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The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...


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Critical mass is actually more about 'the right number of nuclei in a specified space'. As we are talking about solid matter this equivalently translates to a given number of atoms (or molecules depending on your matter). And this furthermore translates to our everyday mass. But it's 'not about mass', it is just a practical way to specify the quantity. So ...


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Actually I think "Meaningful" or "Physically Meaningful" in many cases is as good as you're going to get, although the word can split up into finer meanings. If we think of mathematics as a language, then think of words that describe how well the description meets its intended purpose. Does the mathematical description evoke the "right" ideas? So words you ...


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1 - It is false! If $E = mc^{2}$ is true only for an object that isn’t moving, the mass never changes (is a "Lorentz invariant"). 2 - Can you rephrase it, please? 3 - Energy and mass are not at all the same thing; an object’s energy can change when its motion changes, but its mass remains the same. 4 - In Special Relativity, time can be variable, its ...


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The bomb doesn't "care" what its mass might look like to an observer in another frame. If you calculate critical mass you don't worry how big it might seem to observers located in billions of other possible frames of reference. Local frame of reference is the only one valid for making calculations concerning the occurence of local phenomena.


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The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...



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