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0

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.


2

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 ...


1

Let's use a spring scale in water. Putting both the feather and the iron ball of same masses 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 iron are same. Now let's repeat this experiment in air(or oil or some other material). We notice that the spring scale ...


1

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 ...


2

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 = ...


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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 ...


0

Some times, we write different equations for force, like $$F=\frac{1}{4\pi\epsilon}.\frac{q}{r^2}.q_{o}$$ $$F=Eq_o$$, $$F=m.\frac{GM}{r^2}$$$$F=ma$$, etc. When the force is due to the property of static charge, we say it as electrostatic force (Coulombs law), when the force is due to mass, we say it as gravitational force (Newton's law of Gravitation). ...


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.


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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 ...


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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.) ...


0

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 ...


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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.


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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 ...


1

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 ...


1

You can think of acceleration from a purely mathematical context - it's the rate of change in velocity. (If you're familiar with calculus, you can say that the acceleration is the derivative of velocity.) Because of this, you don't need any mechanics to determine acceleration, so the mass is irrelevant. More concretely, a mathematician could calculate the ...


1

The definition of acceleration is rate of change of velocity. If you know velocity as a function of time, then you know acceleration. No information concerning mass is required.


1

Mass and acceleration are two independent variables. You need to consider them together to arrive at a force with the direct relationship F=ma. In other words, if it's mass doubles, so does the force and the acceleration is the same. Gravity works exactly this way. If you are considering a fixed force from something other than gravity (a force of constant ...


0

This is a partial answer that may get you thinking in the right direction. Three rods connected in a triangle are rigid. 4 our more rods connected in a square or larger polygon are flexible. This is why high voltage power lines are supported by structures built entirely of triangles. However, triangles are not enough to ensure stability. E.G. Two ...


0

The answers posted by others encouraged me to examine the assumptions I was making in trying to solve the problem. Unfortunately I think none of the answers were ultimately heading in the right direction so I can't accept any of them. I believe the answer is that the friction between block B and the plane will always be at its maximum value once the plane ...


2

The rope on the left pulls the weight and the pulley towards each other. The rope on the right pulls the floor and the pulley towards each other. Each of these pull downwards on the pulley with force $T$. The total downward force on the pulley is $2T$. The pulley doesn't move because the ceiling pulls upward on it hard enough to keep it motionless. The ...


1

If there is no air drag, the pulley and the rope are frictionless and massless, the rope is not slacking anywhere, the rope is unstretchable and the rope is attached at exactly the centre of mass of the object so that no torque is produced, yes. How is it counter-intuitive? It may seem like that it would require an exact $mg$ force on the top too but don't ...


2

The quote is taken from just above eq. (1.32) in Ref. 1: [...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance ...


-1

The relation is: $\sum F=0$ comes from $\sum F=ma$ If $\sum F=0$, so that either $m$ or $a$ must be $0$. Currently, there is no substance of mass $0$. So, $a=0$. Means that no force is applied to the object. If it is moving, it will continuously move with constant velocity. If the $v=0$ , the object will stay stationary. Hope you understand.


2

There is no law that says the sum of forces on a given object must be $0$, that is simply the condition for mechanical equilibrium. If an object has constant $0$ velocity (or, more generally, any constant velocity), then its acceleration ($\frac{dv}{dt}$) is $0$ and, by Newton's second law as you have it, the net force acting on it is $0$. However, if all ...


1

There is enough information. Think about the tension in the rope on either side of the pulley - it's the same. Then look at the vertical forces.


0

$f_{AB}(t)$ will be on the left side and $f_{BF}(t)$ on the right. Maximum $f_{BF}(t)=\mu (m_A+m_B)g=(0.6)(30)(10)=180N$ So, $F(t)$ will have to be greater than $180N$ so that $B$ can move. When it does, $A$ will experience a pseudo force say $F'(t)$ in the left direction. If $B$ accelerates at $a_B$, then $$F'(t)=m_Aa_B=10(180/30)=60N$$ But, Maximum ...


0

The strong law of action and reaction says that the forces that two bodies exert on each other have the same magnitude, opposite direction and act along the line joining the particles. When you want that last bit to be true and you want to write the force on particle $i$ as $-\nabla_i V_{ij}$, then the potential has to be a function of the relative distance. ...


1

I can only speak for steel and aluminum which I have had engineering experience with. The answer is that a solid rod is going to be stronger than a stranded one of the same area. There are a few effects to consider, and most of them work against the stranded cables. Geometry Effects. Strands are not straight and so pulling on them causes contact pressure ...


1

I assume the total cross section of the wires in the rope is equal to the cross section of the steel bar. Since there is some space between the fibers, the rope will have a larger diameter. If the fibers in the rope were all straight and you were careful to pull equally on each fiber, the bar and rope should be equally strong. But they typically won't be. ...



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