 m/s^2 (up is my choice of positive direction) we get that v_0= \pm 4.43  m/s. We choose the positive solution. That means my hand must exert ... 1 While your object is in motion its acceleration is g, or -9.81 m/sec^2 (we'll take the upwards direction to be positive). This constant acceleration is why the velocity decreases from its initial value of +v when you throw it to -v when it lands. So far so good. But I think the force you are talking about is the force required to stop the object ... 1 I don't see the paradox you are describing and i have some issues with your text: "it will always fall with an acceleration of 9.8m/s": If there is air resistance, it will reach a terminal velocity at which the object is no longer accelerating. Let's assume no air resistance to make things easy. "If I throw it very high up in the sky, it falls with a ... 0 For one constant acceleration a_{\mathrm{Gravitation}} is an approximation, that holds because the radius of the earth is so big compared to the height of your thrown mass. That being said, what you probably "feel" is, that something with greater height gains more momentum on his way back to the ground (as the distance is greater so is the time needed to ... 2 The situation is exactly the same as if the water was stationary and the body was moving (well, assuming you're far enough from the walls for edge effects not to matter). In that case the equation of motion for the body will be:$$ \frac{dv}{dt} = A(v) $$where A will be given by something like the quadratic drag equation:$$ mA = \frac{1}{2} \rho ...

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The work involved in accelerating the ping-pong ball is the integral of P(dV). If you further use the work-kinetic energy theorem, you arrive at the velocity of the ping-pong ball. Obviously, for the long barrel, this method doesn't work. In my opinion, this is because there is a hidden assumption that the ping-pong ball fits tightly against the tube that ...

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Well, most importantly, this question is fundamentally wrong... The units for acceleration are meters/(second**2). Either you made a typo, or you don't fully understand what you are trying to ask. I will assume it was just a typo. The first detail that must be addressed is the type of system we are analyzing. If it is something like a tube attached to an ...

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Sort of. You are correct in saying (with some caveats) that gravity and acceleration are equivalent. According to general relativity, gravity is manifested as curvature of spacetime. As we know from special relativity and Einstein's famous equation $E = mc^2$, energy and mass are equivalent. As a result, any type of energy contributes to gravity (i.e. to the ...

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Gravity and acceleration are not totally equivalent, only locally so. That's why you get tidal forces, so in some cases, you can detect if you are accelerating or in a gravitational field. From Wikipedia Equivalence Principle What is now called the "Einstein equivalence principle" states that the weak equivalence principle holds, and that: The ...

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It is because general theory of relativity. As it says, one person is in a lift suddenly someone cut the cable, then he will experience no gravity as both the lift and person are falling at same rate.

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You can define energy in an accelerating frame, and you do it every day. The surface of the earth is an accelerating frame. Sometimes you say a frame is close enough to inertial and just treat like it is inertial even though it isn't inertial and hope for the best. Other times you just have to sit down and learn how to do physics in a noninertial frame. ...

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It is true that accelerating charges radiate energy in the form of electromagnetic radiation. However, the relationship is not a naive ratio between the acceleration and the amplitude of the electromagnetic wave. Rather, you need to invoke Maxwell's equations. Specifically, there is Ampere's law: \nabla\times{\bf B}={\bf J}+\frac{\partial{\bf ...

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Deceleration for the situation you describe is acceleration in the opposite direction of the initial velocity vector. Let's say it takes 2 seconds for the boat to stop, and let's ignore the upward movement of the boat when it hits the tires and attempts to ride over them. Say the boat's initial velocity when it hit the tires was 4 meters/sec. Suppose it ...

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Additional possibility: when light travels through a medium such as water, it slows down substantially. If light leaves water and enters air, it does in fact speed up when it enters the air. I have no idea how to calculate the time required for this "speed up" to happen, but if the time interval can be calculated or measured, you can calculate an ...

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The speed of light is constant in a vacuum. However, it can change direction in the presence of gravity so in a sense it does accelerate.

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I think you are imagining the earth as one giant rigid inertial frame and that creates problems. Let's look at the atmosphere, a giant doldrum over the pole to make it simple. What keeps the air up there a certain height. Well there is a stronger pressure from the air below it than from the air above it. Newtonian gravity would say the air stays at rest ...

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