# Tag Info

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The things I learnt in years of pinewood derby racing: use the maximum weight keep it at the back make sure the car tracks straight focus on stability The weight is your "engine". Since you start at a slope, mass at the back has further to drop than mass at the front (really!). You can think about it like this: if the weight of the car is evenly ...

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Disclaimer: This answer was written before I found out that a pinewood derby is for miniature wooden cars that run in a track. Therefore, only part of it will apply to a pinewood derby. As I understand your situation, you will be traveling down an incline and then on to a horizontal surface. You are saying you want to maximize the effect of gravitational ...

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well you see the lift has got to do with the mass. Because of the design of the wings of the airplane the speed of wind going above the wing is more than the speed of air going from the bottom of the wing. Therefore from the Bernoulli's equation the pressure above the wing is less than the pressure below the wing. this pressure difference causes the lift. ...

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It is true that the heavier plane needs a greater lift and this is seen in practice. If two planes at equal altitude loose power at the same time and one weighs more than the other they will be able to glide the ...... same distance! One of them descends faster than the other but it glides forward faster to generates more lift. It seems odd, but one ...

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The reason for stability can be explained in terms of potential energy. Systems want to reach a state of minimum potential energy. Assume a system that is in static equilibrium undergoes a small displacement from that position. If such a displacement results in the potential energy increases the system will try and go towards the stable equilibrium position ...

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The experiment involves sticking the tines of the forks into the cork so that the long heavy handles of the forks extend downward. Take a look at the photo in this link: https://www.kecksci.claremont.edu/physics/demo/corkfork.htm. Now, the cork and the forks are bound together as one object, and the center of mass of that object is down toward the middle ...

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Since most the universe is empty space The short answer to this surprise is to see that the density $\rho = \frac M V = \frac M {\frac 4 3 {{\frac{2 G M}{c^2}}^{3}} \pi} = \frac{positive-constant} {M^2}$ decreases quickly as the mass increases. Now let's play with the values of size and mass that one can find on Wikipedia and other publications. ...

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Although the eV is defined with respect to a particle of unit charge (an electron) in an electric field, it is simply a unit of energy. There are simple relations between thr eV and everyday units of energy, such as the Joule or calorie. Thus energies of all sorts of things, in fact any energy, can be expressed in eV, even if it has nothing whatsoever to ...

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One electron-volt $=1.6 \times 10^{-19} joules$ and is a unit of energy that is equal to the energy acquired by an electron falling across a 1 volt potential difference. The particle (neutrino) doesn't need a charge to have some energy. Instead of expressing the mass of a particle in kg, we can express it as $mc^2$ which is an energy (joules or eV... your ...

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As the definition of "rest mass", when the item is relatively static in your frame, the mass you observed is the rest mass.

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When the particle is stationary in your frame, the mass you observe is the rest mass.

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You are totally correct! Yes, velocity is relative, and therefore "relativistic mass" $m = \gamma ~m_0$ is relative: different people see different values. However, here's the crucial part, anyone who sees you travelling at speed $v$ with mass $m$ agrees on this number $m_0 = m / \gamma$. It is what's called a "Lorentz scalar": every coordinate system ...

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The value of velocity depends on the observer. Therefore the value of $\gamma$ depends on the observer. Therefore (since the value of $m_0$ does not depend on the observer) the value of $m=\gamma m_0$ depends on the observer. It's not entirely clear what's confusing you, but you appear to be assuming that $m$ should be observer-independent. It's not. ...

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Seeing how the other answers mention things like Lagrangian and quite a few details of the Standard Model, I feel that a more popular level explanation is needed. This comes with the pitfall of not being accurate, but for that you have other sources. I will pull a rabbit out of my hat and answer without a single (real) formula. What does "mass" mean to a ...

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The mass always means the same thing – but in different theories, one uses different equations and other tools to express the mass. Inertial mass $m$ is the quantity expressing "resistance of the object with respect to acceleration", i.e. the coefficient that enters Newton's $F=ma$. The gravitational mass is what enters the formula for the gravitational ...

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Particle physics has a set of elementary particles , some of which have zero mass. In this table the general public has heard of the electron and maybe the photon. The mass of the electron has been measured consistent with classical definition of mass. The mathematical theory of elementary particles is called the standard model and the table has the ...

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The volume of the disk is $z\frac{\pi r^2}{2}$ not the weight. Otherwise you have done everything correctly! :) Here's how I solved your problem, it's very similar to your solution but keeps track of the directions involved (and doesn't require any integration). Apologies if this does not clarify things. The rotational force or torque ($\tau$) is ...

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This is a direct consequence of what comes out when you go from equations of motion for single point particles to collective motion of extensive bodies. You start with the second Newton's law: $$m_i \vec{a}_i = \vec{F}_{i}$$ Summation over the entire system cancels out pairs of mutually opposite internal forces, leaving us with external force and ...

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Ok, I thought about it again and with the help of your comments I would explain it this way (please let me know what you think about it): The top quark mass can be determined via the formula $m_t^2 = p_t^2 = (p_W + p_b)^2 = p_W^2 + p_b^2 + 2p_Wp_b = m_W^2 + m_b^2 + 2 E_W E_b(1-\beta_W\beta_bcos(\gamma))$ with $p$ denoting the four-momenta and $\gamma$ ...

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Here's a hint, We know $I=MR^2$ (considering point mass) from the axis of rotation where $R$ is the perpendicular distance from axis of rotation. Since each point mass is moving with a constant velocity in the same direction, it means that perpendicular distance from axis of rotation remains same. So you can calculate moment of inertia of each point mass ...

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Time to brush up on our physics: (1) There are two types of Energy in the Universe: Electromagnetic and Scalar. Scalar waves travel faster than light. (2) A straw does NOT get blown in to a tree during a tornado. There is a warping of space/time within the vortices of tornados. The tree and straw simply occupy the same space but at different times. When ...

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Thanks, that helped me work it out, both the mathematics and the concept hold. I thought about "all the other factors should be constant on the equation". So, say with constant energy E=1, travel 1 metre at 1 metre/second and this takes 1 second and you weigh 1 kilogram. If time slows down then on this journey you would actually experience more seconds. ...

2

Mass doesn't depend on time. As in your equation, if the difference in time increases or decreases, the difference in velocity will increase or decrease, so the mass will be constant. As velocity is a variable you cannot come to a conclusion like that. To get a proportional relationship all the other factors should be constant on the equation.

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Please bare in mind that although your relations are mathematically correct, this is not the case conceptually. Mass is a concept which in Newton's framework, is intrinsic to bodies and independent of its motion or time, or any other conditions. Whereas in Einstein framework, mass as the inertial property, does depend on the state of motion, but the mass as ...

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The effective masses are different because they are on different bands with different curvatures.

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You will need a box ( or a bathtub ) with a plain control of the volume of water and some inflatable balloons whose volume can be changed, controlled and measured 1 ) ( exactly ) immerse the body part to be measured in a liquid at rest. 2 ) Attach the balloons to the feets, 3 ) adjust the volume of the inflatable balloons to cancel the buoyancy , as you ...

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With some prior measurements such as total mass, volume of lower and upper body. I think you can use these techniques to figure the mass of just lower body. Computerized Tomography (CT) and Magnetic Resonance Imaging (MRI) These two imaging techniques are now considered to be the most accurate methods for measuring tissue, organ, and whole-body fat mass as ...

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This link summarizes the measurements of the speed of light. The first measurement of c that didn't make use of the heavens was by Armand Fizeau in 1849. He used a beam of light reflected from a mirror 8 km away. The beam was aimed at the teeth of a rapidly spinning wheel. The speed of the wheel was increased until its motion was such that the light's ...

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$\frac{1}{2} mv^2$= K.E. If mass is doubled then energy also doubles. Thank you I hope my answer is correct .

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Simple questions are good questions, and often do not have simple answers. Gravity will always effect your rocket, although if you're far away enough from the gravity's source it may be negligible (not noticeable to you). So how far is "enough"? It depends on your limit of precision. Suppose you were making a trip that took 10 days, and you wanted to get ...

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Mass affects the acceleration through $f=ma$, The higher the mass, the lower the acceleration for a given force. Drag is not affected by mass, only by the speed and the density of the atmosphere. Mass never ceases to influence the rocket's behaviour. Any change in speed or direction (i.e. any change in velocity) is affected by mass through the above ...

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If I understand what the book pretends to show, the theory identified by the existence of $X^{\mu}(\tau)$ in an Lorentz-Poincaré symmetric spacetime can construct physical scalars only by some functionality of $\dot{X}^{\mu}\dot{X}_{\mu}$. The $\dot{}=d/d\tau$ came from the translations in space-time and the product $A_{\mu}B^{\mu}=g_{\mu\nu}A^{\mu}B^{\nu}$, ...

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Relativistic mass is indeed the time component of the 4-momentum, and that is exactly why is it NOT invariant.

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Per Danu's comment, if $m_0/\sqrt{1-v^2}$ were Lorentz invariant, it would follow (because $m_0$ is Lorentz invariant) that $v$ is Lorentz invariant. So neither relativistic mass nor anything proportional to it can be Lorentz invariant. Of course the notion, elsewhere in the comments, that it's somehow not possible to talk meaningfully about the ...

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