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The spring in the biro will twist one way as you compress, the other way as you decompress. Any twist on compress, no twist on decompres is due to a chinge in the coupling to the pressure button. Another way is a bidirectional twist with a ratchet that favours the rotation of the decompress direction. BTW, a pump screwdriver is an example of a mechanism ...


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This is one disassembled , any stationary store or office centre should have them. They just spin a curved ratchet up and down. Best of luck with your ideas


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Damping The density of atmospheric air is approximately 1.225 kg/m$^2$. At 3000 atm, the density would be 3675 kg/m$^2$ compared to the density of water of 1000 kg/m$^2$. I don't know the viscosity of high density air, but as @DirkBruere answered, it could be a significant factor in the damping of the spring decompression. The relative difference in ...


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In 3000atm it's speed of decompression will be slower because it is facing greater air density and the expanding spring has to move it. There will also be less "resonance" as the denser air damps the spring movement.


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You're both half right. Your teacher is thinking of the invariant mass or rest mass(see footnote [1]), which is thinking in the more modern and much simpler paradigm of naming properties of a particle-like "thing" in a frame of reference at rest relative to that thing and you're thinking of the transformations of the properties to effective values as seen ...


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For inelastic scattering, the initial momentum is $m_b v_{b_i}$. After collision, both $m_b$ and $m_c$ move together, with a velocity $v_{b_f}=v_{c_f}=v_{cm}$. By conservation of momentum $m_b v_{b_i}=m_b v_{b_f}+m_c v_{c_f}=(m_b +m_c)v_{cm}$, whichyield the equation that you are looking for


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The systematic way to set up the equations is to draw a free body diagram for each mass. The FBD shows all the forces acting on that particular mass. You can then use Newton's second law to get the acceleration from the resultant force. The weight of mass 2 acts on mass 2, not on mass 1. Of course the weight of mass 2 will affect the motion of the system, ...


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Yes, there is an error in your reasoning. The gravitational force on an object is always proportional to the mass of that object. So if you want the gravitational force on $m_1$, you use $m_1 g$. If you want the gravitational force on $m_2$, you use $m_2 g$. If you want the gravitational force on the combined system of $m_1$ and $m_2$, you use $(m_1 + m_2)g$,...


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You have two unknowns $A$ and $\phi_o$ so you need two equations. You are given the displacement $y$ at $t=0$ which gives you one of the equations. If you differentiate the equation for the displacement $y$ with respect to time $t$ this will give you an expression for the velocity which is you second equation.


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I assume that the spring’s mass is distributed homogeneously across the spring. In this case, you have to use a partial differential equation (PDE) to perfectly describe the spring, whose solution would for example be a function which describes how each infinitesimal segment of the spring is shifted from its resting position (when no external forces and ...


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I think with every bounce some energy is transferred to the spring. This means that after every bounce the height of the (undeformable) sphere is equal or less than the height it fell from. The sequence of successive heights is highly dependent on the height, the masses and the spring constant, and therefore chaotic (no friction involved).


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On the top of Figure we have $\:n+1\:$ ideal springs and $\:n\:$ particles in equilibrium. The constants of the springs are $\:k_{\rho}\: (\rho=1,2,\cdots, n+1) \:$ with equilibrium lengths $\:\ell_{\rho}\:(\rho=1,2,\cdots, n+1 )\:$ and the particle masses $\:m_{\rho}\:(\rho=1,2,\cdots, n)$. Disturbing the system from this equilibrium, the equation ...


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The results you get in the experiment depend on the dimensions of the sample. Are T and W much smaller than L? The shape of the sample is also important in another way. Is it a dog bone shape, with a short middle section length being used as the sample length? If T and W are small compared to L, then the cross sectional area used in calculating the true ...


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according to me you have erred in assuming that change in KE is 0 as the block will have some velocity when it is compressed.since total mechanical energy remains constant loss in P.E will result in gain of K.E


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$A \approx \frac{W_0*L_0}{L}*\frac{T_0*L_0}{L}=\frac{W_0*T_0*L_0^2}{L^2}$ If my interpretation is correct, you are assuming that $W*L=W_0*L_0$ and $T*L=T_0*L_0$ That would make the volume: $W_0*T_0*L_0^2/L$, which decreases when you stretch the material. For small strain, the Poisson ratio would approach 1. Poisson ratio should be between -1 and 0.5 for a ...


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When the spring reaches maximum compression, the mass is instantaneously at rest but it is not in static equilibrium. The net force on it is not zero : $kx \ne mg$. Like a pendulum at the end of each swing, there is a net force on the mass causing it to accelerate towards the equilibrium position - at which the net force on it is then (for an instant) zero....



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