# Tag Info

4

Estimating is always a fun aspect of physics - so let's do some, without looking up any values. What is the kinetic energy of a plane? We need to know the mass of a plane and its speed. I am going to use seriously rounded numbers - let's see how close we get. We "know" a full size car is about 1000 kg, and can carry 5 passengers of 100 kg. That means a car ...

4

Assuming kerosene is C8H18, has 25 chemical bonds, each of which releases 1eV when burned, gives an energy in fuel of 20 MJ/kg. The weight of a plane shortly after take off is significantly fuel. If I'd guessed I would have said 20 tonnes per plane; (Floris' comment suggests more like 30). This gives 400 GJ per plane, equivalent to 100 tons of TNT per ...

3

Negative energies are totally fine, because you had to pick a zero-point for energy. In your calculation you picked it to be at infinity. You could have chosen the zero-point for potential energy in such a way that your system had zero energy, or whatever. Only changes in energy are meaningful, in general. Consider this: what happens if you add energy to ...

3

In general, when working out the Lagrangian, start in coordinates that you know and then rewrite in generalized coordinates. Kinetic energy in this case is proportional to $v^2 = \dot x^2 + \dot y^2 + \dot v_z^2$. In your spherical coordinates $$x = r \sin \alpha \cos \theta;~ y = r \sin \alpha \sin \theta;~ z = r \cos \alpha.$$ Take full time derivatives ...

2

It is up to you where to define the zero potential (potential energy undefined up to a constant, so by adding any constant, the zero becomes anywhere you want it), but let's consider the point where the potential itself is minimum and subtract this value so that the potential is everywhere positive except at it's minimum (just a convention) $$\vec{F} = ... 2 There's no such thing as an imaginary point. In other words, you can certainly plug imaginary numbers into a formula, but those imaginary numbers don't represent points in space, and thus the results you get will not represent the conditions at any actual point. However, you don't need an actual point in space for this purpose. The potential energy field ... 2 I think what he's saying is that$$F_{net} = F_{nc} + \nabla U,$$which is pretty standard. f^a is your net force, which is the sum of your conservative and nonconservative forces. Conservative forces can be written as the gradient of some potential, which is where you get your \nabla U from. f^e, then, are your nonconservative forces. 2 How is it possible that I "invest" a (constant) teacup worth of potential energy, yet "gain" the potential energy of rising an arbitrary big mass that is floating in my basin by the constant amount caused by the water from the tea cup? The mass cannot be "arbitrarily big". Since it is floating, it has a net density that is less than that of the water. ... 1 What happened with V\left(\sqrt{x^2+y^2+z^2}\right)? You mean, why does V(r) disappear from the \frac{\partial }{\partial \dot q_j} term, right? It's because V(r) is a function only of q_j not \dot q_j. Those variables are treated as independent and so \frac{\partial V}{\partial \dot q_j}=0. and why \partial\dot q_{j} = \partial\dot ... 1 The \frac{1}{2} is missing in their derivation because the potential you are both calculating is different. You integrate from 0 to y. In here is the implicit assumption that at the equilibrium height there is zero potential. In the derivation you linked, they use the formula U = \Delta mgy. What they are doing is taking a column of fluid of height ... 1 The kinetic energy of the particle depends just on the path it is following - if you imagine the cone is suddenly invisible, the particle continues to go around in a circle. That means that there is no reason to add \sin\theta in your expression for the kinetic energy if you used r to mean (as drawn) the distance from the axis of rotation. Note - your ... 1 If we assume that mechanical energy (K+U) is conserved in both the earth frame and the initially co-moving, constant velocity frame, then it's not the differences in velocities which are the same; it's the differences in the squares of the velocities which are the same.$$\frac{1}{2}v_{1e}^2=\frac{1}{2}v_{0e}^2+2gh$$and ... 1 In this case it's because of who's doing the work. The derivation you cite regards the work as being done by the gravitational field. This means in bringing the object from infinity, the gravitational field has lost that energy to the object. Conversely, the work done on the object is the positive value. Everything else works out and when we take the ... 1 When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the \vec v_+, q_+ of the positron, the effect on the electron of its \vec E and \vec B field. Fortunately, 5 keV (kinetic ... 1 The potential energy of the ball is given as E = mah_1 where m is the mass of the ball, h_1 is the height over the point you set as zero potential and a is the acceleration due to gravity (which is different between Mars and Earth). If you assume there is no energy lost due to heating of the ball or other inelasticities, you have an elastic collision ... 1 Since \epsilon is a symmetric tensor, it has 6 independent component that determine it. Hence use a multi-index I\in\{(i,j)|1\leq i\leq j\leq 3\} to denote them. The strain energy density then becomes (perhaps one has to be careful with "diagonal" terms here in order to get the right coefficients)$$\psi = C_{IJ}\epsilon_I\epsilon_J$$where summation is ... 1 It depends on how you define the potential. In mechanics one has the convention$$ F = -\nabla U Since the electric field exerts a force via $F=qE$ it is only natural to apply this convention in electrostatics too. In this way the electric potential $V$ can be directly interpreted as mechanical potential energy $U=qV$. Option number (2) is therefor the ...

Only top voted, non community-wiki answers of a minimum length are eligible