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It is actually an example of continuity equation, which might help you to understand the process. The continuity equation of course implies mass conservation, in a given case. I find this example very useful as it contrasts typical consideration of continuity equation in which there is a flow of liquid through a pipe of variable diameter and speed of liquid ...


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By "it is a case of conservation of mass. " Book means , that it uses Conservation of mass and is not an example of it. Or more clearly, when water falls down from a tap, it is a case which uses conservation of mass.


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Well, if you want an answer at the 9-grader level, it's probably this: We don't know, and it's mostly irrelevant to how the universe behaves now. In particular, the Big Bang Theory doesn't care about what happened before the Big Bang. According to many interpretations of different branches of physics, the question doesn't even make sense, e.g. what happened ...


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The simplest way to look at this is that energy can also reveal itself in negative forms. Don't think of it as something only positive, but also there's a negative part of it in the universe that's not directly visible. For example, we have good reasons to believe that the total energy in the universe adds up to zero. We also have experiments that show ...


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If we observe the path of the ball as radius reduces, we can see that the path followed is not circular. So between the radius and tangent, angle formed will not be 90 degrees.. This creates a component of tension along the path of the ball thus increasing its velocity


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The weak decay is not forbidden. You can have $q\overline q \to ZZ \to e^+e^-$ using a loop like in the EM decay. You need a loop to conserve angular momentum, because a single $Z$ or $\gamma$ has spin $1$ while the $\pi^0$ has spin $0$. If an EM decay is possible, it is so much faster that it will dominate. You can see this in the overall $\pi^0$ decay ...


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Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe. Conservation means that these quantities cannot spontaneously change. Let's ...


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In a collision momentum is conserved if there are no external forces. enefgy is also cionserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones when kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in ...


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Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...


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This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...


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It is not about momentum. The total momentum is conserved: the fingernail flies in one direction, and you (plus the nail clipper) suffer a recoil to the opposite direction. The thing is that your piece of fingernail has a very small mass compared to your body mass, and the firing speed is actually not so high, so the recoil is negligible. About why the ...


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This is essentially the same question as https://astronomy.stackexchange.com/questions/13302/ In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum is $\text{m1} \text{v1}$ and the initial kinetic energy is $\frac{\text{m1} \text{v1}^2}{2}$, since m2 is at rest. Let u1 and u2 be the velocities of the ...


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The reason the Newton's law is referred to as the Newton's laws is because the are applicable to newtonian range objects only. The world of quantum mechanics has no relation with the laws of motion at all. The world of quantum mechanics is random and is not predictable, which is a direct violation of the newton's laws. The point that you make about the ...


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Newton's third law says the the force on A due to B is equal and opposite to the force on B due to A. This in turn means that the changes of momentum of A and B are the same in magnitude but opposite in direction. This is how the momentum becomes rearranged. B loses some momentum and A gained an equal amount. So when two atoms collide you can think of ...


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I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ...


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The behavior of the balls in Newton's cradle does provide a clue to understanding this phenomenon. Because of the spherical symmetry of the balls, and the fact that they make contact at a point, the propagation of compression pulses through the array is not like regular sound propagation through a solid medium. Sound in any medium is usually subject to ...


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Ultimately, the third law expresses in Newton's formalism the principle of conservation of momentum. In advanced physics you will understand that the laws of physics of the world can be defined in terms of "Lagrangians" and "action principles". An action principle takes a bunch of trajectories of particles from some starting positions to their ending ...


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Symmetry, no direction is in any way special so the particles cannot decide which direction the torque can it be, spontaneous symmetry breaking is impossible because this torque apparently is large before anything moves. Also, this will violate conservation of angular momentum because the angular momentum around the centre of mass will change (if one mass ...


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It relies on conservation of energy and momentum and the equation for energy in special relativity: $E^2 = (pc)^2 + (mc^2)^2$. Here you go. Energy of photon: $E_\gamma = \hbar\omega = p_\gamma c$, where $p_\gamma$ is the momentum of the photon. Assume the electron is initially at rest, so it's energy is simply $m_ec^2$. By conservation of energy, the ...


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As another hint to your problem, you'll also want to consider conservation of energy (there's a big clue in your problem that conservation of energy is important --there is no friction between the cart and the path). To start you off, here are the important conservation of energy equations for your problem: $$E_k = \frac{1}{2}mv^2= ...


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I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem: $$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$ Here, you're asking about the line integral around a closed path. When using this equation, you end up making ...


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Conservation of momentum is a little inconvenient under gravity. Anyway, the last sentence in the question means that $p = mv$, throwing each sand bag gives the same momentum boost. Let's name it $\Delta p$. (you might need to work that out since your teacher might ask for a detailed explanation; I'll leave that to you) Now your question asks, how to ...


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The values presented in the bar chart aren't really energies, they are changes in energy. Using the frame of reference shown ($y$-axis) the kinetic ($K$) and potential ($U$) energies at point 1 are: $$K_1=0$$ $$U_1=mgy$$ Assuming no friction or air-drag acts on the car, then Conservation of Energy applies, so: $$K+U=\text{constant}=mgy$$ When the car ...


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An isolated system can experience internal forces like from the car's engine. It is no longer an isolated system when you count friction in, which is an external force on the system. Friction is namely the force that pulls the car forward to make it accelerate. The system consisting of the car can only accelerate when experiencing a non-zero net external ...



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