# Tag Info

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Imagine that the air in the atmosphere was just somehow sitting there unpressurised. What would happen? Well, Earth's gravity would be attracting all that air towards the centre. So the air would start to fall downwards. The very bottom layer of air would be prevented from falling through the solid surface, as the air molecules rebound off the molecules of ...

9

I think it is misleading to explain air pressure as the weight of the air column above our head acting per unit area. Actually your head only feels those air molecules which are directly hitting the surface of your head. You don't feel any air molecules which are further away. As explained in Wikipedia - Kinetic theory of gases - Pressure and kinetic ...

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TL;DR: The planets would not maintain a closed orbit. Bertrand's Theorem states that the only possible central force potentials that will give rise to bound & closed orbits are $F(r) \propto r^{+1}$ (simple harmonic oscillator) and $F(r) \propto r^{-2}$ (gravity, Coulomb force). The proof is slightly involved and I will not reproduce it here. Orbits ...

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The question Which conservation laws relate to the motion of the asteroid? implies that it is the asteroid which is the system under consideration. The gravitational force is an attractive central force which acts along the line joining the centre of masses of the star and the asteroid. The asteroid has an external force acting on it, the ...

5

Since energy is conserved, the total kinetic energy after the yo-yo has fallen a distance $h$ is simply the initial energy, $E_1=mgh$. However, gravitational potential energy is converted into two forms of kinetic energy: translational kinetic energy ($\frac{1}{2}mv^2$) and rotational kinetic energy ($\frac{1}{2}I\omega^2$). Here, $I$ is the moment of ...

4

Tesseract's answer is correct. More conceptually, in general dimensions the angular momentum is what is known as a "two-form", which you can think of a plane with a magnitude and orientation of in-plane rotation. This works in any dimension, and in any dimension the conservation of this plane means that motion under a central force is confined to that plane. ...

4

Rotating reference frame In the rotating reference frame the centrifugal force explains the sinking of dense particles, just as gravity does on the surface of earth. Unfortunately, the terminology “fictitious” makes it seem as though it cannot do anything. For that reason I prefer the term “inertial force” instead of “fictitious force”. In a non-inertial ...

4

What you have described is a double tackle assembly, which gives a W/4 (4 to 1) lift advantage. Frictionless, ideal rope and bars would work the same as ideal pulleys, assuming the ropes do not slip off of the bars. see https://en.wikipedia.org/wiki/Block_and_tackle

4

Consider blocks of equal mass, stacked on each other. The block at the top ‘feel’ the force of gravity and normal force from the block below. The block just below feels force of gravity (it’s weight) plus the force acting from the box above by Newton’s $3^\mathrm{rd}$ Law (if it ‘provides’ normal force on box A, box A acts on it with equal magnitude, but ...

4

The best way to think about this is in terms of the effective potential. For a central-force problem in a potential $V(r)$, it turns out that we can always look at the radial motion of a particle with angular momentum $\ell$ and mass $\mu$ as being equivalent to 1-D motion in an effective potential given by $$V_\text{eff}(r) = V(r) + \frac{\ell^2}{2 \mu r^... 3 As answered elsewhere, yes, the motion is confined to a plane. For me, analogy to cross products is less intuitive in higher dimensions. What makes it planar? Force dictates the second derivative. You always have a second order differential equation. All solutions to "nice" second order differential equations are defined by two functions. Below is a kinda-... 3 Yes. Conservation of angular momentum under the action of a central force allows both the restriction of the orbit to a plane and the elimination of the angular variable in that plane. So central force problems reduce to one-dimensional problems in the radial variable, with an effective potential that includes the “centrifugal” term L^2/2mr^2. With this ... 3 The conservation laws have nothing to do with actual trajectory, but with the nature of forces acting on the object. Take for example conservation of energy of some object in gravitational field. The only force acting on the object is (for simplicity, let us assume radial motion):$$F=-G\frac{mM}{r^2}$$In very short time, this will produce the change in ... 3 Kepler's Second Law is the same for all two-body systems with a potential energy that depends on the distance between them V(r). The Orbital Equation is the differential equation that gives the possible orbits, which is \frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = - \frac{\mu}{L_z^2} \frac{f(u)}{u^2}, where u = \frac{1}{r}, \mu is the reduced mass ... 2 A more mathematical way to think of it is that there are two relevant vectors: the initial velocity \vec v and the initial acceleration \vec a. These span a two dimensional subspace (regardless of the number of dimensions in the full space). The solution will stay within this subspace. There is the degenerate case when \vec v and \vec a are not ... 2 Definition of Conservative Force :- The work done by the force doesn’t depend on the path taken but only on endpoints. Consider the image at bottom :- If the work done (electrical work, gravitational work etc.) in moving something from the point A to point B through the path shown (purple path) is represented as$$ W_{A~to~B} = \int_{A}^{B} \mathbf{F} \...

2

Intuitively it might help if you think about it this way. If the angle to the left is greater than the the angle to right and the wedge is stationary the rope will slide down on the left side. When the wedge accelerates to the left if the component of that acceleration down the incline on the left exactly equals the downward acceleration of the rope that ...

2

You don't feel the air pressure of the air around you (and hence don't feel the "weight" of air above you) because the pressure inside your body is pushing out with the same force (assuming you're breathing normally). This is different from, say, what you would experience if you held your breath and dove 20 feet down into the ocean -- then you'd be able to ...

2

Machinists call this a "press fit" when you need a pin or cylinder to remain tight in a hole. Specifications for press fits vary as to uses, or materials. A .002 inch press fit would mean that the cylinder diameter is .002 inch larger than the hole. Freezing the cylinder as cold as possible will make it contract somewhat to make installation easier. It will ...

2

The pressure is impulse the air molecules give you in collision on unit area and therefore depends on their speed and density. For simplicity, i will consider only density (this would mean that air column has everywhere the same temperature. This is not true for our atmosphere, but such scenario could be achieved in a lab). Because of gravitation, the ...

2

Please read up on pressure here, and how it is derived in the kinetic theory of gases here , third page. This illustrates the motion of molecules in a gas in a closed container, the kinetic energy of all those molecules impinging on the walls and bouncing off each other and back scattering gives the pressure on the walls for the given temperature and ...

2

Your question is too broad and too basic to answer it directly in any way. But I will try to answer it by using some very important checkpoints you will have to pay attention if you decide to study physics and in particular classical mechanics. Gravity and magnetism. These two are completely different phenomena of physics. Gravity acts as a force between ...

2

Answers in words have already been posted, but perhaps an answer with equations may also help, so I will add this. Consider a thin slice of fluid, whether liquid or gas, in a uniform gravitational field. Say the bottom of the slice is at height $z$ and the top at height $z + w$ where the width $w$ will be small (you can call it $\delta z$ if you like). The ...

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However, the concept that I am struggling with is that this procedure means that the engine's power / energy is dependent on the mass of the rocket? Much more troublesome is that the power is also dependent on the speed of the rocket. The faster the rocket is going, the more power you calculate is coming from the engine! This reason is why rocket engine ...

2

Try this: take the definition $$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i},$$ multiply both sides by $M$, and differentiate the equation twice with respect to time.

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I wasn't going to post an answer to this, but the other answers seem very insufficient to me and may give the wrong idea. The work required to lift an object off the ground is not affected by velocity directly. The work is proportional to the product of the force acting on the object, and it's displacement in the direction of that force. This means that ...

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Something that confuse many people on this experiments is that you need to understand that just measuring the s photons will not show an interference pattern. The interference pattern appears only after you have selected the specific photons in s that correspond to the photons in p whose polarization has been erased. So you need this after-measurement in p ...

2

All of the links given are good answers. Since sometimes a different explanation may work best for someone, I offer the following. Heat is energy transfer between objects (solids, liquids or gases) due solely to a temperature difference between the objects. Work is energy transfer between objects due to a net force exerted by one object on another through ...

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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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