# Tag Info

21

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

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 ...

14

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: ...

12

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, ...

11

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 ...

10

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 ...

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 $$\hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ... 9 The way you convert between units is really just multiplying by several factors of 1. But it's 1 written in a slightly unusual way. Think about this: you're probably familiar with conversion factors in the form$$(\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 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 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) ... 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 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 ...

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.

7

Actually, this is the basic stuff every mechanics textbook should have. Center of mass is the most basic it needs just Newton laws: Second: $m_i\ddot{\vec{r}_i}= \vec{F}_i$ And third: $\sum_i\vec{F}_i = 0$ Summing over i one obtains: $\frac{d^2}{dt^2}\left(\sum_i m_ir_i\right) = \sum_i\vec{F}_i = 0$ For the total energy you need those forces to be ...

7

By lines of force, do you mean the magnetic field? If so, you could determine the direction of the B-field by right-hand rule. The reason is that a current element flowing up will create a B-field circulating out of the screen on the wire's left, and into the screen on the right's right. This direction is chosen by convention. Wrap the wire into a ...

7

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 ...

7

Trigonometric functions don't "preserve" units. The expression under a trigonometric function must be dimensionless and so is the value of a trigonometric function. Thus, C2 in your equations is in units of frequency: Hz or 1/s. There is an error in one of the equations, perhaps a missing constant.

7

Starting with your given equation, we add $p^2 c^2$ to both sides to get $$E^2=m^2 c^4 + p^2 c^2$$ now using the definition of relativistic momentum $p=\gamma m v$ we substitute that in above to get $$E^2 = m^2 c^4 +(\gamma m v)^2 c^2=m^2 c^4 +\gamma^2 m^2 v^2 c^2$$ Now, factoring out a common $m^2 c^4$ from both terms on the RHS in anticipation of the ...

7

To a very good approximation the transmission of a metal film falls exponentially with thickness i.e.: $$T = e^{-\alpha t}$$ where $\alpha$ is the absorption coefficient given on the web site Alexander mentioned, http://refractiveindex.info/?group=METALS&material=Copper, and at 500nm wavelength this gives $\alpha = 6.4297\times 10^5/cm$. So you just ...

7

The three capacitors are connected in parallel. There are only two nodes in this circuit. A series connection requires at least three. The equivalent capacitance is just the sum of the three capacitances. UPDATE: The circuit can be redrawn such that the parallel connection is manifest.

7

If you solve for $t$ in Eq. (5.1), and plug that into equation (1.1), you'll see that the solution looks like $x_B \propto v_A^2 sin(\theta) cos(\theta)$. The function on the right is symmetric about $\pi/4$, thus, as long as $\theta$ doesn't equal $\pi/4$, there will be two solutions (symmetrically about $\pi/4$). Of course, in general, there could be ...

7

Alternatively I would look around the lab for an infrared thermometer. There exist in the market close focus ones that go down to 6mm in close focus option ( so as not to advertise, google space accurate infrared thermometers microscopes where I found the number in a one of the first hits). I would choose a large ant, or attract more by a spot of honey ...

7

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 ...

7

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} $[A,B] \neq ... 7 Well, this certainly is an evil trick to play on first year students! Escape velocity isn't actually a velocity at all. It's a speed, i.e., it's scalar quantity as opposed to a vector quantity. Note that when the escape "velocity" at r was calculated, the only assumption made was conservation of mechanical energy, and then magnitude of v is isolated from ... 6 The equation describes parabolic motion, if$a\neq 0$is a non-zero constant acceleration, which I will assume from now on. If you think about it, your solution provides an answer to the question: at what time does the object is in the position$s$? [A note on notation: Traditionally, the letter$s$denotes distance (I guess from the German word "Strecke"), ... 6 For any given$n$, you can work it out via the rules for series and parallel resistors, but to get a general formula, valid for all$n$, doesn't look easy to me. The best way I know of is to get a recursive relationship giving the resistance of an$n$-step ladder in terms of an$(n-1)$-step ladder. If I'm not mistaken, the$n$-step ladder can be thought of ... 6 I'll take it step by step here. First I'll write the answer for the first few cases with circuit analysis. Then I'll apply a reduction to show the pattern that the problem arrives at. N=1 $$Z = R+R=2R$$ N=2 $$Z = R+\frac{1}{\frac{1}{R}+\frac{1}{R + R}} = R \left( 1+\frac{1}{1+\frac{1}{1 + 1}} \right)=\frac{5}{3} R$$ N=3$\$Z = ...

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