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14

Let's draw a graph of velocity against time for the two particles $A$ and $B$. For convenience I've made the total time $2t$: The red line shows the velocity for particle $A$ while the green line shows the velocity for particle $B$. When we draw a velocity:time graph the distance travelled is the area under the line. More precisely it is the integral of ...


14

It's very common to get mixed up about signs. The only recommendation I can give is to establish a clear sign convention and stick carefully to it. To show what I mean let's consider your skater: I'm going to use the convention that positive is to the right and negative is to the left. remember that quantities like velocity and acceleration are vectors, ...


13

I don't think this sounds unreasonable as an estimate at all. Let's check it. One designs a building as a compromise between two competing factors: One needs all of the load bearing materials to be well mildly loaded - working in their linear region so that there is no danger of their undergoing plastic (irreversible) deformation, creeping then ...


7

There IS a potential, and all three bulbs will be on. First, you need a reference point to measure voltages. Let's take the wire right of the right battery, and define its potential as 0V. If both batteries generate a voltage of 5V, the wire between the batteries will have a potential of 5V, and the wire left of the batteries will have 10V: So, bulb 1 ...


6

A very brief Google search gets you the number 1,192,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 (approximately $10^{57}$ atoms)- but in fact this is wrong. That value is derived from the mass of the objects of the solar system (mostly the Sun) divided by the mass of a proton (which is what most of the Sun is made of). But the ...


5

Energy kinematics I have this question, because typically problems that can be solve using conservation of energy or just energy-related principles, can usually be solved sing kinematic equations. Yep. In fact, there are two profound pieces of math, Hamiltonian and Lagrangian dynamics, which say that you can use energies to derive the actual kinematic ...


4

Applying $\partial_1\overline{\partial}_1$, we have the following. First term:$$\partial_1\overline\partial_1 G_1 = -\pi\alpha' \eta^{\mu_1\mu_2} \delta^2 (z_1-z_2,\overline z_1-\overline z_2) X^{\mu_3} (z_3,\overline z_3) X^{\mu_4} (z_4,\overline z_4) $$$$+\text{ } {\rm permutations~of~indices~} (2,3,4).$$In the second term, $z_1$ can be the logarithm or ...


3

It is well known that adding a total time derivative to the Lagrangian does not change equations of motion. The Lagrangian above adds a term $$-q\dot q=-\frac{1}{2}\frac{\mathrm{d}q^2}{\mathrm{d}t}$$ (a total time derivative) to the free particle Lagrangian $\dot q^2$. It is thus fully equivalent to to the standard free particle Lagrangian (up to an ...


3

This paper is interesting. It uses the method of calculating the number of nucleons in the neutron star, $N$, based on the radius, $r$, the number density as a function of radius, $n(r)$, and the metric function $\lambda$, which comes from the equations of general relativity: $$N=\int_0^R 4\pi r^2e^{\lambda/2}n(r)dr=\int_o^R4\pi r^2 ...


3

Peskin & Schroeder, An Intro to QFT, are using that$^1$ $$i\Delta(x-y)~:=~\langle 0 | [\phi(x), \phi(y)] |0\rangle \tag{K} $$ vanishes for space-like vectors, see below eq. (2.53) on p. 28. In particular for equal times $x^0=y^0$, we have $$i\Delta(0,{\bf x}-{\bf y})~=~0.\tag{L}$$ Therefore at the physics level of rigor ...


3

Yes, this is correct. Also there are field lines going out from the top of the top plate and the bottom of the bottom plate. And far away, the field lines should look like those of a single point charge, with charge equal to the total charge on the two plates.


2

We can write total energy $E$ two ways: \begin{equation} E^2=p^2c^2+m^2c^4 \\ E=T+mc^2, \end{equation} where $T$ is kinetic energy. Eliminating $E$ from those two equations will give you the desired result.


2

Hint: $T = E - E_0 = m\gamma c^2 - mc^2 = mc^2(\gamma -1)$ and $p = |\vec p| = m\gamma |\vec v| = m\gamma v$


2

The gravitational mass of a neutron star is quite a lot less than its baryonic rest mass (plus the mass associated with the kinetic energy of its contents), because a bound neutron star, by definition, must have a total energy (the sum of its internal energy and gravitational potential energy) that is less than zero. In a “normal star” this is also true, ...


2

What matters is the surface of the object (in this case sphere). On the sphere a charge would distribute uniformly. In the situation that you described what matters is the relative size of the surfaces of two spheres. Since they are equal the charge on both of the them is the same (and hence is the half of the original charge).


2

First a couple general results: given a function $f(t)$ and an interval $T = [t_1,t_2]$, the square of the mean of $f$ on $T$ is $$\langle f\rangle_T^2 = \biggl(\frac{1}{t_2 - t_1}\int_{t_1}^{t_2}f(t)\,\mathrm{d}t\biggr)^2$$ and the mean of the square of $f$ on $T$ is $$\langle f^2\rangle_T = \frac{1}{t_2 - ...


2

Hints: By mean/average Feynman means temporal mean/average defined as $$\tag{1} \langle f \rangle ~:=~\frac{ \int_{t_i}^{t_f}\! dt~ f(t)}{t_f-t_i} . $$ Inequality: The mean square is always greater than the square of the mean $$\tag{2}\langle f^2 \rangle ~\geq ~ \langle f \rangle^2.$$ There are several proofs of ineq. (2), e.g. the variance is always ...


2

When the skater is on the ice, friction stops him/her in 3.52 seconds as you said. The molecules in the skates rub against the molecules in the ice, and the ice molecules absorb some of the skate molecules's energy, slowing the skater down. The reason the force is negative is because the friction is acting in the opposite direction of the skater's motion. ...


2

Lorentz invariance refers to the action $S=\int\mathcal{L}(x)\,\mathrm{d}x$, not to the Lagrangian. To determine the condition on the Lagrangian which we must have, we make the coordinate change $x\to \Lambda x=:x'$ (a Lorentz transformation) and use the general fact that the Jacobian of a Lorentz transformation is unity, so ...


2

Giving that most of the solar system's mass is concentrated in the sun, you may say that the order of magnitude of the number of atoms in the sun and in the solar system is the same. Thus, we may find this number by using the sun's mass and dividing it by the hydrogen's mass, because the sun is composed of it almost entirely: ...


2

No, Newton's Second Law of Motion does not require the acceleration to be constant. However, for any given acceleration (and mass) there can only be one value that the force can yield, namely $m \cdot a$. In other words, if the force and mass stays the same, then the acceleration stays the same as well. Contrary, if the acceleration and mass stays the same, ...


1

What everyone said, but also: the way the question is phrased, the model it uses, is a 1-dimensional model. Imagine a point travelling along the real number line/the x-axis on a cartesian plane. The model in the problem assumes there is no vertical movement, or 'side to side', but only 'pure' 1-D movement along a straight horizontal line. There is only that, ...


1

$f(v)$ is a probability density, so it is convenient to set $$\int_0^{2V{_0}} \mathrm{d}v \, f(v) = 1$$ In your case this will give you an expression for $A$ in terms of $V_0$. Now, e.g. $$\int_{V_0}^{2V{_0}} \mathrm{d}v \, f(v)$$ is the probability that a particle has velocity between $V_0$ and $2V_0$. The average value of any function of $v$ can then ...


1

Other people have answered your questions about the signs; I want to address your comment, "why does this sound like the normal force?" By Newton's third law, pairs of forces always come in equal and opposite pairs. If the ground exerts an upward normal force on you, you exert a downward normal force on the ground. If sliding on the ground exerts a ...


1

In the end this is physics, so maybe you should try it less formal? (and not confuse eulerian and langrangian methods) Initially the mass element at $a$ had mass ( I use $\delta$ to make clear that we really should consider finite differences and then perform a limit at the end) $$m = \rho(a,0) \delta a$$ now, we follow its motion and after some time t we ...


1

According to the wiki page the 747 has a max takeoff "weight" of about 350,000 kg (depending on the model) and a wing surface of about 500 m2. That means that a force of 3.5 MN must be carried by 5 million square cm, or 0.7 N per square cm. If you can somehow split this evenly between the top and bottom surface, then you need to come up with a paper surface ...


1

Let us make an estimate. Let the skyscraper be 400 m tall, each storey 4 m high, 500 people per storey, 20 m2 per person, 10000 m2 per storey, let us assume that the building is a 100x100 m2 square in the plan and that it only has 10 cm thick structural walls in a 25x25m2 grid. So the cross-section area of the structural walls is 2x5x100x0.1 m2=100 m2. Let ...


1

Young's modulus is the ratio of tensile stress to tensile strain for a material: E = (F/A)/(∆L/L) = (F * L) / (A * ∆L) F/A is force per area, and (∆L/L) is change in length per original length For structural steel, Young's modulus is 200 gigapascals. This quantity can be used to predict how much the steel will compress under a given weight per unit area. ...


1

Try to think of this problem using a polar coordinate system. $x$ is essentially the radius $r$ or $\rho$, measured from pivotal. $w$ is simply the angular velocity. So the position vector of the object is $x\hat{\vec r}+\theta\hat{\vec \theta}$ So the velocity vector is $\dot x\hat{\vec r}+w\hat{\vec \theta}$ The hatted vectors are unit. So the ...


1

The work done by the painter is $1.93$ kJ, which represents the force he applied to the rope times the length of the pull. The work applied to the barrel is $1.47$ kJ. The rest went into friction. I believe the question is ambiguous between the two values, but that is an English question, not a physics one.



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