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The reason is because the time taken for the two trips are different, so the average speed is not simply $\frac{v_1 + v_2}{2}$ We should go back to the definition. The average speed is always (total length) ÷ (total time). In your case, the total time can be calculated as \begin{align} \text{time}_1 &= \frac{120 \mathrm{miles}}{40 \mathrm{mph}} ...

22

The best way to solve it would be experimentally, by doing the run several times, with calibrated instrumentation by the roadside to measure your speed. The acceleration won't have been constant, so that's not an assumption we can use. Knowing the 0-60 time capability won't really help; it could be different when accelerating up hill, compared to on the ...

16

Your calculation is incorrect. $\text{Work} = \text{Force} \cdot \text{displacement} = F \cdot s$ The above product is a "dot" or "scalar" product, which means we only consider the displacement that occurs in the direction of the Force, which in the case of gravity is downwards. Can we set this vertical displacement to 0? No we cannot, and here is why: ...

16

This scientific problem – well, a more general one – has been solved in the following paper: http://arxiv.org/abs/1204.0162 Because it's legal in my country to move backwards in time, I remember the future event – one minute from now – in which Andrew Gibson will mention that he has this paper hanging in his physics lounge. He will curse me. 11 minutes ...

15

If you write down the formulas for kinetic energy, $$E_k = \frac{1}{2} m v^2$$ and momentum $$p = mv$$ you see that you can write the energy in terms of momentum via $$E_k = \frac{p^2}{2m}$$ So, if two objects have the same energy $E_k$, they only have the same momentum if they also have the same mass. Since the bull has a much larger mass than the bullet, ...

12

You basically just need to be careful about the distinction between velocity and speed. In particular, you say that Won't the particles change velocity when exposed to the magnetic field, and therefore change KE? A change in velocity is not necessarily accompanied by a change in speed, and it's the speed that determines the kinetic energy. The ...

11

I agree with @Ron Maimon that these ETS questions are problematic. But this is (i think) the reasoning they go with. Unlike @Mike's assumption you should not take the normal average, but as stated in the question the weighted average. A weighted average assigns to each measurement $x_i$ a weight $w_i$ and the average is then $$\frac{\sum_iw_ix_i}{\sum_i ... 11 Another way of solving such problems is to go to another reference frame, where you obviously don't have enough energy. For example you've got a 5 MeV photon, so you think that there is plenty of energy to make e^-e^+ pair. Now you make a boost along the direction of the photon momentum with v=0.99\,c and you get a 0.35 MeV photon. That is not ... 11 A common mistake when students begin the study of the quantum harmonic oscillator is to try to convert everything to integrals. The thing is, in most curricula, the QHO is also used as a way to secretly acquaint you with bra-ket notation, and all the conveniences it offers. In reality, you shouldn't need any integrals at all here. \lvert n \rangle is a ... 11 If the cage is completely closed, it doesn't make a difference if the bird is hovering inside it or if it sits on the ground. When flying, the bird pushes air to the ground which will exert a downward force on the cage exactly equal to the weight of the bird. This is a direct consequence of the conservation of momentum and Newton's second & third law. ... 11 It seems that the question (v1) is caused by the fact that there are two different notions of the commutator: One for group theory:$$\tag{1} [A,B] ~:=~ ABA^{-1}B^{-1}$$(or sometimes [A,B] := A^{-1}B^{-1}AB, depending on convention), which is relatively seldom used in physics. One for rings/associative algebras:$$\tag{2} [A,B]:=AB-BA,$$which is ... 10 Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator:$$ \hat{H} \psi = E \psi. $$As such, we expect the solutions to be determined only up to scaling. Clearly if \psi_n is a solution with eigenvalue E_n, then$$ \hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ...

10

These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have. You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) ...

10

Batteries do not behave in such an ideal way across all conditions. The simplest model of a battery as a circuit element is the one you describe - a pure voltage source. A slightly-more sophisticated model is as a voltage source connected to a fixed resistor, called the battery's internal resistance. A typical battery has an internal resistance of between 1 ...

9

Electric monopoles exist. Magnetic monopoles don't* exist. This is the reason that Maxwell's laws governing electricity and magnetism aren't symmetric. Maxwell's laws say that $\nabla \cdot B = 0$ and $\nabla \cdot E = 4\pi\rho_{e}$, but if magnetic monopoles existed this would be $\nabla \cdot B = 4\pi\rho_{m}$ and $\nabla \cdot E = 4\pi\rho_{e}$. A ...

9

The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form $$(\text{number})(\text{unit}) = (\text{other number})(\text{other unit})$$ But of course, you can divide both sides of any equation by the same ...

9

Here is a mathematical derivation. We use the sign convention $(+,-,-,-)$ for the Minkowski metric $\eta_{\mu\nu}$. I) First recall the fact that $SL(2,\mathbb{C})$ is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$. This follows partly because: There is a bijective isometry from the Minkowski space ...

9

Recall that a force perpendicular to the direction of motion does no work but simply changes the direction of the velocity vector. The same thing happens here: Initially the ball's motion is perpendicular to the force of gravity and hence at this very moment, gravity does no work but slightly "rotates" this velocity vector towards the downward direction; as ...

9

Hints: Prove that the angular momentum $L^{ij}:=x^ip^j-x^jp^i$ is conserved for a central force law in $d$ spatial dimensions, $i,j\in\{1,2,\ldots ,d\}.$ Choose a 2D plane $\pi$ through the origin that is parallel to the initial position and momentum vectors. Deduce (from the equations of motion $\dot{\bf x} \parallel {\bf p}$ and $\dot{\bf p} \parallel ... 9 This problem is generally called propagation of error / uncertainty. You can google it and find a lot of info (I'd also recommend Taylor's "Introduction to Error Analysis"). Here's the gist of it, though. If you have independent measured quantities$x, y, z, \ldots$with errors$ \sigma_x, \sigma_y, \sigma_z, \ldots$, then the error on a function ... 9 The superscript$^2$in$1750\text{ mm}^2$refers to a squaring of the units, not the number$1750$. A more transparent way to write this is$1750\text{ mm}\cdot\text{mm}$. The idea is now to multiply by$1$, but$1$written in a clever way: $$1=\frac{1\text{ m}}{1000\text{ mm}}$$. Can you see how that number is conceptually equal to$1$? The top and ... 9 In addition to Ali's answer, here are some pictures which may be helpful in convincing people that the origin is not the only point inside the polygon where$\mathbf{E}=\mathbf{0}$. Letting the charges be located at$(\cos(2\pi k/N),\sin(2\pi k/N))$for$k\in\{1,2,...,N\}$, we can generate plots of$|\mathbf{E}|^{-1}$for various$N$. The zeros of ... 8 According to what I understand you will show as weighing more on a carpet than on a hard floor. From what I understand it is due to the way the hard floor affects the feet of the scales. Here is a article that explores that question: http://www.newscientist.com/article/dn2462-people-weigh-less-on-a-hard-surface.html 8 You're confusing independent and dependent variables. When you propogate from uncertainties in the$x_{i}$to some$f(x_{1},x_{2}...)$, the formula$\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$assumes that each of the$x_{i}$is an independently measured variable and that$f$is a dependent variable to be calculated ... 8 Jerry Schirmer's right about why solving for$r$first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right. But the main thing I wanted to point out is that, if the measurements of$V$and$h$are independent, and if the "errors" mean standard deviations as ... 8 1) I assume that OP means that$q=q(t)$is a single dynamical variable for a classical Lagrangian function $$L~=~ \frac{1}{2}m\dot{q}^2 + f(q)\dot{q},$$ where$f=f(q)$is a given function. I also assume that the problem is well-posed, i.e., that the problem has consistent boundary conditions. Say, the$q$variable is fixed at initial and final time, ... 8 I) OP is given a problem of the form $$\tag{1} \dot{q}~=~f(q,p), \qquad \dot{p}~=~g(q,p),$$ where$f$and$gare two given smooth functions. OP is asked to derive the integrability condition for the eqs. (1) to be Hamilton's eqs. $$\tag{2} \dot{q}~=~\frac{\partial H}{\partial p}, \qquad \dot{p}~=~-\frac{\partial H}{\partial q}.$$ OP correctly ... 8 You use the total amount of movement over time. So here that is|: 80km plus 60km equals 140km Which gives you the correct answer. Displacement, using Pythagoras, would be 100km, but you travelled 140km in that hour! You didn't travel along that hypoteneuse, so it is irrelevant here. 8 There is no 'only if' because it is not true: \begin{align} e^{A+B} = e^A e^B \end{align} does not necessarily imply[A,B] = 0. One can easily find an example of this using matrices. Here's one: \begin{align} A= \begin{pmatrix} 0 & 0 \\ 0 & 2\pi i \end{pmatrix}, B=\begin{pmatrix} 0 & 1 \\ 0 & 2 \pi i \end{pmatrix}. \end{align}[A,B] \neq ...

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