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45

I am sorry to say, but your colleague is right. Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity, $$ m g = k v^2 $$ And thus $$ v=\sqrt{\frac{mg}{k}}$$ So, the terminal velocity of a ball 10 times as ...


43

Start by considering what is seen by the people watching you from the Earth. Nothing can travel faster than the speed of light, $c$, so the quickest you could get to Kepler 186f would be if you were travelling at $c$ in which case it would take 490 years. In practice it would take longer than this because you have to accelerate from rest when you leave the ...


31

like even when light gets on the moon why does the space appears dark from the moon? For the same reason it appears dark from the Earth (when flying at an altitude of 80,000 feet or so): Image credit: View from the SR-71 Blackbird. The fact is, we can't 'see space' from the Earth's surface during the day because the atmosphere is 'in the way'- the ...


28

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked. To calculate ...


19

As you have probably noticed, the moon is tidally locked with the earth so that we always see the same side. You can look up in the sky and watch sunlight move across the moon's face. From the surface of the moon this change in illumination would look just like the day/night cycle on Earth ... except that it's roughly a month long. Until the advent of radar ...


19

Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum. In vacuum, there is only the gravitational force on each ball. That force is proportional to mass. The accelleration of a object due to a force is inversely proportional to its mass, so the mass cancels out. Each ball will accellerate the same, which is the ...


14

The factor of two is correct as far as the integral goes; it comes from the unphysical situation of having your test mass exactly on the thin shell. Intuitively, you get the average of the "just outside" result (as if mass is concentrated at the centre) and the "just inside" result of zero. A more physical thing to do would be to `regulate' the calculation ...


13

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 ...


11

In a nutshell, the problem with OP's choice of operators $\hat{p}_j$ and $\hat{H}$ is that they are not selfadjoint wrt. to the pertinent measure $\mu$. In other words, the usual integration by part method to prove selfadjointness does not work. Here are some more details. Let us put the constants $m=1=R$ for simplicity. Then the Lagrangian reads ...


11

Other answers & comments cover the difference in acceleration due to friction, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, ...


10

tl;dr: Velocity required: 1680 m/s Time to hit you: 6500 seconds Part 1: Velocity required (Using Google search values) Radius of moon = 1737.4 kilometers Mass of moon = 7.34767309E22 kilograms Assuming perfectly circular motion of the bullet, and no air resistance, and ignoring gravitational effects of other planets / objects in space, and using simple ...


10

Let's simplify. Let's eliminate the Moon. Let's get rid of the Sun temporarily. Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way. We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical ...


10

If you redraw your diagram as: It should be clear which capacitors are in parallel and which are in series.


10

The ways of rearranging a system is related to the entropy by $W=e^S$. The entropy is given by $S=\sum_i p_i \log p_i$ where $p_i$ is the probability of realising a given microstate. Hence $W = \prod_i \, p_i^{\,p_i}$


10

To figure this out, you need to know about momentum ($p$). That's a combination of how fast something is moving ($v$, for velocity) and how much it weighs ($m$, for mass). You'll also need to understand algebra, which is just using a letter to mean some number you don't know yet. $$ p = m\cdot v $$ Momentum is conserved, which means the momentum from both ...


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 ...


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

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 mistake you're making is that you're looking at the full acceleration when you should look at the relative one. At distance $R=1\mathrm{au}$ from the sun, the gravitational acceleration is given by $$ a_0 = \frac{GM_\odot}{R^2} $$ Assuming a spherical cow earth (in vacuum), at midday at the equator, we're one earth-radius $r$ closer to the sun, ie $$ ...


9

Yes your weight will change. The moon will have a bigger impact than the sun, so you need to look at the position of the moon to decide when you will be heaviest (basically - you are lighter when the moon is overhead, or on the opposite side of the earth; and heaviest when it is on the horizon. So a full moon rising makes you fat...) The effect (the ...


8

$$M_{gal}=Rv^2/G$$ ($G=6.67\times10^{-11} (N*[m/kg]^2) $. Units of $v$ and $R$ are km/sec. and km., respectively) You gave G in MKS, then: R and v are m, m/s, $ m= (\frac {1}{10^3}) km$, that's why you got a wrong result: $ 10^3 * (10^3) ^2 = 10^9 $ that's the order of magnitude you are missing $$ 1.5*10^{11} *(3*10^4)^2/(6.6*10^{-11}) = ...


8

This vector potential can be written in every point on the plane except the origin as: $$ A_x = -\frac{\partial \psi}{\partial y}$$ $$ A_y = \frac{\partial \psi}{\partial x} $$ with $$\psi = \frac{1}{2}\mathrm{log}(x^2+y^2)$$ $A$ is not exact, because $\psi$ is singular at the origin. But this means that the magnetic field is zero at every point except ...


8

You cannot use the second kinematical equation because it is valid only when the acceleration due to gravity, $g$ , is constant. This is incorrect for distances comparable to the radius of the earth, and velocities comparable to the escape velocity. The first correctly assumes a $\frac{1}{R^2}$ fall-off of the gravitational attraction on the body due to ...


8

When quoting results, there are a few good rules to follow: Avoid rounding errors in intermediate calculations. Write your error to 1 significant figure if your data set is smaller than $10^2$, 2 if it's smaller than $10^4$ etc. Write your estimate and its error with the same number of decimal places. Rules 1. and 3. are simple to understand. Rule 2. ...


8

To show that this measure is Lorentz invariant you first need to explicitly write your integral as an integral over mass shell in 4D k-space. This could be done by inserting Dirac delta function $\delta[k^\mu k_\mu-m^2]$ and integrating over the whole 4D space. Then you could apply the following transformations: \begin{align} \theta(k_0)\cdot\delta[k^\mu ...


8

One can do the calculation(expand the potential to the second order around the center) and show that the center of the polygon is a minimum of potential. We are free to choose $V(\infty)=0$, if we do so, then it would be easy to show that the potential at the center of the polygon is positive. Combining the results above with the fact that the potential is ...


7

You say $C,\lambda$ are constants of integration, but that gives $\ddot{x}-\lambda\dot{x} = 0$ instead. Since your followup would be inconsistent, I will assume you meant $C,\omega$ are constants of integration. You should not have three components to the geodesic equation, but rather two: $$\ddot{y} + \Gamma^y_{xx}\dot{x}\dot{x} + ...


7

Freely-moving charges placed on a line will tend to fly away from each other - with no equilibrium position possible - unless there is some potential that confines them to a specific region. Enforcing the charges to lie within an interval $[0,L]$ will always mean one charge is at either end, so you might as well consider $n-2$ charges confined by the ...


7

This problem has been solved by Griffiths in Charge density of a conducting needle. David J. Griffiths and Ye Li. Am. J. Phys. 64 no. 6 (1996), p. 706. PDF from colorado.edu. The problem is nontrivial.



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