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31

Yes, photons can. See https://en.wikipedia.org/wiki/Radiation_pressure (and photons are certainly massless). PS In fact, any massless particle has momentum (which has a fixed value since they can only travel at the speed of light) and if it is scattered on a body, it changes its own and the body's momentum, which is what a force does.


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Newton's 2nd Law of Motion gives the impressed force as $F=dp/dt$, so a physical theory for a massless particle exerting a force requires that the particle have momentum, $p$. First we will discuss mass, momentum, the force law, and Special Relativity. In Newtonian physics mass is identified in two ways: by it's inertia, or as the quantity of matter. The ...


4

For a given mass the gravitational attraction remains the same -- but only if you are far away. For example, the surface gravity of Sol, our sun, is $274$ $ m/s^2$, about 28 times the surface gravity of Terra, which is $9.8$ $ m/s^2$. But as the material is compacted, the surface gravity increases: this is because the effective mass can be treated as ...


4

I'll give you a couple of ways to think about this. First, geometrically, the circle you are thinking about drawing should contain the entire circular path of the car. If we're assuming that the car is remaining at a constant "elevation" on the banked surface, then the center of that circle has to be at the same elevation: otherwise you'd be drawing a cone ...


3

Concerning your wording "force is transmitted (and maybe decreases because of loss of energy)" - no, no, the decrease of force is not easily connected to the loss of energy. Force can be decreased because there is friction, but this does not imply a loss of energy (not if nothing moves). And also energy can be lost (plastic deformation of the rope) without a ...


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At the instantaneous moment shown in the diagram, we can write: $$2R\alpha_{ring}=a_{disc}$$ as both are in pure rolling. This also tells us that the point on the ring where the thread is attached has an acceleration $=2R\alpha_{ring}=2a_{ring}$ so we find that: $$a_{disc}=2a_{ring}$$ Note that when the string moves to another position this will not be true, ...


3

Because the springs are considered massless. So if there were a force difference between the ends, you would get infinite acceleration. And this is valid not only for the ends, but for any two points, if there is no mass between them. The issue is somehow explained in this post: Is the tension in both ends the same (on a massed string)? And indeed, if the ...


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Gravity doesn't have a horizontal component. The component of gravity normal to the plane in your diagram can be said to have a horizontal component, sure (and a vertical component of magitude $mg\cos^{2}\theta$). But there is also a component of gravity parallel to the plane of magnitude $mg\sin{\theta}$. That component can be resolved into a vertical and ...


3

Your confusion might be coming from not clearly understanding that you need to define a system, and then all of your quantities are referenced to the system you have defined. If your system is A and B, then the force that A applies to B is by definition an internal force, and no work is done by that force on the system. But you can choose your system ...


2

If there was no air resistance, the object would still slow down, stop momentarily, then return to the ground - because it is pulled downwards (towards the centre of the Earth) by the force of gravity (ie the object's weight). More than one force can act on an object at the same time. Here, both gravity and air resistance act on the object. Gravity always ...


2

I think you are asking why this argument remains valid when the mass is oscillating, and does not only apply when it is static. I think the answer to this is that the forces in the springs depend only on their extensions (F=kx), not how quickly the extension is changing (F=kx+bx'+cx"). So at any given extension x the tension is the same regardless of the ...


2

There isn't a "centripetal force" vector. As the car goes around the banked curve, the normal force on the car increases relative to what it would be on an un-banked straight road. The vertical component of the normal force supports the weight of the car, and the horizontal component of the normal force provides the centripetal force necessary to cause the ...


1

On page 68 the paragraph headed Calculating displacement given a time and acceleration includes the text: Assume that you’re on your traditional weekend physics data-gathering expedition. Walking around with your clipboard and white lab coat, you happen upon a football game. Very interesting, you think. In a certain situation, you observe that the ...


1

To understand this, use the definition of force $\frac{d{\bf p}}{dt} = {\bf F}$, namely the force is equal to the rate of change of momentum. Something like a collision can be very complicated to model, but the average force is approximately given by ${\bf F}_{average} = \frac{\text{change in momentum}}{\text{time taken}}$. Typically, in a collision, the ...


1

You will need to use two free body diagrams, one for each mass, one FBD will include the force of A on B and the other one the force of B on A. Both mass have the same acceleration (do they), so you end up with a little system of three equations and three unknowns.


1

The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...


1

Okay, so I figured it out myself. Here's what I think: Take $P$ to be the point on top of the ring, where the string is attached. Now, two things contribute to $P$'s acceleration : the acceleration of the centre of mass of the ring, and the acceleration due to the angular motion. So, $a_P = a_{ring} + \alpha \times R$, where $\alpha$ is the angular ...


1

The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ...


1

I can do it by considering $m_1$ and $m_2$ to be a system, which would give me $a=F/(m_1+m_2)$. How can I use a free-body diagram instead to calculate the acceleration? But this is by the use of a free-body diagram. Otherwise, how would you know that it is the force $F$ you should include? Because, your acceleration expression comes from Newton's 2nd ...


1

The net force on the string is not F. You are pulling the string forward with force F but I think you are forgetting that the block is pulling the string backwards with a force that is almost equal to F. If the masses of the string and block are m and M then for the whole system (string plus block) F=(M+m)a. The net force on the string is F '=ma=mF/(M+m). ...


1

The whole point in components is that when you add them, they must must give the original vector. The two components you've drawn don't. Their sum is not the original gravity vector. Remember that components are supposed to follow coordinate axes, so they are perpendicular to each other (in that way they take care of distinct directions so we can treat ...


1

Even though capillary forces may be the dominant force in this situation, hydrostatic (gravitational body) forces are still there. And that results in the equilibrium seen in the left most figure as you start this experiment. So as you bend the tube over, the head (force due to height of the fluid) is reduced and that would allow the surface tension of the ...


1

If you measure a force (weight) for a given acceleration (gravity) in order to determine the mass of an object and you haven't started measuring then the mass is undefined. As soon as you apply an acceleration $a>0$ and you measure corresponding force $F>0$ you can determine the mass. Equations are useful only when they can be used to measure things, ...


1

I disagree with both other answers. It's not enough that the sum of forces is zero This is not a vocabulary problem The work-energy theorem applies also to systems of interacting bodies, where the total work of all forces (internal and external) equals the variation of the sum of the kinetic energies of all bodies. Now to answer the question. ...


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The force of friction is defined as $F_f = \mu N$, where $N$ is the normal force. In the case of a flat surface free of external forces, you can use Newton's laws to determine that $N = mg$, where $m$ is the mass of the object. Notice that we have made no reference to the objects size, or area of contact. This is because in these examples we have ...


1

Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$ \frac12 mv^2=mgh $$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...



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