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5

The canonical example of this is the apparent singularity that arises in spherical coordinates when you pass around the earth only to find that your longitude has gone from $0$ to $180$. Or for example the singularity that arises in the Laplacian with spherical coordinates. These are all non-physical and are a consequence of choosing a coordinate system. A ...


5

Since the Schwarzschild metric is invariant under time translations $t \mapsto t + \Delta t$, we get a Killing field $\partial_t$ and thus we know $g(V, \partial_t)$ is conserved along a geodesic with velocity field $V$. This quantity can be written $$ -Q = g(V, \partial_t) = V^i g_{it} = -(1-2M/r) V^t,$$ and can be thought of as an energy of some kind. We ...


3

It seems that you forgot that $\theta$ depends on $y^0$ when you computed $\dfrac{\partial x^1}{\partial y^0}$. The correct answer should be $$\frac{\partial x^1}{\partial y^0} = -\omega(y^1\sin\theta+y^2\cos\theta),$$ and then $$\frac{\partial^2 x^1}{\partial y^1\partial y^0} = -\omega\sin\theta,$$ as you desired.


3

You misread a greek nu, $\nu$, for a Latin 'v'. The first equation is $E=h\nu$ with a nu, which denotes $f$. (As Winther rightly pointed out in a comment, the second occurrence, in $c=\nu\lambda$, is also a nu that denotes the frequency $f$. Only the third occurrence, in $v=f\lambda$, is actually a $v$ for velocity.)


2

It is important to first note that $x^\mu$ is a scalar function, not a vector, so that we must use the scalar wave operator. We will raise and lower indices with the background metric, $\eta_{ab}$. Working in the $x^\mu$ coordinate system, we then have\begin{align*} g^{ab} \nabla_a \nabla_b x^\mu & = g^{ab} \partial_a \partial_b x^\mu - g^{ab} {\Gamma^c}...


2

Unfortunately it is unclear what is the expression $$\Omega_i^2 x_i^2$$ as there are indeed $4$ i's in the expression. But after reading previous page of the book, this term corresponds to $$\Omega ^2 r^2 = (\Omega \cdot \Omega)(r\cdot r)$$ so the expression should be written as $$ \Omega_i^2 x_l^2$$ instead.


2

suppose $1/R_1 = x$ and $1/R_2=y$ and $1/R_{eq}=z$ now we know that $z=x+y$ and $x>0,y>0$ therefore $z>x$ and $z>y$ and hence $1/R_{eq}>1/R_1 $ and $1/R_{eq}>1/R_2 $ and therefore $R_{eq}<R_1$ and $R_{eq}<R_2$


2

$$a(x)=\dfrac{dv}{dt}=\dfrac{dv}{dx}\dfrac{dx}{dt}=v \dfrac{dv}{dx}$$ $$a(x)dx=vdv$$ $$\int_{x_0=0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}(v(x)^2-v_0^2)$$ $$\int_{0}^{x} a(\bar{x}) d\bar{x}=\frac{1}{2}v(x)^2$$ $$v(x)=\pm\sqrt{2\int_{0}^{x} a(\bar{x}) d\bar{x}}$$


2

This question is a bit vague, so let me rephrase it first: Let $f_T(n)$ be the least natural number not describable by $T$ by a expression of length $n$. If $T, S$ are both "reasonable" (e.g. true computably axiomatizable theories of arithmetic extending $PA$), does $f_T=f_{S}$? OK, first off we need to make this a bit more precise: what does it mean ...


2

As you stated, $F=-\frac{GMm}{r^2}.$ If we proceed very far away from the planet, i.e. take $r\rightarrow \infty$, then $F\rightarrow 0$ (this is what your problem assumes with $r_1$). Suppose we stop at a point "infinitely" far away from the planet and move towards the planet some minor distance $\Delta r$. Because $F=0$, $W=0$. But if we then approach the ...


2

For a function $f(x)$, its average value on an interval $[a, b]$ is given as; $$f_{avg} =\frac{1}{b-a}\int_a^b f(x)dx$$ Suppose the given distance function respect to time ($t$) is given by $s(t)$. Then the velocity would be given as $v(t)=s'(t)$ Now if we were to calculate the average velocity, we would compute; $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$$ ...


2

Discontinuous functions are fairly common. What's the magnitude of the force between two point charges, or particles which can be considered point charges, $$F=\frac{kq_1 q_2}{r^2}$$ Where $q$'s are the charges, $k$ is constant, and $r$ is the distance between them. This is quite clearly discontinuous when the distance is is zero, and diverges as the ...


1

Being continuous or not is often a matter of referential: origin, orientation of axes and units, or scale. When discontinuity appears, the source is often in the simplification of the model, think about reducing a moving object to its center of mass. But would physics exists without discontinuity? If an equation is continuous, and differentiable, it ...


1

This is a cleaned up version of some comments of mine on the original question. Some bad behavior is removable, some is not. The behavior that is removable in some sense "already is", from the physicists' perspective. For example, sinc can be thought of as (a multiple of) the Fourier transform of the indicator function of some interval symmetric about zero. ...


1

The lorry is moving at constant speed so the forces are in equilibrium. Therefore you have $$\frac Pv=mg\sin\theta$$ Plugging in the values, you get $v=5$


1

It looks like you replaced $\beta_p\beta_{-p}|1\rangle$ by $0$ in your calculation of $A_p|1\rangle$. I think that's were you lost the missing term. More generally speaking, I'd suggest to try to make more use of symmetry and modularisation – your calculations seem unnecessarily lengthy and could probably be exhibited a lot more succinctly if you just ...


1

Subtracting the first equation from the second yields: $$ m_Bg - m_Ag \sin\theta = (m_A + m_B)a $$ Note that $m_A = 2m_B$, so we have: $$ m_Bg - 2m_Bg \sin\theta = 3m_Ba $$ Now if the blocks remain at rest when released, then $a = 0$, so: \begin{align*} m_Bg - 2m_Bg \sin\theta &= 0 \\ m_Bg &= 2m_Bg \sin\theta \\ \frac{1}{2} &= \sin\theta \\ \...


1

Hint: Without arguing by symmetry: the electric potential of a configuration of the charges is given by $$V(x_1,\ldots,x_n) = \sum_{i<j}\frac{1}{\|x_i-x_j\|}$$ where $x_1,\ldots,x_n$ are distinct points in $S^1\subset\mathbb{R}^2$. The system is in equilibrium if, and only if $V$ is at a critical point. Use Lagrange multipliers.


1

The stick is balanced if the weights are in inverse proportion to the lever lengths. Thus if the shorter end has length $x$, the probability that it goes down is $x$, so the overall probability is $$ 2\int_0^\frac12x\mathrm dx=\frac14\;. $$


1

Yes, the motion still takes place in a plane (the plane spanned by $\mathbf{r}(0)$ and $\mathbf{\dot r}(0)$). One way you can see it is by expanding $\mathbf{r}(t)=\mathbf{r}(0)+\mathbf{\dot r}(0)t+...$ in a series and using the differential equation to see that $\mathbf{r}^{(k)}(0)\in $ span$\{ \mathbf{r}(0), \mathbf{\dot r}(0)\}$. The other way is the ...


1

The equations and idea in your calculation is correct. However, you used the wrong distance $D$. In a bench press, the weight does not move a distance of your wingspan, but rather the distance from your chest to however far your arms reach. It should be somewhat less than half of your wingspan. So you need to measure that distance for both of $A$ and $B$, ...


1

I am going to give you a sketch on how you can find the solution. As I stated in the comments, notice that this is not an RLC circuit, because C and L are in parallel whereas they are in series in an RLC circuit. Since C and L are in parallel, we have $$v_C(t) = v_L(t)$$ and $$i(t) = i_R(t) = i_L(t) + i_C(t)$$ From Kirchhoff's voltage law: $$v_R(t)+...


1

I don't know if I've got your question right. If you are asking where to attach a thread so that your cuboid would be rotated of a certain angle under the influence of its weight, then the answer is very simple: at equilibrium the center of mass (that is the center of the cuboid, I suppose) is directly below the suspension point (or pivot point, as you call ...


1

Mass is proportional to volume not surface area. In general surface area is proportional to the square of a linear measurement and volume is equal to the cube of a linear measurement. So volume is proportional to the 3/2 power of surface area. The ratio of areas is 50/18 so the ratio of volumes is $(50/18)^{3/2}= 4.63. Since A has mass 500 grams, B has ...


1

That is more trivial. If $$ \frac{1}{R_{eq}}=\sum_{k=1}^{n}\frac{1}{R_k} $$ obviously $$ \frac{1}{R_{eq}}> \frac{1}{R_k} $$ for any $k\in\{1,2,\ldots,n\}$, hence $R_{eq}< R_k$, so $$ R_{eq} < \min_{k} R_k.$$


1

First you calculate how long before the driver reacts. 22ft/s times 0.7s = feet traveled before the driver reacts. Then, the driver reacts. How long does it take for the driver to completely stop (forgetting about jerk and higher derivatives)? Eyeballing it, about 1.8 seconds. This is the $\Delta t$ you use in formula 1, with $v = 22$ initially.


1

We have $2 \zeta \omega= \frac{b}{m}$ by definition, $\zeta$ is just defined to be this quantity.


1

The torque due to gravity and centripetal force must be equal. Let's call the coordinate along the rod $x$, varying from 0 to $L$. A small piece, of length $dx$ has a mass of $\frac{m}{L} dx$. The torque due to gravity of this small piece is $$d\tau_g=\frac{m}{L} dx g x \sin \theta$$ The total torque due to gravity is given by $$\tau_g=\int_0^L\frac{m}{L} dx ...


1

Here is the physics: In one dimension $x(t)=x_0+v_0t+at^2/2$. $x_0$ is the initial coordinate, $v_0$ is the initial velocity. For your problem, you have two directions (along $x$ and along $y$) and two objects (target will be denoted by subscript T, missile by subscript M) $$ x_T(t)=x_{T,0}+v_{T,x,0} t+a_{T,x}t^2/2\\ y_T(t)=y_{T,0}+v_{T,y,0} t+a_{T,y}t^2/2\\...



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