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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 ...


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 ...


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 ...


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I was confused by this too -- pop descriptions of the equivalence principle don't mention the problem where the gravitational field points in different directions in different places. It is true that gravity is equivalent to acceleration, and that as a result, if you are freely falling, you feel like you're in an inertial frame. But this frame is only ...


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 ...


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

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 ...


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

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 ...


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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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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. ...


1

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 ...


1

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 ...


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 ...


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 $a=5 m/s^2$ ...


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 ...


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

If you consider that gravity is weak compared to the electromagnetic force because $G \approx 6.67 \times 10^{-11} Nm^2 kg^{-2} $ and $k_e \approx 8,987 \times10^9 N m^2 C^{-2} $ it would require very small distances in order for the gravitational force to be effective, but at this distances the electromagnetic force would be several times higher, ...


1

If you work through the numbers, you will find that all of the air on earth has a mass that is less than 1 millionth the mass of the earth. You can't get more than a tiny fraction of that air close to your falling object, so the effects of wind and other disturbances would FAR outweigh any effects due to gravity, because G is sooooooo small.


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Assuming that the pressure change across the cone is small (e.g., no significant density changes for the flowing gas), use the continuity equation. With constant density, this simplifies to $A_1 \cdot v_1 = A_2 \cdot v_2$, where $A$ is the cross sectional area of the flow stream and $v$ is the velocity of the flow stream. If you additionally need the ...



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