# Tag Info

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First, water at seabed pressure will also be between the sand grains in the bucket, so the water pressure does not add up to the force that will oppose your pull. This is already at equilibrium. So away go your 9800 N. I'm not sure either about the 231 N, as the sand will probably mostly stay with the seabed as you pull out the bucket. So your hope is in ...

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When the majority of the turn has completed the racer cranks up the throttle, allowing the anti-wheelie control and stability control to do their job. They are drifting outward. I believe the reason is to on a fully open throttle, the slippage allowing the engine to spin at higher RPM's making more horsepower. At some point they run out of headroom and ...

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Folks are looking for non-Newtonian gravity. The experimental gravity group at U. Washington have really been the leaders in the field over the past ten years; they have some nice review papers available for free. Because of the way that short-range forces work in quantum mechanics, we expect that a short-range gravitational interaction would produce a ...

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Speed is always the speed of one thing relative to another. For example, by speed of the carrier, you mean the speed of the carrier through the ocean. If the wind speed is 0, this means the wind is still compared to the ocean. But someone on the moving carrier will feel a headwind. For a jet to fly, the wind must flow over its wings. It doesn't matter if ...

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What may confuses you is the fact that there are two opposing forces doing work against each other if the body is moving and no forces at all if the body is at rest. This is a actually a nice example of the principle of relativity: Whether the body is at rest or whether it is moving with constant velocity shouldn't change the physics. :-) 'Two equal ...

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Because the forces' magnitudes are equal, I would expect no net acceleration of $m$ Correct. Even if $m$ is moving, there will be no acceleration, since there is no net force. if $m$ is moving, the forces are doing work on $m$ ($F_1$'s work being the inverse of $F_2$'s work), whereas if $m$ is stationary, no work is done by either $F_1$ or $F_2$; ...

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Schrödinger's Wave Equation is an application of Hamiltonian Mechanics. Unlike Newtonian Mechanics, Hamiltonian Mechanics relies on knowing about the things that contribute to the energy of the system. If you know the things which contribute to the energy of a system, then you can determine things like forces, accelerations, and positions. (All through the ...

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As said before, the answer is no(t always), but there is a simple law which can help you predict whether it will be the case or not, and how the torque is distributed across your solid*. Using simple algebra and $\times$ distributivity, one can easily prove that $$\vec{\tau}_{\vec{p}}=\vec{\tau}_{\vec{o}}+\vec{op}\times\vec{R}$$ where $\vec{R}$ is the ...

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No, you wouldn't. The moment (which is called torque, by the way - at least, the kind of moment you're talking about) of a force around a reference point is $$\vec{\tau} = \vec{r}\times\vec{F}$$ where $\vec{r}$ is the vector from the reference point to where the force is applied. If the force is zero, then you can tell that the moment (the torque) will be ...

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Molten glass is suspended (and spun) on air jets to form "pre-forms" for molding glass optics. I suspect that the same must be possible for metals.

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Yes, it is possible to magnetically levitate molten metal. This is not due to ferromagnetism however. As seen in the below references, the metal sample is placed within a tapered conducting coil, which carries alternating electric current in the ~400kilohertz range. This sets up a magnetic field gradient inside the coil and causes eddy currents in the ...

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Let us consider there is no external force and the tyres are rolling smoothly without slipping. Here friction doesn't come into play let me explain you how. Even if friction is present, this friction is not the answer. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer. The correct answer is rolling ...

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The tyres of the cycle are rolling and the remaining cycle moves with a velocity same as that the centre of mass of the tyres have. Now the question is which force is responsible to bring the cycle at rest. The answer is Air-friction and Rolling-friction. It should be noted that the static and kinetic friction does not come into the picture because the point ...

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Based on your comments (that the exam question said "switch A is pressed"), the question can be answered - and the tutor was correct. The key is to look closely at the diagram, and observe that the lower halves of the compartments are connected together, as are the upper halves. In this diagram, $p_1$ represents the driving pressure. Now across the ...

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You are correct. Each cylinder has a hydraulic force applied at the bottom that is sufficient to accelerate its mass upward. The one with the lighter mass will accelerate more rapidly and reach the top of travel more quickly, but the other will already be moving.

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A very tempting mental model of an atom, reinforced by many illustrations in books, has protons and neutrons as "large" spheres in the nucleus and electrons as "small" spheres somewhere near the nucleus. If you assume that all of these particles are made of some "stuff" that has roughly the same density (which is the case for everyday solid and liquid ...

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A note about the statement that a maximal gravitational force would occur at r=0: we can at least exclude the case for two distinct fermions with identical quantum numbers , by the Pauli Exclusion Principle.

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You're absolutely correct - objects do not need to ever reach earth's escape velocity of 11.2 km/s, and many spacecraft that leave orbit, don't. That being said, note that escape velocity depends on where you are: the velocity that a cannonball 1000 km above the earth's surface would need to escape is substantially lower than that needed by a cannonball on ...

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Let's use a spring scale in water s.t the spring scale is sealed in a air tight glass box containing only vacuum so as to remove the buoyancy. Putting both the feather and the iron ball of same masses on the scale we notice that the spring shows the same reading(say $a_0$) for both feather and iron ball. Hence the gravitational force on both the feather and ...

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It is true that the deformation wave travels at the speed of sound, but you have to get away from thinking of objects as rigid if you ask about deformation waves. One good image is striking a foam ball with your fist. The overall shape of the ball will change as the ball wraps around your fist. Some of the deformation will travel across the diameter of ...

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The deformation wave will travel in both directions - there's no way for it to "know" the shortest path. And the resulting set of vibrations will interfere with each other in interesting ways, causing complicated resonances. So, let's look at a simpler example: just a thin, large torus. We'll look at two points $90°$ apart from one another; one at $\theta = ... 0 It takes zero force to push an object along a frictionless plane with a constant velocity. In fact a force can be defined as the rate of change of linear momentum, and zero force applied to a body yields no change on momentum. When friction is added, only when the applied force exactly matches the friction force the resulting motion is uniform. In ... 2 A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ... 2 Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ $$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$ Second possibility : If your box is spherical, By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where$\eta$is coefficient of viscosity. Hence, your ball attains terminal velocity. $$F=6\pi\eta rv$$ $$v=\frac{F}{6\pi\eta ... 0 If you apply a force on the box, and see no acceleration, then the force you apply is equal to the friction force. Friction is velocity dependent, you cannot say "the friction force is so much" independently of the force you are applying. 2 By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing F_{app} otherwise there will be an acceleration. 2 The formula F=G \frac{m_1 \cdot m_2}{r^2} is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ... 4 During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ... 1 I have always thought that electrons can and do spend some (tiny) fraction of their time in the nucleus, depending on the orbital they occupy - the same quantum mechanics that says "they must remain in orbit" does in fact allow for orbitals that, while strictly speaking having zero probability at r=0, have a very small probability at a radius comparable to ... 1 So firstly, it is not the strong nuclear force that keeps electrons in fixed orbits around the nucleus. The strong nuclear force, that is, the Quantum Chromodyanmic interactions, hold the protons and neutrons together in the nucleus, as well as holding the quarks and/ or antiquarks together in other hadrons and mesons. These interactions do not come into ... 1 The equation for gravitational force F=Gm1m2/r^2 gives the force of attraction b/n any 2 bodies with point mass m1 and m2 and separated by a distance 'r'..it means both the objects are attracted towards each other by a force F=Gm1m2/r^2..It is also coherent with newtons 3rd law i.e action and reaction forces are equal.. The expression F=ma or a=F/m ... 0 Yes, this is all correct so far. What you need to remember here is that Force is a vector quantity. That is, it has a direction associated with it. A force pushing you into the ground is the not same as one pushing you up into the sky, like the seat of a flying airplane. So here you need to lable your force \vec{F}_1, say, and this would be the force ... 1 If a is the acceleration of object 1 (should write as a_1), then m should be m_1. Vice versa. 0 If it is the mechanical damage just after impact that is of interest, and not the recovery, you are interested in what is felt locally at the scale of a single cell e.g. Then the quantities you may want to calculate are also local: e.g., the energy dissipated in the tissue per unit volume. The energy dissipated in the sample is the kinetic energy of the ... 1 This is because it is assumed that the test charge does not produce any electric field of its own and its magnitude is negligibly small, so it doesn't apply any force on the test charge. 1 The question seems to be wrong. If they want to know tangential acceleration, they should have given angular acceleration. From given things , 'r' , 'v', and "mu" we can only find centripetal acceleration. And as you said you are getting answer 5 m/s². How it is possible? 0 Sit in the frame of car if you are having problems. Apply tangential and centrifugal pseudo forces. As we are at rest, friction has to act of same magnitude of their resultant and in opposite direction. The answer will be 4ms^{-2} 0.5 \times 10=\sqrt{3^2+a_t^2} a_t=4ms^{-2} 5 Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower. You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or ... 1 Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: \frac{1}{2}mv_A^2>2mgR In this case the pearl has a velocity v>0 in the top point and will continue its movement. \frac{1}{2}mv_A^2<2mgR In this case the pearl won't reach the top and will oscillate around point A. ... 2 I'm going to say the same thing as the author but explain more with words. I am also going to take the external force to be zero and ignore it, since the part you're confused about isn't independent of the external force. The force on particle \alpha, due to all the other particles in the system is$$f_{\alpha} = \sum_{\textrm{all other particles ... 0 Lots of good answers here, but most of them are pretty mathematical and not very intuitive. Lets consider a realistic example. You're on the moon with a six shooter and some extra bullets. You are in a uniform field, and you make two point masses travel through the same distance by dropping a bullet with one hand and firing at the lunar surface with the ... 0 This is all just terminology. 'Force' is a term from Classical mechanics really. 'Fundamental Force' is a term for any one of the set of four theories, gravity, and the three Standard Model interactions, Strong nuclear, weak nuclear and electromagnetic. The strong nuclear interactions (plural) for example could be said to be eight 'forces'. It's just that ... 0 It had a tangential velocity and its weight was enough to provide the centripetal force for motion. 1 Work done is also defined as change in kinetic energy of the body. Since F is constant force so F/m=a is a constant acceleration of m. So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force. The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is ... 1 Without any math and considering only Newtonian model here, I would say that if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system. 6 Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass$m$lifted a distance$h$against gravity with an acceleration$g$is given by:$$W=F\times h=mgh$$ If you are told that someone is going to drop a$1$kilogram mass on your head from a height of$10$... 4 Well, the reason it doesn't matter is that work is defined as $$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$ so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity. Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a ... 2 As you describe, the definition of work is just:$W=F d$. What you are confusing maybe is the rate of work$P$and the force$F$. When you move fast,$P=Fv$is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then: $$W=Pt=Fvt=Fd$$ Independent of velocity. 2 As you note, for a constant force acting on an object which moves in one direction, the work done is equal to$Fd\$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe. When work is applied by an external ...

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Consider two masses M and m in circular motion with same velocity,v. Both has acceleration v^2/R. The forces acting on the two masses are different. Force will become more on the greater mass. But acceleration of both are same. Because, if you put M and m in the following relation, you get same v^2/R. $$(mv^2/R)/m=v^2/R$$ since we know $$F/m=a$$ where ...

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