Hot answers tagged

7

TL; DR The material with the greater effusivity will be more likely to burn you upon contact. Analysis of a simplified case First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = ...


5

Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae. For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to ...


4

I'm guessing you mean this sort of thing: The light follows the arcs of water because it is kept inside by total internal reflection. The stream of water is behaving like an optic fibre. Most of the light travels along the stream of water and the light you see coming from the arc is scattering due to impurities in the water or ripples at the surface of ...


3

First, when calculating force, the unit produced is newtons (N), not kg. You can use kg(force) if you're willing to risk getting confused, and your calculation of the falcon weight shows that you did get confused. In a 1g environment, 1 kg of mass produces 1 kg of force, so there is no multiplication by 9.8. The gravitational attraction between any two ...


3

I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static ...


2

You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ...


2

Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over $m$ is really the integral over the volume: $$\int_m r \cdot dm = \int_V r \cdot\rho(r')~ dV$$ Here I deliberately distinguish between $r$, the vector from the origin of the coordinate ...


2

I fear you are overthinking it. I am not sure why the original references, Kubo (1964) and Royer (1976) and their proofs are not adequate for you. Working in Fock space with creation and annihilation operators is a bit self-defeating, unless you were suitably adroit. Sticking to standard phase space operators $\hat{x}, \hat{p}$ and using the standard CBH ...


2

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in your body ...


2

Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are.


2

We know that the Levi-Civita connection satisfies $\nabla_a g_{bc} = 0$ and the product rule. The definition of the inverse metric $g^{ab}$ is $g^{ab}g_{bc} = \delta^a_c$. Therefore, we have: $$\begin{align} 0 &= \nabla_a \delta^b_c \\ &= \nabla_a (g^{bd}g_{dc}) \\ &= (\nabla_a g^{bd}) g_{dc} + g^{bd} \nabla_a g_{dc} \\ &= (\nabla_a g^{bd}) ...


2

In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $ V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$


2

I think you have all the right pieces to answer the question, here are a few hints that should be of some use. You say that you picked coordinates $ \{v^{\mu} \}$. It seems to me that they should instead be called $ \{ x^{\mu} \}$, as that is what you're taking partial derivatives with respect to. As you correctly pointed out, you are working with ...


2

You can't get an exact number for this without some assumptions. However, you can develop a relationship. What we do know is that one full period corresponds to some amount of time (my niave guess would be 1 second but I believe I have heard faster clocks which might be half seconds). So we introduce some constant C which is the amount of time reported on ...


2

There many configurations that statisfy your assumptions. But you had forgot about many other constraints like potential difference, charge density, end effects etc. When you consider PD you come to know, ignoring end effects, that only one configuration is possible. Since 1st plate has charge Q its surface charge density of 1st side is Q/2A 2nd side again ...


2

Dip773, you wrote the most general solution already. As stated by Floris, you can obtain a generalized formula ($k_{eff} = k_p\cdot\phi$, where $\phi = \frac{1+\sqrt{5}}{2}=1.61803...$ is the golden ratio) if all spring constants are equal ($k_n = k_p$ for each $n$). In order to obtain this, you also need the hypotesis that $n$ is large, and that adding one ...


2

Do we feel less weight on surface of Mount Everest? (Or have I mixed some wrong values?) The answer to both of these questions is "yes". One would weight a tiny bit less on Mount Everest, but not as much less as the question poses. You have used some incorrect values and assumptions. If you use the numbers you yourself used to compute the gravitational ...


1

Since the springs are connected in series, they will experience an equal amount of force as the tension is same inside both the springs in equilibrium. If the tension is non-zero, then some part of spring will accelerate and therefore equilibrium will be destroyed. In case of parallel springs, the force exerted by them will be the sum of the two forces as ...


1

As we know, for a mechanical system being in equilibrium means having total force equal to zero. Let's look now at the second spring on a picture b): to be in equilibrium, force from the first spring $F_{k1}$ (action of the first spring on the second) should equalise external force $F_{out}$. Thus, $F_{out}=-F'_{k1}$. This, in turn, after applying Newton's ...


1

You're only a half-step away. You listed conservation of energy and linear momentum, both of which are due to there being no external forces on the three-charge system. But with no external forces, you know that the center of mass of the system won't accelerate. Since the COM starts at rest, this means that the COM will remain stationary. Think about what ...


1

The magnitude of centripetal acceleration is $\frac{v^2}{r}$ instantaneously. It applies no matter the speed on your circular path. (Technically it's true for any curve, but $r$ would be changing on non-circular curves, making calculations more difficult.) The tangential acceleration is constant, so you can write a function for $v$. Then you have two ...


1

Assume that each swing advances the second hand by one second on the dial. You have a pendulum which is supposed to have a period of exactly one second. If the pendulum has got longer in length it will have a longer period and so it will take longer for the pendulum to advance the second hand by one second on the dial. So the clock will run slow. Just ...


1

You are on the right track. For a mechanical pendulum, the relationship is linear. You don't need to know how many swings of the pendulum corresponds to how many seconds. If the pendulum is x% slower, it will report x% fewer seconds per day. Now since length goes as $\ell = \ell_0(1+\alpha \Delta T)$ and period of pendulum as $$T = ...


1

I have an old pendulum clock (probably over 100 years old) still in working order. I'm pretty sure that the hands advance linearly with the number of swings of the pendulum. I keep my house about 10 degrees Fahrenheit cooler in the winter than in the summer, but don't really notice a difference in its timekeeping. The bob is suspended by a wire made of ...


1

I'll answer since you've given it the old college try. Start by defining a coordinate system in which x is towards the center of the circle and y is upwards. The components of the normal force are $$R_x=R\sin\alpha,\qquad R_y=R\cos\alpha$$ Then Newton's second law gives you $$\sum F_x=R\sin\alpha=ma_c=m\frac{v^2}{r}$$ $$\sum F_y=R\cos\alpha-mg=0$$ The ...


1

TL;DR: $R \neq mg\cos(\alpha)$, because there is a component of $\frac{mv^2}{r}$ in the same direction. Relative to horizontal and vertical The horizontal force must be equal to resultant force towards the center of the circle (for circular motion). $$ R \sin(\alpha) = \frac{mv^2}{r} \\ $$ The particle is in equilibrium in the vertical direction, so $$ ...


1

Yes, it is a sort of contradiction. More precisely, it says that you cannot model water's liquid-to-gas phase transition with an ideal gas model, and any attempt to do so will be fraught with contradictions. Now as to some misunderstandings that you may have: first, be careful about thinking of potential energy as infinite at infinite separation: that's ...


1

If this article is to be believed, you would have no problem at all - in fact you could feel where the cities are, let alone the mountains.


1

You have$\frac E{m_0}$, the energy divided by the rest mass is $\gamma=\sqrt{1-\frac{v^2}{c^2}}$. The proper time is lab time divided by $\gamma$. Since you have a fixed $E$, as $m_0 \to 0, \gamma \to \infty$ and the proper time goes to $0$. For the last part, you are supposed to assume that an $11$ MeV neutrino arrived $7$ seconds before a $7$ MeV ...


1

Think about the problem from the point where you neglect the wire and just analyse the motion. You have gravity and at any other point the force has to balance gravity that you end up in a circular motion. In A its towards the origin whereas in B there is no force (except gravity). Different from the classical space-orbit or car-in-a-curve problem is ...



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